This preview shows page 1. Sign up to view the full content.
Unformatted text preview: y vR2 vR1
u vD2 x V #2 #1 y A1B
0.230 m v50
x y
T arad
x mg a (m/s2)
100 v (m/s)
4000 80 3000 60 2000 40 1000 20 0 0 20 40 60 80 100 120 t (s) 0 0 20 (a) 40 60 80 100 120 (b) t (s) y y dy y
y x x x dx 2x
(b) (a) ELECTRIC CHARGE AND ELECTRIC FIELD 21.1. 21 (a) IDENTIFY and SET UP:! Use the charge of one electron ( !1.602 " 10 !19 C) to find the number of electrons
required to produce the net charge.
EXECUTE:! The number of excess electrons needed to produce net charge q is
q
!3.20 " 10!9 C
#
# 2.00 " 1010 electrons.
!e !1.602 " 10!19 C/electron
(b) IDENTIFY and SET UP:! Use the atomic mass of lead to find the number of lead atoms in 8.00 " 10 !3 kg of
lead. From this and the total number of excess electrons, find the number of excess electrons per lead atom.
EXECUTE:! The atomic mass of lead is 207 " 10!3 kg/mol, so the number of moles in 8.00 " 10 !3 kg is mtot
8.00 " 10!3 kg
#
# 0.03865 mol. N A (Avogadro’s number) is the number of atoms in 1 mole, so the
M
207 " 10!3 kg/mol
number of lead atoms is N # nN A # (0.03865 mol)(6.022 " 1023 atoms/mol) = 2.328 " 1022 atoms. The number of
n# 2.00 " 1010 electrons
# 8.59 " 10!13.
2.328 " 1022 atoms
EVALUATE:! Even this small net charge corresponds to a large number of excess electrons. But the number of
atoms in the sphere is much larger still, so the number of excess electrons per lead atom is very small.
IDENTIFY:! The charge that flows is the rate of charge flow times the duration of the time interval.
SET UP:! The charge of one electron has magnitude e # 1.60 " 10!19 C.
EXECUTE:! The rate of charge flow is 20,000 C/s and t # 100 $ s # 1.00 " 10!4 s.
Q
# 1.25 " 1019.
Q # (20,000 C/s)(1.00 " 10 !4 s) # 2.00 C. The number of electrons is ne #
1.60 " 10!19 C
EVALUATE:! This is a very large amount of charge and a large number of electrons.
IDENTIFY:! From your mass estimate the number of protons in your body. You have an equal number of electrons.
SET UP:! Assume a body mass of 70 kg. The charge of one electron is !1.60 " 10!19 C.
EXECUTE:! The mass is primarily protons and neutrons of m # 1.67 " 10!27 kg. The total number of protons and
70 kg
neutrons is np and n #
# 4.2 " 10 28. About onehalf are protons, so np # 2.1 " 1028 # ne . The number of
1.67 " 10!27 kg
electrons is about 2.1" 1028. The total charge of these electrons is
Q # ( !1.60 " 10 !19 C/electron)(2.10 " 1028 electrons) # !3.35 " 109 C.
EVALUATE:! This is a huge amount of negative charge. But your body contains an equal number of protons and
your net charge is zero. If you carry a net charge, the number of excess or missing electrons is a very small fraction
of the total number of electrons in your body.
IDENTIFY:! Use the mass m of the ring and the atomic mass M of gold to calculate the number of gold atoms.
Each atom has 79 protons and an equal number of electrons.
SET UP:! N A # 6.02 " 1023 atoms/mol . A proton has charge +e.
excess electrons per lead atom is 21.2. 21.3. 21.4. EXECUTE:! The mass of gold is 17.7 g and the atomic weight of gold is 197 g mol. So the number of atoms
% 17.7 g &
22
is N A n # (6.02 " 1023 atoms/mol) '
( # 5.41" 10 atoms . The number of protons is
) 197 g mol *
np # (79 protons/atom)(5.41" 1022 atoms) # 4.27 " 1024 protons . Q # (np )(1.60 " 10!19 C/proton) # 6.83 " 105 C .
(b) The number of electrons is ne # np # 4.27 " 1024.
EVALUATE:! The total amount of positive charge in the ring is very large, but there is an equal amount of negative
charge. 211 212 21.5. Chapter 21 IDENTIFY:! Apply F # k q1q2
r2 and solve for r. SET UP:! F # 650 N . k q1q2
(8.99 " 109 N + m 2 /C2 )(1.0 C) 2
#
# 3.7 " 103 m # 3.7 km
650 N
F
EVALUATE:! Charged objects typically have net charges much less than 1 C.
IDENTIFY:! Apply Coulomb's law and calculate the net charge q on each sphere.
EXECUTE:! r # 21.6. SET UP:! The magnitude of the charge of an electron is e # 1.60 " 10!19 C . 1 q2
. This gives q # 4, P0 Fr 2 # 4, P0 (4.57 " 10!21 N)(0.200 m) 2 # 1.43 " 10 !16 C. And
4, P0 r 2
therefore, the total number of electrons required is n # q /e # (1.43 "10!16 C)/(1.60 "10!19 C/electron) # 890 electrons.
EXECUTE:! F # 21.7. EVALUATE:! Each sphere has 890 excess electrons and each sphere has a net negative charge. The two like
charges repel.
IDENTIFY:! Apply Coulomb’s law.
SET UP:! Consider the force on one of the spheres.
(a) EXECUTE:! q1 # q2 # q F# 1 q1q2
q2
F
0.220 N
#
so q # r
# 0.150 m
# 7.42 " 10!7 C (on each)
4, P0 r 2
4, P0 r 2
(1/4, P0 )
8.988 " 109 N + m 2 /C2 (b) q2 # 4q1 F# 21.8. 1 q1q2
4q12
F
F
#
so q1 # r
# 1r
# 1 (7.42 " 10!7 C) = 3.71 " 10 !7 C.
2
4, P0 r
4, P0r 2
4(1/4, P0 ) 2 (1/4, P0 ) 2 And then q2 # 4q1 # 1.48 " 10 !6 C.
EVALUATE:! The force on one sphere is the same magnitude as the force on the other sphere, whether the sphere
have equal charges or not.
IDENTIFY:! Use the mass of a sphere and the atomic mass of aluminum to find the number of aluminum atoms in
one sphere. Each atom has 13 electrons. Apply Coulomb's law and calculate the magnitude of charge q on each
sphere.
SET UP:! N A # 6.02 " 10 23 atoms/mol . q # nee , where ne is the number of electrons removed from one sphere
and added to the other.
EXECUTE:! (a) The total number of electrons on each sphere equals the number of protons.
%
&
0.0250 kg
24
ne # np # (13)( N A ) '
( # 7.25 " 10 electrons .
) 0.026982 kg mol *
(b) For a force of 1.00 " 104 N to act between the spheres, F # 1.00 " 104 N # 1 q2
. This gives
4, P0 r 2 q # 4, P0 (1.00 " 10 4 N)(0.0800 m) 2 # 8.43 " 10 !4 C . The number of electrons removed from one sphere and added to the other is ne # q / e # 5.27 "1015 electrons. 21.9. (c) ne / ne # 7.27 " 10 !10 .
EVALUATE:! When ordinary objects receive a net charge the fractional change in the total number of electrons in
the object is very small .
qq
IDENTIFY:! Apply F # ma , with F # k 1 2 2 .
r
SET UP:! a # 25.0 g # 245 m/s 2 . An electron has charge !e # !1.60 " 10!19 C. EXECUTE:! F # ma # (8.55 " 10!3 kg)(245 m/s 2 ) # 2.09 N . The spheres have equal charges q, so F # k q2
and
r2 q 2.29 " 10!6 C
F
2.09 N
# (0.150 m)
# 2.29 " 10!6 C . N # #
# 1.43 " 1013 electrons . The
8.99 " 109 N + m 2 /C 2
e 1.60 " 10!19 C
k
charges on the spheres have the same sign so the electrical force is repulsive and the spheres accelerate away from
each other.
q #r Electric Charge and Electric Field! ! 213 21.10. EVALUATE:! As the spheres move apart the repulsive force they exert on each other decreases and their
acceleration decreases.
(a) IDENTIFY:! The electrical attraction of the proton gives the electron an acceleration equal to the acceleration
due to gravity on earth.
SET UP:! Coulomb’s law gives the force and Newton’s second law gives the acceleration this force produces. ma # 1 q1q2
e2
.
and r #
4, P0 r 2
4, P0 ma . 9.00 "10 N + m /C /.1.60 "10 C /
. 9.11" 10 kg /. 9.80 m/s /
9 EXECUTE:! r = 2 !31 2 !19 2 2 = 5.08 m EVALUATE:! The electron needs to be about 5 m from a single proton to have the same acceleration as it receives
from the gravity of the entire earth.
(b) IDENTIFY:! The force on the electron comes from the electrical attraction of all the protons in the earth.
SET UP:! First find the number n of protons in the earth, and then find the acceleration of the electron using
Newton’s second law, as in part (a). n = mE/mp = (5.97 " 1024 kg)/(1.67 " 10!27 kg) = 3.57 " 1051 protons.
1 qp qe
1
ne2
2
4, P0 RE
4, P0
a = F/m =
#
.
2
me
me RE 21.11. EXECUTE:! a = (9.00 " 109 N + m2/C2)(3.57 " 1051)(1.60 " 10!19 C)2/[(9.11 " 10!31 kg)(6.38 " 106 m)2] = 2.22 "
1040 m/s2. One can ignore the gravitation force since it produces an acceleration of only 9.8 m/s2 and hence is much
much less than the electrical force.
EVALUATE:! With the electrical force, the acceleration of the electron would nearly 1040 times greater than with
gravity, which shows how strong the electrical force is.
IDENTIFY:! In a space satellite, the only force accelerating the free proton is the electrical repulsion of the other
proton.
SET UP:! Coulomb’s law gives the force, and Newton’s second law gives the acceleration: a = F/m =
(1/ 4, P0 ) (e2/r2)/m.
EXECUTE:! (a) a = (9.00 " 109 N + m2/C2)(1.60 " 1019 C)2/[(0.00250 m)2(1.67 " 1027 kg)] = 2.21 " 104 m/s2.
(b) The graphs are sketched in Figure 21.11.
EVALUATE:! The electrical force of a single stationary proton gives the moving proton an initial acceleration
about 20,000 times as great as the acceleration caused by the gravity of the entire earth. As the protons move
farther apart, the electrical force gets weaker, so the acceleration decreases. Since the protons continue to repel, the
velocity keeps increasing, but at a decreasing rate. Figure 21.11
21.12. IDENTIFY:! Apply Coulomb’s law.
SET UP:! Like charges repel and unlike charges attract.
EXECUTE:! (a) F # !6
1 q1q2
1 (0.550 " 10 C) q2
. This gives 0.200 N #
and q2 # 03.64 " 10!6 C . The
2
4, P0 r
4, P0
(0.30 m) 2 force is attractive and q1 1 0 , so q2 # 03.64 " 10!6 C .
(b) F # 0.200 N. The force is attractive, so is downward.
EVALUATE:! The forces between the two charges obey Newton's third law. 214 Chapter 21 21.13. IDENTIFY:! Apply Coulomb’s law. The two forces on q3 must have equal magnitudes and opposite directions.
SET UP:! Like charges repel and unlike charges attract.
!
!
qq
EXECUTE:! The force F2 that q2 exerts on q3 has magnitude F2 # k 2 2 3 and is in the +x direction. F1 must be in
r2 the ! x direction, so q1 must be positive. F1 # F2 gives k
2 21.14. q1 q3
qq
#k 2 2 3 .
2
r1
r2 2 %r &
% 2.00 cm &
q1 # q2 ' 1 ( # (3.00 nC) '
( # 0.750 nC .
) 4.00 cm *
) r2 *
EVALUATE:! The result for the magnitude of q1 doesn’t depend on the magnitude of q2 .
IDENTIFY:! Apply Coulomb’s law and find the vector sum of the two forces on Q.
SET UP:! The force that q1 exerts on Q is repulsive, as in Example 21.4, but now the force that q2 exerts is
attractive.
EXECUTE:! The xcomponents cancel. We only need the ycomponents, and each charge contributes equally.
1 (2.0 " 10!6 C) (4.0 " 10!6 C)
sin 2 # !0.173 N (since sin2 # 0.600). Therefore, the total force is
F1 y # F2 y # !
4, P0
(0.500 m) 2
2 F # 0.35 N, in the ! y direction .
EVALUATE:! If q1 is !2.0 $ C and q2 is 02.0 $ C , then the net force is in the +ydirection. 21.15. IDENTIFY:! Apply Coulomb’s law and find the vector sum of the two forces on q1 .
!
!
SET UP:! Like charges repel and unlike charges attract, so F2 and F3 are both in the +xdirection.
EXECUTE:! F2 # k 21.16. q1q2
qq
# 6.749 " 10!5 N, F3 # k 1 2 3 # 1.124 " 10!4 N . F # F2 0 F3 # 1.8 " 10!4 N .
2
r12
r13 F # 1.8 " 10!4 N and is in the +xdirection.
!
!
EVALUATE:! Comparing our results to those in Example 21.3, we see that F1 on 3 # ! F3 on 1 , as required by
Newton’s third law.
IDENTIFY:! Apply Coulomb’s law and find the vector sum of the two forces on q2 .
!
SET UP:! F2 on 1 is in the +ydirection.
EXECUTE:! F2on 1 # (9.0 " 109 N + m 2 C2 ) (2.0 " 10!6 C) (2.0 " 10!6 C) . F2 on 1 / y # 00.100 N . .F / Q on 1 y . 0.60 m / 2 # 0.100 N . . F2 on 1 / x # 0 and FQ on 1 is equal and opposite to F1 on Q (Example 21.4), so . FQ on 1 / # !0.23N and
x # 0.17 N . Fx # . F2 on 1 / x 0 . FQ on 1 / # !0.23 N . Fy # . F2 on 1 / y 0 . FQ on 1 / # 0.100 N 0 0.17 N # 0.27 N .
x y !
0.23
# 403 , so F is
0.27
40! counterclockwise from the +y axis, or 130! counterclockwise from the +x axis.
EVALUATE:! Both forces on q1 are repulsive and are directed away from the charges that exert them.
The magnitude of the total force is F # 21.17. . 0.23 N / 2 0 . 0.27 N / # 0.35 N. tan !1
2 IDENTIFY and SET UP:! Apply Coulomb’s law to calculate the force exerted by q2 and q3 on q1. Add these forces as vectors to get the net force. The target variable is the xcoordinate of q3 .
!
EXECUTE:! F2 is in the xdirection.
F2 # k q1q2
# 3.37 N, so F2 x # 03.37 N
2
r12 Fx # F2 x 0 F3 x and Fx # !7.00 N
F3 x # Fx ! F2 x # !7.00 N ! 3.37 N # !10.37 N For F3 x to be negative, q3 must be on the ! x axis. F3 # k q1q3
k q1q3
, so x #
# 0.144 m, so x # !0.144 m
2
x
F3 EVALUATE:! q2 attracts q1 in the 0 x direction so q3 must attract q1 in the ! x direction, and q3 is at negative x. Electric Charge and Electric Field! ! 215 21.18. IDENTIFY:! Apply Coulomb’s law.
!
!
SET UP:! Like charges repel and unlike charges attract. Let F21 be the force that q2 exerts on q1 and let F31 be the force that q3 exerts on q1 .
EXECUTE:! The charge q3 must be to the right of the origin; otherwise both q2 and q3 would exert forces in the
0 x direction. Calculating the two forces:
1 q1q2 (9 " 109 N + m 2 C2 )(3.00 " 10!6 C)(5.00 " 10!6 C)
F21 #
#
# 3.375 N , in the +x direction.
2
4, P0 r12
(0.200 m) 2 F31 # (9 " 109 N + m 2 C2 ) (3.00 " 10!6 C) (8.00 " 10!6 C) 0.216 N + m 2
#
, in the ! x direction.
2
2
r13
r13 We need Fx # F21 ! F31 # !7.00 N , so 3.375 N ! 0.216 N + m 2
0.216 N + m 2
# !7.00 N . r13 #
# 0.144 m . q3
2
r13
3.375 N 0 7.00 N is at x # 0.144 m .
EVALUATE:! F31 # 10.4 N. F31 is larger than F21 , because q3 is larger than q2 and also because r13 is less than r12 .
21.19. IDENTIFY:! Apply Coulomb’s law to calculate the force each of the two charges exerts on the third charge. Add
these forces as vectors.
SET UP:! The three charges are placed as shown in Figure 21.19a. Figure 21.19a
EXECUTE:! Like charges repel and unlike attract, so the freebody diagram for q3 is as shown in Figure 21.19b.
F1 # 1 q1q3
2
4, P0 r13 F2 # 1 q2 q3
2
4, P0 r23 Figure 21.19b
F1 # (8.988 " 109 N + m 2 /C2 ) (1.50 " 10!9 C)(5.00 " 10!9 C)
# 1.685 " 10!6 N
(0.200 m) 2 (3.20 " 10!9 C)(5.00 " 10!9 C)
# 8.988 " 10!7 N
(0.400 m)2
!!!
The resultant force is R # F1 0 F2 .
F2 # (8.988 " 109 N + m 2 /C2 ) Rx # 0.
Ry # F1 0 F2 # 1.685 " 10!6 N +8.988 " 10!7 N = 2.58 " 10!6 N.
The resultant force has magnitude 2.58 " 10!6 N and is in the –ydirection.
EVALUATE:! The force between q1 and q3 is attractive and the force between q2 and q3 is replusive.
21.20. IDENTIFY:! Apply F # k qq to each pair of charges. The net force is the vector sum of the forces due to q1 and q2 .
r2
SET UP:! Like charges repel and unlike charges attract. The charges and their forces on q3 are shown in Figure 21.20. 216 Chapter 21 EXECUTE:! F1 # k
F2 # k q2 q3
2
2 r q1q3
(4.00 " 10!9 C)(0.600 " 10!9 C)
# (8.99 " 109 N + m 2 /C2 )
# 5.394 " 10!7 N .
2
r1
(0.200 m) 2 # (8.99 " 109 N + m 2 /C 2 ) (5.00 " 10!9 C)(0.600 " 10!9 C)
# 2.997 " 10!7 N .
(0.300 m) 2 Fx # F1x 0 F2 x # 0 F1 ! F2 # 2.40 " 10!7 N . The net force has magnitude 2.40 " 10!7 N and is in the 0 x direction.
EVALUATE:! Each force is attractive, but the forces are in opposite directions because of the placement of the
charges. Since the forces are in opposite directions, the net force is obtained by subtracting their magnitudes. Figure 21.20
21.21. IDENTIFY:! Apply Coulomb’s law to calculate each force on !Q .
!
!
SET UP:! Let F1 be the force exerted by the charge at y # a and let F2 be the force exerted by the charge at y # ! a.
EXECUTE:! (a) The two forces on !Q are shown in Figure 21.21a. sin 4 # a
and r # ( a 2 0 x 2 )1/ 2 is the
( a 2 0 x 2 )1/ 2 distance between q and !Q and between ! q and !Q .
(b) Fx # F1x 0 F2 x # 0 . Fy # F1 y 0 F2 y # 2
(c) At x # 0, Fy # 1
qQ
1
2qQa
sin 4 #
.
2
2
2
4, P0 (a 0 x )
4, P0 (a 0 x 2 )3 2 1 2qQ
, in the +y direction.
4, P0 a 2 (d) The graph of Fy versus x is given in Figure 21.21b.
EVALUATE:! Fx # 0 for all values of x and Fy 5 0 for all x. Figure 21.21
21.22. IDENTIFY:! Apply Coulomb’s law to calculate each force on !Q .
!
!
SET UP:! Let F1 be the force exerted by the charge at y # a and let F2 be the force exerted by the charge at y # ! a . The distance between each charge q and Q is r # . a 2 0 x 2 / 1/ 2 . cos4 # x .a 2 0 x2 / 1/ 2 . EXECUTE:! (a) The two forces on !Q are shown in Figure 21.22a.
(b) When x 5 0 , F1x and F2 x are negative. Fx # F1x 0 F2 x # !2 1
qQ
1
!2qQx
cos4 #
. When
4, P0 ( a 2 0 x 2 )
4, P0 (a 2 0 x 2 )3 / 2 x 1 0 , F1x and F2 x are positive and the same expression for Fx applies. Fy # F1 y 0 F2 y # 0 .
(c) At x # 0 , Fx # 0 .
(d) The graph of Fx versus x is sketched in Figure 21.22b. Electric Charge and Electric Field! ! 217 EVALUATE:! The direction of the net force on !Q is always toward the origin. Figure 21.22
21.23. IDENTIFY:! Apply Coulomb’s law to calculate the force exerted on one of the charges by each of the other three
and then add these forces as vectors.
(a) SET UP:! The charges are placed as shown in Figure 21.23a. q1 # q2 # q3 # q4 # q Figure 21.23a Consider forces on q4 . The freebody diagram is given in Figure 21.23b. Take the yaxis to be parallel to the
!
diagonal between q2 and q4 and let 0 y be in the direction away from q2 . Then F2 is in the 0 y direction.
EXECUTE:! F3 # F1 # F2 # 1 q2
4, P0 L2 1 q2
4, P0 2 L2 F1x # ! F1 sin 453 # ! F1/ 2 F1 y # 0 F1 cos 453 # 0 F1/ 2
F3 x # 0 F3 sin 453 # 0 F3 / 2 F3 y # 0 F3 cos 453 # 0 F3 / 2
F2 x # 0, F2 y # F2 Figure 21.23b
(b) Rx # F1x 0 F2 x 0 F3 x # 0 Ry # F1 y 0 F2 y 0 F3 y # (2/ 2)
R# 1 q2
1 q2
q2
(1 0 2 2)
0
#
4, P0 L2 4, P0 2 L2 8, P0 L2 q2
(1 0 2 2). Same for all four charges.
8, P0 L2 218 21.24. Chapter 21 EVALUATE:! In general the resultant force on one of the charges is directed away from the opposite corner. The
forces are all repulsive since the charges are all the same. By symmetry the net force on one charge can have no
component perpendicular to the diagonal of the square.
k qqIDENTIFY:! Apply F # 2 to find the force of each charge on 0 q . The net force is the vector sum of the
r
individual forces.
SET UP:! Let q1 # 02.50 $ C and q2 # !3.50 $ C . The charge 0 q must be to the left of q1 or to the right of q2 in
order for the two forces to be in opposite directions. But for the two forces to have equal magnitudes, 0 q must be
closer to the charge q1 , since this charge has the smaller magnitude. Therefore, the two forces can combine to give
zero net force only in the region to the left of q1 . Let 0 q be a distance d to the left of q1 , so it is a distance
d 0 0.600 m from q2 .
EXECUTE:! F1 # F2 gives kq q1
d 2 # kq q2
( d 0 0.600 m) 2 . d #6 q1
q2 (d 0 0.600 m) # 6 (0.8452)(d 0 0.600 m) . d must (0.8452)(0.600 m)
# 3.27 m . The net force would be zero when 0 q is at x # !3.27 m .
1 ! 0.8452
!
!
EVALUATE:! When 0 q is at x # !3.27 m , F1 is in the ! x direction and F2 is in the +x direction. be positive, so d # 21.25. IDENTIFY:! F # q E . Since the field is uniform, the force and acceleration are constant and we can use a constant acceleration equation to find the final speed.
SET UP:! A proton has charge +e and mass 1.67 " 10!27 kg .
EXECUTE:! (a) F # (1.60 " 10!19 C)(2.75 " 103 N/C) # 4.40 " 10!16 N
(b) a # 21.26. (c) vx # v0 x 0 axt gives v # (2.63 " 1011 m/s 2 )(1.00 " 10!6 s) # 2.63 " 105 m/s
EVALUATE:! The acceleration is very large and the gravity force on the proton can be ignored.
q
IDENTIFY:! For a point charge, E # k 2 .
r
!
SET UP:! E is toward a negative charge and away from a positive charge.
EXECUTE:! (a) The field is toward the negative charge so is downward.
3.00 " 10!9 C
E # (8.99 " 109 N + m 2 /C2 )
# 432 N/C .
(0.250 m) 2 (8.99 " 109 N + m 2 /C2 )(3.00 " 10!9 C)
# 1.50 m
E
12.0 N/C
EVALUATE:! At different points the electric field has different directions, but it is always directed toward the
negative point charge.
IDENTIFY:! The acceleration that stops the charge is produced by the force that the electric field exerts on it.
Since the field and the acceleration are constant, we can use the standard kinematics formulas to find acceleration
and time.
(a) SET UP:! First use kinematics to find the proton’s acceleration. vx # 0 when it stops. Then find the electric
field needed to cause this acceleration using the fact that F = qE.
2
2
EXECUTE:! vx # v0 x 0 2ax ( x ! x0 ) . 0 = (4.50 " 106 m/s)2 + 2a(0.0320 m) and a = 3.16 " 1014 m/s2. Now find the
(b) r # 21.27. F 4.40 " 10!16 N
#
# 2.63 " 1011 m/s 2
m 1.67 " 10!27 kg kq # electric field, with q = e. eE = ma and E = ma/e = (1.67 " 10!27 kg)(3.16 " 1014 m/s2)/(1.60 " 10!19 C) = 3.30 "
106 N/C, to the left.
(b) SET UP:! Kinematics gives v = v0 + at, and v = 0 when the electron stops, so t = v0/a.
EXECUTE:! t = v0/a = (4.50 " 106 m/s)/(3.16 " 1014 m/s2) = 1.42 " 10!8 s = 14.2 ns
(c) SET UP:! In part (a) we saw that the electric field is proportional to m, so we can use the ratio of the electric
fields. Ee / Ep # me / mp and Ee # . me / mp / Ep .
EXECUTE:! Ee = [(9.11 " 10!31 kg)/(1.67 " 10!27 kg)](3.30 " 106 N/C) = 1.80 " 103 N/C, to the right
EVALUATE:! Even a modest electric field, such as the ones in this situation, can produce enormous accelerations
for electrons and protons. Electric Charge and Electric Field! ! 219 21.28. !
!
IDENTIFY:! Use constant acceleration equations to calculate the upward acceleration a and then apply F # qE to
calculate the electric field.
SET UP:! Let +y be upward. An electron has charge q # !e .
EXECUTE:! (a) v0 y # 0 and a y # a , so y ! y0 # v0 yt 0 1 a yt 2 gives y ! y0 # 1 at 2 . Then
2
2
2 2( y ! y0 )
2(4.50 m)
F ma (9.11 " 10!31 kg) (1.00 " 1012 m s )
2
#
# 1.00 " 1012 m s . E # #
#
# 5.69 N C
q
q
1.60 " 10!19 C
t2
(3.00 " 10!6 s)2
The force is up, so the electric field must be downward since the electron has negative charge.
(b) The electron’s acceleration is ~ 1011 g , so gravity must be negligibly small compared to the electrical force.
EVALUATE:! Since the electric field is uniform, the force it exerts is constant and the electron moves with
constant acceleration.
(a) IDENTIFY:! Eq. (21.4) relates the electric field, charge of the particle, and the force on the particle. If the
particle is to remain stationary the net force on it must be zero.
SET UP:! The freebody diagram for the particle is sketched in Figure 21.29. The weight is mg, downward. For
the net force to be zero the force exerted by the electric field must be upward. The electric field is downward. Since
the electric field and the electric force are in opposite directions the charge of the particle is negative.
a# 21.29. mg # q E Figure 21.29 mg (1.45 " 10!3 kg)(9.80 m/s2 )
#
# 2.19 " 10!5 C and q # !21.9 $ C
E
650 N/C
(b) SET UP:! The electrical force has magnitude FE # q E # eE. The weight of a proton is w # mg . FE # w so EXECUTE:! q# eE # mg
mg (1.673 "10!27 kg)(9.80 m/s2 )
#
# 1.02 " 10!7 N/C.
e
1.602 "10!19 C
This is a very small electric field.
EVALUATE:! In both cases q E # mg and E # (m / q ) g . In part (b) the m / q ratio is much smaller (" 10!8 ) than
EXECUTE:! E # 21.30. in part (a) (" 10!2 ) so E is much smaller in (b). For subatomic particles gravity can usually be ignored compared to
electric forces.
1q
IDENTIFY:! Apply E #
.
4, P0 r 2
SET UP:! The iron nucleus has charge 0 26e. A proton has charge 0e .
1 (26)(1.60 " 10!19 C)
# 1.04 " 1011 N/C.
EXECUTE:! (a) E #
4, P0 (6.00 " 10!10 m)2
1 (1.60 " 10!19 C)
# 5.15 " 1011 N/C.
4, P0 (5.29 " 10!11 m)2
EVALUATE:! These electric fields are very large. In each case the charge is positive and the electric fields are
directed away from the nucleus or proton.
q
IDENTIFY:! For a point charge, E # k 2 . The net field is the vector sum of the fields produced by each charge. A
r
!
!
!
charge q in an electric field E experiences a force F # qE .
SET UP:! The electric field of a negative charge is directed toward the charge. Point A is 0.100 m from q2 and
0.150 m from q1. Point B is 0.100 m from q1 and 0.350 m from q2.
EXECUTE:! (a) The electric fields due to the charges at point A are shown in Figure 21.31a.
q1
6.25 " 10!9 C
# 2.50 " 103 N/C
E1 # k 2 # (8.99 " 109 N + m 2 /C2 )
(0.150 m)2
rA1
(b) Eproton # 21.31. E2 # k q2
2
A2 r # (8.99 " 109 N + m 2 /C2 ) 12.5 " 10!9 C
# 1.124 " 104 N/C
(0.100 m) 2 2110 Chapter 21 Since the two fields are in opposite directions, we subtract their magnitudes to find the net field.
E # E2 ! E1 # 8.74 " 103 N/C, to the right.
(b) The electric fields at points B are shown in Figure 21.31b.
q1
6.25 " 10!9 C
# 5.619 " 103 N/C
E1 # k 2 # (8.99 " 109 N + m 2 /C2 )
(0.100 m)2
rB1
E2 # k q2
rB22 # (8.99 " 109 N + m 2 /C 2 ) 12.5 " 10!9 C
# 9.17 " 102 N/C
(0.350 m) 2 Since the fields are in the same direction, we add their magnitudes to find the net field. E # E1 0 E2 # 6.54 "103 N/C,
to the right.
(c) At A, E # 8.74 " 103 N/C , to the right. The force on a proton placed at this point would be
F # qE # (1.60 " 10!19 C)(8.74 " 103 N/C) # 1.40 " 10!15 N, to the right.
EVALUATE:! A proton has positive charge so the force that an electric field exerts on it is in the same direction as
the field. 21.32. Figure 21.31
!
!
IDENTIFY:! The electric force is F # qE .
SET UP:! The gravity force (weight) has magnitude w # mg and is downward.
EXECUTE:! (a) To balance the weight the electric force must be upward. The electric field is downward,
so for an upward force the charge q of the person must be negative. w # F gives mg # q E and mg (60 kg)(9.80 m/s 2 )
#
# 3.9 C .
E
150 N/C
qq(3.9 C) 2
# 1.4 " 107 N . The repulsive force is immense and this is not a
(b) F # k 2 # (8.99 " 109 N + m 2 /C2 )
r
(100 m)2
feasible means of flight.
EVALUATE:! The net charge of charged objects is typically much less than 1 C.
IDENTIFY:! Eq. (21.3) gives the force on the particle in terms of its charge and the electric field between the
plates. The force is constant and produces a constant acceleration. The motion is similar to projectile motion; use
constant acceleration equations for the horizontal and vertical components of the motion.
(a) SET UP:! The motion is sketched in Figure 21.33a.
q# 21.33. For an electron q # !e.
Figure 21.33a
!
!
!
!
!
F # qE and q negative gives that F and E are in opposite directions, so F is upward. The freebody diagram
for the electron is given in Figure 21.33b.
EXECUTE:! 7F y # ma y eE # ma
Figure 21.33b Solve the kinematics to find the acceleration of the electron: Just misses upper plate says that x ! x0 # 2.00 cm
when y ! y0 # 00.500 cm.
xcomponent
v0 x # v0 # 1.60 " 106 m/s, ax # 0, x ! x0 # 0.0200 m, t # ? x ! x0 # v0 xt 0 1 axt 2
2 Electric Charge and Electric Field! ! 2111 x ! x0
0.0200 m
#
# 1.25 " 10!8 s
v0 x
1.60 " 106 m/s
In this same time t the electron travels 0.0050 m vertically:
ycomponent
t # 1.25 " 10!8 s, v0 y # 0, y ! y0 # 00.0050 m, a y # ?
t# y ! y0 # v0 yt 0 1 a yt 2
2
ay # 2( y ! y0 )
2(0.0050 m)
#
# 6.40 " 1013 m/s 2
t2
(1.25 " 10!8 s) 2 (This analysis is very similar to that used in Chapter 3 for projectile motion, except that here the acceleration is
upward rather than downward.) This acceleration must be produced by the electricfield force: eE # ma
E# ma (9.109 " 10!31 kg)(6.40 " 1013 m/s2 )
#
# 364 N/C
e
1.602 " 10!19 C Note that the acceleration produced by the electric field is much larger than g, the acceleration produced by
gravity, so it is perfectly ok to neglect the gravity force on the elctron in this problem.
eE (1.602 " 10!19 C)(364 N/C)
#
# 3.49 " 1010 m/s 2
(b) a #
mp
1.673 " 10!27 kg
This is much less than the acceleration of the electron in part (a) so the vertical deflection is less and the
proton won’t hit the plates. The proton has the same initial speed, so the proton takes the same time
t # 1.25 " 10!8 s to !
travel horizontally the length of the plates. The force on the proton is downward (in the
same direction as E , since q is positive), so the acceleration is downward and a y # !3.49 " 1010 m/s 2 .
y ! y0 # v0 y t 0 1 a y t 2 # 1 ( !3.49 " 1010 m/s 2 )(1.25 " 10!8 s) 2 # !2.73 " 10!6 m. The displacement is 2.73 " 10!6 m,
2
2 21.34. downward.
(c) EVALUATE:! The displacements are in opposite directions because the electron has negative charge and the
proton has positive charge. The electron and proton have the same magnitude of charge, so the force the electric
field exerts has the same magnitude for each charge. But the proton has a mass larger by a factor of 1836 so its
acceleration and its vertical displacement are smaller by this factor.
IDENTIFY:! Apply Eq.(21.7) to calculate the electric field due to each charge and add the two field vectors to find
the resultant field.
!
ˆ
ˆj
ˆ
SET UP:! For q1 , r = ˆ . For q2 , r = cos4 i + sin 4 ˆ , where 4 is the angle between E2 and the +xaxis.
j
!
q1 ˆ (9.0 " 109 N + m 2 /C2 )(!5.00 " 10!9 C)
# ( !2.813 " 104 N/C) ˆ .
EXECUTE:! (a) E1 #
j#
j
2
4, P0 r12
. 0.0400 m /
!
!
q2
(9.0 " 109 N + m 2 /C2 )(3.00 " 10!9 C)
E2 #
#
# 1.080 " 104 N/C . The angle of E2 , measured from the
2
2
2
4, P0 r2
. 0.0300 m / 0 (0.0400 m) . 21.35. / xaxis, is 180! ! tan !1 4.00 cm # 126.93 Thus
3.00 cm
!
4
ˆ cos126.93 0 ˆ sin126.93) # ( ! 6.485 " 103 N/C) i 0 (8.64 " 103 N/C) ˆ
ˆ
E2 # (1.080 " 10 N/C)(i
j
j
!
!
3
ˆ 0 (!2.813 " 104 N/C 0 8.64 " 103 N/C) ˆ .
(b) The resultant field is E1 + E 2 # (!6.485 " 10 N/C) i
j
!
!
3
ˆ ! (1.95 " 104 N/C) ˆ .
E1 + E 2 # (!6.485 " 10 N/C) i
j
!
!
EVALUATE:! E1 is toward q1 since q1 is negative. E2 is directed away from q2 , since q2 is positive.
IDENTIFY:! Apply constant acceleration equations to the motion of the electron.
SET UP:! Let +x be to the right and let 0 y be downward. The electron moves 2.00 cm to the right and 0.50 cm
downward.
EXECUTE:! Use the horizontal motion to find the time when the electron emerges from the field.
x ! x0 # 0.0200 m, ax # 0, v0 x # 1.60 " 106 m s . x ! x0 # v0 xt 0 1 axt 2 gives t # 1.25 " 10!8 s . Since ax # 0 ,
2 % v 0 vy
vx # 1.60 " 106 m s . y ! y0 # 0.0050 m, v0y # 0, t # 1.25 " 10!8 s . y ! y0 # ' 0 y
)2 &
5
( t gives v y # 8.00 " 10 m s .
* 2
2
Then v # vx 0 v y # 1.79 " 106 m s . EVALUATE:! v y # v0 y 0 a yt gives a y # 6.4 " 1013 m/s 2 . The electric field between the plates is E# ma y
e # (9.11" 10!31 kg)(6.4 " 1013 m/s2 )
# 364 V/m . This is not a very large field.
1.60 " 10!19 C 2112 21.36. Chapter 21 !
!
IDENTIFY:! Use the components of E from Example 21.6 to calculate the magnitude and direction of E . Use
!
!
F = qE to calculate the force on the !2.5 nC charge and use Newton's third law for the force on the
!8.0 nC charge.
!
ˆ
j
SET UP:! From Example 21.6, E # (!11 N/C) i + (14 N/C) ˆ .
% Ey &
2
( # tan !1 (14 11) # 51.83 , so
EXECUTE:! (a) E # Ex2 0 E y # (!11 N/C) 2 0 (14 N/C)2 # 17.8 N/C . tan !1 '
' Ex (
)
*
4 # 1283 counterclockwise from the +xaxis.
!!
(b) (i) F = Eq so F # (17.8 N C)(2.5 " 10!9 C) # 4.45 " 10!8 N , at 52! below the +xaxis. 21.37. (ii) 4.45 " 10!8 N at 128! counterclockwise from the +xaxis.
EVALUATE:! The forces in part (b) are repulsive so they are along the line connecting the two charges and in each
case the force is directed away from the charge that exerts it.
IDENTIFY and SET UP:! The electric force is given by Eq. (21.3). The gravitational force is we # me g . Compare
these forces.
(a) EXECUTE:! we # (9.109 " 10!31 kg)(9.80 m/s2 ) # 8.93 " 10!30 N
In Examples 21.7 and 21.8, E # 1.00 " 104 N/C, so the electric force on the electron has magnitude
FE # q E # eE # (1.602 " 10!19 C)(1.00 " 104 N/C) # 1.602 " 10!15 N.
we 8.93 " 10!30 N
#
# 5.57 " 10!15
FE 1.602 " 10!15 N
The gravitational force is much smaller than the electric force and can be neglected.
(b) mg # q E
m # q E / g # (1.602 " 10!19 C)(1.00 " 104 N/C)/(9.80 m/s2 ) # 1.63 " 10!16 kg
m 1.63 " 10!16 kg
#
# 1.79 " 1014 ; m # 1.79 " 1014 me .
me 9.109 " 10!31 kg
EVALUATE:! m is much larger than me . We found in part (a) that if m # me the gravitational force is much smaller than the electric force. q is the same so the electric force remains the same. To get w large enough to equal FE , 21.38. the mass must be made much larger.
(c) The electric field in the region between the plates is uniform so the force it exerts on the charged object is
independent of where between the plates the object is placed.
IDENTIFY:! Apply constant acceleration equations to the motion of the proton. E # F / q .
SET UP:! A proton has mass mp # 1.67 " 10!27 kg and charge 0e . Let +x be in the direction of motion of the proton.
EXECUTE:! (a) v0 x # 0 . a # E# 2(0.0160 m)(1.67 " 10!27 kg)
# 148 N C.
(1.60 " 10!19 C)(1.50 " 10!6 s)2 (b) vx # v0 x 0 axt # 21.39. eE
1
1 eE 2
t . Solving for E gives
. x ! x0 # v0 xt 0 1 axt 2 gives x ! x0 # axt 2 #
2
mp
2
2 mp eE
t # 2.13 " 104 m s.
mp EVALUATE:! The electric field is directed from the positively charged plate toward the negatively charged plate
and the force on the proton is also in this direction.
ˆ
ˆ
ˆ
IDENTIFY:! Find the angle 4 that r makes with the +xaxis. Then r = (cos4 ) i + (sin 4 ) ˆ .
j
SET UP:! tan 4 # y / x ,
% !1.35 &
ˆ
ˆ
EXECUTE:! (a) tan !1 '
( # ! rad . r = ! j .
0*
2
)
2ˆ
2ˆ
% 12 & ,
ˆ
(b) tan !1 ' ( # rad . r #
i+
j.
2
2
) 12 * 4
% 2.6 &
ˆ
ˆ
ˆ
(c) tan !1 '
( # 1.97 rad # 112.93 . r = !0.39i + 0.92 j (Second quadrant).
) 01.10 *
ˆ
ˆˆ
EVALUATE:! In each case we can verify that r is a unit vector, because r + r = 1 . Electric Charge and Electric Field! ! 2113 21.40. IDENTIFY:! The net force on each charge must be zero.
SET UP:! The force diagram for the !6.50 $ C charge is given in Figure 21.40. FE is the force exerted on the
charge by the uniform electric field. The charge is negative and the field is to the right, so the force exerted by the
field is to the left. Fq is the force exerted by the other point charge. The two charges have opposite signs, so the
force is attractive. Take the +x axis to be to the right, as shown in the figure.
EXECUTE:! (a) F # q E # (6.50 " 10!6 C)(1.85 " 108 N/C) # 1.20 " 103 N
Fq # k 7F x q1q2
r2 # (8.99 " 109 N + m 2 /C 2 ) (6.50 " 10!6 C)(8.75 " 10!6 C)
# 8.18 " 102 N
(0.0250 m) 2 # 0 gives T 0 Fq ! FE # 0 and T # FE ! Fq # 382 N . (b) Now Fq is to the left, since like charges repel.
7 Fx # 0 gives T ! Fq ! FE # 0 and T # FE 0 Fq # 2.02 " 103 N . EVALUATE:! The tension is much larger when both charges have the same sign, so the force one charge exerts on
the other is repulsive. 21.41. Figure 21.40
!
!!
!
!
IDENTIFY and SET UP:! Use E in Eq. (21.3) to calculate F , F # ma to calculate a , and a constant acceleration
equation to calculate the final velocity. Let +x be east.
(a) EXECUTE:! Fx # q E # (1.602 " 10!19 C)(1.50 N/C) = 2.403 " 10!19 N ax # Fx /m # (2.403 " 10!19 N)/(9.109 " 10!31 kg) = +2.638 " 1011 m/s2
v0 x # 04.50 " 105 m/s, ax # 02.638 " 1011 m/s2 , x ! x0 # 0.375 m, vx # ?
2
2
vx # v0 x 0 2ax ( x ! x0 ) gives vx # 6.33 " 105 m/s
!
!
EVALUATE:! E is west and q is negative, so F is east and the electron speeds up.
(b) EXECUTE:! Fx # ! q E # !(1.602 " 10!19 C)(1.50 N/C) = ! 2.403 " 10!19 N ax # Fx / m # (!2.403 " 10!19 N)/(1.673 " 10!27 kg) # !1.436 " 108 m/s2
v0 x # 01.90 " 104 m/s, ax # !1.436 " 108 m/s 2 , x ! x0 # 0.375 m, vx # ? 21.42. 21.43. 2
2
vx # v0 x 0 2ax ( x ! x0 ) gives vx # 1.59 " 104 m/s
!
EVALUATE:! q 5 0 so F is west and the proton slows down.
IDENTIFY:! Coulomb’s law for a single pointcharge gives the electric field.
(a) SET UP:! Coulomb’s law for a pointcharge is E # (1/ 4, P0 )q / r 2 . EXECUTE:! E = (9.00 " 109 N + m2/C2)(1.60 " 10!19 C)/(1.50 " 10!15 m)2 = 6.40 " 1020 N/C
(b) Taking the ratio of the electric fields gives
E/Eplates = (6.40 " 1020 N/C)/(1.00 " 104 N/C) = 6.40 " 1016 times as strong
EVALUATE:! The electric field within the nucleus is huge compared to typical laboratory fields!
IDENTIFY:! Calculate the electric field due to each charge and find the vector sum of these two fields.
SET UP:! At points on the xaxis only the x component of each field is nonzero. The electric field of a point
charge points away from the charge if it is positive and toward it if it is negative.
EXECUTE:! (a) Halfway between the two charges, E # 0.
(b) For  x  1 a , Ex # For x 5 a , Ex # 1
4!P0 1
4!P0 %
q
q&
4q
ax
!
.
'
(#!
(a 0 x) 2 ( a ! x) 2 *
4, P0 ( x 2 ! a 2 ) 2
) %
q
q & 2q x 2 0 a 2
.
0
'
(#
2
(a ! x) 2 * 4, P0 ( x 2 ! a 2 ) 2
) (a 0 x) %
q
q&
2q x 2 0 a 2
0
#!
.
'
2
2(
(a ! x) *
4, P0 ( x 2 ! a 2 ) 2
) (a 0 x)
The graph of Ex versus x is sketched in Figure 21.43. For x 1 ! a , Ex # !1
4!P0 2114 Chapter 21 EVALUATE:! The magnitude of the field approaches infinity at the location of one of the point charges. Figure 21.43
21.44. IDENTIFY:! For a point charge, E # k q
2 !
!
. For the net electric field to be zero, E1 and E2 must have equal r
magnitudes and opposite directions.
!
SET UP:! Let q1 # 00.500 nC and q2 # 08.00 nC. E is toward a negative charge and away from a positive charge.
EXECUTE:! The two charges and the directions of their electric fields in three regions are shown in Figure 21.44.
Only in region II are the two electric fields in opposite directions. Consider a point a distance x from q1 so a 0.500 nC
8.00 nC
#k
. 16 x 2 # (1.20 m ! x )2 . 4 x # 6 (1.20 m ! x )
(1.20 ! x )2
x2
and x # 0.24 m is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from
the 0.500 nC charge and 0.96 m from the 8.00 nC charge.
EVALUATE:! There is only one point along the line connecting the two charges where the net electric field is zero.
This point is closer to the charge that has the smaller magnitude.
distance 1.20 m ! x from q2 . E1 # E2 gives k Figure 21.44
21.45. IDENTIFY:! Eq.(21.7) gives the electric field of each point charge. Use the principle of superposition and add the
electric field vectors. In part (b) use Eq.(21.3) to calculate the force, using the electric field calculated in part (a).
(a) SET UP:! The placement of charges is sketched in Figure 21.45a. Figure 21.45a The electric field of a point charge is directed away from the point charge if the charge is positive and toward the
1q
point charge if the charge is negative. The magnitude of the electric field is E #
, where r is the distance
4, P0 r 2
between the point where!the field is calculated and the point charge.
!
(i) At point a the fields E1 of q1 and E 2 of q2 are directed as shown in Figure 21.45b. Figure 21.45b Electric Charge and Electric Field! ! 2115 EXECUTE:! E1 #
E2 # !9 1 q1
2.00 " 10 C
# (8.988 " 109 N + m 2 /C 2 )
# 449.4 N/C
4, P0 r12
(0.200 m) 2 1 q2
5.00 " 10 !9 C
# (8.988 " 109 N + m 2 /C 2 )
# 124.8 N/C
4, P0 r22
(0.600 m) 2 E1x # 449.4 N/C, E1 y # 0
E2 x # 124.8 N/C, E2 y # 0
Ex # E1x 0 E2 x # 0449.4 N/C 0 124.8 N/C # 0574.2 N/C E y # E1 y 0 E2 y # 0
The resultant field at point a has magnitude 574 N/C and is in the +xdirection.
!
!
(ii) SET UP:! At point b the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45c. Figure 21.45c
EXECUTE:! E1 #
E2 # 1 q1
2.00 " 10!9 C
# (8.988 " 109 N + m 2 /C 2 )
# 12.5 N/C
2
4, P0 r12
.1.20 m / 1 q2
5.00 " 10!9 C
# . 8.988 " 109 N + m 2 /C2 /
# 280.9 N/C
2
2
4, P0 r2
. 0.400 m / E1x # 12.5 N/C, E1 y # 0
E2 x # !280.9 N/C, E2 y # 0
Ex # E1x 0 E2 x # 012.5 N/C ! 280.9 N/C # !268.4 N/C E y # E1 y 0 E2 y # 0
The resultant field at point b has magnitude 268 N/C and is in the ! x direction.
!
!
(iii) SET UP:! At point c the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45d. Figure 21.45d
EXECUTE:! E1 #
E2 # 1 q1
2.00 " 10!9 C
# (8.988 " 109 N + m 2 /C2 )
# 449.4 N/C
2
2
4, P0 r1
. 0.200 m / 1 q2
5.00 " 10!9 C
# . 8.988 " 109 N + m 2 /C 2 /
# 44.9 N/C
2
4, P0 r2
(1.00 m) 2 E1x # !449.4 N/C, E1 y # 0
E2 x # 044.9 N/C, E2 y # 0
Ex # E1x 0 E2 x # !449.4 N/C 0 44.9 N/C # !404.5 N/C
E y # E1 y 0 E2 y # 0
The resultant field at point b has magnitude 404 N/C and is in the ! x direction.
!
!
!
(b) SET UP:! Since we have calculated E at each point the simplest way to get the force is to use F # !eE .
EXECUTE:! (i) F # (1.602 " 10!19 C)(574.2 N/C) # 9.20 " 10!17 N, ! x direction
(ii) F # (1.602 " 10!19 C)(268.4 N/C) # 4.30 " 10!17 N, 0 x direction
(iii) F # (1.602 " 10!19 C)(404.5 N/C) # 6.48 " 10!17 N, 0 x direction
EVALUATE:! The general rule for electric field direction is away from positive charge and toward negative
charge. Whether the field is in the 0 x  or ! x direction depends on where the field point is relative to the charge
that produces the field. In part (a) the field magnitudes were added because the fields were in the same direction
and in (b) and (c) the field magnitudes were subtracted because the two fields were in opposite directions. In
part (b) we could have used Coulomb's law to find the forces on the electron due to the two charges and then
added these force vectors, but using the resultant electric field is much easier. 2116 Chapter 21 21.46. IDENTIFY:! Apply Eq.(21.7) to calculate the field due to each charge and then require that the vector sum of the
two fields to be zero.
SET UP:! The field of each charge is directed toward the charge if it is negative and away from the charge if it is
positive .
EXECUTE:! The point where the two fields cancel each other will have to be closer to the negative charge,
because it is smaller. Also, it can’t be between the two charges, since the two fields would then act in the same
direction. We could use Coulomb’s law to calculate the actual values, but a simpler way is to note that the 8.00 nC
charge is twice as large as the ! 4.00 nC charge. The zero point will therefore have to be a factor of 2 farther
from the 8.00 nC charge for the two fields to have equal magnitude. Calling x the distance from the –4.00 nC 21.47. charge: 1.20 0 x # 2 x and x # 2.90 m .
EVALUATE:! This point is 4.10 m from the 8.00 nC charge. The two fields at this point are in opposite directions
and have equal magnitudes.
q
IDENTIFY:! E # k 2 . The net field is the vector sum of the fields due to each charge.
r
SET UP:! The electric field of a negative charge is directed toward the charge. Label the charges q1, q2 and q3, as
shown in Figure 21.47a. This figure also shows additional distances and angles. The electric fields at point P are
shown in Figure 21.47b. This figure also shows the xy coordinates we will use and the x and y components of the
!!
!
fields E1 , E2 and E3 .
EXECUTE:! E1 # E3 # (8.99 " 109 N + m 2 /C 2 ) E2 # (8.99 " 109 N + m 2 /C2 ) 5.00 " 10!6 C
# 4.49 " 106 N/C
(0.100 m) 2 2.00 " 10!6 C
# 4.99 " 106 N/C
(0.0600 m)2 E y # E1 y 0 E2 y 0 E3 y # 0 and Ex # E1x 0 E2 x 0 E3 x # E2 0 2 E1 cos53.1! # 1.04 " 107 N/C
E # 1.04 " 107 N/C , toward the !2.00 $ C charge.
EVALUATE:! The xcomponents of the fields of all three charges are in the same direction. Figure 21.47
21.48. IDENTIFY:! A positive and negative charge, of equal magnitude q, are on the xaxis, a distance a from the origin.
Apply Eq.(21.7) to calculate the field due to each charge and then calculate the vector sum of these fields.
!
SET UP:! E due to a point charge is directed away from the charge if it is positive and directed toward the charge
if it is negative.
1 2q
EXECUTE:! (a) Halfway between the charges, both fields are in the ! x direction and E #
, in the
4, P0 a 2
! xdirection .
1 % !q
q&
1 % !q
q&
(b) E x #
!
0
'
( for  x  1 a . E x #
'
( for x 5 a .
4, P0 ) (a 0 x) 2 (a ! x) 2 *
4, P0 ) (a 0 x) 2 ( a ! x ) 2 *
Ex # 1 % !q
q&
!
'
( for x 1 ! a . Ex is graphed in Figure 21.48.
2
4, P0 ) (a 0 x) (a ! x) 2 * Electric Charge and Electric Field! ! 2117 EVALUATE:! At points on the x axis and between the charges, Ex is in the ! x direction because the fields from
both charges are in this direction. For x 1 !a and x 5 0 a , the fields from the two charges are in opposite
directions and the field from the closer charge is larger in magnitude. Figure 21.48
21.49. IDENTIFY:! The electric field of a positive charge is directed radially outward from the charge and has magnitude
1q
E#
. The resultant electric field is the vector sum of the fields of the individual charges.
4, P0 r 2
SET UP:! The placement of the charges is shown in Figure 21.49a. Figure 21.49a
EXECUTE:! (a) The directions of the two fields are shown in Figure 21.49b.
E1 # E2 # 1q
with r # 0.150 m.
4, P0 r 2 E # E2 ! E1 # 0; Ex # 0, E y # 0
Figure 21.49b
(b) The two fields have the directions shown in Figure 21.49c. E # E1 0 E2 , in the 0 x direction
Figure 21.49c
E1 # 1 q1
6.00 " 10!9 C
# (8.988 " 109 N + m 2 /C 2 )
# 2396.8 N/C
4, P0 r12
(0.150 m) 2 E2 # 1 q2
6.00 " 10!9 C
# (8.988 " 109 N + m 2 /C2 )
# 266.3 N/C
2
4, P0 r2
(0.450 m)2 E # E1 0 E2 # 2396.8 N/C 0 266.3 N/C # 2660 N/C; Ex # 02260 N/C, E y # 0 2118 Chapter 21 (c) The two fields have the directions shown in Figure 21.49d. 0.400 m
# 0.800
0.500 m
0.300 m
cos4 #
# 0.600
0.500 m sin 4 # Figure 21.49d
E1 # 1 q1
4, P0 r12 E1 # (8.988 " 109 N + m 2 /C2 )
E2 # 6.00 " 10!9 C
# 337.1 N/C
(0.400 m)2 1 q2
4, P0 r22 E2 # (8.988 " 109 N + m 2 /C 2 ) 6.00 " 10!9 C
# 215.7 N/C
(0.500 m) 2 E1x # 0, E1 y # ! E1 # !337.1 N/C
E2x # 0 E2 cos4 # 0 (215.7 N/C)(0.600) # 0129.4 N/C
E2y # ! E2 sin 4 # !(215.7 N/C)(0.800) # !172.6 N/C
Ex # E1x 0 E2 x # 0129 N/C
Ey # E1 y 0 E2 y # !337.1 N/C ! 172.6 N/C # !510 N/C
2
E # Ex2 0 E y # (129 N/C)2 0 (!510 N/C)2 # 526 N/C
!
E and its components are shown in Figure 21.49e. tan 2 # Ey
Ex !510 N/C
# !3.953
0129 N/C
2 # 2843C, counterclockwise from 0 x axis
tan 2 # Figure 21.49e
(d) The two fields have the directions shown in Figure 21.49f. sin 4 # Figure 21.49f 0.200 m
# 0.800
0.250 m Electric Charge and Electric Field! ! 2119 The components of the two fields are shown in Figure 21.49g.
E1 # E2 # 1q
4, P0 r 2 E1 # (8.988 "109 N + m 2 /C2 ) 6.00 "10!9 C
(0.250 m)2 E1 # E2 # 862.8 N/C
Figure 21.49g E1x # ! E1 cos4 , E2 x # 0 E2 cos4
Ex # E1x 0 E2 x # 0
E1 y # 0 E1 sin 4 , E2 y # 0 E2 sin 4 21.50. E y # E1 y 0 E2 y # 2 E1 y # 2 E1 sin 4 # 2 . 862.8 N/C / . 0.800 / # 1380 N/C
E # 1380 N/C, in the 0 y direction.
EVALUATE:! Point a is symmetrically placed between identical charges, so symmetry tells us the electric field
must be zero. Point b is to the right of both charges and both electric fields are in the +xdirection and the resultant
field is in this direction. At point c both fields have a downward component and the field of q2 has a component to
!
the right, so the net E is in the 4th quadrant. At point d both fields have an upward component but by symmetry
they have equal and opposite xcomponents so the net field is in the +ydirection. We can use this sort of reasoning
to deduce the general direction of the net field before doing any calculations.
IDENTIFY:! Apply Eq.(21.7) to calculate the field due to each charge and then calculate the vector sum of those fields.
SET UP:! The fields due to q1 and to q2 are sketched in Figure 21.50.
!
1 (6.00 " 10!9 C) ˆ
ˆ
EXECUTE:! E2 #
( !i ) # !150i N/C .
4, P0
(0.6 m) 2
!
%
&
1
1
1
ˆ
ˆ
E1 #
j
j
(4.00 " 10!9 C) '
(0.600) i 0
(0.800) ˆ ( # (21.6 i + 28.8 ˆ ) N C .
4, P0
(1.00 m)2
(1.00 m) 2
)
*
!!
!
ˆ
E = E + E # (!128.4 N/C) i + (28.8 N/C) ˆ . E # (128.4 N/C) 2 0 (28.8 N/C)2 # 131.6 N/C at
j
1 2 % 28.8 &
4 # tan '
( # 12.63 above the ! x axis and therefore 196.2! counterclockwise from the +x axis.
) 128.4 *
!
!
EVALUATE:! E1 is directed toward q1 because q1 is negative and E 2 is directed away from q2 because q2 is positive.
!1 Figure 21.50 21.51. !
!
IDENTIFY:! The resultant electric field is the vector sum of the field E1 of q1 and E 2 of q2 .
SET UP:! The placement of the charges is shown in Figure 21.51a. Figure 21.51a 2120 Chapter 21 EXECUTE:! (a) The directions of the two fields are shown in Figure 21.51b.
1 q1
E1 # E2 #
4, P0 r12 E1 # (8.988 " 109 N + m 2 /C2 ) 6.00 " 10!9 C
(0.150 m)2 E1 # E2 # 2397 N/C Figure 21.51b E1x # !2397 N/C, E1 y # 0 E2x # !2397 N/C, E2 y # 0
E x # E1x 0 E2 x # 2(!2397 N/C) # !4790 N/C
E y # E1 y 0 E2 y # 0 The resultant electric field at point a in the sketch has magnitude 4790 N/C and is in the ! x direction.
(b) The directions of the two fields are shown in Figure 21.51c. Figure 21.51c 1 q1
6.00 " 10!9 C
# (8.988 " 109 N + m 2 /C2 )
# 2397 N/C
2
4, P0 r1
(0.150 m)2
1 q2
6.00 " 10!9 C
# (8.988 " 109 N + m 2 /C2 )
# 266 N/C
E2 #
4, P0 r22
(0.450 m)2
E1x # 02397 N/C, E1 y # 0 E2 x # !266 N/C, E2 y # 0
E1 # E x # E1x 0 E2 x # 02397 N/C ! 266 N/C # 02130 N/C
E y # E1 y 0 E2 y # 0 The resultant electric field at point b in the sketch has magnitude 2130 N/C and is in the 0 x direction.
(c) The placement of the charges is shown in Figure 21.51d.
0.300 m
# 0.600
0.500 m
0.400 m
# 0.800
cos4 #
0.500 m sin 4 # Figure 21.51d The directions of the two fields are shown in Figure 21.51e. E1 # 1 q1
4, P0 r12 E1 # (8.988 " 109 N + m 2 /C2 ) 6.00 " 10!9 C
(0.400 m) 2 E1 # 337.0 N/C
1 q2
E2 #
4, P0 r22 E2 # (8.988 " 109 N + m 2 /C2 )
E2 # 215.7 N/C Figure 21.51e E1x # 0, E1 y # ! E1 # !337.0 N/C
E2 x # ! E2 sin 4 # !(215.7 N/C)(0.600) # !129.4 N/C
E2 y # 0 E2 cos4 # 0(215.7 N/C)(0.800) # 0172.6 N/C 6.00 " 10!9 C
(0.500 m)2 Electric Charge and Electric Field! ! 2121 Ex # E1x 0 E2 x # !129 N/C
E y # E1 y 0 E2 y # !337.0 N/C 0 172.6 N/C # !164 N/C
2
E # Ex2 0 E y # 209 N/C
!
The field E and its components are shown in Figure 21.51f. tan 2 # Ey
Ex !164 N/C
# 01.271
!129 N/C
2 # 2323, counterclockwise from 0 x axis
tan 2 # Figure 21.51f
(d) The placement of the charges is shown in Figure 21.51g. 0.200 m
# 0.800
0.250 m
0.150 m
# 0.600
cos4 #
0.250 m sin 4 # Figure 21.51g The directions of the two fields are shown in Figure 21.51h. E1 # E2 # 1q
4, P0 r 2 E1 # (8.988 " 109 N + m 2 /C2 ) 6.00 " 10!9 C
(0.250 m)2 E1 # 862.8 N/C
E 2 # E1 # 862.8 N/C Figure 21.51h E1x # ! E1 cos4 , E2 x # ! E2 cos4 Ex # E1x 0 E2 x # !2 . 862.8 N/C /. 0.600 / # !1040 N/C
E1 y # 0 E1 sin 4 , E2 y # ! E2 sin 4 E y # E1 y 0 E2 y # 0
E # 1040 N/C, in the ! x direction.
EVALUATE: The electric field produced by a charge is toward a negative charge and away from a positive charge.
As in Exercise 21.45, we can use this rule to deduce the direction of the resultant field at each point before doing
any calculations.
21.52. IDENTIFY:! For a long straight wire, E #
SET UP:! 8
2, P0 r . 1
# 4.49 " 109 N + m 2 /C2 .
2, P0 EXECUTE:! r # 1.5 " 10!10 C m
# 1.08 m
2, P0 (2.50 N C) EVALUATE:! For a point charge, E is proportional to 1/r 2 . For a long straight line of charge, E is proportional to 1/r . 2122 Chapter 21 21.53. IDENTIFY:! Apply Eq.(21.10) for the finite line of charge and E #
for the infinite line of charge.
2, P0
!
SET UP:! For the infinite line of positive charge, E is in the +x direction.
EXECUTE:! (a) For a line of charge of length 2a centered at the origin and lying along the yaxis, the electric field
!
1
8
ˆ
is given by Eq.(21.10): E =
i.
2, P0 x x 2 a 2 0 1
!
8ˆ
(b) For an infinite line of charge: E =
i . Graphs of electric field versus position for both distributions of
2, P0 x
charge are shown in Figure 21.53.
EVALUATE:! For small x, close to the line of charge, the field due to the finite line approaches that of the infinite
line of charge. As x increases, the field due to the infinite line falls off more slowly and is larger than the field of
the finite line. 8 Figure 21.53
21.54. (a) IDENTIFY:! The field is caused by a finite uniformly charged wire.
SET UP:! The field for such a wire a distance x from its midpoint is
%1&
1
8
8
E#
# 2'
.
(
2
2, P0 x ( x / a ) 0 1
4, P0 * x ( x / a) 2 0 1
) .18.0 "10 9 EXECUTE:! E = N + m 2 / C2 / .175 " 10!9 C/m /
2 = 3.03 " 104 N/C, directed upward. % 6.00 cm &
(0.0600 m) '
( 01
) 4.25 cm *
(b) IDENTIFY:! The field is caused by a uniformly charged circular wire.
SET UP:! The field for such a wire a distance x from its midpoint is E # 1
Qx
. We first find the radius
4, P0 ( x 2 0 a 2 )3/ 2 of the circle using 2!r = l.
EXECUTE:! Solving for r gives r = l/2! = (8.50 cm)/2! = 1.353 cm
The charge on this circle is Q = 8l = (175 nC/m)(0.0850 m) = 14.88 nC
The electric field is E# 21.55. .9.00 " 109 N + m2 /C2 / .14.88 " 10!9 C/m / . 0.0600 m /
1
Qx
=
3/ 2
3/ 2
4, P0 . x 2 0 a 2 /
9(0.0600 m) 2 0 (0.01353 m) 2 :
;
< E = 3.45 " 104 N/C, upward.
EVALUATE:! In both cases, the fields are of the same order of magnitude, but the values are different because the
charge has been bent into different shapes.
!
!
IDENTIFY:! For a ring of charge, the electric field is given by Eq. (21.8). F = qE . In part (b) use Newton's third
law to relate the force on the ring to the force exerted by the ring.
SET UP:! Q # 0.125 " 10!9 C, a # 0.025 m and x # 0.400 m .
!
1
Qx
ˆ
ˆ
EXECUTE:! (a) E =
i = (7.0 N/C) i .
2
4, P0 ( x 0 a 2 )3/ 2
!
!
!
ˆ
ˆ
= ! F # !qE # !(!2.50 " 10!6 C) (7.0 N/C) i # (1.75 " 10!5 N) i
(b) F
on ring on q EVALUATE:! Charges q and Q have opposite sign, so the force that q exerts on the ring is attractive. Electric Charge and Electric Field! ! 2123 21.56. IDENTIFY:! We must use the appropriate electric field formula: a uniform disk in (a), a ring in (b) because all the
charge is along the rim of the disk, and a pointcharge in (c).
(a) SET UP:! First find the surface charge density (Q/A), then use the formula for the field due to a disk of charge,
:
=9
1
Ex #
>1 !
?.
2
2P0 >
( R / x) 0 1 ?
;
<
EXECUTE:! The surface charge density is = # Q
Q
6.50 " 10!9 C
# 2#
= 1.324 " 10!5 C/m2.
, (0.0125 m)2
A ,r The electric field is
Ex # =9 :
1
1.324 " 10!5 C/m 2
1
9
:
1!
>1 !
?=
?
2
2
2P0 >
2(8.85 " 10!12 C2 /N + m 2 ) >
( R / x) 0 1 ?
% 1.25 cm &
;
<
>
01?
'
(
>
?
) 2.00 cm *
;
< Ex = 1.14 " 105 N/C, toward the center of the disk.
(b) SET UP:! For a ring of charge, the field is E # 1
Qx
.
2
4, P0 . x 0 a 2 /3 / 2 EXECUTE:! Substituting into the electric field formula gives E# (9.00 " 109 N + m 2 /C2 )(6.50 " 10!9 C)(0.0200 m)
1
Qx
=
3/ 2
3/ 2
4, P0 . x 2 0 a 2 /
9(0.0200 m) 2 0 (0.0125 m) 2 :
;
< E = 8.92 " 104 N/C, toward the center of the disk.
(c) SET UP:! For a point charge, E # .1/ 4, P0 / q / r 2 . 21.57. EXECUTE:! E = (9.00 " 109 N + m2/C2)(6.50 " 10!9 C)/(0.0200 m)2 = 1.46 " 105 N/C
(d) EVALUATE:! With the ring, more of the charge is farther from P than with the disk. Also with the ring the
component of the electric field parallel to the plane of the ring is greater than with the disk, and this component
cancels. With the point charge in (c), all the field vectors add with no cancellation, and all the charge is closer to
point P than in the other two cases.
IDENTIFY:! By superposition we can add the electric fields from two parallel sheets of charge.
SET UP:! The field due to each sheet of charge has magnitude = / 2P0 and is directed toward a sheet of negative
charge and away from a sheet of positive charge.
(a) The two fields are in opposite directions and E # 0.
(b) The two fields are in opposite directions and E # 0.
(c) The fields of both sheets are downward and E # 2 21.58. = # = , directed downward.
2P0 P0
EVALUATE:! The field produced by an infinite sheet of charge is uniform, independent of distance from the sheet.
IDENTIFY and SET UP:! The electric field produced by an infinite sheet of charge with charge density = has
magnitude E # = . The field is directed toward the sheet if it has negative charge and is away from the sheet if it
2P0
has positive charge.
EXECUTE:! (a) The field lines are sketched in Figure 21.58a.
(b) The field lines are sketched in Figure 21.58b.
EVALUATE:! The spacing of the field lines indicates the strength of the field. In part (a) the two fields add
between the sheets and subtract in the regions to the left of A and to the right of B. In part (b) the opposite is true. Figure 21.58 2124 Chapter 21 21.59. IDENTIFY:! The force on the particle at any point is always tangent to the electric field line at that point.
SET UP:! The instantaneous velocity determines the path of the particle.
EXECUTE:! In Fig.21.29a the field lines are straight lines so the force is always in a straight line and velocity and
acceleration are always in the same direction. The particle moves in a straight line along a field line, with
increasing speed. In Fig.21.29b the field lines are curved. As the particle moves its velocity and acceleration are
not in the same direction and the trajectory does not follow a field line.
EVALUATE:! In twodimensional motion the velocity is always tangent to the trajectory but the velocity is not
always in the direction of the net force on the particle.
IDENTIFY:! The field appears like that of a point charge a long way from the disk and an infinite sheet close to the
disk’s center. The field is symmetrical on the right and left.
SET UP:! For a positive point charge, E is proportional to 1/r 2 and is directed radially outward. For an infinite
sheet of positive charge, the field is uniform and is directed away from the sheet.
EXECUTE:! The field is sketched in Figure 21.60.
EVALUATE:! Near the disk the field lines are parallel and equally spaced, which corresponds to a uniform field.
Far from the disk the field lines are getting farther apart, corresponding to the 1/r 2 dependence for a point charge. 21.60. Figure 21.60
21.61. IDENTIFY:! Use symmetry to deduce the nature of the field lines.
(a) SET UP:! The only distinguishable direction is toward the line or away from the line, so the electric field lines
are perpendicular to the line of charge, as shown in Figure 21.61a. Figure 21.61a
(b) EXECUTE and EVALUATE:! The magnitude of the electric field is inversely proportional to the spacing of the
field lines. Consider a circle of radius r with the line of charge passing through the center, as shown in
Figure 21.61b. Figure 21.61b 21.62. The spacing of field lines is the same all around the circle, and in the direction perpendicular to the plane of the
circle the lines are equally spaced, so E depends only on the distance r. The number of field lines passing out
through the circle is independent of the radius of the circle, so the spacing of the field lines is proportional to the
reciprocal of the circumference 2, r of the circle. Hence E is proportional to 1/r.
IDENTIFY:! Field lines are directed away from a positive charge and toward a negative charge. The density of
field lines is proportional to the magnitude of the electric field.
SET UP:! The field lines represent the resultant field at each point, the net field that is the vector sum of the fields
due to each of the three charges.
EXECUTE:! (a)! Since field lines pass from positive charges and toward negative charges, we can deduce that the
top charge is positive, middle is negative, and bottom is positive.
(b) The electric field is the smallest on the horizontal line through the middle charge, at two positions on either
side where the field lines are least dense. Here the ycomponents of the field are cancelled between the positive
charges and the negative charge cancels the xcomponent of the field from the two positive charges.
EVALUATE:! Far from all three charges the field is the same as the field of a point charge equal to the algebraic
sum of the three charges. Electric Charge and Electric Field! ! 2125 21.63. (a) IDENTIFY and SET UP:! Use Eq.(21.14) to relate the dipole moment to the charge magnitude and the
separation d of the two charges. The direction is from the negative charge toward the positive charge.
!
EXECUTE:! p # qd # (4.5 " 10!9 C)(3.1 " 10!3 m) # 1.4 " 10!11 C + m; The direction of p is from q1 toward q2 .
(b) IDENTIFY and SET UP:! Use Eq. (21.15) to relate the magnitudes of the torque and field.
EXECUTE:! @ # pE sin A , with A as defined in Figure 21.63, so E#
E# @
p sin A
7.2 " 10!9 N + m
# 860 N/C
(1.4 " 10!11 C + m)sin 36.93 Figure 21.63
EVALUATE:! Eq.(21.15) gives the torque about an axis through the center of the dipole. But the forces on the two
charges form a couple (Problem 11.53) and the torque is the same for any axis parallel to this one. The force on
each charge is q E and the maximum moment arm for an axis at the center is d /2, so the maximum torque is 2( q E )(d/2) # 1.2 " 10!8 N + m. The torque for the orientation of the dipole in the problem is less than this 21.64. maximum.
(a) IDENTIFY:! The potential energy is given by Eq.(21.17).
!
!!
!
SET UP:! U (A ) # ! p + E # ! pE cosA , where A is the angle between p and E .
EXECUTE:! parallel: A # 0 and U . 03 / # ! pE perpendicular: A # 903 and U . 903 / # 0 BU # U . 903 / ! U . 03 / # pE # . 5.0 " 10!30 C + m / .1.6 " 106 N/C / # 8.0 " 10!24 J.
(b) 21.65. 3
2 kT # BU so T # 2 . 8.0 " 10!24 J /
2BU
#
# 0.39 K
3k
3 .1.381 " 10!23 J/K / EVALUATE:! Only at very low temperatures are the dipoles of the molecules aligned by a field of this strength. A
much larger field would be required for alignment at room temperature.
IDENTIFY:! Follow the procedure specified in part (a) of the problem.
SET UP:! Use that y 55 d .
EXECUTE:! (a) ! 1
1
( y 0 d 2) 2 ! ( y ! d 2)2
2 yd
!
#
#2
. This gives
2
2
2
2
2
( y ! d 2) ( y 0 d 2)
( y ! d 4)
( y ! d 2 4)2 q
2 yd
qd
y
p
#
. Since y 2 55 d 2 / 4 , E y C
.
4, P0 ( y 2 ! d 2 4) 2 2, P0 ( y 2 ! d 2 4) 2
2, P0 y 3
!
!
(b) For points on the ! y axis , E! is in the +y direction and E0 is in the ! y direction. The field point is closer to
Ey # ! q , so the net field is upward. A similar derivation gives E y C p
. E y has the same magnitude and direction
2, P0 y 3 at points where y 55 d as where y 11 !d . 21.66. EVALUATE:! E falls off like 1/ r 3 for a dipole, which is faster than the 1/ r 2 for a point charge. The total charge of
the dipole is zero.
!
!
IDENTIFY:! Calculate the electric field due to the dipole and then apply F = qE .
SET UP:! From Example 21.15, Edipole ( x) #
EXECUTE:! Edipole # 21.67. p
.
2, P0 x3 6.17 " 10!30 C + m
# 4.11 " 106 N C . The electric force is F # qE #
2, P0 (3.0 " 10!9 m)3 (1.60 " 10!19 C)(4.11" 106 N/C) # 6.58 " 10!13 N and is toward the water molecule (negative xdirection).
!
!
!
EVALUATE:! Edipole is in the direction of p ¸ so is in the +x direction. The charge q of the ion is negative, so F is
!
directed opposite to E and is therefore in the ! x direction.
IDENTIFY:! Like charges repel and unlike charges attract. The force increases as the distance between the charges
decreases.
SET UP:! The forces on the dipole that is between the slanted dipoles are sketched in Figure 21.67a. 2126 Chapter 21 EXECUTE:! The forces are attractive because the + and ! charges of the two dipoles are closest. The forces are
toward the slanted dipoles so have a net upward component. In Figure 21.67b, adjacent dipoles charges of opposite
sign are closer than charges of the same sign so the attractive forces are larger than the repulsive forces and the
dipoles attract.
EVALUATE:! Each dipole has zero net charge, but because of the charge separation there is a nonzero force
between dipoles. Figure 21.67
21.68. IDENTIFY:! Find the vector sum of the fields due to each charge in the dipole.
SET UP:! A point on the xaxis with coordinate x is a distance r # (d / 2) 2 0 x 2 from each charge.
EXECUTE:! (a) The magnitude of the field the due to each charge is E # &
1q
q%
1
#
'
(,
4, P0 r 2 4, P0 ) (d 2) 2 0 x 2 * where d is the distance between the two charges. The xcomponents of the forces due to the two charges
are equal and oppositely directed and so cancel each other. The two fields have equal ycomponents,
&
d2
2q %
1
so E # 2 E y #
'
( sin 4 , where 4 is the angle below the xaxis for both fields. sin 4 #
4, P0 ) (d 2) 2 0 x 2 *
(d 2) 2 0 x 2
% 2q & %
&%
1
d2
'
and Edipole # '
('
2
2(
2
2
) 4, P0 * ) (d 2) 0 x * ' ( d 2) 0 x
) 21.69. &
qd
(#
. The field is the ! y direction.
( 4, P0 ((d 2) 2 0 x 2 )3 2
*
qd
(b) At large x, x 2 55 ( d 2) 2 , so the expression in part (a) reduces to the approximation Edipole C
.
4, P0 x 3
qd
EVALUATE:! Example 21.15 shows that at points on the +y axis far from the dipole, Edipole C
. The
2, P0 y 3
expression in part (b) for points on the x axis has a similar form.
!!!
IDENTIFY:! The torque on a dipole in an electric field is given by @ = p × E .
!
!
SET UP:! @ # pE sin A , where A is the angle between the direction of p and the direction of E .
!
!
EXECUTE:! (a) The torque is zero when p is aligned either in the same direction as E or in the opposite
direction, as shown in Figure 21.69a.
!
!
(b) The stable orientation is when p is aligned in the same direction as E . In this case a small rotation of the
!
!
!
dipole results in a torque directed so as to bring p back into alignment with E . When p is directed opposite to
!
!
!
E , a small displacement results in a torque that takes p farther from alignment with E .
(c) Field lines for Edipole in the stable orientation are sketched in Figure 21.69b.
EVALUATE:! The field of the dipole is directed from the + charge toward the ! charge. Figure 21.69
21.70. IDENTIFY:! The plates produce a uniform electric field in the space between them. This field exerts torque on a
dipole and gives it potential energy.
SET UP:! The electric field between the plates is given by E # = / P0 , and the dipole moment is p = ed. The
!!
potential energy of the dipole due to the field is U # ! p + E # ! pE cos A , and the torque the field exerts on it is @ = pE sin A. Electric Charge and Electric Field! ! 2127
!!
EXECUTE:! (a) The potential energy, U # ! p + E # ! pE cos A , is a maximum when A = 180°. The field between the plates is E # = / P0 , giving Umax = (1.60 " 10–19 C)(220 " 10–9 m)(125 " 10–6 C/m2)/(8.85 " 10–12 C2/N + m2) = 4.97 " 10–19 J
The orientation is parallel to the electric field (perpendicular to the plates) with the positive charge of the dipole
toward the positive plate.
(b) The torque, @ = pE sin A, is a maximum when A = 90° or 270°. In this case @ max # pE # p= / P0 # ed= / P0 @ max # .1.60 " 10!19 C / . 220 " 10!9 m / .125 " 10!6 C/mD / . 8.85 " 10!12 C2 / N + m 2 /
@ max # 4.97 " 10!19 N + m 21.71. The dipole is oriented perpendicular to the electric field (parallel to the plates).
(c) F = 0.
EVALUATE:! When the potential energy is a maximum, the torque is zero. In both cases, the net force on the
dipole is zero because the forces on the charges are equal but opposite (which would not be true in a nonuniform
electric field).
(a) IDENTIFY:! Use Coulomb's law to calculate each force and then add them as vectors to obtain the net force.
Torque is force times moment arm.
SET UP:! The two forces on each charge in the dipole are shown in Figure 21.71a. sin 4 # 1.50 / 2.00 so 4 # 48.63
Opposite charges attract and like charges repel.
Fx # F1x 0 F2 x # 0 Figure 21.71a
EXECUTE:! F1 # k qq(5.00 "10!6 C)(10.0 " 10!6 C)
#k
# 1.124 " 103 N
2
r
(0.0200 m) 2 F1 y # ! F1 sin 4 # !842.6 N
F2 y # !842.6 N so Fy # F1 y 0 F2 y # !1680 N (in the direction from the 05.00$ C charge toward the !5.00$ C
charge).
EVALUATE: The xcomponents cancel and the ycomponents add.
(b) SET UP:! Refer to Figure 21.71b. The ycomponents have zero moment arm and
therefore zero torque.
F1x and F2 x both produce clockwise torques. Figure 21.71b
EXECUTE:! F1 X # F1 cos4 # 743.1 N @ # 2( F1x )(0.0150 m) # 22.3 N + m, clockwise
EVALUATE:! The electric field produced by the ! 10.00$ C charge is not uniform so Eq. (21.15) does not apply. 2128 Chapter 21 21.72. IDENTIFY:! Apply F # k qqfor each pair of charges and find the vector sum of the forces that q1 and q2 exert on q3 .
r2
SET UP:! Like charges repel and unlike charges attract. The three charges and the forces on q3 are shown in
Figure 21.72. Figure 21.72
q1q3
(5.00 " 10!9 C)(6.00 " 10!9 C)
# (8.99 " 109 N + m 2 /C2 )
# 1.079 " 10!4 C .
2
(0.0500 m) 2
r1
4 # 36.9! . F1x # 0 F1 cos4 # 8.63 " 10!5 N . F1 y # 0 F1 sin 4 # 6.48 " 10!5 N . EXECUTE: (a) F1 # k q2 q3
(2.00 " 10!9 C)(6.00 " 10!9 C)
# (8.99 " 109 N + m 2 /C2 )
# 1.20 " 10!4 C .
2
r2
(0.0300 m) 2
F2 x # 0 , F2 y # ! F2 # !1.20 " 10!4 N . Fx # F1x 0 F2 x # 8.63 " 10!5 N .
F2 # k Fy # F1 y 0 F2 y # 6.48 " 10!5 N 0 (!1.20 " 10!4 N) # !5.52 " 10!5 N . (b) F # Fx2 0 Fy2 # 1.02 " 10!4 N . tan A # 21.73. Fy
Fx # 0.640 . A # 32.6! , below the 0 x axis. EVALUATE: The individual forces on q3 are computed from Coulomb’s law and then added as vectors, using
components.
(a) IDENTIFY: Use Coulomb's law to calculate the force exerted by each Q on q and add these forces as vectors to
find the resultant force. Make the approximation x 55 a and compare the net force to F # ! kx to deduce k and then f # (1/ 2, ) k / m .
SET UP:! The placement of the charges is shown in Figure 21.73a. Figure 21.73a
EXECUTE:! Find the net force on q. Fx # F1x 0 F2 x and F1x # 0 F1 , F2 x # ! F2
Figure 21.73b
F1 # 1
qQ
1
qQ
, F2 #
4, P0 . a 0 x /2
4, P0 . a ! x /2 Fx # F1 ! F2 # Fx # qQ 9 1
1:
!
4, P0 > . a 0 x /2 . a ! x /2 ?
;
< !2
!2
qQ 9 %
x&
% x& :
> 0 '1 0 ( ! '1 ! ( ?
4, P0 a 2 > ) a *
) a* ?
;
< Electric Charge and Electric Field! ! 2129 Since x 11 a we can use the binomial expansion for (1 ! x/a) !2 and (1 0 x/a ) !2 and keep only the first two terms: (1 0 z ) n C 1 0 nz. For (1 ! x/a) !2 , z # ! x/a and n # !2 so (1 ! x/a) !2 C 1 0 2 x/a. For (1 0 x/a )!2 , z # 0 x/a and
n # !2 so (1 0 x/a) !2 C 1 ! 2 x/a. Then F C % qQ &
qQ 9% 2 x & % 2 x & :
x. For simple harmonic
'1 ! ( ! '1 0 ( ? # ! '
3(
4, P0 a 2 >)
a* )
a *<
;
) , P0 a * motion F # !kx and the frequency of oscillation is f # .1/ 2, / k / m . The net force here is of this form, with qQ
.
, P0 ma 3
(b) The forces and their components are shown in Figure 21.73c. k # qQ / , P0 a 3 . Thus f # 1
2, Figure 21.73c The xcomponents of the forces exerted by the two charges cancel, the ycomponents add, and the net force is in
the +ydirection when y > 0 and in the ! y direction when y < 0. The charge moves away from the origin on the
yaxis and never returns.
EVALUATE:! The directions of the forces and of the net force depend on where q is located relative to the other
two charges. In part (a), F # 0 at x # 0 and when the charge q is displaced in the +x or –xdirection the net force is
a restoring force, directed to return q to x # 0. The charge oscillates back and forth, similar to a mass on a spring.
21.74. IDENTIFY:! Apply 7F x # 0 and 7F y # 0 to one of the spheres. SET UP:! The freebody diagram is sketched in Figure 21.74. Fe is the repulsive Coulomb force between the
spheres. For small 4 , sin 4 C tan 4 .
EXECUTE:! 7 Fx # T sin 4 ! Fe # 0 and 7 Fy # T cos4 ! mg # 0 . So mg sin 4
kq 2
# Fe # 2 . But tan 4 C sin 4 # d ,
cos4
2L
d 1/3 % q2L &
2kq 2 L
and d # '
(.
mg
) 2, P0 mg *
EVALUATE:! d increases when q increases. so d 3 # Figure 21.74
21.75. IDENTIFY:! Use Coulomb's law for the force that one sphere exerts on the other and apply the 1st condition of
equilibrium to one of the spheres.
(a) SET UP:! The placement of the spheres is sketched in Figure 21.75a. Figure 21.75a 2130 Chapter 21 The freebody diagrams for each sphere are given in Figure 21.75b. Figure 21.75b
Fc is the repulsive Coulomb force exerted by one sphere on the other.
(b) EXECUTE:! From either force diagram in part (a): 7 Fy # ma y
mg
T cos 25.03 ! mg # 0 and T #
cos 25.03
7 Fx # max T sin 25.03 ! Fc # 0 and Fc # T sin 25.03 Use the first equation to eliminate T in the second: Fc # . mg/ cos 25.03 / . sin 25.03 / # mg tan 25.03
Fc # 1 q1q2
1 q2
1
q2
#
#
2
2
4, P0 r
4, P0 r
4, P0 [2(1.20 m)sin 25.03]2 Combine this with Fc # mg tan 25.03 and get mg tan 25.03 #
q # . 2.40 m / sin 25.03 q # . 2.40 m / sin 25.03 1
q2
4, P0 [2(1.20 m)sin 25.03]2 mg tan 25.03
.1/ 4, P0 / .15.0 "10 !3 kg /. 9.80 m/s 2 / tan 25.03 # 2.80 " 10!6 C
8.988 " 109 N + m 2 /C2
(c) The separation between the two spheres is given by 2L sin4 . q # 2.80$ C as found in part (b).
Fc # .1/ 4, P0 / q 2 / . 2 L sin 4 / and Fc # mg tan4 . Thus .1/ 4, P0 / q 2 / . 2 L sin 4 / # mg tan 4 .
2 . sin 4 / 2 . 2.80 "10 C /
1
q2
tan 4 #
# . 8.988 " 109 N + m 2 / C2 /
# 0.3328.
2
2
4, P0 4 L mg
4 . 0.600 m / .15.0 " 10!3 kg /. 9.80 m/s 2 /
!6 2 2 Solve this equation by trial and error. This will go quicker if we can make a good estimate of the value of 4 that
solves the equation. For 4 small, tan 4 C sin 4 . With this approximation the equation becomes sin 3 4 # 0.3328
and sin 4 # 0.6930, so 4 # 43.93. Now refine this guess: 4
45.03
40.03
39.63
39.53
39.43 21.76. sin 2 4 tan 4
0.5000
0.3467
0.3361
0.3335
0.3309 so 4 # 39.53 EVALUATE:! The expression in part (c) says 4 E 0 as L E F and 4 E 903 as L E 0. When L is decreased from
the value in part (a), 4 increases.
IDENTIFY:! Apply 7 Fx # 0 and 7 Fy # 0 to each sphere.
SET UP:! (a) Free body diagrams are given in Figure 21.76. Fe is the repulsive electric force that one sphere
exerts on the other.
kq q
EXECUTE:! (b) T # mg cos 203 # 0.0834 N , so Fe # T sin 203 # 0.0285 N # 12 2 . (Note:
r1 r1 # 2(0.500 m)sin 203 # 0.342 m.)
(c) From part (b), q1q2 # 3.71 " 10!13 C2 . Electric Charge and Electric Field! ! 2131 (d) The charges on the spheres are made equal by connecting them with a wire, but we still have
q 0q
Q2
, where Q # 1 2 . But the separation r2 is known:
Fe # mg tan 4 # 0.0453 N # 1
2
4, P0 r22
q1 0 q2
# 4, P0 Fe r22 # 1.12 " 10 !6 C. This equation, along
2
with that from part (c), gives us two equations in q1 and q2 : q1 0 q2 # 2.24 " 10 !6 C and q1q2 # 3.71 " 10 !13 C 2 . r2 # 2(0.500 m)sin 303 # 0.500 m. Hence: Q # By elimination, substitution and after solving the resulting quadratic equation, we find: q1 # 2.06 " 10 !6 C and
q2 # 1.80 " 10 !7 C .
EVALUATE:! After the spheres are connected by the wire, the charge on sphere 1 decreases and the charge on
sphere 2 increases. The product of the charges on the sphere increases and the thread makes a larger angle with the
vertical. Figure 21.76
21.77. IDENTIFY and SET UP:! Use Avogadro's number to find the number of Na + and Cl ! ions and the total positive
!
!
and negative charge. Use Coulomb's law to calculate the electric force and F # ma to calculate the acceleration.
(a) EXECUTE:! The number of Na + ions in 0.100 mol of NaCl is N # nN A . The charge of one ion is +e, so the total charge is q1 # nN A e # (0.100 mol)(6.022 " 1023 ions/mol)(1.602 " 10!19 C/ion) # 9.647 " 103 C
There are the same number of Cl! ions and each has charge !e, so q2 # !9.647 " 103 C. F# 1 q1q2
(9.647 " 103 C)2
# (8.988 " 109 N + m2 /C2 )
# 2.09 " 1021 N
2
4, P0 r
(0.0200 m)2 (b) a # F/m. Need the mass of 0.100 mol of Cl ! ions. For Cl, M # 35.453 " 10 !3 kg/mol, so
F
2.09 " 1021 N
m # . 0.100 mol / 35.453 " 10!3 kg/mol # 35.45 " 10!4 kg. Then a # #
# 5.90 " 1023 m/s 2 .
m 35.45 " 10!4 kg
(c) EVALUATE:! Is is not reasonable to have such a huge force. The net charges of objects are rarely larger than 1 $C;
a charge of 104 C is immense. A small amount of material contains huge amounts of positive and negative charges.
IDENTIFY:! For the acceleration (and hence the force) on Q to be upward, as indicated, the forces due to q1 and q2
must have equal strengths, so q1 and q2 must have equal magnitudes. Furthermore, for the force to be upward, q1
must be positive and q2 must be negative.
SET UP:! Since we know the acceleration of Q, Newton’s second law gives us the magnitude of the force on it.
We can then add the force components using F # FQq1 cos4 0 FQq2 cos4 # 2 FQq1 cos4 . The electrical force on Q is . 21.78. / 1 Qq1
(for q1) and likewise for q2.
4, P0 r 2
EXECUTE:! First find the net force: F = ma = (0.00500 kg)(324 m/s2) = 1.62 N. Now add the force
components, calling 4 the angle between the line connecting q1 and q2 and the line connecting q1 and Q.
F
1.62 N
#
F # FQq1 cos4 0 FQq2 cos4 # 2 FQq1 cos4 and FQq1 #
= 1.08 N. Now find the charges by
% 2.25 cm &
2cos4
2'
(
) 3.00 cm *
solving for q1 in Coulomb’s law and use the fact that q1 and q2 have equal magnitudes but opposite signs.
r 2 FQq1
(0.0300 m)2 (1.08 N)
1 Qq1
#
FQq1 #
and q1 #
# 6.17 " 10!8 C.
2
1
4, P0 r
. 9.00 "109 N + m2 /C2 /.1.75 "10!6 C /
Q
4, P0 given by Coulomb’s law, FQq1 # q2 # ! q1 # !6.17 " 10!8 C. 2132 21.79. Chapter 21 EVALUATE:! Simple reasoning allows us first to conclude that q1 and q2 must have equal magnitudes but opposite
signs, which makes the equations much easier to set up than if we had tried to solve the problem in the general
case. As Q accelerates and hence moves upward, the magnitude of the acceleration vector will change in a
complicated way.
IDENTIFY:! Use Coulomb's law to calculate the forces between pairs of charges and sum these forces as vectors to
find the net charge.
(a) SET UP:! The forces are sketched in Figure 21.79a. !!
!!
EXECUTE:! F1 0 F3 = 0, so the net force is F # F2 . F# 1 q(3q )
6q 2
#
, away from the vacant corner.
4, P0 ( L/ 2) 2 4, P0 L2 Figure 21.79a
(b) SET UP:! The forces are sketched in Figure 21.79b. EXECUTE:! F2 #
F1 # F3 # 1 q . 3q /
3q 2
#
4, P0 2 L 2 4, P0 . 2 L2 / . / 1 q . 3q /
3q
#
4, P0 L2
4, P0 L2
2 The vector sum of F1 and F3 is F13 # F12 0 F32 . Figure 21.79b F13 # 2 F1 # !
3 2q 2 !
; F13 and F2 are in the same direction.
2
4, P0 L 3q 2 %
1&
' 2 0 ( , and is directed toward the center of the square.
4, P0 L2 )
2*
EVALUATE:! By symmetry the net force is along the diagonal of the square. The net force is only slightly larger
when the !3q charge is at the center. Here it is closer to the charge at point 2 but the other two forces cancel.
IDENTIFY:! Use Eq.(21.7) for the electric field produced by each point charge. Apply the principle of
superposition and add the fields as vectors to find the net field.
(a) SET UP:! The fields due to each charge are shown in Figure 21.80a.
F # F13 0 F2 # 21.80. cos4 # Figure 21.80a x
x 0 a2
2 Electric Charge and Electric Field! ! 2133 EXECUTE:! The components of the fields are given in Figure 21.80b. E1 # E2 # E3 # 1% q &
'
(
4, P0 ) a 2 0 x 2 * 1 % 2q &
'(
4, P0 ) x 2 * Figure 21.80b E1 y # ! E1 sin 4 , E2 y # 0 E2 sin 4 so E y # E1 y 0 E2 y # 0.
&
1 % q &%
x
( , E3 x # ! E3
'2
2 ('
2
2
4, P0 ) a 0 x * ) x 0 a *
% 1 % q &%
&&
x
2q
E x # E1x 0 E2 x 0 E3 x # 2 '
!
'2
2 ('
' 4, P ) a 0 x * x 2 0 a 2 ( ( 4, P x 2
(
0
0
)
**
)
2q % 1
x
2q %
1
&
&
Ex # !
1!
#!
' x2 ! 2
2'
2 3/ 2 (
2
2 3/ 2 (
4, P0 '
. a 0 x / ( 4, P0 x ' .1 0 a / x / (
)
*
)
*
2q %
1
&
Thus E #
1!
, in the ! xdirection.
4, P0 x 2 ' .1 0 a 2 / x 2 /3 / 2 (
'
(
)
*
E1x # E2 x # 0 E1 cos4 # (b) x 55 a implies a 2 / x 2 11 1 and .1 0 a 2 / x 2 / Thus E C 21.81. 21.82. 21.83. !3 / 2 C 1 ! 3a 2 / 2 x 2 . 2q % % 3a 2 & & 3qa 2
.
'1 ! '1 !
(( #
4, P0 x 2 ' ) 2 x 2 * ( 4, P0 x 4
)
* EVALUATE:! E " 1/ x 4 . For a point charge E " 1/ x 2 and for a dipole E " 1/ x 3 . The total charge is zero so !
at
large distances the electric field should decrease faster with distance than for a point charge. By symmetry E must
lie along the xaxis, which is the result we found in part (a).
IDENTIFY:! The small bags of protons behave like pointmasses and pointcharges since they are extremely far apart.
SET UP:! For pointparticles, we use Newton’s formula for universal gravitation (F = Gm1m2/r2) and Coulomb’s
law. The number of protons is the mass of protons in the bag divided by the mass of a single proton.
EXECUTE:! (a) (0.0010 kg) /(1.67 " 10!27 kg) # 6.0 " 1023 protons
(b) Using Coulomb’s law, where the separation is twice is the radius of the earth, we have
Felectrical = (9.00 " 109 N + m2/C2)(6.0 " 1023 " 1.60 " 10–19 C)2/(2 " 6.38 " 106 m)2 = 5.1 " 105 N
Fgrav = (6.67 " 10–11 N + m2/kg2)(0.0010 kg)2/(2 " 6.38 " 106 m)2 = 4.1 " 10–31 N
(c) EVALUATE:! The electrical force ("200,000 lb!) is certainly large enough to feel, but the gravitational force
clearly is not since it is about 1036 times weaker.
IDENTIFY:! We can treat the protons as pointcharges and use Coulomb’s law.
SET UP:! (a) Coulomb’s law is F = (1/ 4, P0 ) q1q2 / r 2 .
EXECUTE:! F = (9.00 " 109 N + m2/C2)(1.60 " 10–19 C)2/(2.0 " 1015 m) = 58 N = 13 lb, which is certainly large
enough to feel.
(b) EVALUATE:! Something must be holding the nucleus together by opposing this enormous repulsion. This is
the strong nuclear force.
IDENTIFY:! Estimate the number of protons in the textbook and from this find the net charge of the textbook.
Apply Coulomb's law to find the force and use Fnet # ma to find the acceleration.
SET UP:! With the mass of the book about 1.0 kg, most of which is protons and neutrons, we find that the number
of protons is 1 (1.0 kg) (1.67 " 10!27 kg) # 3.0 " 10 26 .
2
EXECUTE:! (a) The charge difference present if the electron’s charge was 99.999% of the proton’s is
Bq # (3.0 " 10 26 )(0.00001)(1.6 " 10 !19 C) # 480 C . (b) F # k (Bq ) 2 r 2 # k (480 C) 2 (5.0 m) 2 # 8.3 " 1013 N , and is repulsive.
2 a # F m # (8.3 " 1013 N) (1 kg) # 8.3 " 1013 m s .
EXECUTE:! (c) Even the slightest charge imbalance in matter would lead to explosive repulsion! 2134 Chapter 21 21.84. IDENTIFY:! The electric field exerts equal and opposite forces on the two balls, causing them to swing away from
each other. When the balls hang stationary, they are in equilibrium so the forces on them (electrical, gravitational,
and tension in the strings) must balance.
SET UP:! (a) The force on the left ball is in the direction of the electric field, so it must be positive, while the force
on the right ball is opposite to the electric field, so it must be negative.
(b) Balancing horizontal and vertical forces gives qE = T sin 4/2 and mg = T cos 4/2.
EXECUTE:! Solving for the angle 4 gives: 4 = 2 arctan(qE/mg).
(c) As E E #, 4 E 2 arctan(#) = 2 (!/2) = ! = 180°
EVALUATE:! If the field were large enough, the gravitational force would not be important, so the strings would
be horizontal.
IDENTIFY and SET UP:! Use the density of copper to calculate the number of moles and then the number of atoms.
Calculate the net charge and then use Coulomb's law to calculate the force.
3
%4
&
%4 &
EXECUTE:! (a) m # GV # G ' , r 3 ( # . 8.9 " 103 kg/m3 / ' , ( .1.00 " 10!3 m / # 3.728 " 10!5 kg
3
3*
)
*
) 21.85. n # m / M # . 3.728 " 10!5 kg / / . 63.546 " 10!3 kg/mol / # 5.867 " 10!4 mol
N # nN A # 3.5 " 1020 atoms (b) N e # . 29 / . 3.5 " 1020 / # 1.015 " 1022 electrons and protons qnet # eN e ! . 0.99900 / eN e # . 0.100 " 10!2 / .1.602 " 10!19 C / .1.015 " 1022 / # 1.6 C
q2
.1.6 C / # 2.3 " 1010 N
F #k 2 #k
2
r
.1.00 m /
2 21.86. EVALUATE:! The amount of positive and negative charge in even small objects is immense. If the charge of an
electron and a proton weren't exactly equal, objects would have large net charges.
IDENTIFY:! Apply constant acceleration equations to a drop to find the acceleration. Then use F # ma to find the
force and F # q E to find q .
SET UP:! Let D # 2.0 cm be the horizontal distance the drop travels and d # 0.30 mm be its vertical
displacement. Let +x be horizontal and in the direction from the nozzle toward the paper and let +y be vertical, in
the direction of the deflection of the drop. ax # 0 and a y # a .
!6
m)3 &
3 % 4, (15.0 " 10
!11
EXECUTE:! First, the mass of the drop: m # GV # (1000 kg m ) '
( # 1.41" 10 kg . Next, the
3
)
*
1
2d 2(3.00 " 10!4 m)
2
# 600 m s .
time of flight: t # D v # (0.020 m) (20 m/s) # 0.00100 s . d # at 2 . a # 2 #
2
t
(0.001 s) 2
2 (1.41 " 10!11 kg)(600 m s )
# 1.06 " 10!13 C .
8.00 " 104 N C
EVALUATE:! Since q is positive the vertical deflection is in the direction of the electric field.
IDENTIFY:! Eq. (21.3) gives the force exerted by the electric field. This force is constant since the electric field is
uniform and gives the proton a constant acceleration. Apply the constant acceleration equations for the x and
ycomponents of the motion, just as for projectile motion.
(a) SET UP:! The electric field is upward so the electric force on the positively charged proton is upward and has
!
!
magnitude F = eE. Use coordinates where positive y is downward. Then applying 7 F # ma to the proton gives
Then a # F m # qE m gives q # ma E # 21.87. that ax # 0 and a y # !eE / m. In these coordinates the initial velocity has components vx # 0 v0 cos 2 and
v y # 0v0 sin 2 , as shown in Figure 21.87a. Figure 21.87a Electric Charge and Electric Field! ! 2135 EXECUTE:! Finding hmax : At y # hmax the ycomponent of the velocity is zero. v y # 0, v0 y # v0 sin 2 , a y # !eE / m, y ! y0 # hmax # ?
2
2
v y # v0 y 0 2a y . y ! y0 / y ! y0 # hmax # 2
2
v y ! v0 y 2a y 2
2
!v0 sin 2 2 mv0 sin 2 2
#
2 . !eE / m /
2eE (b) Use the vertical motion to find the time t: y ! y0 # 0, v0 y # v0 sin 2 , a y # !eE / m, t # ?
y ! y0 # v0 yt 0 1 a y t 2
2 With y ! y0 # 0 this gives t # ! 2v0 y
ay #! 2 . v0 sin 2 / 2mv0 sin 2
#
!eE / m
eE Then use the xcomponent motion to find d: ax # 0, v0 x # v0 cos2 , t # 2mv0 sin 2 / eE , x ! x0 # d # ?
2
2
% 2mv0 sin 2 & mv0 2sin 2 cos 2 mv0 sin 22
#
x ! x0 # v0 xt 0 1 axt 2 gives d # v0 cos 2 '
(#
2
eE
eE
eE
)
*
(c) The trajectory of the proton is sketched in Figure 21.87b. Figure 21.87b
9. 4.00 " 105 m/s / . sin 30.03 / : .1.673 " 10!27 kg /
<
#;
# 0.418 m
2 .1.602 " 10!19 C / . 500 N/C /
2 (d) Use the expression in part (a): hmax .1.673 "10 kg /. 4.00 " 10 m/s / sin 60.03 # 2.89 m
Use the expression in part (b): d #
.1.602 "10 C / . 500 N/C/
!27 5 2 !19 EVALUATE:! In part (a), a y # !eE / m # !4.8 " 1010 m/s 2 . This is much larger in magnitude than g, the acceleration
due to gravity, so it is reasonable to ignore gravity. The motion is just like projectile motion, except that the
acceleration is upward rather than downward and has a much different magnitude. hmax and d increase when 2 or v0 increase and decrease when E increases.
21.88. IDENTIFY:! Ex # E1x 0 E2 x . Use Eq.(21.7) for the electric field due to each point charge.
!
SET UP:! E is directed away from positive charges and toward negative charges.
EXECUTE:! (a) Ex # 050.0 N/C . E1x # 1 q1
4.00 " 10!9 C
# (8.99 " 109 N + m 2 /C 2 )
# 099.9 N/C .
2
4, P0 r1
(0.60 m) 2 Ex # E1x 0 E2 x , so E2 x # Ex ! E1x # 050.0 N/C ! 99.9 N/C # !49.9 N/C . Since E2 x is negative, q2 must be
negative. q2 # E2 x r22
(49.9 N/C)(1.20 m)2
#
# 7.99 " 10!9 C . q2 # !7.99 " 10!9 C
(1/ 4, P0 ) 8.99 " 109 N + m 2 /C2 (b) Ex # !50.0 N/C . E1x # 099.9 N/C , as in part (a). E2 x # Ex ! E1x # !149.9 N/C . q2 is negative.
q2 # E2 x r22
(149.9 N/C)(1.20 m)2
#
# 2.40 " 10!8 C . q2 # !2.40 " 10!8 C .
(1/ 4, P0 ) 8.99 " 109 N + m 2 /C2 EVALUATE:! q2 would be positive if E2 x were positive. 2136 Chapter 21 21.89. IDENTIFY:! Divide the charge distribution into infinitesimal segments of length dx . Calculate Exand Eu due to a
segment and integrate to find the total field.
SET UP:! The charge dQ of a segment of length dx is dQ # (Q / a )dx . The distance between a segment at x and
the charge q is a 0 r ! x . (1 ! y ) !1 C 1 0 y when y 11 1 .
a
1
dQ
1
Qdx
1 Q%1
1&
so Ex #
2
H a(a 0 r ! x)2 # 4, P0 a ' r ! a 0 r ( .
P0 ( a 0 r ! x)
4,
4, P0 0
)
*
1 Q% 1
1&
a 0 r # x , so Ex #
! ( . Ey # 0 .
'
4, P0 a ) x ! a x *
!
!
1 qQ % 1
1& ˆ
(b) F = qE =
! (i.
'
4, P0 a ) x ! a x *
kqQ
kqQ
kqQ
1 qQ
((1 ! a x) !1 ! 1) #
(1 0 a x 0 +++ ! 1) C 2 C
EVALUATE:! (c) For x 55 a, F #
. (Note that for
ax
ax
x
4, P0 r 2
x 55 a , r # x ! a C x .) The charge distribution looks like a point charge from far away, so the force takes the form
of the force between a pair of point charges.
IDENTIFY:! Use Eq. (21.7) to calculate the electric field due to a small slice of the line of charge and integrate as
!
in Example 21.11. Use Eq. (21.3) to calculate F .
SET UP:! The electric field due to an infinitesimal segment of the line of charge is sketched in Figure 21.90. EXECUTE:! (a) dEx # 21.90. sin 4 #
cos4 # y
x 0 y2
2 x
x 0 y2
2 Figure 21.90
Slice the charge distribution up into small pieces of length dy. The charge dQ in each slice is dQ # Q (dy/a ). The
!
electric field this produces at a distance x along the xaxis is dE. Calculate the components of dE and then integrate
over the charge distribution to find the components of the total field.
1 % dQ &
Q % dy &
EXECUTE:! dE #
'
(#
'
(
4, P0 ) x 2 0 y 2 * 4, P0 a ) x 2 0 y 2 * Qx %
dy
&
4, P0 a ' . x 2 0 y 2 /3 / 2 (
'
(
)
*
Q%
ydy
&
dE y # !dE sin 4 # !
'2
2 3/ 2 (
4, P0 a ' . x 0 y / (
)
*
Qx a
dy
Qx 9 1
#
Ex # H dEx # !
4, P0 a H 0 ( x 2 0 y 2 )3/ 2 4, P0 a > x 2
;
dEx # dE cos4 # E y # H dEy # ! !
!
(b) F # q0 E
Fx # !qEx # a Q
:
#
2
2?
4, P0 x
x 0 y <0
y a x 0 a2 a
Q
ydy
Q9
1
Q %1
1
:
&
#!
!
#!
!
2
2?
2
2(
4, P0 a H 0 ( x 2 0 y 2 )3/ 2
4, P0 a >
4, P0 a ' x
x 0 y <0
x 0a *
)
; ! qQ
1
qQ % 1
1
&
!
; Fy # !qE y #
(
4, P0 x x 2 0 a 2
4, P0 a ' x
x2 0 a2 *
)
!1 / 2 1 % a2 &
1%
a2 & 1 a2
# '1 0 2 (
# '1 ! 2 ( # ! 3
x*
x ) 2x * x 2x
x2 0 a2 x )
2
qQ
qQ % 1 1 a & qQa
Fx C !
, Fy C
'!0
(#
4, P0 x 2
4, P0 a ) x x 2 x3 * 8, P0 x3 (c) For x >> a, 1
2 1 Electric Charge and Electric Field! ! 2137 EVALUATE:! For x 55 a, Fy 11 Fx and F C Fx #
21.91. !
qQ
and F is in the –xdirection. For x >> a the charge
2
4, P0 x distribution Q acts like a point charge.
IDENTIFY:! Apply Eq.(21.9) from Example 21.11.
!
SET UP:! a # 2.50 cm . Replace Q by Q . Since Q is negative, E is toward the line of charge and !
Q
1
ˆ
E=!
i.
4, P0 x x 2 0 a 2
!
Q
1
1
9.00 " 10 !9 C
ˆ
ˆ
ˆ
EXECUTE:! E = !
i #!
i # ( !7850 N/C) i .
4, P0 x x 2 0 a 2
4, P0 (0.100 m) (0.100 m) 2 0 (0.025 m) 2 21.92. (b) The electric field is less than that at the same distance from a point charge (8100 N/C). For large x,
&
1
1%
1 Q%
a2
a2 &
( x 0 a ) !1/ 2 # (1 0 a 2 / x 2 ) !1/ 2 C '1 ! 2 ( . E x EF #
1 ! 2 0 +++ ( . The first correction term to the point
2'
4, P0 x ) 2 x
x
x ) 2x *
*
charge result is negative.
(c) For a 1% difference, we need the first term in the expansion beyond the point charge result to be less than
a2
0.010:
C 0.010 I x C a 1 (2(0.010)) # 0.025 1 0.020 I x C 0.177 m .
2x2
EVALUATE:! At x # 10.0 cm (part b), the exact result for the line of charge is 3.1% smaller than for a point
charge. It is sensible, therefore, that the difference is 1.0% at a somewhat larger distance, 17.7 cm.
!
!
kQ 2
IDENTIFY:! The electrical force has magnitude F # 2 and is attractive. Apply 7 F # ma to the earth.
r v2
2, r
. The period T is
. The mass of the earth is mE # 5.97 " 10 24 kg , the
r
v
orbit radius of the earth is 1.50 " 1011 m and its orbital period is 3.146 " 107 s .
SET UP:! For a circular orbit, a # EXECUTE:! F # ma gives
Q# 21.93. kQ
v2
4, 2 r 2
, so
# mE . v 2 #
r2
r
T2 (5.97 " 1024 kg)(4)(, 2 )(1.50 " 1011 m)3
mE 4, 2 r 3
#
# 2.99 " 1017 C .
(8.99 " 109 N + m 2 /C 2 )(3.146 " 107 s) 2
kT 2 EVALUATE:! A very large net charge would be required.
IDENTIFY:! Apply Eq.(21.11).
SET UP:! = # Q / A # Q / , R 2 . (1 0 y 2 ) !1/ 2 C 1 ! y 2 / 2 , when y 2 11 1 .
EXECUTE:! (a) E # = 91 ! . R 2 / x 2 0 1/
;
2P0 > !1 2 :.
?
< &
4.00 pC , (0.025 m) 2 9 % (0.025 m)2
>1 ! '
E#
0 1(
2
2P0
> ) (0.200 m)
*
; (b) For x 55 R E # 21.94. =
2P0 [1 ! (1 ! R 2 2 x 2 0 +++)] C !1 2 :
? # 0.89 N/C , in the +x direction.
?
< = R2
2P0 2 x 2 # =, R 2
Q
.
#
4, P0 x 2 4, P0 x 2 (c) The electric field of (a) is less than that of the point charge (0.90 N/C) since the first correction term to the point
charge result is negative.
(0.90 ! 0.89)
# 0.01 # 1% . For x # 0.100 m ,
(d) For x # 0.200 m , the percent difference is
0.89
(3.60 ! 3.43)
# 0.047 C 5%.
Edisk # 3.43 N C and Epoint # 3.60 N C , so the percent difference is
3.60
EVALUATE:! The field of a disk becomes closer to the field of a point charge as the distance from the disk
increases. At x # 10.0 cm , R / x # 25% and the percent difference between the field of the disk and the field of a
point charge is 5%.
IDENTIFY:! Apply the procedure specified in the problem.
SET UP:! H x2
x1 x1 f ( x)dx # !H f ( x) dx .
x2 2138 Chapter 21 EXECUTE:! (a) For f ( x ) # f (! x ) , replace ! x with y. This gives
(b) For g ( x) # ! g (! x ) , H a
!a H a
!a H a
!a f ( x)dx # H 0
!a a !a 0 0 f ( x)dx 0 H f ( x)dx # H a a 0 H a
0 f ( x)dx . Now a 0 f ( ! x) d ( ! x ) 0 0 f ( x)dx # H f ( y )dy 0 H f ( x)dx # 2 H f ( x)dx .
0 a !a !a 0 0 g ( x)dx # H g ( x)dx 0 H g ( x)dx # ! H a a 0 0 a !a a ! g (! x)(! d (! x)) 0 H g ( x) dx . Now replace 0 ! x with y. This gives H g ( x) dx # ! H g ( y )dy 0 H g ( x)dx # 0.
(c) The integrand in E y for Example 21.11 is odd, so E y =0. 21.95. 21.96. 21.97. EVALUATE:! In Example 21.11, Ey # 0 because for each infinitesimal segment in the upper half of the line of charge,
there is a corresponding infinitesimal segment in the bottom half of the line that has Ey in the opposite direction.
!
!
IDENTIFY:! Find the resultant electric field due to the two point charges. Then use F # qE to calculate the force
on the point charge.
SET UP:! Use the results of Problems 21.90 and 21.89.
EXECUTE:! (a) The ycomponents of the electric field cancel, and the xcomponent from both charges, as given in
!
1 !2Q % 1
1
1 !2Qq % 1
1
&
&ˆ
Problem 21.90, is Ex #
' y ! ( y 2 0 a 2 )1/ 2 ( . Therefore, F = 4, P a ' y ! ( y 2 0 a 2 )1/ 2 ( i . If y 55 a
4, P0 a )
*
)
*
0
!
1 !2Qq
1 Qqa ˆ
ˆ
(1 ! (1 ! a 2 /2 y 2 0 +++))i = !
FC
i.
4, P0 ay
4, P0 y 3
(b) If the point charge is now on the xaxis the two halves of the charge distribution provide different forces,
!
!
1 Qq % 1
1&ˆ
! (i
though still along the xaxis, as given in Problem 21.89: F0 = qE0 #
'
4, P0 a ) x ! a x *
!
!
!!
!
1 Qq % 1
1 &ˆ
1 Qq % 1
2
1 &ˆ
!0
and F! = q E! = !
'!
( i . Therefore, F = F0 0 F! #
'
( i . For x 55 a ,
4, P0 a ) x x 0 a *
4, P0 a ) x ! a x x 0 a *
!
&
% a a2
&& ˆ
1 Qq % % a a 2
1 2Qqa ˆ
FC
i.
' '1 0 0 2 0 . . . ( ! 2 0 '1 ! 0 2 ! . . .( ( i #
(
4, P0 ax ' )
4, P0 x3
xx
xx
*
)
**
)
EVALUATE:! If the charge distributed along the xaxis were all positive or all negative, the force would be
proportional to 1/ y 2 in part (a) and to 1/ x 2 in part (b), when y or x is very large.
!
IDENTIFY:! Divide the semicircle into infinitesimal segments. Find the electric field dE due to each segment and
integrate over the semicircle to find the total electric field.
SET UP:! The electric fields along the xdirection from the left and right halves of the semicircle cancel. The
remaining ycomponent points in the negative ydirection. The charge per unit length of the semicircle is
Q
k 8 dl k 8 d4
and dE #
.
8#
#
a2
a
,a
k 8 sin 4 d4
2k 8 , 2
2k 8
2k 8 2kQ
,2
EXECUTE:! dE y # dE sin 4 #
. Therefore, E y #
H0 sin4 d4 # a [!cos4 ]0 # a # , a 2 , in
a
a
the ! y direction .
EVALUATE:! For a full circle of charge the electric field at the center would be zero. For a quartercircle of
charge, in the first quadrant, the electric field at the center of curvature would have nonzero x and y components.
The calculation for the semicircle is particularly simple, because all the charge is the same distance from point P.
IDENTIFY:! Divide the charge distribution into small segments, use the point charge formula for the electric field
due to each small segment and integrate over the charge distribution to find the x and y components of the total field.
SET UP:! Consider the small segment shown in Figure 21.97a.
EXECUTE:! A small segment that
subtends angle d4 has length a d4 and
% ad4 &
2Q
contains charge dQ # ' 1
d4 .
(Q #
,a *
,
)2
( 1 , a is the total length of the charge
2
distribution.)
Figure 21.97a Electric Charge and Electric Field! ! 2139 The charge is negative, so the field at the origin is directed toward the small segment. The small segment is located at
angle 4 as shown in the sketch. The electric field due to dQ is shown in Figure 21.97b, along with its components.
dE # 1 dQ
4, P0 a 2 dE # Q
d4
2, 2P0 a 2 Figure 21.97b
dEx # dE cos4 # . Q / 2, 2P0 a 2 / cos4 d4
Ex # H dEx # . , /2
Q
Q
cos4 d4 # 2 2 sin 4
2
2 H0
2, P0 a
2, P0 a , /2
0 2 2 0 dE y # dE sin 4 # . Q / 2, 2P0 a 2 / sin 4 d4
E y # H dE y # / # 2, Q a
P . , /2
Q
Q
sin 4 d4 # 2 2 ! cos4
2, 2P0 a 2 H 0
2, P0 a , /2
0 / # 2, Q a
P
2 2 0 EVALUATE:! Note that Ex # E y , as expected from symmetry.
21.98. IDENTIFY:! Apply 7F x # 0 and 7F y # 0 to the sphere, with x horizontal and y vertical. !
SET UP:! The freebody diagram for the sphere is given in Figure 21.98. The electric field E of the sheet is directed away from the sheet and has magnitude E #
EXECUTE:! 7F y # 0 gives T cos 2 # mg and T # Combining these two equations we have = 2P0 (Eq.21.12). mg
.
cos 2 7F x # 0 gives T sin 2 # q=
q=
and T #
.
2P0
2P0 sin 2 % q= &
mg
q=
q=
#
and tan 2 #
. Therefore, 2 # arctan '
(.
cos2 2P0 sin 2
2P0 mg
) 2P0 mg * EVALUATE:! The electric field of the sheet, and hence the force it exerts on the sphere, is independent of the
distance of the sphere from the sheet. Figure 21.98
21.99. IDENTIFY:! Each wire produces an electric field at P due to a finite wire. These fields add by vector addition.
1
Q
SET UP:! Each field has magnitude
. The field due to the negative wire points to the left, while
4, P0 x x 2 0 a 2 the field due to the positive wire points downward, making the two fields perpendicular to each other and of equal
1
Q
magnitude. The net field is the vector sum of these two, which is Enet = 2E1 cos 45° = 2
cos 453 . In
4, P0 x x 2 0 a 2
part (b), the electrical force on an electron at P is eE.
1
Q
EXECUTE:! (a) The net field is Enet = 2
cos 453 .
4, P0 x x 2 0 a 2
Enet = 2 . 9.00 " 109 N + m 2 /C2 / . 2.50 " 10!6 C / cos 453
(0.600 m) (0.600 m)2 0 (0.600 m)2 = 6.25 " 104 N/C. The direction is 225° counterclockwise from an axis pointing to the right through the positive wire. 2140 Chapter 21 (b) F = eE = (1.60 " 1019 C)(6.25 " 104 N/C) = 1.00 " 10–14 N, opposite to the direction of the electric field, since
the electron has negative charge.
EVALUATE:! Since the electric fields due to the two wires have equal magnitudes and are perpendicular to each
other, we only have to calculate one of them in the solution.
21.100. IDENTIFY:! Each sheet produces an electric field that is independent of the distance from the sheet. The net field
is the vector sum of the two fields.
SET UP:! The formula for each field is E # = /2P0 , and the net field is the vector sum of these,
Enet # =B
2P0 6 =A
2P0 # =B 6= A
2P0 , where we use the + or – sign depending on whether the fields are in the same or opposite directions and = B and = A are the magnitudes of the surface charges.
EXECUTE:! (a) The two fields oppose and the field of B is stronger than that of A, so
Enet = =B
2P0 ! =A
2P0 # = B != A
2P0 = 11.6 µC/m 2 ! 9.50 µC/m 2
= 1.19 " 105 N/C, to the right.
2 . 8.85 " 10!12 C 2 /N + m 2 / (b) The fields are now in the same direction, so their magnitudes add.
Enet = (11.6 µC/m2 + 9.50 µC/m2)/2 P0 = 1.19 " 106 N/C, to the right (c) The fields add but now point to the left, so Enet = 1.19 " 106 N/C, to the left.
EVALUATE:! We can simplify the calculations by sketching the fields and doing an algebraic solution first.
21.101. IDENTIFY:! Each sheet produces an electric field that is independent of the distance from the sheet. The net field
is the vector sum of the two fields.
SET UP:! The formula for each field is E # = / 2P0 , and the net field is the vector sum of these,
Enet # =B
2P0 6 =A
2P0 # =B 6= A
2P0 , where we use the + or – sign depending on whether the fields are in the same or opposite directions and = B and = A are the magnitudes of the surface charges.
EXECUTE:! (a) The fields add and point to the left, giving Enet = 1.19 " 106 N/C.
(b) The fields oppose and point to the left, so Enet = 1.19 " 105 N/C.
(c) The fields oppose but now point to the right, giving Enet = 1.19 " 105 N/C.
EVALUATE:! We can simplify the calculations by sketching the fields and doing an algebraic solution first.
21.102. IDENTIFY:! The sheets produce an electric field in the region between them which is the vector sum of the fields
from the two sheets.
SET UP:! The force on the negative oil droplet must be upward to balance gravity. The net electric field between
the sheets is E # = / P0 , and the electrical force on the droplet must balance gravity, so qE = mg.
EXECUTE:! (a) The electrical force on the drop must be upward, so the field should point downward since the
drop is negative.
(b) The charge of the drop is 5e, so qE # mg. (5e)(= / P0 ) # mg and =# mgP0 . 324 " 10
#
5e !9 kg / . 9.80 m/s 2 / . 8.85 " 10!12 C2 /N + m2 /
5 .1.60 " 10 !19 C/ = 35.1 C/m2 EVALUATE:! Balancing oil droplets between plates was the basis of the Milliken OilDrop Experiment which
produced the first measurement of the mass of an electron.
21.103. IDENTIFY and SET UP:! Example 21.12 gives the electric field due to one infinite sheet. Add the two fields as
vectors.
!
ˆ
EXECUTE:! The electric field due to the first sheet, which is in the xyplane, is E1 # .= / 2P0 / k for z 5 0 and
!
!
ˆ
ˆ
E1 # ! .= /2P0 / k for z 1 0. We can write this as E1 # .= /2P0 / . z/ z / k , since z/ z # 01 for z 5 0 and z/ z # ! z/z # !1
!
ˆ
for z 1 0. Similarly, we can write the electric field due to the second sheet as E 2 # ! .= /2P0 / . x/ x / i , since its
!!
!
ˆ
ˆ
charge density is != . The net field is E # E1 0 E2 # .= /2P0 / ! . x/ x / i 0 . z/ z / k . . / EVALUATE:! The electric field is independent of the ycomponent of the field point since displacement in the
6 y  direction is parallel to both planes. The field depends on which side of each plane the field is located.
21.104. IDENTIFY:! Apply Eq.(21.11) for the electric field of a disk. The hole can be described by adding a disk of charge
density != and radius R1 to a solid disk of charge density 0= and radius R2 . Electric Charge and Electric Field! ! 2141
2
SET UP:! The area of the annulus is , ( R2 ! R12 )= . The electric field of a disk, Eq.(21.11) is E# =9
1!1 ( R x)2 0 1: .
< 2P0 ; 2
EXECUTE:! (a) Q # A= # , ( R2 ! R12 )=
!
xˆ !
xˆ
=9
!=
(b) E ( x) #
i . E ( x) #
i.
1 ! 1/ ( R2 /x ) 2 0 1 : ! 91 ! 1/ ( R1/x )2 0 1:
1/ ( R1/x ) 2 0 1 ! 1/ ( R2 /x )2 0 1
<;
<x
x
2P0 ;
2P0
The electric field is in the +x direction at points above the disk and in the ! x direction at points below the disk, and
xˆ
i specifies these directions.
the factor
x
2
!
x
x
= %x x&x ˆ = %1 1& ˆ
(c) Note that 1 ( R1 / x) 2 0 1 = (1 0 ( x/R1 ) 2 ) !1/ 2 C . This gives E ( x) =
i=
'!(
' ! (x i.
R1
R1
2P0 ) R1 R2 * x
2P0 ) R1 R2 *
Sufficiently close means that ( x R1 ) 2 11 1. / . (d) Fx # qEx # !
f# 1
2, . / q= % 1 1 &
q= % 1 1 &
' ! ( x . The force is in the form of Hooke’s law: Fx # ! kx , with k #
' ! (.
2P0 ) R1 R2 *
2P0 ) R1 R2 * 1
k
#
m 2, 1&
q= % 1
' ! (.
2P0 m ) R1 R2 * EVALUATE:! The frequency is independent of the initial position of the particle, so long as this position is
sufficiently close to the center of the annulus for ( x / R1 ) 2 to be small.
21.105. IDENTIFY:! Apply Coulomb’s law to calculate the forces that q1 and q2 exert on q3 , and add these force vectors
to get the net force.
SET UP:! Like charges repel and unlike charges attract. Let +x be to the right and +y be toward the top of the page.
EXECUTE:! (a) The four possible force diagrams are sketched in Figure 21.105a.
Only the last picture can result in a net force in the –xdirection.
(b) q1 # !2.00 $ C, q3 # 04.00 $ C, and q2 5 0.
!
!
(c) The forces F1 and F2 and their components are sketched in Figure 21.105b.
q1 q3
q2 q3
1
1
sin 41 0
sin 4 2 . This gives
Fy # 0 # !
2
4, P0 (0.0400 m)
4, P0 (0.0300 m) 2
9
sin 41 9
3 5 27
#
#
q2 #
q1
q1
q1 # 0.843 $ C .
16 sin 4 2 16
4 5 64
q1
q2
1%
4
3&
0
'
( # 56.2 N .
4, P0 ) (0.0400 m)2 5 (0.0300 m)2 5 *
!
EVALUATE:! The net force F on q3 is in the same direction as the resultant electric field at the location of q3 due
to q1 and q2 . (d) Fx # F1x 0 F2 x and Fy # 0 , so F # q3 Figure 21.105
21.106. IDENTIFY:! Calculate the electric field at P due to each charge and add these field vectors to get the net field.
SET UP:! The electric field of a point charge is directed away from a positive charge and toward a negative
charge. Let +x be to the right and let +y be toward the top of the page.
EXECUTE:! (a) The four possible diagrams are sketched in Figure 21.106a.
The first diagram is the only one in which the electric field must point in the negative ydirection.
(b) q1 # !3.00 $ C, and q2 1 0 . 2142 Chapter 21 !
!
5
12
(c) The electric fields E1 and E2 and their components are sketched in Figure 24.106b. cos41 # , sin 41 #
,
13
13
k q1
k q2
k q2
k q1
5
12
5
12
5
cos4 2 # and sin 4 2 # . Ex # 0 # !
. This gives
.
0
#
2
2
2
13
13
(0.050 m) 13 (0.120 m) 13
(0.120 m)
(0.050 m)2 12
Solving for q2 gives q2 # 7.2 $C , so q2 # !7.2 $C . Then k q1
kq2
12
5
!
# !1.17 " 107 N/C . E # 1.17 " 107 N/C .
2
2
(0.050 m) 13 (0.120 m) 13
!
EVALUATE:! With q1 known, specifying the direction of E determines both q2 and E.
Ey # ! Figure 21.106
21.107. IDENTIFY:! To find the electric field due to the second rod, divide that rod into infinitesimal segments of length
dx, calculate the field dE due to each segment and integrate over the length of the rod to find the total field due to
!
!
the rod. Use dF # dq E to find the force the electric field of the second rod exerts on each infinitesimal segment of
the first rod.
SET UP:! An infinitesimal segment of the second rod is sketched in Figure 21.107. dQ # (Q / L)dx .
EXECUTE:! (a) dE #
L Ex # H dEx #
0 k dQ
kQ
dx#
.
2
( x 0 a / 2 0 L ! x)
L ( x 0 a / 2 0 L ! x) 2
L kQ L
dxkQ 9
1
kQ % 1
1
:
&
#
#
!
'
(.
L H 0 ( x 0 a / 2 0 L ! x) 2
L > x 0 a / 2 0 L ! x ? 0
L ) x 0 a/2 x 0 a/20 L *
;
< 2kQ % 1
1
&
!
'
(.
L ) 2x 0 a 2L 0 2x 0 a *
(b) Now consider the force that the field of the second rod exerts on an infinitesimal segment dq of the first rod.
This force is in the +xdirection. dF # dq E .!
Ex # F # H E dq # H L0a 2
a2 EQ
2kQ 2 L 0 a 2% 1
1
&
!
dx # 2 H
'
( dx .
a2
L
L
) 2x 0 a 2L 0 2x 0 a * 2 F# 2kQ 1
.[ln (a 0 2x)]aL 0 a2
L2 2 F# 2 % % a 0 2 L 0 a & % 2 L 0 2a & &
1n ' '
('
(( .
2a
* ) 4 L 0 2a * *
)) kQ 2 % (a 0 L) 2 &
1n '
(.
L2
) a ( a 0 2 L) * (c) For a 55 L , F # 2 L
! [ln(2 L 0 2 x 0 a )]a 0 a
2 2 / # kQ
L
2 kQ 2 % a 2 (1 0 L a) 2 & kQ 2
1n ' 2
( # 2 (21n (1 0 L a ) ! ln(1 0 2 L a )) .
L2
) a (1 0 2 L a ) * L For small z, ln(1 0 z ) C z ! & % 2 L 2 L2
& & kQ 2
kQ 2 % % L L2
z2
. Therefore, for a 55 L , F C 2 ' 2 ' ! 2 0 +++ ( ! '
! 2 0 +++ ( ( C 2 .
(
2
L ' ) a 2a
a
* )a
** a
) EVALUATE:! The distance between adjacent ends of the rods is a. When a 55 L the distance between the rods is
much greater than their lengths and they interact as point charges. Figure 21.107 22 GAUSS’S LAW 22.1. ˆ
(a) IDENTIFY and SET UP: J E # H E cos A dA, where A is the angle between the normal to the sheet n and the
!
electric field E .
EXECUTE: In this problem E and cosA are constant over the surface so
J E # E cos A H dA # E cos A A # .14 N/C /. cos 603 / . 0.250 m 2 / # 1.8 N + m 2 /C. (b) EVALUATE:
(c) EXECUTE: J E is independent of the shape of the sheet as long as A and E are constant at all points on the sheet.
(i) J E # E cosA A. J E is largest for A # 03, so cosA # 1 and J E # EA. (ii) J E is smallest for A # 903, so cosA # 0 and J E # 0. J E is 0 when the surface is parallel to the field so no electric field lines pass through the surface. EVALUATE:
22.2. The field is uniform and the surface is flat, so use J E # EA cosA . IDENTIFY:
SET UP: ! A is the angle between the normal to the surface and the direction of E , so A # 70! . EXECUTE: J E # (75.0 N/C)(0.400 m)(0.600 m)cos70! # 6.16 N + m2 /C If the field were perpendicular to the surface the flux would be J E # EA # 18.0 N + m 2 /C. The flux in
!
this problem is much less than this because only the component of E perpendicular to the surface contributes to the
flux.
IDENTIFY: The electric flux through an area is defined as the product of the component of the electric field
perpendicular to the area times the area.
(a) SET UP: In this case, the electric field is perpendicular to the surface of the sphere, so J E # EA # E (4, r 2 ) .
EXECUTE: Substituting in the numbers gives
EVALUATE: 22.3. J E # .1.25 " 106 N/C / 4, . 0.150 m / # 3.53 " 105 N + m 2 /C
2 We use the electric field due to a point charge.
1q
SET UP: E #
4, P0 r 2
EXECUTE: Solving for q and substituting the numbers gives
1
2
q # 4, P0 r 2 E #
0.150 m / .1.25 " 106 N/C / # 3.13 " 10!6 C
9
2
2.
9.00 " 10 N + m /C
EVALUATE: The flux would be the same no matter how large the circle, since the area is proportional to r2 while
the electric field is proportional to 1/r2.
IDENTIFY: Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate the total enclosed charge.
!
ˆ
ˆ
SET UP: E = (!5.00 N/C + m) x i + (3.00 N/C + m) z k . The area of each face is L2 , where L # 0.300 m .
!
ˆ
ˆ
EXECUTE: ns1 = ! ˆ I J1 # E + nS1 A # 0 .
j
!
ˆ I J # E + n A # (3.00 N C + m)(0.300 m)2 z # (0.27 (N C) + m) z .
ˆ
ˆ S2
nS2 = 0 k
2
(b) IDENTIFY: 22.4. J 2 # (0.27 (N/C)m)(0.300 m) # 0.081 (N/C) + m2 .
!
ˆ
ˆ
nS3 = 0 ˆ I J 3 # E + nS3 A # 0 .
j
!
ˆ I J # E + n A # !(0.27 (N/C) + m) z # 0 (since z # 0).
ˆ
ˆ S4
nS4 = !k
4
!
ˆ I J # E + n A # (!5.00 N/C + m)(0.300 m)2 x # !(0.45 (N/C) + m) x.
ˆ
ˆ S5
nS5 = 0 i
5
J 5 # !(0.45 (N/C) + m)(0.300 m) # !(0.135 (N/C) + m2 ).
!
ˆ
ˆ
ˆ
n = !i I J # E + n A # 0(0.45 (N/C) + m) x # 0 (since x # 0).
S6 6 S6 221 222 Chapter 22 (b)Total flux:! 22.5. 22.6. J # J 2 0 J 5 # (0.081 ! 0.135)(N/C) + m2 # !0.054 N + m2 /C. Therefore, q # P0J # !4.78 " 10!13 C.
!
EVALUATE: Flux is positive when E is directed out of the volume and negative when it is directed into the
volume.
IDENTIFY: The flux through the curved upper half of the hemisphere is the same as the flux through the flat circle
defined by the bottom of the hemisphere because every electric field line that passes through the flat circle also must
pass through the curved surface of the hemisphere.
SET UP: The electric field is perpendicular to the flat circle, so the flux is simply the product of E and the area of
the flat circle of radius r.
EXECUTE: JE = EA = E( , r 2 ) = , r 2 E
EVALUATE: The flux would be the same if the hemisphere were replaced by any other surface bounded by the flat
circle.
IDENTIFY: Use Eq.(22.3) to calculate the flux for each surface.
!
!!
ˆ
SET UP: J # E + A # EA cos A where A = An .
ˆ (left) . J # !(4 " 103 N/C)(0.10 m)2 cos(90! ! 36.93) # !24 N + m2 /C.
ˆ
EXECUTE: (a) n = ! j
S1 S1 ˆ
ˆ
nS2 = 0 k (top) . J S2 # !(4 " 103 N/C)(0.10 m)2 cos903 # 0 .
ˆ
nS3 = 0 ˆ (right) . J S3 # 0 (4 " 103 N/C)(0.10 m) 2 cos(903 ! 36.93) # 024 N + m2 /C .
j
ˆ
ˆ
nS4 = !k (bottom) . J S4 # (4 " 103 N/C)(0.10 m)2 cos903 # 0 .
ˆ
ˆ
nS5 = 0 i (front) . J S5 # 0 (4 " 103 N/C)(0.10 m) 2 cos36.93 # 32 N + m 2 /C .
ˆ
ˆ
nS6 = !i (back) . J S6 # !(4 " 103 N/C)(0.10 m)2 cos36.93 # !32 N + m 2 /C .
22.7. EVALUATE: (b) The total flux through the cube must be zero; any flux entering the cube must also leave it, since
the field is uniform. Our calculation gives the result; the sum of the fluxes calculated in part (a) is zero.
(a) IDENTIFY: Use Eq.(22.5) to calculate the flux through the surface of the cylinder.
SET UP: The line of charge and the cylinder are sketched in Figure 22.7. Figure 22.7
EXECUTE: The area of the curved part of the cylinder is A # 2, rl .
!!
The electric field is parallel to the end caps of the cylinder, so E + A # 0 for the ends and the flux through the
cylinder end caps is zero.
The electric field is normal to the curved surface of the cylinder and has the same magnitude E # 8 / 2, P0 r at all
points on this surface. Thus A # 03 and J E # EA cos A # EA # . 8 / 2, P0 r /. 2, rl / # 8l . 6.00 "10 !6 C/m / . 0.400 m / # 2.71 " 105 N + m 2 / C
P0
8.854 " 10!12 C2 / N + m 2
(b) In the calculation in part (a) the radius r of the cylinder divided out, so the flux remains the same,
J E # 2.71 " 105 N + m 2 / C.
(c) J E # 22.8. 8l # . 6.00 "10 !6 # C/m / . 0.800 m / # 5.42 " 105 N + m 2 / C (twice the flux calculated in parts (b) and (c)).
P0
8.854 " 10!12 C 2 / N + m 2
EVALUATE:! The flux depends on the number of field lines that pass through the surface of the cylinder.
IDENTIFY:! Apply Gauss’s law to each surface.
SET UP:! Qencl is the algebraic sum of the charges enclosed by each surface. Flux out of the volume is positive and
flux into the enclosed volume is negative.
EXECUTE:! (a) J S1 # q1/P0 # (4.00 " 10!9 C)/P0 # 452 N + m 2 /C.
(b) J S2 # q2 /P0 # (!7.80 " 10!9 C)/P0 # !881 N + m 2 /C.
(c) J S3 # (q1 0 q2 )/P0 # ((4.00 ! 7.80) " 10!9 C)/P0 # !429 N + m 2 /C.
(d) J S4 # (q1 0 q3 )/P0 # ((4.00 0 2.40) " 10!9 C)/P0 # 723 N + m 2 /C.
(e) J S5 # (q1 0 q2 0 q3 )/P0 # ((4.00 ! 7.80 0 2.40) " 10!9 C)/P0 # !158 N + m 2 /C. Gauss’s Law 22.9. 223 EVALUATE:! (f ) All that matters for Gauss’s law is the total amount of charge enclosed by the surface, not its
distribution within the surface.
IDENTIFY:! Apply the results in Example 21.10 for the field of a spherical shell of charge.
q
SET UP:! Example 22.10 shows that E # 0 inside a uniform spherical shell and that E # k 2 outside the shell.
r
EXECUTE:! (a) E # 0
15.0 " 10!6 C
# 3.75 " 107 N/C
(b) r # 0.060 m and E # (8.99 " 109 N + m 2 /C2 )
(0.060 m)2 15.0 " 10!6 C
# 1.11" 107 N/C
(0.110 m) 2
EVALUATE:! Outside the shell the electric field is the same as if all the charge were concentrated at the center of
the shell. But inside the shell the field is not the same as for a point charge at the center of the shell, inside the shell
the electric field is zero.
IDENTIFY:! Apply Gauss’s law to the spherical surface.
SET UP:! Qencl is the algebraic sum of the charges enclosed by the sphere.
EXECUTE:! (a) No charge enclosed so J # 0 .
q
!6.00 " 10!9 C
# ! 678 N + m 2 C.
(b) J # 2 #
P0 8.85 " 10!12 C2 N + m 2
(c) r # 0.110 m and E # (8.99 " 109 N + m 2 /C2 ) 22.10. q1 0 q2 (4.00 ! 6.00) " 10!9 C
#
# !226 N + m 2 C.
P0
8.85 " 10!12 C 2 N + m 2
EVALUATE:! Negative flux corresponds to flux directed into the enclosed volume. The net flux depends only on the
net charge enclosed by the surface and is not affected by any charges outside the enclosed volume.
IDENTIFY:! Apply Gauss’s law.
SET UP:! In each case consider a small Gaussian surface in the region of interest.
!
EXECUTE:! (a) Since E is uniform, the flux through a closed surface must be zero. That is:
!!q
J # úE + dA # # 1 H "dV # 0 I H "dV # 0. But because we can choose any volume we want, " must be zero if
P0 P0
the integral equals zero.
(b) If there is no charge in a region of space, that does NOT mean that the electric field is uniform. Consider a
closed volume close to, but not including, a point charge. The field diverges there, but there is no charge in that
region.
EVALUATE:! The electric field within a region can depend on charges located outside the region. But the flux
through a closed surface depends only on the net charge contained within that surface.
IDENTIFY:! Apply Gauss’s law.
SET UP:! Use a small Gaussian surface located in the region of question.
EXECUTE:! (a) If " 5 0 and uniform, then q inside any closed surface is greater than zero. This implies J 5 0 , so
!!
úE + dA 5 0 and so the electric field cannot be uniform. That is, since an arbitrary surface of our choice encloses a (c) J # 22.11. 22.12. 22.13. nonzero amount of charge, E must depend on position.
(b) However, inside a small bubble of zero charge density within the material with density " , the field can be
uniform. All that is important is that there be zero flux through the surface of the bubble (since it encloses no
charge). (See Problem 22.61.)
EVALUATE:! In a region of uniform field, the flux through any closed surface is zero.
!!
(a) IDENTIFY and SET UP:! It is rather difficult to calculate the flux directly from J # úE + dA since the magnitude
!
!
of E and its angle with dA varies over the surface of the cube. A much easier approach is to use Gauss's law to
calculate the total flux through the cube. Let the cube be the Gaussian surface. The charge enclosed is the point
charge.
9.60 " 10!6 C
# 1.084 " 106 N + m 2 / C.
EXECUTE:! J E # Qencl / P0 #
8.854 " 10!12 C2 / N + m 2
By symmetry the flux is the same through each of the six faces, so the flux through one face is
1
1.084 " 106 N + m 2 / C / # 1.81" 105 N + m 2 / C.
6.
(b) EVALUATE:! In part (a) the size of the cube did not enter into the calculations. The flux through one face
depends only on the amount of charge at the center of the cube. So the answer to (a) would not change if the size of
the cube were changed. 224 Chapter 22 22.14. IDENTIFY:! Apply the results of Examples 22.9 and 22.10.
q
SET UP:! E # k 2 outside the sphere. A proton has charge +e.
r
q
92(1.60 " 10!19 C)
# 2.4 " 1021 N/C
EXECUTE:! (a) E # k 2 # (8.99 " 109 N + m 2 /C2 )
(7.4 " 10!15 m) 2
r
2 % 7.4 " 10!15 m &
13
(b) For r # 1.0 " 10 m , E # (2.4 " 10 N/C) '
( # 1.3 " 10 N/C
!10
) 1.0 " 10 m *
(c) E # 0 , inside a spherical shell.
EVALUATE:! The electric field in an atom is very large.
IDENTIFY:! The electric fields are produced by point charges.
1q
SET UP:! We use Coulomb’s law, E #
, to calculate the electric fields.
4, P0 r 2
!10 22.15. 21 EXECUTE:! (a) E # . 9.00 " 109 N + m 2 /C2 /
(b) E # . 9.00 " 109 N + m 2 /C2 / 22.16. 22.17. 5.00 " 10!6 C 5.00 " 10!6 C .1.00 m / 2 # 4.50 " 104 N/C # 9.18 " 102 N/C . 7.00 m /
(c) Every field line that enters the sphere on one side leaves it on the other side, so the net flux through the surface is
zero.
EVALUATE:! The flux would be zero no matter what shape the surface had, providing that no charge was inside the
surface.
IDENTIFY:! Apply the results of Example 22.5.
SET UP:! At a point 0.100 m outside the surface, r # 0.550 m .
1q
1 (2.50 " 10!10 C)
EXECUTE:! (a) E #
#
# 7.44 N C.
4!P0 r 2 4!P0 (0.550 m)2
(b) E # 0 inside of a conductor or else free charges would move under the influence of forces, violating our
electrostatic assumptions (i.e., that charges aren’t moving).
EVALUATE:! Outside the sphere its electric field is the same as would be produced by a point charge at its center,
with the same charge.
IDENTIFY:! The electric field required to produce a spark 6 in. long is 6 times as strong as the field needed to
produce a spark 1 in. long.
1q
SET UP:! By Gauss’s law, q # P0 EA and the electric field is the same as for a pointcharge, E #
.
4, P0 r 2
2 EXECUTE:! (a) The electric field for 6inch sparks is E # 6 " 2.00 " 104 N/C # 1.20 " 105 N/C
The charge to produce this field is
q # P0 EA # P0 E (4, r 2 ) # (8.85 " 10!12 C2 /N + m 2 )(1.20 " 105 N/C)(4, )(0.15 m)2 # 3.00 " 10!7 C . 3.00 " 10!7 C
# 1.20 " 105 N/C .
(0.150 m)2
EVALUATE:! It takes only about 0.3 $ C to produce a field this strong.
IDENTIFY:! According to Exercise 21.32, the Earth’s electric field points towards its center. Since Mars’s electric
field is similar to that of Earth, we assume it points toward the center of Mars. Therefore the charge on Mars must
be negative. We use Gauss’s law to relate the electric flux to the charge causing it.
q
SET UP:! Gauss’s law is J E #
and the electric flux is J E # EA .
P0
EXECUTE:! (a) Solving Gauss’s law for q, putting in the numbers, and recalling that q is negative, gives
q # !P0J E # !(3.63 " 1016 N + m 2 /C)(8.85 " 10!12 C2 /N + m2 ) # !3.21" 105 C .
(b) Use the definition of electric flux to find the electric field. The area to use is the surface area of Mars.
J
3.63 " 1016 N + m 2 /C
E# E #
# 2.50 " 102 N/C
A
4, (3.40 " 106 m) 2
(b) Using Coulomb’s law gives E # (9.00 " 109 N + m 2 /C2 ) 22.18. !3.21 " 105 C
# !2.21" 10!9 C/m 2
AMars 4, (3.40 " 106 m) 2
EVALUATE:! Even though the charge on Mars is very large, it is spread over a large area, giving a small surface
charge density. (c) The surface charge density on Mars is therefore = # q # Gauss’s Law 22.19. 225 IDENTIFY and SET UP:! Example 22.5 derived that the electric field just outside the surface of a spherical
1q
. Calculate q and from this the number of excess electrons.
conductor that has net charge q is E #
4, P0 R 2
R2E
. 0.160 m / .1150 N/C / # 3.275 " 10!9 C.
#
1/ 4, P0 /
8.988 " 109 N + m 2 /C2
.
2 EXECUTE:! q # Each electron has a charge of magnitude e # 1.602 " 10!19 C, so the number of excess electrons needed is 22.20. 22.21. 3.275 " 10!9 C
# 2.04 " 1010 .
1.602 " 10!19 C
EVALUATE:! The result we obtained for q is a typical value for the charge of an object. Such net charges
correspond to a large number of excess electrons since the charge of each electron is very small.
IDENTIFY:! Apply Gauss’s law.
SET UP:! Draw a cylindrical Gaussian surface with the line of charge as its axis. The cylinder has radius 0.400 m
and is 0.0200 m long. The electric field is then 840 N/C at every point on the cylindrical surface and is directed
perpendicular to the surface.
!!
EXECUTE:! úE + dA # EAcylinder # E (2, rL) # (840 N/C)(2! )(0.400 m)(0.0200 m) # 42.2 N + m 2 /C.
!!
The field is parallel to the end caps of the cylinder, so for them úE + dA # 0. From Gauss’s law,
q # P0J E # (8.854 " 10!12 C2 /N + m 2 )(42.2 N + m2 /C) # 3.74 " 10!10 C.
EVALUATE:! We could have applied the result in Example 22.6 and solved for 8 . Then q # 8 L.
IDENTIFY:! Add the vector electric fields due to each line of charge. E(r) for a line of charge is given by
Example 22.6 and is directed toward a negative line of chage and away from a positive line.
SET UP:! The two lines of charge are shown in Figure 22.21. E# 18
2, P0 r Figure 22.21
!
!
EXECUTE:! (a) At point a, E1 and E 2 are in the 0 y direction (toward negative charge, away from positive
charge).
E1 # .1/ 2, P0 / 9. 4.80 " 10!6 C/m / / . 0.200 m / : # 4.314 " 105 N/C
;
<
E2 # .1/ 2, P0 / 9. 2.40 " 10!6 C/m / / . 0.200 m / : # 2.157 " 105 N/C
;
< E # E1 0 E2 # 6.47 " 105 N/C, in the ydirection.
!
!
(b) At point b, E1 is in the 0 y direction and E2 is in the ! y direction.
E1 # .1/ 2, P0 / 9. 4.80 " 10!6 C/m / / . 0.600 m / : # 1.438 " 105 N/C
;
< E2 # .1/ 2, P0 / 9. 2.40 " 10!6 C/m / / . 0.200 m / : # 2.157 " 105 N/C
;
< 22.22. E # E2 ! E1 # 7.2 " 104 N/C, in the ! y direction.
EVALUATION:! At point a the two fields are in the same direction and the magnitudes add. At point b the two fields
are in opposite directions and the magnitudes subtract.
IDENTIFY:! Apply the results of Examples 22.5, 22.6 and 22.7.
SET UP:! Gauss’s law can be used to show that the field outside a long conducting cylinder is the same as for a line
of charge along the axis of the cylinder.
EXECUTE:! (a) For points outside a uniform spherical charge distribution, all the charge can be considered to be
concentrated at the center of the sphere. The field outside the sphere is thus inversely proportional to the square of
the distance from the center. In this case,
2 % 0.200 cm &
E # (480 N C) '
( # 53 N C
) 0.600 cm * 226 Chapter 22 22.23. (b) For points outside a long cylindrically symmetrical charge distribution, the field is identical to that of a long line
%
of charge: E #
, that is, inversely proportional to the distance from the axis of the cylinder. In this case
2!P0 r
% 0.200 cm &
E # (480 N/C) '
( # 160 N/C
) 0.600 cm *
(c) The field of an infinite sheet of charge is E # #/2P0 ; i.e., it is independent of the distance from the sheet. Thus in
this case E # 480 N/C.
EVALUATE:! For each of these three distributions of charge the electric field has a different dependence on distance.
IDENTIFY:! The electric field inside the conductor is zero, and all of its initial charge lies on its outer surface. The
introduction of charge into the cavity induces charge onto the surface of the cavity, which induces an equal but
opposite charge on the outer surface of the conductor. The net charge on the outer surface of the conductor is the
sum of the positive charge initially there and the additional negative charge due to the introduction of the negative
charge into the cavity.
(a) SET UP:! First find the initial positive charge on the outer surface of the conductor using qi # = A, where A is
the area of its outer surface. Then find the net charge on the surface after the negative charge has been introduced
into the cavity. Finally use the definition of surface charge density.
EXECUTE:! The original positive charge on the outer surface is
qi # = A # = (4, r 2 ) # (6.37 " 10!6 C/m 2 )4, (0.250 m2 ) # 5.00 " 10!6 C/m2 After the introduction of !0.500 $ C into the cavity, the outer charge is now
5.00 $ C ! 0.500 $ C # 4.50 $ C q
q
4.50 " 10!6 C
#
#
# 5.73 " 10!6 C/m2
A 4, r 2 4, (0.250 m) 2
J
q
q
#
(b) SET UP:! Using Gauss’s law, the electric field is E # E #
A P0 A P0 4, r 2
EXECUTE:! Substituting numbers gives
4.50 " 10!6 C
E#
# 6.47 " 105 N/C.
(8.85 " 10!12 C 2 /N + m 2 )(4, )(0.250 m) 2
The surface charge density is now = # (c) SET UP:! We use Gauss’s law again to find the flux. J E #
EXECUTE:! Substituting numbers gives q
.
P0 !0.500 " 10!6 C
# !5.65 " 104 N + m 2 /C2 .
8.85 " 10!12 C 2 /N + m 2
EVALUATE:! The excess charge on the conductor is still 0 5.00 $ C, as it originally was. The introduction of the
!0.500 $ C inside the cavity merely induced equal but opposite charges (for a net of zero) on the surfaces of the
conductor.
IDENTIFY:! We apply Gauss’s law, taking the Gaussian surface beyond the cavity but inside the solid.
q
SET UP:! Because of the symmetry of the charge, Gauss’s law gives us E # total , where A is the surface area of a
P0 A
sphere of radius R = 9.50 cm centered on the pointcharge, and qtotal is the total charge contained within that sphere.
This charge is the sum of the !2.00 $ C point charge at the center of the cavity plus the charge within the solid
between r = 6.50 cm and R = 9.50 cm. The charge within the solid is qsolid = GV = G( [4 / 3], R 3 ! [4 / 3], r 3 ) =
( [4, / 3] G)(R3 – r3)
EXECUTE:! First find the charge within the solid between r = 6.50 cm and R = 9.50 cm:
4,
qsolid #
(7.35 " 10!4 C/m 3 ) 9(0.0950 m)3 ! (0.0650 m)3 : # 1.794 " 10!6 C,
;
<
3
Now find the total charge within the Gaussian surface:
qtotal # qsolid 0 qpoint # !2.00 $ C 0 1.794 $ C # !0.2059 $ C
JE # 22.24. Now find the magnitude of the electric field from Gauss’s law:
0.2059 " 10!6 C
q
q
#
#
# 2.05 " 105 N/C .
E#
2
!12
P0 A P0 4, r
(8.85 " 10 C 2 /N + m 2 )(4, )(0.0950 m) 2
The fact that the charge is negative means that the electric field points radially inward.
EVALUATE:! Because of the uniformity of the charge distribution, the charge beyond 9.50 cm does not contribute
to the electric field. Gauss’s Law 22.25. 227 IDENTIFY:! The magnitude of the electric field is constant at any given distance from the center because the charge
density is uniform inside the sphere. We can use Gauss’s law to relate the field to the charge causing it.
q
q
q
(a) SET UP:! Gauss’s law tells us that EA # , and the charge density is given by G # #
.
P0
V (4 / 3), R 3
EXECUTE:! Solving for q and substituting numbers gives
q # EAP0 # E (4, r 2 )P0 # (1750 N/C)(4, )(0.500 m)2 (8.85 " 10!12 C2 /N + m2 ) # 4.866 " 10!8 C . Using the formula for q
q
4.866 " 10!8 C
#
#
# 2.60 " 10!7 C/m3 .
3
3
V (4 / 3), R
(4 / 3), . 0.355 m /
(b) SET UP:! Take a Gaussian surface of radius r = 0.200 m, concentric with the insulating sphere. The charge
%4
&
enclosed within this surface is qencl # GV # G ' , r 3 ( , and we can treat this charge as a pointcharge, using
)3
*
1 qencl
. The charge beyond r = 0.200 m makes no contribution to the electric field.
Coulomb’s law E #
4, P0 r 2
EXECUTE:! First find the enclosed charge:
charge density we get G # 3:
%4
&
94
qencl # G ' , r 3 ( # . 2.60 " 10!7 C/m3 / > , . 0.200 m / ? # 8.70 " 10!9 C
3
3
)
*
;
< Now treat this charge as a pointcharge and use Coulomb’s law to find the field:
E # . 9.00 " 109 N + m 2 /C2 / 22.26. 8.70 " 10!9 C . 0.200 m / 2 # 1.96 " 103 N/C EVALUATE:! Outside this sphere, it behaves like a pointcharge located at its center. Inside of it, at a distance r
from the center, the field is due only to the charge between the center and r.
IDENTIFY:! Apply Gauss’s law and conservation of charge.
SET UP:! Use a Gaussian surface that lies wholly within he conducting material.
EXECUTE:! (a) Positive charge is attracted to the inner surface of the conductor by the charge in the cavity. Its
magnitude is the same as the cavity charge: qinner # 0 6.00 nC, since E # 0 inside a conductor and a Gaussian
surface that lies wholly within the conductor must enclose zero net charge.
(b) On the outer surface the charge is a combination of the net charge on the conductor and the charge “left behind”
when the 06.00 nC moved to the inner surface: qtot # qinner 0 qouter I qouter # qtot ! qinner # 5.00 nC ! 6.00 nC # ! 1.00 nC.
22.27. 22.28. EVALUATE:! The electric field outside the conductor is due to the charge on its surface.
IDENTIFY:! Apply Gauss’s law to each surface.
SET UP:! The field is zero within the plates. By symmetry the field is perpendicular to the plates outside the plates
and can depend only on the distance from the plates. Flux into the enclosed volume is positive.
EXECUTE:! S2 and S3 enclose no charge, so the flux is zero, and electric field outside the plates is zero. Between the plates, S 4 shows that ! EA # ! q P0 # !# A P0 and E # # P0 .
EVALUATE:! Our result for the field between the plates agrees with the result stated in Example 22.8.
IDENTIFY:! Close to a finite sheet the field is the same as for an infinite sheet. Very far from a finite sheet the field
is that of a point charge.
=
1q
SET UP:! For an infinite sheet, E #
. For a point charge, E #
.
2P0
4, P0 r 2
EXECUTE:! (a) At a distance of 0.1 mm from the center, the sheet appears “infinite,” so
EC q
7.50 " 10!9 C
#
# 662 N/C .
2P0 A 2P0 (0.800 m) 2 (b) At a distance of 100 m from the center, the sheet looks like a point, so: EC 1q
1 (7.50 " 10!9 C)
#
# 6.75 " 10!3 N C.
2
4!P0 r
4!P0 (100 m) 2 (c) There would be no difference if the sheet was a conductor. The charge would automatically spread out evenly
over both faces, giving it half the charge density on either face as the insulator but the same electric field. Far away,
they both look like points with the same charge.
EVALUATE:! The sheet can be treated as infinite at points where the distance to the sheet is much less than the
distance to the edge of the sheet. The sheet can be treated as a point charge at points for which the distance to the
sheet is much greater than the dimensions of the sheet. 228 Chapter 22 22.29. IDENTIFY:! Apply Gauss’s law to a Gaussian surface and calculate E.
(a) SET UP:! Consider the charge on a length l of the cylinder. This can be expressed as q # 8l . But since the
surface area is 2, Rl it can also be expressed as q # = 2, Rl . These two expressions must be equal, so 8l # = 2, Rl
and 8 # 2, R= .
(b) Apply Gauss’s law to a Gaussian surface that is a cylinder of length l, radius r, and whose axis coincides with
the axis of the charge distribution, as shown in Figure 22.29.
EXECUTE:!
Qencl # = . 2, Rl / J E # 2, rlE
Figure 22.29
JE # E# = . 2, Rl /
Qencl
gives 2, rlE #
P0
P0 =R
P0 r (c) EVALUATE:! Example 22.6 shows that the electric field of an infinite line of charge is E # 8 / 2, P0 r. = # EA # =R 1
(5 $ C m 2 0 2 $ C m 2 0 4 $ C m 2 ! 6 $ C m 2 ) # 2.82 " 105 N C to the left.
2P0 (b)! EB # EB # #1
#
#
#
1
0 30 4! 2#
( #1 0 # 3 0 # 4 ! # 2 ) .
2P0 2P0 2P0 2P0 2P0 1
(6 $ C m 2 0 2 $ C m 2 0 4 $ C m 2 ! 5 $ C m 2 ) # 3.95 " 105 N C to the left.
2P0 (c) EC # #4
#
#
#
1
0 1! 2! 3#
( # 2 0 # 3 ! # 4 ! #1 ) .
2P0 2P0 2P0 2P0 2P0 1
(4 $ C m 2 0 6 $ C m 2 ! 5 $ C m 2 ! 2 $ C m 2 ) # 1.69 " 105 N C to the left
2P0
EVALUATE:! The field at C is not zero. The pieces of plastic are not conductors.
IDENTIFY:! Apply Gauss’s law and conservation of charge.
SET UP:! E # 0 in a conducting material.
EXECUTE:! (a) Gauss’s law says +Q on inner surface, so E # 0 inside metal.
(b) The outside surface of the sphere is grounded, so no excess charge.
(c) Consider a Gaussian sphere with the –Q charge at its center and radius less than the inner radius of the metal.
This sphere encloses net charge –Q so there is an electric field flux through it; there is electric field in the cavity.
(d) In an electrostatic situation E # 0 inside a conductor. A Gaussian sphere with the !Q charge at its center and
radius greater than the outer radius of the metal encloses zero net charge (the !Q charge and the 0Q on the inner
surface of the metal) so there is no flux through it and E # 0 outside the metal.
(e) No, E # 0 there. Yes, the charge has been shielded by the grounded conductor. There is nothing like positive
and negative mass (the gravity force is always attractive), so this cannot be done for gravity.
EVALUATE:! Field lines within the cavity terminate on the charges induced on the inner surface.
EC # 22.31. , R% 8 &
8
, the same as for an infinite line of charge that is along the axis of the cylinder.
#
'
(#
P0 r P0 r ) 2, R * 2, P0 r
IDENTIFY:! The net electric field is the vector sum of the fields due to each of the four sheets of charge.
SET UP:! The electric field of a large sheet of charge is E # = / 2P0 . The field is directed away from a positive sheet
and toward a negative sheet.
#
#
#
#
1
EXECUTE:! (a) At A : E A # 2 0 3 0 4 ! 1 #
( # 2 0 #3 0 # 4 ! #1 ) .
2P0 2P0 2P0 2P0 2P0
so E # 22.30. 8
2, R Gauss’s Law 22.32. 22.33. 229 ˆ
IDENTIFY and SET UP:! Eq.(22.3) to calculate the flux. Identify the direction of the normal unit vector n for each
surface.
!
ˆ
ˆ
j
EXECUTE:! (a) E # ! Bi 0 Cˆ ! Dk ; A # L2
ˆ
face S1: n # ! ˆ
j
!! !
ˆ
ˆ
ˆ
J E # E + A # E + ( An) # (! Bi 0 Cˆ ! Dk ) + (! Aˆ) # !CL2 .
j
j
ˆ
ˆ # 0k
face S2 : n
!! !
ˆ
ˆ
ˆ
ˆ
J E # E + A # E + ( An) # (! Bi 0 Cˆ ! Dk ) + ( Ak ) # ! DL2 .
j
ˆ
face S3: n # 0 ˆ
j
!! !
ˆ
ˆ
ˆ
J E # E + A # E + ( An) # (! Bi 0 Cˆ ! Dk ) + ( Aˆ) # 0CL2 .
j
j
ˆ
ˆ
face S4 : n # !k
!! !
ˆ
ˆ
ˆ
ˆ
J E # E + A # E + ( An) # (! Bi 0 Cˆ ! Dk ) + (! Ak ) # 0 DL2 .
j
ˆ
ˆ
face S5: n # 0 i
!! !
ˆ
ˆ
ˆ
ˆ
J E # E + A # E + ( An) # (! Bi 0 Cˆ ! Dk ) + ( Ai ) # ! BL2 .
j
ˆ
ˆ
face S6 : n # !i
!! !
ˆ
ˆ
ˆ
ˆ
J E # E + A # E + ( An) # (! Bi 0 Cˆ ! Dk ) + (! Ai ) # 0 BL2 .
j
(b) Add the flux through each of the six faces: J E # !CL2 ! DL2 0 CL2 0 DL2 ! BL2 0 BL2 # 0
The total electric flux through all sides is zero.
EVALUATE:! All electric field lines that enter one face of the cube leave through another face. No electric field
lines terminate inside the cube and the net flux is zero.
IDENTIFY:! Use Eq.(22.3) to calculate the flux through each surface and use Gauss’s law to relate the net flux to
the enclosed charge.
SET UP:! Flux into the enclosed volume is negative and flux out of the volume is positive.
EXECUTE:! (a) J # EA # (125 N C)(6.0 m 2 ) # 750 N + m 2 C.
(b)! Since the field is parallel to the surface, J # 0.
(c) Choose the Gaussian surface to equal the volume’s surface. Then 750 N + m 2 /C ! EA # q / P0 and
1 (2.40 " 10!8 C/P 0 750 N + m 2 /C) # 577 N/C , in the positive xdirection. Since q 1 0 we must have some
0
6.0 m 2
net flux flowing in so the flux is ! E A on second face.
E# 22.34. EVALUATE:! (d) q 1 0 but we have E pointing away from face I. This is due to an external field that does not
affect the flux but affects the value of E.
IDENTIFY:! Apply Gauss’s law to a cube centered at the origin and with side length 2L.
SET UP:! The total surface area of a cube with side length 2L is 6(2 L)2 # 24 L2 .
EXECUTE:! (a) The square is sketched in Figure 22.34.
(b) Imagine a charge q at the center of a cube of edge length 2L. Then: J # q / P0 . Here the square is one 24th of the surface area of the imaginary cube, so it intercepts 1/24 of the flux. That is, J # q 24P0 .
EVALUATE:! Calculating the flux directly from Eq.(22.5) would involve a complicated integral. Using Gauss’s law
and symmetry considerations is much simpler. Figure 22.34 2210 Chapter 22 22.35. (a) IDENTIFY:! Find the net flux through the parallelepiped surface and then use that in Gauss’s law to find the net
charge within. Flux out of the surface is positive and flux into the surface is negative.
!
SET UP:! E1 gives flux out of the surface. See Figure 22.35a.
EXECUTE:! J1 # 0 E1K A A # (0.0600 m)(0.0500 m) # 3.00 " 10!3 m2
E1K # E1 cos 603 # (2.50 " 10 4 N/C) cos 603 E1K # 1.25 " 104 N/C
Figure 22.35a
J E1 # 0 E1K A # 0 (1.25 " 10 4 N/C)(3.00 " 10!3 m 2 ) # 37.5 N + m 2 / C
!
SET UP:! E2 gives flux into the surface. See Figure 22.35b.
EXECUTE:! J 2 # ! E2 K A A # (0.0600 m)(0.0500 m) # 3.00 " 10!3 m2
E2 K # E2 cos 603 # (7.00 " 104 N/C)cos 603 E2 K # 3.50 " 104 N/C
Figure 22.35b
J E2 # ! E2 K A # !(3.50 " 104 N/C)(3.00 " 10!3 m 2 ) # !105.0 N + m 2 /C The net flux is J E # J E1 0 J E2 # 037.5 N + m 2 /C ! 105.0 N + m2 /C # !67.5 N + m2 /C. 22.36. The net flux is negative (inward), so the net charge enclosed is negative.
Q
Apply Gauss’s law: J E # encl
P0
2
Qencl # J E P0 # ( !67.5 N + m /C)(8.854 " 10!12 C 2 /N + m 2 ) # !5.98 " 10!10 C.
(b) EVALUATE:! If there were no charge within the parallelpiped the net flux would be zero. This is not the case, so
there is charge inside. The electric field lines that pass out through the surface of the parallelpiped must terminate on
charges, so there also must be charges outside the parallelpiped.
IDENTIFY:! The 2 particle feels no force where the net electric field due to the two distributions of charge is zero.
SET UP:! The fields can cancel only in the regions A and B shown in Figure 22.36, because only in these two
regions are the two fields in opposite directions.
%
#
50 $ C/m
EXECUTE:! Eline # Esheet gives
and r # % ,= #
# 0.16 m # 16 cm .
#
2!P0 r 2P0
, (100 $ C/m 2 )
The fields cancel 16 cm from the line in regions A and B.
EVALUATE:! The result is independent of the distance between the line and the sheet. The electric field of an
infinite sheet of charge is uniform, independent of the distance from the sheet. Figure 22.36
22.37. (a) IDENTIFY:! Apply Gauss’s law to a Gaussian cylinder of length l and radius r, where a 1 r 1 b, and calculate E
on the surface of the cylinder.
SET UP:! The Gaussian surface is sketched in Figure 22.37a.
EXECUTE:! J E # E . 2, rl / Qencl # 8l (the charge on the
length l of the inner conductor
that is inside the Gaussian
surface).
Figure 22.37a
JE #
E# Qencl
8l
gives E . 2, rl / #
P0
P0 !
8
. The enclosed charge is positive so the direction of E is radially outward.
2, P0 r Gauss’s Law 2211 (b) SET UP:! Apply Gauss’s law to a Gaussian cylinder of length l and radius r, where r > c, as shown in
Figure 22.37b.
EXECUTE:! J E # E . 2, rl / Qencl # 8l (the charge on the length l
of the inner conductor that is inside
the Gaussian surface; the outer
conductor carries no net charge).
Figure 22.37b Qencl
8l
gives E . 2, rl / #
P0
P0
!
8
E#
. The enclosed charge is positive so the direction of E is radially outward.
2, P0 r
(c) E = 0 within a conductor. Thus E = 0 for r < a;
JE # 8
for a 1 r 1 b; E # 0 for b 1 r 1 c;
2, P0 r
8
E#
for r 5 c. The graph of E versus r is sketched in Figure 22.37c.
2, P0 r
E# Figure 22.37c 22.38. EVALUATE:! Inside either conductor E = 0. Between the conductors and outside both conductors the electric field
is the same as for a line of charge with linear charge density 8 lying along the axis of the inner conductor.
(d) IDENTIFY and SET UP:! inner surface: Apply Gauss’s law to a Gaussian cylinder with radius r, where
b 1 r 1 c. We know E on this surface; calculate Qencl .
EXECUTE:! This surface lies within the conductor of the outer cylinder, where E # 0, so J E # 0. Thus by Gauss’s
law Qencl # 0. The surface encloses charge 8 l on the inner conductor, so it must enclose charge ! 8 l on the inner
surface of the outer conductor. The charge per unit length on the inner surface of the outer cylinder is ! 8 .
outer surface: The outer cylinder carries no net charge. So if there is charge per unit length ! 8 on its inner surface
there must be charge per until length 0 8 on the outer surface.
EVALUATE:! The electric field lines between the conductors originate on the surface charge on the outer surface of
the inner conductor and terminate on the surface charges on the inner surface of the outer conductor. These surface
charges are equal in magnitude (per unit length) and opposite in sign. The electric field lines outside the outer
conductor originate from the surface charge on the outer surface of the outer conductor.
IDENTIFY:! Apply Gauss’s law.
SET UP:! Use a Gaussian surface that is a cylinder of radius r, length l and that has the line of charge along its axis.
The charge on a length l of the line of charge or of the tube is q # 2 l .
Q
%l
%
EXECUTE:! (a) (i)! For r 1 a , Gauss’s law gives E (2!rl ) # encl #
and E #
.
P0
P0
2!P0 r
(ii)! The electric field is zero because these points are within the conducting material.
Q
2% l
%
(iii) For r 5 b , Gauss’s law gives E (2!rl ) # encl #
and E #
.
P0
P0
!P0 r
The graph of E versus r is sketched in Figure 22.38. 2212 Chapter 22 (b) (i) The Gaussian cylinder with radius r, for a 1 r 1 b , must enclose zero net charge, so the charge per unit length
on the inner surface is !2 . ! (ii) Since the net charge per length for the tube is 02 and there is !2 on the inner
surface, the charge per unit length on the outer surface must be 022 .
EVALUATE:! For r 5 b the electric field is due to the charge on the outer surface of the tube. Figure 22.38
22.39. (a) IDENTIFY:! Use Gauss’s law to calculate E(r).
(i) SET UP:! r 1 a: Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where r 1 a, as
sketched in Figure 22.39a.
EXECUTE:! J E # E . 2, rl / Qencl # 2 l (the charge on the length l
of the line of charge)
Figure 22.39a
JE # Qencl
2l
gives E . 2, rl / #
P0
P0 !
2
. The enclosed charge is positive so the direction of E is radially outward.
2, P0 r
(ii) a 1 r 1 b: Points in this region are within the conducting tube, so E = 0.
(iii) SET UP:! r 5 b: Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where r 5 b, as
sketched in Figure 22.39b. E# EXECUTE:! J E # E . 2, rl / Qencl # 2 l (the charge on length l of the
line of charge) !2 l (the charge on
length l of the tube) Thus Qencl # 0.
Figure 22.39b
JE # Qencl
gives E . 2, rl / # 0 and E # 0. The graph of E versus r is sketched in Figure 22.39c.
P0 Figure 22.39c
(b) IDENTIFY:! Apply Gauss’s law to cylindrical surfaces that lie just outside the inner and outer surfaces of the
tube. We know E so can calculate Qencl .
(i) SET UP:! inner surface
Apply Gauss’s law to a cylindrical Gaussian surface of length l and radius r, where a 1 r 1 b. Gauss’s Law 2213 EXECUTE:! This surface lies within the conductor of the tube, where E = 0, so J E # 0. Then by Gauss’s law 22.40. Qencl # 0. The surface encloses charge 2 l on the line of charge so must enclose charge !2 l on the inner surface of
the tube. The charge per unit length on the inner surface of the tube is !2 .
(ii) outer surface
The net charge per unit length on the tube is !2 . We have shown in part (i) that this must all reside on the inner
surface, so there is no net charge on the outer surface of the tube.
EVALUATE:! For r < a the electric field is due only to the line of charge. For r > b the electric field of the tube is
the same as for a line of charge along its axis. The fields of the line of charge and of the tube are equal in magnitude
and opposite in direction and sum to zero. For r < a the electric field lines originate on the line of charge and
terminate on the surface charge on the inner surface of the tube. There is no electric field outside the tube and no
surface charge on the outer surface of the tube.
IDENTIFY:! Apply Gauss’s law.
SET UP:! Use a Gaussian surface that is a cylinder of radius r and length l, and that is coaxial with the cylindrical
charge distributions. The volume of the Gaussian cylinder is , r 2l and the area of its curved surface is 2, rl . The
charge on a length l of the charge distribution is q # 8l , where 8 # G, R 2 . EXECUTE:! (a) For r 1 R , Qencl # G, r 2l and Gauss’s law gives E (2!rl ) # Qencl "!r 2l
"r
and E #
, radially
#
2P0
P0
P0 outward.
(b) For! r 5 R , Qencl =8l # "!R2l and Gauss’s law gives E (2!rl ) # q " !R 2 l
"R 2
and E #
#
# % , radially
2P0 r 2!P0 r
P0
P0 outward.
(c) At r # R , the electric field for BOTH regions is E # "R
, so they are consistent.
2P0 (d) The graph of E versus r is sketched in Figure 22.40.
EVALUATE:! For r 5 R the field is the same as for a line of charge along the axis of the cylinder. Figure 22.40
22.41. IDENTIFY:! First make a freebody diagram of the sphere. The electric force acts to the left on it since the electric
field due to the sheet is horizontal. Since it hangs at rest, the sphere is in equilibrium so the forces on it add to zero,
by Newton’s first law. Balance horizontal and vertical force components separately.
SET UP:! Call T the tension in the thread and E the electric field. Balancing horizontal forces gives T sin 4 = qE.
Balancing vertical forces we get T cos 4 = mg. Combining these equations gives tan 4 = qE/mg, which means that
4 = arctan(qE/mg). The electric field for a sheet of charge is E # = 2P0 .
EXECUTE:! Substituting the numbers gives us E # =
2P0 # 2.50 " 10!7 C/m 2
# 1.41 " 104 N/C . Then
2 . 8.85 " 10!12 C2 /N + m 2 / 9 . 5.00 " 10!8 C / .1.41" 104 N/C / :
? # 19.83
!2
2
> . 2.00 " 10 kg /. 9.80 m/s / ?
;
< 4 # arctan > EVALUATE:! Increasing the field, or decreasing the mass of the sphere, would cause the sphere to hang at a larger
angle. 2214 Chapter 22 22.42. IDENTIFY:! Apply Gauss’s law.
SET UP:! Use a Gaussian surface that is a sphere of radius r and that is concentric with the conducting spheres.
EXECUTE:! (a) For r 1 a, E # 0, since these points are within the conducting material. q
For a 1 r 1 b, E # 1 2 , since there is +q inside a radius r.
4, P0 r
For b 1 r 1 c, E # 0, since since these points are within the conducting material
1q
, since again the total charge enclosed is +q.
For r 5 c, E #
4, P0 r 2
(b) The graph of E versus r is sketched in Figure 22.42a.
(c) Since the Gaussian sphere of radius r, for b 1 r 1 c , must enclose zero net charge, the charge on inner shell
surface is –q.
(d) Since the hollow sphere has no net charge and has charge !q on its inner surface, the charge on outer shell
surface is +q.
(e) The field lines are sketched in Figure 22.42b. Where the field is nonzero, it is radially outward.
EVALUATE:! The net charge on the inner solid conducting sphere is on the surface of that sphere. The presence of
the hollow sphere does not affect the electric field in the region r 1 b . Figure 22.42
22.43. IDENTIFY:! Apply Gauss’s law.
SET UP:! Use a Gaussian surface that is a sphere of radius r and that is concentric with the charge distributions.
EXECUTE:! (a) For r 1 R, E # 0, since these points are within the conducting material. For R 1 r 1 2R , 1 2Q
Q
E # 1 2 , since the charge enclosed is Q. For r 5 2 R , E #
since the charge enclosed is 2Q.
4!P0 r
4, P0 r 2
(b) The graph of E versus r is sketched in Figure 22.43.
EVALUATE:! For r 1 2 R the electric field is unaffected by the presence of the charged shell. Figure 22.43
22.44. IDENTIFY:! Apply Gauss’s law and conservation of charge.
SET UP:! Use a Gaussian surface that is a sphere of radius r and that has the point charge at its center.
1Q
, radially outward, since the charge enclosed is Q, the charge of the point
EXECUTE:! (a) For r 1 a , E #
4, P0 r 2
2Q
charge. For a 1 r 1 b , E # 0 since these points are within the conducting material. For r 5 b , E # 1
,
4!P0 r 2
radially inward, since the total enclosed charge is –2Q. Gauss’s Law 2215 (b) Since a Gaussian surface with radius r, for a 1 r 1 b , must enclose zero net charge, the total charge on the inner
Q
surface is !Q and the surface charge density on inner surface is # # !
.
4!a 2
(c) Since the net charge on the shell is !3Q and there is !Q on the inner surface, there must be !2Q on the outer
2Q
.
4, b 2
(d) The field lines and the locations of the charges are sketched in Figure 22.44a.
(e) The graph of E versus r is sketched in Figure 22.44b. surface.! The surface charge density on the outer surface is = # ! Figure 22.44 22.45. EVALUATE:!! For r 1 a the electric field is due solely to the point charge Q. For r 5 b the electric field is due to
the charge !2Q that is on the outer surface of the shell.
IDENTIFY:! Apply Gauss’s law to a spherical Gaussian surface with radius r. Calculate the electric field at the
surface of the Gaussian sphere.
(a) SET UP:! (i) r 1 a : The Gaussian surface is sketched in Figure 22.45a.
EXECUTE:! J E # EA # E (4, r 2 ) Qencl # 0; no charge is enclosed
JE # Qencl
says E (4, r 2 ) # 0 and E # 0.
P0 Figure 22.45a (ii) a 1 r 1 b: Points in this region are in the conductor of the small shell, so E = 0.
(iii) SET UP:! b 1 r 1 c: The Gaussian surface is sketched in Figure 22.45b.
Apply Gauss’s law to a spherical Gaussian surface with radius b 1 r 1 c. EXECUTE:! J E # EA # E (4, r 2 )
The Gaussian surface encloses all of the small
shell and none of the large shell, so Qencl # 02q. Figure 22.45b Qencl
2q
2q
gives E (4, r 2 ) #
so E #
. Since the enclosed charge is positive the electric field is radially
4, P0 r 2
P0
P0
outward.
(iv) c 1 r 1 d : Points in this region are in the conductor of the large shell, so E # 0.
JE # 2216 Chapter 22 (v) SET UP:! r 5 d : Apply Gauss’s law to a spherical Gaussian surface with radius r > d, as shown in Figure 22.45c. EXECUTE:! J E # EA # E . 4, r 2 / The Gaussian surface encloses all of the small shell
and all of the large shell, so Qencl # 02q 0 4q # 6q. Figure 22.45c Qencl
6q
gives E . 4, r 2 / #
P0
P0
6q
E#
. Since the enclosed charge is positive the electric field is radially outward.
4, P0 r 2
The graph of E versus r is sketched in Figure 22.45d.
JE # Figure 22.45d
(b) IDENTIFY and SET UP:! Apply Gauss’s law to a sphere that lies outside the surface of the shell for which we
want to find the surface charge.
EXECUTE:! (i) charge on inner surface of the small shell: Apply Gauss’s law to a spherical Gaussian surface with
radius a 1 r 1 b. This surface lies within the conductor of the small shell, where E = 0, so J E # 0. Thus by Gauss’s 22.46. law Qencl # 0, so there is zero charge on the inner surface of the small shell.
(ii) charge on outer surface of the small shell: The total charge on the small shell is 02q. We found in part (i) that
there is zero charge on the inner surface of the shell, so all 0 2q must reside on the outer surface.
(iii) charge on inner surface of large shell: Apply Gauss’s law to a spherical Gaussian surface with radius c 1 r 1 d .
The surface lies within the conductor of the large shell, where E = 0, so J E # 0. Thus by Gauss’s law Qencl # 0. The
surface encloses the 02q on the small shell so there must be charge !2q on the inner surface of the large shell to
make the total enclosed charge zero.
(iv) charge on outer surface of large shell: The total charge on the large shell is 0 4q . We showed in part (iii) that
the charge on the inner surface is !2q , so there must be 06q on the outer surface.
EVALUATE:! The electric field lines for b 1 r 1 c originate from the surface charge on the outer surface of the
inner shell and all terminate on the surface charge on the inner surface of the outer shell. These surface charges have
equal magnitude and opposite sign. The electric field lines for r 5 d originate from the surface charge on the outer
surface of the outer sphere.
IDENTIFY:! Apply Gauss’s law.
SET UP:! Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells.
EXECUTE:! (a)! (i)!For r 1 a, E # 0, since the charge enclosed is zero.! (ii) For a 1 r 1 b, E # 0, since the points
1 2q
, outward, since charge enclosed is 0 2q. !
4, P0 r 2
(iv) For c 1 r 1 d , E # 0, since the points are within the conducting material.! (v) For r 5 d , E # 0, since the net
charge enclosed is zero. The graph of E versus r is sketched in Figure 22.46. are within the conducting material.! (iii) For b 1 r 1 c, E # Gauss’s Law 2217 (b)! (i) small shell inner surface: Since a Gaussian surface with radius r, for a 1 r 1 b , must enclose zero net
charge, the charge on this surface is zero.! (ii)!small shell outer surface: 02q .! (iii)!large shell inner surface: Since
a Gaussian surface with radius r, for c 1 r 1 d , must enclose zero net charge, the charge on this surface is !2q .!
(iv)!large shell outer surface: Since there is !2q on the inner surface and the total charge on this conductor is !2q ,
the charge on this surface is zero.
EVALUATE:! The outer shell has no effect on the electric field for r 1 c . For r 5 d the electric field is due only to
the charge on the outer surface of the larger shell. Figure 22.46
22.47. IDENTIFY:! Apply Gauss’s law
SET UP:! Use a Gaussian surface that is a sphere of radius r and that is concentric with the charged shells.
EXECUTE:! (a)! (i) For r 1 a, E # 0, since charge enclosed is zero.! (ii) a 1 r 1 b, E # 0, since the points are
2q
within the conducting material.! (iii) For b 1 r 1 c, E # 1 2 , outward, since charge enclosed is +2q.!
4!P0 r
2q
(iv) For c 1 r 1 d , E # 0, since the points are within the conducting material.! (v)!For r 5 d , E # 1 2 , inward,
4!P0 r
since charge enclosed is –2q. The graph of the radial component of the electric field versus r is sketched in
Figure 22.47, where we use the convention that outward field is positive and inward field is negative.
(b)! (i)!small shell inner surface: Since a Gaussian surface with radius r, for a < r < b, must enclose zero net
charge, the charge on this surface is zero.! (ii)!small shell outer surface: +2q.! (iii)!large shell inner surface: Since
a Gaussian surface with radius r, for c < r < d, must enclose zero net charge, the charge on this surface is –2q.!
(iv) large shell outer surface: Since there is –2q on the inner surface and the total charge on this conductor is –4q,
the charge on this surface is –2q.
EVALUATE:! The outer shell has no effect on the electric field for r 1 c . For r 5 d the electric field is due only to
the charge on the outer surface of the larger shell. Figure 22.47
22.48. IDENTIFY:! Apply Gauss’s law.
SET UP:! Use a Gaussian surface that is a sphere of radius r and that is concentric with the sphere and shell. The
4
28, 3
R.
volume of the insulating shell is V # , [2 R]3 ! R3 #
3
3
3Q
28! "R 3
EXECUTE:! (a) Zero net charge requires that !Q #
, so " # !
.
28!R L
3
(b) For r 1 R, E # 0 since this region is within the conducting sphere. For r 5 2 R, E # 0, since the net charge enclosed
Q 4! " 3
(r ! R3 ) and
by the Gaussian surface with this radius is zero. For R 1 r 1 2R , Gauss’s law gives E (4!r 2 ) # 0
P0 3P0
Q
"
Q
Qr
E#
0
(r 3 ! R 3 ) . Substituting " from part (a) gives E # 2 2 !
. The net enclosed charge for
7!P0 r
28!P0 R 3
4!P0 r 2 3P0r 2
each r in this range is positive and the electric field is outward. . / 2218 Chapter 22 (c) The graph is sketched in Figure 22.48. We see a discontinuity in going from the conducting sphere to the
insulator due to the thin surface charge of the conducting sphere. But we see a smooth transition from the uniform
insulator to the surrounding space.
EVALUATE:! The expression for E within the insulator gives E # 0 at r # 2 R . 22.49. Figure 22.48
!
IDENTIFY:! Use Gauss’s law to find the electric field E produced by the shell for r 1 R and r 5 R and then use
!
!
F # qE to find the force the shell exerts on the point charge.
(a) SET UP:! Apply Gauss’s law to a spherical Gaussian surface that has radius r 5 R and that is concentric with
the shell, as sketched in Figure 22.49a.
EXECUTE:! J E # E . 4, r 2 / Qencl # !Q Figure 22.49a JE # Qencl
!Q
gives E . 4, r 2 / #
P0
P0 Q
qQ
and it is directed toward the center of the shell. Then F # qE #
,
4, P0 r 2
4, P0 r 2
!
!
directed toward the center of the shell. (Since q is positive, E and F are in the same direction.)
(b) SET UP:! Apply Gauss’s law to a spherical Gaussian surface that has radius r 1 R and that is concentric with
the shell, as sketched in Figure 22.49b. The magnitude of the field is E # EXECUTE:! J E # E . 4, r 2 / Qencl # 0
Figure 22.49b Qencl
gives E . 4, r 2 / # 0
P0
Then E # 0 so F # 0.
EVALUATE: Outside the shell the electric field and the force it exerts is the same as for a point charge !Q located
at the center of the shell. Inside the shell E # 0 and there is no force.
IDENTIFY:! The method of Example 22.9 shows that the electric field outside the sphere is the same as for a point
charge of the same charge located at the center of the sphere.
SET UP:! The charge of an electron has magnitude e # 1.60 " 10!19 C .
JE # 22.50. Gauss’s Law 2219 q
Er 2 (1150 N/C)(0.150 m)2
. For r # R # 0.150 m , E # 1150 N/C so q #
#
# 2.88 " 10!9 C .
2
k
r
8.99 " 109 N + m 2 /C2
2.88 " 10!9 C
# 1.80 " 1010 electrons .
The number of excess electrons is
1.60 " 10!19 C/electron
EXECUTE:! (a) E # k (b) r # R 0 0.100 m # 0.250 m . E # k 22.51. q
r 2 # (8.99 " 109 N + m 2 /C 2 ) 2.88 " 10!9 C
# 414 N/C .
(0.250 m) 2 EVALUATE:! The magnitude of the electric field decreases according to the square of the distance from the center
of the sphere.
IDENTIFY:! The net electric field is the vector sum of the fields due to the sheet of charge on each surface of the
plate.
SET UP:! The electric field due to the sheet of charge on each surface is E # = / 2P0 and is directed away from the
surface.
EXECUTE:! (a) For the conductor the charge sheet on each surface produces fields of magnitude = / 2P0 and in the same direction, so the total field is twice this, or = / P0 .
(b) At points inside the plate the fields of the sheets of charge on each surface are equal in magnitude and opposite
in direction, so their vector sum is zero. At points outside the plate, on either side, the fields of the two sheets of
charge are in the same direction so their magnitudes add, giving E # = / P0 .
22.52. EVALUATE:! Gauss’s law can also be used directly to determine the fields in these regions.
IDENTIFY:! Example 22.9 gives the expression for the electric field both inside and outside a uniformly charged
!
!
sphere. Use F = !eE to calculate the force on the electron.
SET UP:! The sphere has charge Q # 0 e .
EXECUTE:! (a) Only at r # 0 is E # 0 for the uniformly charged sphere.
(b) At points inside the sphere, Er # er
1 e2r
. The field is radially outward. Fr # !eE # !
. The minus sign
3
4, P0 R 3
4!P0 R denotes that Fr is radially inward. For simple harmonic motion, Fr # ! kr # ! m&2 r , where M # k / m # 2, f .
Fr # ! m&2 r # ! 1 e2r
1 e2
1
1 e2
so & #
and f #
.
3
3
4!P0 R
4!P0 mR
2! 4!P0 mR 3 (c) If f # 4.57 " 1014 Hz # 1
2, 1 e2
1
(1.60 " 10!19 C)2
then R # 3
# 3.13 " 10!10 m.
3
2
4!P0 mR
4!P0 4! (9.11 " 10!31 kg)(4.57 " 1014 Hz) 2 The atom radius in this model is the correct order of magnitude.
e
e2
(d) If r 5 R , Er #
and Fr # !
. The electron would still oscillate because the force is directed toward
4, P0 r 2
4, P0 r 2 22.53. the equilibrium position at r # 0 . But the motion would not be simple harmonic, since Fr is proportional to 1/ r 2 and
simple harmonic motion requires that the restoring force be proportional to the displacement from equilibrium.
EVALUATE:! As long as the initial displacement is less than R the frequency of the motion is independent of the
initial displacement.
IDENTIFY:! There is a force on each electron due to the other electron and a force due to the sphere of charge. Use
Coulomb’s law for the force between the electrons. Example 22.9 gives E inside a uniform sphere and Eq.(21.3)
gives the force.
SET UP:! The positions of the electrons are sketched in Figure 22.53a. If the electrons are in
equilibrium the net force on
each one is zero.
Figure 22.53a 2220 Chapter 22 EXECUTE:! Consider the forces on electron 2. There is a repulsive force F1 due to the other electron, electron 1. F1 # 1
e2
4, P0 . 2d /2 The electric field inside the uniform distribution of positive charge is E # Qr
(Example 22.9), where Q # 02e.
4, P0 R 3 At the position of electron 2, r = d. The force Fcd exerted by the positive charge distribution is Fcd # eE # e . 2e / d
4, P0 R 3 and is attractive.
The force diagram for electron 2 is given in Figure 22.53b. Figure 22.53b
1 e2
2e 2 d
Net force equals zero implies F1 # Fcd and
#
2
4, P0 4d
4, P0 R 3 Thus .1/ 4d 2 / # 2d / R 3 , so d 3 # R 3 /8 and d # R / 2. 22.54. EVALUATE:! The electric field of the sphere is radially outward; it is zero at the center of the sphere and increases
with distance from the center. The force this field exerts on one of the electrons is radially inward and increases as
the electron is farther from the center. The force from the other electron is radially outward, is infinite when d = 0
and decreases as d increases. It is reasonable therefore for there to be a value of d for which these forces balance.
IDENTIFY:! Use Gauss’s law to find the electric field both inside and outside the slab.
SET UP:! Use a Gaussian surface that has one face of area A in the y z plane at x # 0, and the other face at a
general value x. The volume enclosed by such a Gaussian surface is Ax.
EXECUTE:! (a) The electric field of the slab must be zero by symmetry. There is no preferred direction in the y z
plane, so the electric field can only point in the xdirection. But at the origin, neither the positive nor negative
xdirections should be singled out as special, and so the field must be zero.
"A x
"x
Q
xˆ
(b) For x N d , Gauss’s law gives EA # encl #
and E #
, with direction given by
i (away from the
P0
P0
P0
x
center of the slab). Note that this expression does give E # 0 at x # 0. ! Outside the slab, the enclosed charge does
Q
"Ad
"d
not depend on x and is equal to G Ad . For x O d , Gauss’s law gives EA # encl #
and E #
, again with
P0
P0
P0 xˆ
i.
 x
EVALUATE:! At the surfaces of the slab, x # 6 d . For these values of x the two expressions for E (for inside and
outside the slab) give the same result. The charge per unit area = of the slab is given by = A # G A(2d ) and direction given by 22.55. G d # = / 2 . The result for E outside the slab can therefore be written as E # = / 2P0 and is the same as for a thin sheet
of charge.
(a) IDENTIFY and SET UP:! Consider the direction of the field for x slightly greater than and slightly less than zero.
The slab is sketched in Figure 22.55a. G . x / # G0 . x / d / 2 Figure 22.55a
EXECUTE:! The charge distribution is symmetric about x = 0, so by symmetry E . x / # E . ! x / . But for x > 0 the field is in the 0 x direction and for x < 0 the field is in the ! x direction. At x = 0 the field can’t be both in the
0 x and ! x directions so must be zero. That is, Ex . x / # ! Ex . ! x / . At point x = 0 this gives Ex . 0 / # ! Ex . 0 / and
this equation is satisfied only for Ex . 0 / # 0. Gauss’s Law (b) IDENTIFY and SET UP:! 2221 x 5 d (outside the slab) Apply Gauss’s law to a cylindrical Gaussian surface whose axis is perpendicular to the slab and whose end caps
have area A and are the same distance x 5 d from x = 0, as shown in Figure 22.55b. EXECUTE:! J E # 2 EA Figure 22.55b To find Qencl consider a thin disk at
coordinate x and with thickness dx, as
shown in Figure 22.55c. The charge
within this disk is
dq # G dV # G Adx # . G0 A / d 2 / x 2 dx.
Figure 22.55c The total charge enclosed by the Gaussian cylinder is
2
Qencl # 2 H dq # . 2 G0 A / d 2 / H x 2 dx # . 2 G 0 A / d 2 /. d 3 / 3/ # 3 G0 Ad .
d Then J E # d 0 0 Qencl
gives 2 EA # 2 G 0 Ad / 3P0 .
P0 E # G 0 d / 3P0
!
!
ˆ
E is directed away from x # 0, so E # . G 0 d / 3P0 / . x / x / i .
IDENTIFY and SET UP:! x 1 d (inside the slab) Apply Gauss’s law to a cylindrical Gaussian surface whose axis is perpendicular to the slab and whose end caps
have area A and are the same distance x 1 d from x # 0, as shown in Figure 22.55d. EXECUTE:! J E # 2 EA Figure 22.55d Qencl is found as above, but now the integral on dx is only from 0 to x instead of 0 to d.
Qencl # 2 H dq # . 2 G0 A / d 2 / H x 2 dx # . 2 G 0 A / d 2 /. x 3 / 3/ .
x 0 Then J E # x 0 Qencl
gives 2 EA # 2 G 0 Ax3 / 3P0 d 2 .
P0
E # G 0 x 3 / 3P0 d 2 !
!
ˆ
E is directed away from x # 0, so E # . G0 x3 / 3P0 d 2 / i . EVALUATE:! Note that E = 0 at x = 0 as stated in part (a). Note also that the expressions for x 5 d and x 1 d
22.56. agree for x # d .
!
!
IDENTIFY:! Apply F # qE to relate the force on q to the electric field at the location of q.
SET UP:! Flux is negative if the electric field is directed into the enclosed volume. 2222 Chapter 22 EXECUTE:! (a) We could place two charges 0Q on either side of the charge 0 q , as shown in Figure 22.56.
(b) In order for the charge to be stable, the electric field in a neighborhood around it must always point back to the
equilibrium position.
(c) If q is moved to infinity and we require there to be an electric field always pointing in to the region where q
had been, we could draw a small Gaussian surface there. We would find that we need a negative flux into the
surface. That is, there has to be a negative charge in that region. However, there is none, and so we cannot get such a
stable equilibrium.
(d) For a negative charge to be in stable equilibrium, we need the electric field to always point away from the charge
position. The argument in (c) carries through again, this time implying that a positive charge must be in the space
where the negative charge was if stable equilibrium is to be attained.
EVALUATE:! If q is displaced to the left or right in Figure 22.56, the net force is directed back toward the
equilibrium position. But if q is displaced slightly in a direction perpendicular to the line connecting the two charges
Q, then the net force on q is directed away from the equilibrium position and the equilibrium is not stable for such a
displacement. 22.57. Figure 22.56
G . r / # G 0 .1 ! r / R / for r N R where G 0 # 3Q / , R . G . r / # 0 for r O R
3 (a) IDENTIFY:! The charge density varies with r inside the spherical volume. Divide the volume up into thin
concentric shells, of radius r and thickness dr. Find the charge dq in each shell and integrate to find the total charge.
SET UP:! The thin shell is sketched in Figure 22.57a.
EXECUTE:! The volume of such
a shell is dV # 4, r 2 dr
The charge contained within the
shell is
dq # G . r / dV # 4, r 2 G 0 .1 ! r / R / dr
Figure 22.57a The total charge Q in the charge distribution is obtained by integrating dq over all such shells into which the sphere
can be subdivided:
R R 0 0 Q # H dq # H 4, r 2 G0 .1 ! r / R / dr # 4,G 0 H .r 2 ! r 3 / R / dr R 9 r3 r 4 :
% R3 R 4 &
3
3
3
!
Q # 4,G 0 > !
( # 4,G 0 . R /12 / # 4, . 3Q / , R /. R /12 / # Q, as was to be shown.
? # 4,G 0 '
3 4R <0
3 4R *
;
)
(b) IDENTIFY:! Apply Gauss’s law to a spherical surface of radius r, where r > R.
SET UP:! The Gaussian surface is shown in Figure 22.57b.
EXECUTE:! J E # E . 4, r 2 / # Qencl
P0 Q
P0 Figure 22.57b Q
; same as for point charge of charge Q.
4, P0 r 2
(c) IDENTIFY:! Apply Gauss’s law to a spherical surface of radius r, where r < R:
SET UP:! The Gaussian surface is shown in Figure 22.57c.
E# EXECUTE:! J E # J E # E . 4, r 2 /
Figure 22.57c Qencl
P0 Gauss’s Law 2223 The calculate the enclosed charge Qencl use the same technique as in part (a), except integrate dq out to r rather than
R. (We want the charge that is inside radius r.)
r
r%
r3 &
% r &
Qencl # H 4, r 2 G0 '1 ! ( dr  # 4,G 0 H ' r2 ! (dr 0
0
R*
) R*
)
r 9 r 3 r 4 :
% r3 r4 &
r&
3%1
Qencl # 4,G0 > !
( # 4,G 0 r ' !
? # 4,G 0 ' !
(
3 4R <0
3 4R *
) 3 4R *
;
) G0 # % r3 &%
3Q
r3 % 1 r &
r&
# Q ' 3 ( ' 4 ! 3 (.
so Qencl # 12Q 3 ' !
(
3
,R
R ) 3 4R *
R*
) R *) Thus Gauss’s law gives E . 4, r 2 / # Q % r3 &%
r&
' 3 (' 4 ! 3 (
P0 ) R * )
R* 3r &
Qr %
' 4 ! (, r N R
R*
4, P0 R 3 )
(d) The graph of E versus r is sketched in Figure 22.57d.
E# Figure 22.57d
dE
d%
3r 2 &
# 0. Thus
(e) Where the electric field is a maximum,
' 4r !
( # 0 so 4 ! 6r / R # 0 and r # 2 R / 3.
dr
dr )
R* At this value of r, E # 3 2R &
Q % 2 R &%
Q
'
(' 4 !
(#
R 3 * 3, P0 R 2
4, P0 R 3 ) 3 *) EVALUATE:! Our expressions for E . r / for r 1 R and for r 5 R agree at r # R. The results of part (e) for the value of r where E . r / is a maximum agrees with the graph in part (d).
22.58. IDENTIFY:! Apply Gauss’s law.
SET UP:! Use a Gaussian surface that is a sphere of radius r and that is concentric with the spherical distribution of
charge. The volume of a thin spherical shell of radius r and thickness dr is dV # 4, r 2 dr .
F
R
R
9R 2
4r & 2
4 3:
%
EXECUTE:! (a) Q # H "(r )dV # 4! H "(r )r 2 dr # 4!"0 H ' 1 !
r dr ?
( r dr # 4!"0 > H r dr !
3R *
3R H
0
0)
0
;0
< 9 R3 4 R 4 :
Q # 4!"0 > !
? # 0 . The total charge is zero.
; 3 3R 4 <
!!Q
(b) For r O R , ú E + dA # encl # 0 , so E # 0 .
P0
! ! Qencl 4! r
4!" 9 r 2
4 r3 :
2
2
(c) For r N R , ú E + dA #
#
H 0 "(r)r dr . E 4!r # P0 0 > H 0 r dr ! 3R H 0 r dr? and
P0
P0
;
<
"0 1 9 r 3 r 4 : "0 9
r:
!
r >1 ! ? .
?#
2>
P0 r ; 3 3R < 3P0 ; R <
(d) The graph of E versus r is sketched in Figure 22.58.
2" r
" R 9 1: " R
"
R
dE
(e) Where E is a maximum,
# 0 . This gives 0 ! 0 max # 0 and rmax # . At this r, E # 0 >1 ! ? # 0 .
3P0 2 ; 2 < 12P0
2
dr
3P0
3P0 R
E# 2224 Chapter 22 EVALUATE:! The result in part (b) for r N R gives E # 0 at r # R ; the field is continuous at the surface of the
charge distribution. Figure 22.58
22.59. IDENTIFY:! Follow the steps specified in the problem.
!
ˆ
SET UP:! In spherical polar coordinates dA = r 2 sin 4 d4 dA r . ú sin 4 d4 dA # 4, . r sin' d' dA
!!
EXECUTE:! (a) J g # ú g + dA # !Gm ú
# !4!Gm.
r2
(b) For any closed surface, mass OUTSIDE the surface contributes zero to the flux passing through the surface.
Thus the formula above holds for any situation where m is the mass enclosed by the Gaussian surface.
!!
That is, J g # ú g + dA # !4!GM encl .
2 22.60. EVALUATE:! The minus sign in the expression for the flux signifies that the flux is directed inward.
!!
IDENTIFY:! Apply ú g + dA # !4!GM encl .
SET UP:! Use a Gaussian surface that is a sphere of radius r, concentric with the mass distribution. Let J g denote
!!
ú g + dA
EXECUTE:! (a) Use a Gaussian sphere with radius r 5 R , where R is the radius of the mass distribution. g is
constant on this surface and the flux is inward. The enclosed mass is M. Therefore, J g # ! g 4!r 2 # !4!GM and GM
, which is the same as for a point mass.
r2
(b) For a Gaussian sphere of radius r 1 R , where R is the radius of the shell, M encl # 0, so g # 0.
g# %4
&
(c) Use a Gaussian sphere of radius r 1 R , where R is the radius of the planet. Then M encl # G ' , r 3 ( # Mr 3 / R 3 .
)3
* 22.61. % r3 &
GMr
This gives J g # ! g 4!r 2 # ! 4!GM encl # ! 4!G ' M 3 ( and g # 3 , which is linear in r.
R
) R*
!
EVALUATE:!! The spherically synmetric distribution of mass results in an acceleration due to gravity g that is
radical and that depends only on r, the distance from the center of the mass distribution.
!!
(a) IDENTIFY:! Use E . r / from Example (22.9) (inside the sphere) and relate the position vector of a point inside
the sphere measured from the origin to that measured from the center of the sphere.
SET UP:! For an insulating sphere of uniform charge density G and centered at the origin, the electric field inside
!
the sphere is given by E # Qr  / 4, P0 R3 (Example 22.9), where r  is the vector from the center of the sphere to the
point where E is calculated.
!
But G # 3Q / 4, R 3 so this may be written as E # G r / 3P0 . And E is radially outward, in the direction of
!
!
!
r , so E = G r  / 3P0 . Gauss’s Law 2225 !
!
For a sphere whose center is located by vector b , a point inside the sphere and located by r is located by the vector
!
!!
r  # r ! b relative to the center of the sphere, as shown in Figure 22.61. !!
! G r !b
EXECUTE:! Thus E #
3P0 . / Figure 22.61 !!
EVALUATE:! When b = 0 this reduces to the result of Example 22.9. When r # b , this gives E # 0, which is
correct since we know that E # 0 at the center of the sphere.
(b) IDENTIFY:! The charge distribution can be represented as a uniform sphere with charge density G and centered
!!
at the origin added to a uniform sphere with charge density ! G and centered at r = b.
!!
!
!
SET UP:! E = Euniform 0 Ehole , where Euniform is the field of a uniformly charged sphere with charge density G and
!
Ehole is the field of a sphere located at the hole and with charge density ! G . (Within the spherical hole the net
charge density is 0 G ! G # 0. )
!
!
Gr
!
EXECUTE:! E uniform #
, where r is a vector from the center of the sphere.
3P0
!!
!G r ! b
!
, at points inside the hole.
Ehole #
3P0
!!
!
!
! G r % !G r ! b & Gb
(#
Then E #
.
0'
( 3P0
3P0 '
3P0
)
*
!
!
!
EVALUATE:! E is independent of r so is uniform inside the hole. The direction of E inside the hole is in the
!
direction of the vector b , the direction from the center of the insulating sphere to the center of the hole.
IDENTIFY:! We first find the field of a cylinder offaxis, then the electric field in a hole in a cylinder is the
difference between two electric fields: that of a solid cylinder onaxis, and one offaxis, at the location of the hole.
!
!
SET UP:! Let r locate a point within the hole, relative to the axis of the cylinder and let r  locate this point relative
!
to the axis of the hole. Let b locate the axis of the hole relative to the axis of the cylinder. As shown in Figure 22.62,
!
! Gr
!!!
r  = r ! b . Problem 23.48 shows that at points within a long insulating cylinder, E =
.
2P0
!
!
!!
!
!!
!
!
!
"r  " ( r ! b ) !
" r "( r ! b ) " b
EXECUTE:! Eoffaxis =
. Ehole = Ecylinder ! Eoffaxis #
.
#
!
#
2P0
2P0
2P0
2P0
2P0
!
Note that E is uniform.
EVALUATE:! If the hole is coaxial with the cylinder, b # 0 and Ehole # 0 . . / . 22.62. / Figure 22.62 2226 Chapter 22 22.63. IDENTIFY:! The electric field at each point is the vector sum of the fields of the two charge distributions.
Gr
.
SET UP:! Inside a sphere of uniform positive charge, E #
3P0 G# 4
3 Q
3Q
Qr
#
so E #
, directed away from the center of the sphere. Outside a sphere of uniform
3
3
4, P0 R 3
, R 4, R Q
, directed away from the center of the sphere.
4, P0 r 2
EXECUTE:! (a) x # 0. This point is inside sphere 1 and outside sphere 2. The fields are shown in
Figure 22.63a. positive charge, E # E1 # Qr
# 0, since r # 0.
4, P0 R 3 Figure 22.63a Q
Q
with r # 2 R so E2 #
, in the ! x direction.
2
4, P0 r
16, P0 R 2
!!
!
Q
ˆ
Thus E # E1 0 E2 #
i.
16, P0 R 2
(b) x # R / 2. This point is inside sphere 1 and outside sphere 2. Each field is directed away from the center of the
sphere that produces it. The fields are shown in Figure 22.63b.
E2 # E1 # Qr
with r # R / 2 so
4, P0 R 3 E1 # Q
8, P0 R 2 Figure 22.63b E2 # Q
Q
with r # 3R / 2 so E2 #
4, P0 r 2
9, P0 R 2 !
Q
Q
ˆ
, in the 0 xdirection and E #
i
2
72, P0 R
72, P0 R 2
(c) x = R. This point is at the surface of each sphere. The fields have equal magnitudes and opposite directions, so
E = 0.
(d) x = 3R. This point is outside both spheres. Each field is directed away from the center of the sphere that produces
it. The fields are shown in Figure 22.63c. E # E1 ! E2 # E1 # Q
with r # 3R so
4, P0 r 2 E1 # Q
36, P0 R 2 Figure 22.63c E2 # Q
Q
with r # R so E2 #
4, P0 r 2
4, P0 R 2 !
5Q
5Q ˆ
i
, in the 0 x direction and E #
2
18, P0 R
18, P0 R 2
EVALUATE:! The field of each sphere is radially outward from the center of the sphere. We must use the correct
expression for E(r) for each sphere, depending on whether the field point is inside or outside that sphere.
IDENTIFY:! The net electric field at any point is the vector sum of the fields at each sphere.
SET UP:! Example 22.9 gives the electric field inside and outside a uniformly charged sphere. For the positively
charged sphere the field is radially outward and for the negatively charged sphere the electric field is radially
inward. E # E1 0 E2 # 22.64. Gauss’s Law 2227 EXECUTE:!! (a) At this point the field of the lefthand sphere is zero and the field of the righthand sphere is toward
the center of that sphere, in the +xdirection. This point is outside the righthand sphere, a distance r # 2R from its
!
1 Qˆ
i.
center. E = 0
4!P0 4 R 2
(b) This point is inside the lefthand sphere, at r # R / 2 , and is outside the righthand sphere, at r # 3R / 2 . Both
fields are in the +xdirection. !
:ˆ
1 9 Q ( R / 2)
Q
1 9Q
4Q : ˆ
1 17Q ˆ
0
0
E=
i=
i.
>
?i =
4!P0 ; R 3
(3 R 2) 2 <
4!P0 > 2 R 2 9 R 2 ?
4!P0 18R 2
;
<
(c) This point is outside both spheres, at a distance r # R from their centers. Both fields are in the +xdirection.
!
1 9Q Q :ˆ
Qˆ
E=
i=
i.
0
4!P0 > R 2 R 2 ?
2!P0 R 2
;
<
(d) This point is outside both spheres, a distance r # 3R from the center of the lefthand sphere and a distance
r # R from the center of the righthand sphere. The field of the lefthand sphere is in the +xdirection and the field
!
19Q
Q :ˆ
1 9Q
Q : ˆ !1 8Q ˆ
i=
i.
of the righthand sphere is in the ! x direction . E =
! ?i =
!
>
4!P0 ; (3R ) 2 R 2 <
4!P0 > 9 R 2 R 2 ?
4!P0 9 R 2
;
< 22.65. EVALUATE:! At all points on the xaxis the net field is parallel to the xaxis.
IDENTIFY:! Let ! dQ be the electron charge contained within a spherical shell of radius r  and thickness dr  .
Integrate r  from 0 to r to find the electron charge within a sphere of radius r. Apply Gauss’s law to a sphere of
radius r to find the electric field E ( r ) .
SET UP:! The volume of the spherical shell is dV # 4, r 2 dr  .
Q 4, !2 r / a0 2
4Q r
EXECUTE:! (a) Q (r ) # Q ! H "dV # Q !
e
r dr # Q ! 3 H r 2e!2 r  / a0 dr .
3H
!a0
a0 0 Q(r ) # Q ! 4Qe!2r
(2e2r ! % 2 r 2 ! 22r ! 2) # Qe !2 r / a0 [2(r / a0 ) 2 0 2(r / a0 ) 0 1].
a032 3 Note if r E F, Q (r ) E 0 ; the total net charge of the atom is zero.
(b) The electric field is radially outward. Gauss’s law gives E (4, r 2 ) # Q(r )
and
P0 kQe !2 r / a0
(2(r a0 ) 2 0 2(r a0 ) 0 1) .
r2
(c) The graph of E versus r is sketched in Figure 22.65. What is plotted is the scaled E, equal to E / Ept charge , versus
E# kQ
is the field of a point charge.
r2
EVALUATE:! As r E 0 , the field approaches that of the positive point charge (the proton). For increasing r the
electric field falls to zero more rapidly than the 1/ r 2 dependence for a point charge.
scaled r, equal to r / a0 . Ept charge # 22.66. Figure 22.65
IDENTIFY:! The charge in a spherical shell of radius r and thickness dr is dQ # G ( r )4, r 2 dr . Apply Gauss’s law.
SET UP:! Use a Gaussian surface that is a sphere of radius r. Let Qi be the charge in the region r N R / 2 and let Q0 be the charge in the region where R / 2 N r N R . 2228 Chapter 22 4! ( R 2)3 %!R 3
#
and
3
6
R
% ( R 3 ! R 3 8) ( R 4 ! R 4 16) & 11%!R 3
15%!R 3
8Q
. Therefore, Q #
.
Q0 # 4! (2% ) H (r 2 ! r 3 / R )dr # 8%! '
and % #
!
(#
R/2
5!R 3
24
3
4R
24
)
*
%r
8Qr
% 4!r 3
(b) For r N R 2 , Gauss’s law gives E 4!r 2 #
and E #
. For R 2 N r N R ,
#
3P0 15!P0 R 3
3P0
EXECUTE:! (a) The total charge is Q # Qi 0 Q0 , where Qi # 2 E 4 !r 2 # E# % (r 3 ! R 3 8) (r 4 ! R 4 16) & &
Qi 1 %
0 ' 8%! '
!
( ( and
'
(
3
4R
P0 P0 )
)
** %!R 3
kQ
(64(r R)3 ! 48(r R ) 4 ! 1) #
(64(r R )3 ! 48(r R )4 ! 1).
2
24P0 (4!r )
15r 2 For r O R , E (4, r 2 ) #
(c) Q
Q
and E #
.
P0
4!P0 r 2 Qi ( 4Q /15) 4
#
# # 0.267.
Q
Q
15 8eQ
r , so the restoring force depends upon displacement to the first power, and
15!P0 R 3
we have simple harmonic motion. (d) For r N R / 2 , Fr # ! eE # ! (e) Comparing to F # ! kr , k # 22.67. 8eQ
k
8eQ
2!
15!P0 R 3me
. Then & #
and T #
#
# 2!
.
3
3
15!P0 R me
&
8eQ
15!P0 R
me EVALUATE:! (f ) If the amplitude of oscillation is greater than R / 2, the force is no longer linear in r , and is thus
no longer simple harmonic.
IDENTIFY:! The charge in a spherical shell of radius r and thickness dr is dQ # G ( r )4, r 2 dr . Apply Gauss’s law.
SET UP:! Use a Gaussian surface that is a sphere of radius r. Let Qi be the charge in the region r N R / 2 and let Q0 be the charge in the region where R / 2 N r N R .
3%r 3
6!% 1 R 4 3
dr #
# !%R 3 and
2R
R 4 16 32
R
31 & 47
47 &
233
%7
%3
3
Q0 # 4!% H (1 ! (r/R ) 2 ) r 2 dr # 4!%R 3 ' !
!%R 3 . Therefore, Q # ' 0
!%R 3 and
(#
( !%R #
R/2
480
) 24 160 * 120
) 32 120 *
480Q
%#
.
233!R 3
4! r 3%r 3
3!%r 4
6%r 2
180Qr 2
dr  #
#
(b) For r N R 2 , Gauss’s law gives E 4!r 2 #
and E #
. For R 2 N r N R ,
2P0 R
16P0 R 233!P0 R 4
P0 H 0 2 R
EXECUTE:! (a) The total charge is Q # Qi 0 Q0 , where Qi # 4! H R/2 0 E 4!r 2 # Qi 4!% r
Q 4!% % r 3 R 3
r5
R3 &
2
2
0
H R / 2 (1 ! (r / R) )r dr # P0i 0 P0 ' 3 ! 24 ! 5R 2 0 160 ( .
P0
P0
)
* E 4!r 2 # 3 4!%R 3 4!%R 3
0
P0
128 P0 E# 3
5
% 1 % r &3 1 % r &5 17 &
480Q % 1 % r & 1 % r &
23 &
' ' (! ' (!
( . For r O R ,
' ' (! ' (!
( and E #
2'
' 3 ) R * 5 ) R * 480 (
233!P0 r ) 3 ) R * 5 ) R * 1920 (
*
)
* Q
, since all the charge is enclosed.
4!P0 r 2 (c) The fraction of Q between R 2 N r N R is Q0 47 480
#
# 0.807.
Q 120 233 Q
(d) E # 180
using either of the electric field expressions above, evaluated at r # R / 2.
233 4!P0 R 2 EVALUATE:! (e) The force an electron would feel never is proportional to ! r which is necessary for simple
harmonic oscillations. It is oscillatory since the force is always attractive, but it has the wrong power of r to be
simple harmonic motion. 23 ELECTRIC POTENTIAL 23.1. IDENTIFY: Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given
by Eq.(23.9).
SET UP: Let the initial position of q2 be point a and the final position be point b, as shown in Figure 23.1. ra # 0.150 m
rb # (0.250 m) 2 0 (0.250 m) 2
rb # 0.3536 m Figure 23.1
EXECUTE: Wa Eb # U a ! U b Ua # 1 q1q2
(02.40 " 10!6 C)(!4.30 " 10!6 C)
# (8.988 " 109 N + m 2 / C D )
4, P0 ra
0.150 m U a # !0.6184 J
Ub # 1 q1q2
( 02.40 " 10!6 C)(!4.30 " 10!6 C)
# (8.988 " 109 N + m 2 / CD )
4, P0 rb
0.3536 m U b # !0.2623 J
Wa Eb # U a ! U b # !0.6184 J ! (!0.2623 J) # !0.356 J 23.2. EVALUATE: The attractive force on q2 is toward the origin, so it does negative work on q2 when q2 moves to
larger r.
IDENTIFY: Apply Wa E b # U a ! U b .
SET UP: 23.3. U a # 05.4 " 10!8 J. Solve for U b . EXECUTE: Wa Eb # !1.9 " 10!8 J # U a ! U b . U b # U a ! Wa Eb # 1.9 " 10!8 J ! (!5.4 " 10!8 J) # 7.3 " 10!8 J.
EVALUATE: When the electric force does negative work the electrical potential energy increases.
IDENTIFY: The work needed to assemble the nucleus is the sum of the electrical potential energies of the protons
in the nucleus, relative to infinity.
SET UP: The total potential energy is the scalar sum of all the individual potential energies, where each potential
energy is U # (1/ 4, P0 )(qq0 / r ). Each charge is e and the charges are equidistant from each other, so the total potential energy is U #
EXECUTE: 1 % e2 e2 e2 &
3e2
.
0 0 (#
'
4, P0 ) r
r
r * 4, P0 r Adding the potential energies gives
U# 3e 2
3(1.60 " 10!19 C) 2 (9.00 " 109 N + m 2 /C2 )
#
# 3.46 " 10!13 J # 2.16 MeV
4, P0 r
2.00 " 10!15 m EVALUATE: This is a small amount of energy on a macroscopic scale, but on the scale of atoms, 2 MeV is quite a
lot of energy.
231 232 23.4. Chapter 23 IDENTIFY: The work required is the change in electrical potential energy. The protons gain speed after being
released because their potential energy is converted into kinetic energy.
(a) SET UP: Using the potential energy of a pair of point charges relative to infinity, U # (1/ 4, P0 )( qq0 / r ). we have W # BU # U 2 ! U 1 #
EXECUTE: 1 % e2 e2 &
' ! (.
4, P0 ) r2 r1 * Factoring out the e2 and substituting numbers gives
2%
1
1
&
!14
W # . 9.00 " 109 N + m 2 /C 2 /.1.60 " 10!19 C / '
!
( # 7.68 " 10 J
3.00 " 10!15 m 2.00 " 10!15 m *
) The protons have equal momentum, and since they have equal masses, they will have equal speeds
%1
&
and hence equal kinetic energy. BU # K1 0 K 2 # 2 K # 2 ' mv 2 ( # mv 2 .
)2
* (b) SET UP: EXECUTE: 23.5. Solving for v gives v # BU
7.68 " 10!14 J
= 6.78 " 106 m/s
#
1.67 " 10!27 kg
m EVALUATE: The potential energy may seem small (compared to macroscopic energies), but it is enough to give
each proton a speed of nearly 7 million m/s.
(a) IDENTIFY: Use conservation of energy: K a 0 U a 0 Wother # K b 0 U b
U for the pair of point charges is given by Eq.(23.9).
SET UP:
Let point a be where q2 is 0.800 m from
q1 and point b be where q2 is 0.400 m
from q1, as shown in Figure 23.5a.
Figure 23.5a
EXECUTE: Only the electric force does work, so Wother # 0 and U # 1 q1q2
.
4, P0 r 2
K a # 1 mva # 1 (1.50 " 10!3 kg)(22.0 m/s)2 # 0.3630 J
2
2 Ua # 1 q1q2
(!2.80 " 10!6 C)(!7.80 " 10!6 C)
# (8.988 " 109 N + m 2 /C 2 )
# 00.2454 J
4, P0 ra
0.800 m
2
K b # 1 mvb
2 Ub # 1 q1q2
(!2.80 " 10!6 C)(!7.80 " 10!6 C)
# (8.988 " 109 N + m 2 /C2 )
# 00.4907 J
4, P0 rb
0.400 m The conservation of energy equation then gives K b # K a 0 (U a ! U b )
1
2 2
mvb # 00.3630 J 0 (0.2454 J ! 0.4907 J) # 0.1177 J vb # 2(0.1177 J)
# 12.5 m/s
1.50 " 10!3 kg EVALUATE: The potential energy increases when the two positively charged spheres get closer together, so the
kinetic energy and speed decrease.
(b) IDENTIFY: Let point c be where q2 has its speed momentarily reduced to zero. Apply conservation of energy to
points a and c: K a 0 U a 0 Wother # K c 0 U c . Electric Potential SET UP: 233 Points a and c are shown in Figure 23.5b. EXECUTE: K a # 00.3630 J (from part (a)) U a # 00.2454 J (from part (a)) Figure 23.5b K c # 0 (at distance of closest approach the speed is zero) Uc # 1 q1q2
# 00.3630 J 0 0.2454 J # 0.6084 J
4, P0 rc Thus conservation of energy K a 0 U a # U c gives
rc # 1 q1q2
4, P0 rc 1
q1q2
(!2.80 " 10!6 C)( !7.80 " 10!6 C)
# (8.988 " 109 N + m 2 /C2 )
# 0.323 m.
4, P0 0.6084 J
00.6084 J U E F as r E 0 so q2 will stop no matter what its initial speed is.
qq
IDENTIFY: Apply U # k 1 2 and solve for r.
r
!6
SET UP: q1 # !7.2 " 10 C , q2 # 02.3 " 10!6 C
EVALUATE: 23.6. kq1q2 (8.99 " 109 N + m 2 /C2 )(!7.20 " 10!6 C)(02.30 " 10!6 C)
#
# 0.372 m
U
!0.400 J
EVALUATE: The potential energy U is a scalar and can take positive and negative values.
(a) IDENTIFY and SET UP: U is given by Eq.(23.9).
1 qqEXECUTE: U #
4P, 0 r
EXECUTE: 23.7. r# (04.60 " 10!6 C)(01.20 " 10!6 C)
# 00.198 J
0.250 m
EVALUATE: The two charges are both of the same sign so their electric potential energy is positive.
(b) IDENTIFY: Use conservation of energy: K a 0 U a 0 Wother # K b 0 U b
SET UP: Let point a be where q is released and point b be at its final position, as shown in Figure 23.7.
U # (8.988 " 109 N + m 2 /C2 ) EXECUTE: K a # 0 (released from rest) U a # 00.198 J (from part (a))
2
K b # 1 mvb
2 Figure 23.7 Only the electric force does work, so Wother # 0 and U # 1 qQ
.
4, P0 r (i) rb # 0.500 m
Ub # 1 qQ
(04.60 " 10!6 C)(01.20 " 10!6 C)
# (8.988 " 109 N + m 2 /C2 )
# 00.0992 J
4, P0 r
0.500 m Then K a 0 U a 0 Wother # K b 0 U b gives K b # U a ! U b and
vb # 1
2 2
mvb # U a ! U b and 2(U a ! U b )
2(00.198 J ! 0.0992 J)
#
# 26.6 m/s.
m
2.80 " 10!4 kg (ii) rb # 5.00 m rb is now ten times larger than in (i) so U b is ten times smaller: U b # 00.0992 J /10 # 00.00992 J.
vb # 2(U a ! U b )
2(00.198 J ! 0.00992 J)
#
# 36.7 m/s.
m
2.80 " 10!4 kg 234 Chapter 23 (iii) rb # 50.0 m
rb is now ten times larger than in (ii) so Ub is ten times smaller:
U b # 00.00992 J/10 # 00.000992 J.
vb # 23.8. EVALUATE: The force between the two charges is repulsive and provides an acceleration to q. This causes the
speed of q to increase as it moves away from Q.
IDENTIFY: Call the three charges 1, 2 and 3. U # U12 0 U13 + U 23
SET UP: U12 # U 23 # U13 because the charges are equal and each pair of charges has the same separation, 0.500 m. 3kq 2
3k (1.2 " 10!6 C) 2
#
# 0.078 J.
0.500 m
0.500 m
EVALUATE: When the three charges are brought in from infinity to the corners of the triangle, the repulsive
electrical forces between each pair of charges do negative work and electrical potential energy is stored.
%qq
qq
qq &
IDENTIFY: U # k ' 1 2 0 1 3 0 2 3 (
r13
r23 *
) r12
EXECUTE: 23.9. 2(U a ! U b )
2(00.198 J ! 0.000992 J)
#
# 37.5 m/s.
m
2.80 " 10!4 kg SET UP: U# In part (a), r12 # 0.200 m , r23 # 0.100 m and r13 # 0.100 m. In part (b) let particle 3 have coordinate x, so r12 # 0.200 m , r13 # x and r23 # 0.200 ! x.
EXECUTE: % (4.00 nC)(!3.00 nC) (4.00 nC)(2.00 nC) (!3.00 nC)(2.00 nC) &
!7
(a) U # k '
0
0
( # 3.60 " 10 J
(0.200 m)
(0.100 m)
(0.100 m)
)
* %qq qq
qq &
(b) If U # 0 , then 0 # k ' 1 2 0 1 3 0 2 3 ( . Solving for x we find:
r12
x
r12 ! x *
)
8
6
0 # ! 60 0 !
I 60 x 2 ! 26 x 0 1.6 # 0 I x # 0.074 m, 0.360 m. Therefore, x # 0.074 m since it is the only
x 0.2 ! x
value between the two charges.
EVALUATE: U13 is positive and both U 23 and U12 are negative. If U # 0 , then U13 # U 23 0 U12 . For
23.10. x # 0.074 m , U13 # 09.7 " 10!7 J , U 23 # !4.3 " 10!7 J and U12 # !5.4 " 10!7 J. It is true that U # 0 at this x.
IDENTIFY: The work done on the alpha particle is equal to the difference in its potential energy when it is moved
from the midpoint of the square to the midpoint of one of the sides.
SET UP: We apply the formula Wa Eb # U a ! U b . In this case, a is the center of the square and b is the midpoint of
one of the sides. Therefore Wcenter Eside # U center ! U side .
There are 4 electrons, so the potential energy at the center of the square is 4 times the potential energy of a single
alphaelectron pair. At the center of the square, the alpha particle is a distance r1 = 50 nm from each electron. At
the midpoint of the side, the alpha is a distance r2 = 5.00 nm from the two nearest electrons and a distance r2 =
125 nm from the two most distant electrons. Using the formula for the potential energy (relative to infinity) of
two point charges, U # (1/ 4, P0 )(qq0 / r ), the total work is
Wcenter Eside # U center ! U side = 4 1 q2 qe % 1 q2 qe
1 q2 qe &
!'2
02
(
4, P0 r1
4, P0 r3 *
) 4, P0 r2 Substituting qe = e and q2 = 2e and simplifying gives
Wcenter Eside # !4e2
EXECUTE: 1 9 2 % 1 1 &:
> ! ' 0 (?
4, P0 > r1 ) r2 r3 * ?
;
< Substituting the numerical values into the equation for the work gives
29
2
1
1
%
&:
!21
W # !4 .1.60 " 10!19 C / >
!'
0
( ? # 6.08 " 10 J ???
50 m ) 5.00 nm
125 nm * <
; EVALUATE: Since the work is positive, the system has more potential energy with the alpha particle at the center
of the square than it does with it at the midpoint of a side. Electric Potential 23.11. 235 IDENTIFY: Apply Eq.(23.2). The net work to bring the charges in from infinity is equal to the change in potential
energy. The total potential energy is the sum of the potential energies of each pair of charges, calculated from
Eq.(23.9).
SET UP: Let 1 be where all the charges are infinitely far apart. Let 2 be where the charges are at the corners of the
triangle, as shown in Figure 23.11. Let qc be the third, unknown charge. Figure 23.11
EXECUTE: W # !BU # !(U 2 ! U1 ) U1 # 0
U 2 # U ab 0 U ac 0 U bc # 1
(q 2 0 2qqc )
4, P0 d Want W # 0, so W # !(U 2 ! U1 ) gives 0 # !U 2
0# 1
(q 2 0 2qqc )
4, P0 d q 2 0 2qqc # 0 and qc # !q/2.
EVALUATE: The potential energy for the two charges q is positive and for each q with qc it is negative. There are
two of the q, qc terms so must have qc 1 q.
23.12. IDENTIFY: Use conservation of energy U a 0 K a # U b 0 K b to find the distance of closest approach rb . The maximum force is at the distance of closest approach, F # k
SET UP: q1q2
.
rb2 K b # 0. Initially the two protons are far apart, so U a # 0. A proton has mass 1.67 " 10!27 kg and charge q # 0 e # 01.60 " 10!19 C.
EXECUTE: rb # 23.14. q1q2
e2
2
. mva # k
and
rb
rb ke 2 (8.99 " 109 N + m 2 /C2 )(1.60 " 10!19 C)2
#
# 1.38 " 10!13 m.
2
mva
(1.67 " 10!27 kg)(1.00 " 106 m/s)2 e2
(1.60 " 10!19 C) 2
# (8.99 " 109 N + m 2 /C 2 )
# 0.012 N.
rb2
(1.38 " 10!13 C) 2
EVALUATE: The acceleration a # F/m of each proton produced by this force is extremely large.
!
W
IDENTIFY: E points from high potential to low potential. a Eb # Va ! Vb .
q0
!
SET UP: The force on a positive test charge is in the direction of E .
EXECUTE: V decreases in the eastward direction. A is east of B, so VB 5 VA . C is east of A, so VC 1 VA . The force
on a positive test charge is east, so no work is done on it by the electric force when it moves due south (the force and
displacement are perpendicular), and VD # VA .
EVALUATE: The electric potential is constant in a direction perpendicular to the electric field.
Wa Eb
kq
IDENTIFY:
# Va ! Vb . For a point charge, V # .
r
q0
SET UP: Each vacant corner is the same distance, 0.200 m, from each point charge.
EXECUTE: Taking the origin at the center of the square, the symmetry means that the potential is the same at the
two corners not occupied by the 05.00 $ C charges. This means that no net work is done is moving from one corner
to the other.
EVALUATE: If the charge q0 moves along a diagonal of the square, the electrical force does positive work for part
of the path and negative work for another part of the path, but the net work done is zero.
F #k 23.13. 2
K a # U b . 2( 1 mva ) # k
2 236 Chapter 23 23.15. IDENTIFY and SET UP: Apply conservation of energy to points A and B.
EXECUTE: K A 0 U A # K B 0 U B
U # qV , so K A 0 qVA # K B 0 qVB K B # K A 0 q(VA ! VB ) # 0.00250 J 0 (!5.00 " 10!6 C)(200 V ! 800 V) # 0.00550 J
vB # 2 K B /m # 7.42 m/s
It is faster at B; a negative charge gains speed when it moves to higher potential.
W
IDENTIFY: The workenergy theorem says Wa Eb # K b ! K a . a Eb # Va ! Vb .
q
SET UP: Point a is the starting and point b is the ending point. Since the field is uniform,
Wa Eb # Fs cos A # E q s cos A . The field is to the left so the force on the positive charge is to the left. The particle
EVALUATE: 23.16. moves to the left so A # 0! and the work Wa Eb is positive.
EXECUTE: (a) Wa Eb # K b ! K a # 1.50 " 10!6 J ! 0 # 1.50 " 10!6 J (b) Va ! Vb # Wa Eb 1.50 " 10!6 J
#
# 357 V. Point a is at higher potential than point b.
q
4.20 " 10!9 C (c) E q s # Wa Eb , so E # Wa Eb Va ! Vb
357 V
#
#
# 5.95 " 103 V/m.
6.00 " 10!2 m
qs
s A positive charge gains kinetic energy when it moves to lower potential; Vb 1 Va .
!
b!
IDENTIFY: Apply the equation that precedes Eq.(23.17): Wa Eb # qH E + dl .
a
!
SET UP: Use coordinates where 0 y is upward and 0 x is to the right. Then E = Eˆ with E # 4.00 " 104 N/C.
j
(a) The path is sketched in Figure 23.17a.
EVALUATE: 23.17. Figure 23.17a
EXECUTE: !
!!
b!
ˆ
E + dl # ( Eˆ) + (dxi ) # 0 so Wa Eb # qH E + dl # 0.
j
a EVALUATE: The electric force on the positive charge is upward (in the direction of the electric field) and does no
work for a horizontal displacement of the charge.
(b) SET UP: The path is sketched in Figure 23.17b. !
dl = dyˆ
j Figure 23.17b
EXECUTE: !!
E + dl # ( Eˆ) + (dyˆ) # E dy
j
j !
b!
b
Wa Eb # qH E + dl # qE H dy # qE ( yb ! ya )
a a yb ! ya # 00.670 m, positive since the displacement is upward and we have taken 0 y to be upward. Wa Eb # qE ( yb ! ya ) # (028.0 " 10!9 C)(4.00 " 104 N/C)(00.670 m) # 07.50 " 10!4 J.
EVALUATE: The electric force on the positive charge is upward so it does positive work for an upward
displacement of the charge. Electric Potential (c) SET UP: 237 The path is sketched in Figure 23.17c.
ya # 0
yb # ! r sin4 # !(2.60 m)sin 453 # !1.838 m
The vertical component of the 2.60 m
displacement is 1.838 m downward.
Figure 23.17c !
ˆ
j
EXECUTE: dl = dxi + dyˆ (The displacement has both horizontal and vertical components.)
!!
ˆ
E + dl # ( Eˆ) + (dxi + dyˆ) # E dy (Only the vertical component of the displacement contributes to the work.)
j
j
!
b!
b
Wa Eb # qH E + dl # qE H dy # qE ( yb ! ya )
a a Wa Eb # qE ( yb ! ya ) # (028.0 " 10!9 C)(4.00 " 104 N/C)(!1.838 m) # !2.06 " 10!3 J. 23.18. EVALUATE: The electric force on the positive charge is upward so it does negative work for a displacement of the
charge that has a downward component.
IDENTIFY: Apply K a 0 U a # K b 0 U b .
SET UP: Let q1 # 03.00 nC and q2 # 02.00 nC. At point a, r1a # r2 a # 0.250 m . At point b, r1b # 0.100 m and r2b # 0.400 m . The electron has q # !e and me # 9.11 " 10!31 kg . K a # 0 since the electron is released from rest.
EXECUTE: ! keq1 keq2
keq1 keq2 1
2
!
#!
!
0 mevb .
r1a
r2 a
r1b
r2b
2 % (3.00 " 10!9 C) (2.00 " 10!9 C) &
!17
Ea # K a 0 U a # k (! 1.60 " 10!19 C) '
0
( # ! 2.88 " 10 J .
0.250 m *
) 0.250 m
% (3.00 " 10!9 C) (2.00 " 10!9 C & 1
1
2
2
!17
0
Eb # K b 0 U b # k (! 1.60 " 10!19 C) '
( 0 mevb # ! 5.04 " 10 J 0 mevb
0.400 m * 2
2
) 0.100 m
Setting Ea # Eb gives vb # 23.19. 23.20. 2
(5.04 " 10!17 J ! 2.88 " 10!17 J) # 6.89 " 106 m s.
9.11 " 10!31 kg EVALUATE: Va # V1a 0 V2 a # 180 V. Vb # V1b 0 V2b # 315 V. Vb 5 Va . The negatively charged electron gains kinetic
energy when it moves to higher potential.
kq
IDENTIFY and SET UP: For a point charge V #
. Solve for r.
r
kq (8.99 " 109 N + m 2 /C2 )(2.50 " 10!11 C)
EXECUTE: (a) r #
#
# 2.50 " 10!3 m # 2.50 mm
V
90.0 V
%V &
% 90.0 V &
(b) Vr # kq # constant so V1r1 # V2 r2 . r2 # r1 ' 1 ( # (2.50 mm) '
( # 7.50 mm .
) 30.0 V *
) V2 *
EVALUATE: The potential of a positive charge is positive and decreases as the distance from the point charge
increases.
IDENTIFY: The total potential is the scalar sum of the individual potentials, but the net electric field is the vector
sum of the two fields.
SET UP: The net potential can only be zero if one charge is positive and the other is negative, since it is a scalar.
The electric field can only be zero if the two fields point in opposite directions.
EXECUTE: (a) (i) Since both charges have the same sign, there are no points for which the potential is zero.
(ii) The two electric fields are in opposite directions only between the two charges, and midway between them the
fields have equal magnitudes. So E = 0 midway between the charges, but V is never zero.
(b) (i) The two potentials have equal magnitude but opposite sign midway between the charges, so V = 0 midway
between the charges, but E ( 0 there since the fields point in the same direction.
(ii) Between the two charges, the fields point in the same direction, so E cannot be zero there. In the other two
regions, the field due to the nearer charge is always greater than the field due to the more distant charge, so they
cannot cancel. Hence E is not zero anywhere.
EVALUATE: It does not follow that the electric field is zero where the potential is zero, or that the potential is zero
where the electric field is zero. 238 Chapter 23 23.21. IDENTIFY: 1
q
7 ri
4, P0 i i
SET UP: The locations of the changes and points A and B are sketched in Figure 23.21.
V# Figure 23.21
(a) VA # EXECUTE: 1 % q1 q2 &
'0
(
4, P0 ) rA1 rA2 *
% 02.40 " 10 !9 C !6.50 " 10 !9 C &
0
VA # (8.988 " 109 N + m 2 /C 2 ) '
( # !737 V
0.050 m *
) 0.050 m (b) VB # 1 % q1 q2 &
0
'
(
4, P0 ) rB1 rB 2 * % 02.40 " 10!9 C !6.50 " 10!9 C &
0
VB # (8.988 " 109 N + m2 /C2 ) '
( # !704 V
0.060 m *
) 0.080 m 23.22. (c) IDENTIFY and SET UP: Use Eq.(23.13) and the results of parts (a) and (b) to calculate W.
!9
!8
EXECUTE: WB E A # q(VB ! VA ) # (2.50 "10 C)(!704 V ! (!737 V)) # 08.2 " 10 J
EVALUATE: The electric force does positive work on the positive charge when it moves from higher potential
(point B) to lower potential (point A).
kq
IDENTIFY: For a point charge, V #
. The total potential at any point is the algebraic sum of the potentials of the
r
two charges.
SET UP: (a) The positions of the two charges are shown in Figure 23.22a. r # a 2 0 x 2 . Figure 23.22a 1q
.
4, P0 a EXECUTE: (b) V0 # 2 (c) V ( x) # 2 1q
1
#2
4, P0 r
4, P0 q
a 0 x2
2 Electric Potential 239 (d) The graph of V versus x is sketched in Figure 23.22b. Figure 23.22b
1 2q
, just like a point charge of charge 02q. At distances from the charges
4, P0 x
much greater than their separation, the two charges act like a single point charge.
kq
IDENTIFY: For a point charge, V #
. The total potential at any point is the algebraic sum of the potentials of the
r
two charges.
SET UP: (a) The positions of the two charges are shown in Figure 23.23.
kq k ( ! q)
EXECUTE: (b) V #
0
# 0.
r
r
(c) The potential along the xaxis is always zero, so a graph would be flat.
(d) If the two charges are interchanged, then the results of (b) and (c) still hold. The potential is zero.
EVALUATE: The potential is zero at any point on the xaxis because any point on the xaxis is equidistant from the
two charges. EVALUATE: 23.23. (e) When x 55 a, V # Figure 23.23
23.24. IDENTIFY: For a point charge, V # kq
. The total potential at any point is the algebraic sum of the potentials of the
r two charges.
SET UP: Consider the distances from the point on the yaxis to each charge for the three regions !a N y N a
(between the two charges), y 5 a (above both charges) and y 1 !a (below both charges).
2kqy
kq
kq
kq
kq
! 2kqa
.
. y 5 a :V #
!
#2
!
#2
2
(a 0 y ) (a ! y ) y ! a
(a 0 y) y ! a y ! a 2
2kqa
kq
! kq
.
y 1 ! a :V #
!
#
(a 0 y ) (! y 0 a) y 2 ! a 2 EXECUTE: (a)  y  1 a : V # % !q
q&
A general expression valid for any y is V # k '
0
(.
) y ! a   y 0 a *
(b) The graph of V versus y is sketched in Figure 23.24.
!2kqa !2kqa
(c) y 55 a : V # 2
.
C
y ! a2
y2
(d) If the charges are interchanged, then the potential is of the opposite sign. 2310 Chapter 23 V # 0 at y # 0 . V E 0F as the positive charge is approached and V E !F as the negative charge is EVALUATE:
approached. Figure 23.24
23.25. IDENTIFY: For a point charge, V # kq
. The total potential at any point is the algebraic sum of the potentials of the
r two charges.
SET UP: (a) The positions of the two charges are shown in Figure 23.25a. Figure 23.25a kq 2kq ! kq( x 0 a)
kq 2kq
kq(3x ! a)
. 0 1 x 1 a :V #
.
!
#
!
#
x x!a
x( x ! a)
x a!x
x( x ! a )
kq( x 0 a)
! kq 2kq
q
2q &
x 1 0 :V #
. A general expression valid for any y is V # k %
0
#
' x !  x ! a ( .
x
x!a
x( x ! a)
)
*
(c) The potential is zero at x # ! a and a /3.
(d) The graph of V versus x is sketched in Figure 23.25b.
(b) x 5 a : V # Figure 23.25b
! kqx ! kq
#
, which is the same as the potential of a point charge –q. Far from
x2
x
the two charges they appear to be a point charge with a charge that is the algebraic sum of their two charges. EVALUATE: (e) For x 55 a : V C Electric Potential 23.26. For a point charge, V # IDENTIFY: 2311 kq
. The total potential at any point is the algebraic sum of the potentials of the
r two charges.
SET UP: The distance of a point with coordinate y from the positive charge is y and the distance from the
negative charge is r # a 2 0 y 2 .
&
(.
(
*
a2 0 y2
a
(b) V # 0, when y 2 #
I 3 y2 # a2 I y # 6
.
4
3
(c) The graph of V versus y is sketched in Figure 23.26. V E F as the positive charge at the origin is approached.
(a) V # EXECUTE: %1
kq 2kq
2
!
# kq '
!
2
' y
 y
r
a 0 y2
) %1 2&
kq
(d) y 55 a : V C kq ' ! ( # ! , which is the potential of a point charge ! q . Far from the two
y
) y y*
charges they appear to be a point charge with a charge that is the algebraic sum of their two charges.
EVALUATE: Figure 23.26
23.27. K a 0 qVa # K b 0 qVb . IDENTIFY:
SET UP: Let point a be at the cathode and let point b be at the anode. K a # 0 . Vb ! Va # 295 V . An electron has q # !e and m # 9.11 " 10!31 kg .
EXECUTE: vb # 12
K b # q(Va ! Vb ) # !(1.60 " 10!19 C)( ! 295 V) # 4.72 " 10!17 J . K b # mvb , so
2 2(4.72 " 10!17 J)
# 1.01 " 107 m s.
9.11 " 10!31 kg The negatively charged electron gains kinetic energy when it moves to higher potential.
kq
kq
IDENTIFY: For a point charge, E # 2 and V #
.
r
r
SET UP: The electric field is directed toward a negative charge and away from a positive charge.
2
V
kq/r
4.98 V
% kq & % r &
EXECUTE: (a) V 5 0 so q 5 0 .
#
# ' (' ( # r . r #
# 0.415 m .
2
E k q /r
12.0 V/m
) r * ) kq *
EVALUATE: 23.28. (0.415 m)(4.98 V)
rV
#
# 2.30 " 10!10 C
8.99 " 109 N + m 2 /C2
k
(c) q 5 0 , so the electric field is directed away from the charge.
EVALUATE: The ratio of V to E due to a point charge increases as the distance r from the charge increases, because
E falls off as 1/r 2 and V falls off as 1/r .
!
(a) IDENTIFY and SET UP: The direction of E is always from high potential to low potential so point b is at
higher potential.
(b) Apply Eq.(23.17) to relate Vb ! Va to E.
!b
b!
EXECUTE: Vb ! Va # ! H E + dl # H E dx # E ( xb ! xa ).
(b) q # 23.29. a a E# Vb ! Va
0240 V
#
# 800 V/m
xb ! xa 0.90 m ! 0.60 m 2312 23.30. Chapter 23 (c) Wb E a # q (Vb ! Va ) # ( !0.200 " 10!6 C)(0240 V) # !4.80 " 10!5 J.
EVALUATE: The electric force does negative work on a negative charge when the negative charge moves from
high potential (point b) to low potential (point a).
kq
IDENTIFY: For a point charge, V #
. The total potential at any point is the algebraic sum of the potentials of the
r
kq
two charges. For a point charge, E # 2 . The net electric field is the vector sum of the electric fields of the two
r
charges.
!
SET UP: E produced by a point charge is directed away from the point charge if it is positive and toward the
charge if it is negative.
EXECUTE: (a) V # VQ 0 V2Q 5 0, so V is zero nowhere except for infinitely far from the charges. The fields can
cancel only between the charges, because only there are the fields of the two charges in opposite directions. Consider a
kQ
k (2Q)
point a distance x from Q and d ! x from 2Q, as shown in Figure 23.30a. EQ # E2Q E 2 #
E (d ! x)2 # 2 x 2 .
x
(d ! x)2
x # d . The other root, x # d , does not lie between the charges.
10 2
1! 2
(b) V can be zero in 2 places, A and B, as shown in Figure 23.30b. Point A is a distance x from !Q and d ! x from 2Q. B is a distance y from !Q and d 0 y from 2Q. At A : k ( ! Q) k (2Q)
0
#0E x # d 3.
x
d!x k ( ! Q) k (2Q)
0
#0E y#d .
y
d0y
The two electric fields are in opposite directions to the left of !Q or to the right of 2Q in Figure 23.30c. But for the
magnitudes to be equal, the point must be closer to the charge with smaller magnitude of charge. This can be the
kQ
k (2Q)
d
.
and x #
case only in the region to the left of !Q . EQ # E2Q gives 2 #
2
x
(d 0 x)
2 !1
!
EVALUATE: (d) E and V are not zero at the same places. E is a vector and V is a scalar. E is proportional to 1/r 2
!
and V is proportional to 1/r . E is related to the force on a test charge and BV is related to the work done on a test
charge when it moves from one point to another.
At B : Figure 23.30
23.31. IDENTIFY and SET UP: Apply conservation of energy, Eq.(23.3). Use Eq.(23.12) to express U in terms of V.
(a) EXECUTE: K1 0 qV1 # K 2 0 qV2 q (V1 ! V2 ) # K 2 ! K1;
K1 # 1 mev12 # 4.099 " 10!18 J;
2 q # !1.602 " 10!19 C
2
K 2 # 1 mev2 # 2.915 " 10!17 J
2 K 2 ! K1
# !156 V
q
EVALUATE: The electron gains kinetic energy when it moves to higher potential.
(b) EXECUTE: Now K1 # 2.915 " 10!17 J, K 2 # 0
V1 ! V2 # V1 ! V2 #
23.32. K 2 ! K1
# 0182 V
q EVALUATE: The electron loses kinetic energy when it moves to lower potential.
IDENTIFY and SET UP: Expressions for the electric potential inside and outside a solid conducting sphere are
derived in Example 23.8.
kq k (3.50 " 10!9 C)
EXECUTE: (a) This is outside the sphere, so V #
#
# 65.6 V.
r
0.480 m k (3.50 " 10!9 C)
# 131 V .
0.240 m
(c) This is inside the sphere. The potential has the same value as at the surface, 131 V.
EVALUATE: All points of a conductor are at the same potential.
(b) This is at the surface of the sphere, so V # Electric Potential 23.33. 2313 (a) IDENTIFY and SET UP: The electric field on the ring’s axis is calculated in Example 21.10. The force on the
electron exerted by this field is given by Eq.(21.3).
EXECUTE: When the electron is on either side of the center of the ring, the ring exerts an attractive force directed
toward the center of the ring. This restoring force produces oscillatory motion of the electron along the axis of the
ring, with amplitude 30.0 cm. The force on the electron is not of the form F # !kx so the oscillatory motion is not
simple harmonic motion.
(b) IDENTIFY: Apply conservation of energy to the motion of the electron.
SET UP: K a 0 U a # K b 0 U b with a at the initial position of the electron and b at the center of the ring. From Example 23.11, V #
EXECUTE: 1
4, P0 Q , where R is the radius of the ring. x 0 R2
xa # 30.0 cm, xb # 0.
2 K a # 0 (released from rest), K b # 1 mv 2
2
Thus 1
2 mv 2 # U a ! U b
2e(Vb ! Va )
.
m And U # qV # !eV so v #
Va # 1
4, P0 Q
2
xa 0 R 2 # (8.988 " 109 N + m 2 / C2 ) 24.0 " 10!9 C
(0.300 m)2 0 (0.150 m) 2 Va # 643 V
Vb # v# 1
4, P0 Q
2
xb 0 R 2 # (8.988 " 109 N + m 2 / C 2 ) 24.0 " 10!9 C
# 1438 V
0.150 m 2e(Vb ! Va )
2(1.602 " 10!19 C)(1438 V ! 643 V)
#
# 1.67 " 107 m/s
m
9.109 " 10!31 kg EVALUATE: The positively charged ring attracts the negatively charged electron and accelerates it. The electron
has its maximum speed at this point. When the electron moves past the center of the ring the force on it is opposite
to its motion and it slows down.
23.34. IDENTIFY: Example 23.10 shows that for a line of charge, Va ! Vb # 8
ln( rb / ra ) . Apply conservation of energy
2, P0 to the motion of the proton.
SET UP: Let point a be 18.0 cm from the line and let point b be at the distance of closest approach, where K b # 0 .
EXECUTE: (a) K a # 1 mv 2 # 1 (1.67 " 10!27 kg)(1.50 " 103 m/s)2 # 1.88 " 10!21 J .
2
2 (b) K a 0 qVa # Kb 0 qVb . Va ! Vb # 23.35. K b ! K a !1.88 " 10!21 J
% 2, P0 &
#
# !0.01175 V . ln(rb / ra ) # '
( ( !0.01175 V) .
!19
q
1.60 " 10 C
)8* % 2, P0 (0.01175 V) &
% 2, P0 ( !0.01175 V) &
rb # ra exp '
( # 0.158 m .
( # (0.180 m)exp ' !
!12
8
)
*
) 5.00 " 10 C/m *
EVALUATE: The potential increases with decreasing distance from the line of charge. As the positively charged
proton approaches the line of charge it gains electrical potential energy and loses kinetic energy.
IDENTIFY: The voltmeter measures the potential difference between the two points. We must relate this quantity to
the linear charge density on the wire.
SET UP:
EXECUTE: For a very long (infinite) wire, the potential difference between two points is BV # 8 2, P0 ln . rb / ra / . (a) Solving for 8 gives 8# ( BV )2, P0
#
ln . rb / ra / 575 V = 9.49 " 108 C/m
% 3.50 cm &
.18 "10 N + m /C / ln ' 2.50 cm (
)
*
(b) The meter will read less than 575 V because the electric field is weaker over this 1.00cm distance than it was
over the 1.00cm distance in part (a).
(c) The potential difference is zero because both probes are at the same distance from the wire, and hence at the
same potential.
EVALUATE: Since a voltmeter measures potential difference, we are actually given BV, even though that is not
stated explicitly in the problem. We must also be careful when using the formula for the potential difference because
each r is the distance from the center of the cylinder, not from the surface.
9 2 2 2314 Chapter 23 23.36. IDENTIFY: The voltmeter reads the potential difference between the two points where the probes are placed.
Therefore we must relate the potential difference to the distances of these points from the center of the cylinder. For
points outside the cylinder, its electric field behaves like that of a line of charge.
SET UP: 23.37. Using BV # ln . rb / ra / and solving for rb, we have rb # ra e2, P0 BV / 8 . 1
%
&
(175 V)
'
9
2
2(
) 2 " 9.00 " 10 N + m /C *
# 0.648 , which gives
EXECUTE: The exponent is
15.0 " 10!9 C/m
rb = (2.50 cm) e0.648 = 4.78 cm.
The distance above the surface is 4.78 cm – 2.50 cm = 2.28 cm.
EVALUATE: Since a voltmeter measures potential difference, we are actually given BV, even though that is not
stated explicitly in the problem. We must also be careful when using the formula for the potential difference because
each r is the distance from the center of the cylinder, not from the surface.
IDENTIFY: For points outside the cylinder, its electric field behaves like that of a line of charge. Since a voltmeter
reads potential difference, that is what we need to calculate.
SET UP:
EXECUTE: The potential difference is BV # 8 2, P0 ln . rb / ra / . (a) Substituting numbers gives BV # 23.38. 8 2, P0 8
2, P0 % 10.0 cm &
ln . rb / ra / = . 8.50 " 10!6 C/m /. 2 " 9.00 " 109 N + m 2 /C2 / ln '
(
) 6.00 cm *
BV = 7.82 " 104 V = 78,200 V = 78.2 kV (b) E = 0 inside the cylinder, so the potential is constant there, meaning that the voltmeter reads zero.
EVALUATE: Caution! The fact that the voltmeter reads zero in part (b) does not mean that V = 0 inside the
cylinder. The electric field is zero, but the potential is constant and equal to the potential at the surface.
IDENTIFY: The work required is equal to the change in the electrical potential energy of the chargering system.
We need only look at the beginning and ending points, since the potential difference is independent of path for a
conservative field.
%1Q
&
! 0(
SET UP: (a) W = BU # qBV # q .Vcenter ! VF / # q '
) 4,P 0 a
* Substituting numbers gives
BU = (3.00 " 106 C)(9.00 " 109 N + m2/C2)(5.00 " 10–6 C)/(0.0400 m) = 3.38 J
(b) We can take any path since the potential is independent of path.
(c) SET UP: The net force is away from the ring, so the ball will accelerate away. Energy conservation gives
U 0 # K max # 1 mv 2 .
2
EXECUTE: Solving for v gives
2U 0
2(3.38 J)
= 67.1 m/s
v#
#
m
0.00150 kg
EVALUATE: Direct calculation of the work from the electric field would be extremely difficult, and we would need
to know the path followed by the charge. But, since the electric field is conservative, we can bypass all this
calculation just by looking at the end points (infinity and the center of the ring) using the potential.
IDENTIFY: The electric field is zero everywhere except between the plates, and in this region it is uniform and
points from the positive to the negative plate (to the left in Figure 23.32).
SET UP: Since the field is uniform between the plates, the potential increases linearly as we go from left to right
starting at b.
EXECUTE: Since the potential is taken to be zero at the left surface of the negative plate (a in Figure 23.32), it is
zero everywhere to the left of b. It increases linearly from b to c, and remains constant (since E = 0) past c. The
graph is sketched in Figure 23.39.
EVALUATE: When the electric field is zero, the potential remains constant but not necessarily zero (as to the right
of c). When the electric field is constant, the potential is linear.
EXECUTE: 23.39. Figure 23.39 Electric Potential 23.40. 23.41. 2315 IDENTIFY and SET UP: For oppositely charged parallel plates, E # = / P0 between the plates and the potential
difference between the plates is V # Ed .
= 47.0 " 10!9 C m 2
EXECUTE: (a) E # #
# 5310 N C.
P0
P0
(b) V # Ed # (5310 N/C)(0.0220 m) # 117 V.
(c) The electric field stays the same if the separation of the plates doubles. The potential difference between the
plates doubles.
EVALUATE: The electric field of an infinite sheet of charge is uniform, independent of distance from the sheet.
The force on a test charge between the two plates is constant because the electric field is constant. The potential
difference is the work per unit charge on a test charge when it moves from one plate to the other. When the distance
doubles the work, which is force times distance, doubles and the potential difference doubles.
IDENTIFY and SET UP: Use the result of Example 23.9 to relate the electric field between the plates to the potential
difference between them and their separation. The force this field exerts on the particle is given by Eq.(21.3). Use
the equation that precedes Eq.(23.17) to calculate the work.
V
360 V
# 8000 V/m
EXECUTE: (a) From Example 23.9, E # ab #
d
0.0450 m
(b) F # q E # (2.40 " 10!9 C)(8000 V/m) # 01.92 " 10!5 N
(c) The electric field between the plates is shown in Figure 23.41. Figure 23.41 The plate with positive charge (plate a) is at higher potential. The electric field is directed from high potential
!
!
!
toward low potential (or, E is from + charge toward ! charge), so E points from a to b. Hence the force that E
exerts on the positive charge is from a to b, so it does positive work.
!
b!
W # H F + dl # Fd , where d is the separation between the plates.
a W # Fd # (1.92 " 10!5 N)(0.0450 m) # 08.64 " 10!7 J
(d) Va ! Vb # 0360 V (plate a is at higher potential) BU # U b ! U a # q (Vb ! Va ) # (2.40 " 10!9 C)(!360 V) # !8.64 " 10!7 J.
23.42. EVALUATE: We see that Wa Eb # !(U b ! U a ) # U a ! U b .
IDENTIFY: The electric field is zero inside the sphere, so the potential is constant there. Thus the potential at the
center must be the same as at the surface, where it is equivalent to that of a pointcharge.
SET UP: At the surface, and hence also at the center of the sphere, the field is that of a pointcharge,
E # Q /(4, P0 R ).
EXECUTE: (a) Solving for Q and substituting the numbers gives Q # 4, P0 RV # (0.125 m)(1500 V)/(9.00 " 109 N + m2/C2) = 2.08 " 108 C = 20.8 nC 23.43. (b) Since the potential is constant inside the sphere, its value at the surface must be the same as at the center,
1.50 kV.
EVALUATE: The electric field inside the sphere is zero, so the potential is constant but is not zero.
IDENTIFY and SET UP: Consider the electric field outside and inside the shell and use that to deduce the potential.
EXECUTE: (a) The electric field outside the shell is the same as for a point charge at the center of the shell, so the
potential outside the shell is the same as for a point charge: V# q
for r 5 R.
4, P0 r The electric field is zero inside the shell, so no work is done on a test charge as it moves inside the shell and all
q
points inside the shell are at the same potential as the surface of the shell:V #
for r N R.
4, P0 R
kq
RV (0.15 m)( !1200 V)
so q #
#
# !20 nC
R
k
k
(c) EVALUATE: No, the amount of charge on the sphere is very small. Since U # qV the total amount of electric
(b) V # energy stored on the balloon is only (20 nC)(1200 V) # 2.4 " 10!5 J. 2316 Chapter 23 23.44. IDENTIFY: Example 23.8 shows that the potential of a solid conducting sphere is the same at every point inside the
sphere and is equal to its value V # q / 2, P0 R at the surface. Use the given value of E to find q.
SET UP: For negative charge the electric field is directed toward the charge.
For points outside this spherical charge distribution the field is the same as if all the charge were concentrated at the
center.
q
(3800 N/C)(0.200 m) 2
EXECUTE: E #
and q # 4, P0 Er 2 #
# 1.69 " 10!8 C .
4, P0 r 2
8.99 " 109 N + m 2 /C 2
Since the field is directed inward, the charge must be negative. The potential of a point charge, taking F as zero, is
q
(8.99 " 109 N + m 2 /C2 )( !1.69 " 10!8 C)
V#
#
# !760 V at the surface of the sphere. Since the charge all resides
4, P0 r
0.200 m
on the surface of a conductor, the field inside the sphere due to this symmetrical distribution is zero. No work is
therefore done in moving a test charge from just inside the surface to the center, and the potential at the center must
also be ! 760 V.
EVALUATE: Inside the sphere the electric field is zero and the potential is constant.
IDENTIFY: Example 23.9 shows that V ( y ) # Ey , where y is the distance from the negatively charged plate, whose
potential is zero. The electric field between the plates is uniform and perpendicular to the plates.
!
SET UP: V increases toward the positively charged plate. E is directed from the positively charged plated toward
the negatively charged plate.
V
480 V
V
EXECUTE: (a) E # #
# 2.82 " 104 V/m and y # . V # 0 at y # 0 , V # 120 V at y # 0.43 cm ,
E
d 0.0170 m
V # 240 V at y # 0.85 cm , V # 360 V at y # 1.28 cm and V # 480 V at y # 1.70 cm . The equipotential surfaces
are sketched in Figure 23.45. The surfaces are planes parallel to the plates.
(b) The electric field lines are also shown in Figure 23.45. The field lines are perpendicular to the plates and the
equipotential lines are parallel to the plates, so the electric field lines and the equipotential lines are mutually
perpendicular.
EVALUATE: Only differences in potential have physical significance. Letting V # 0 at the negative plate is a
choice we are free to make. 23.45. Figure 23.45
23.46. IDENTIFY: By the definition of electric potential, if a positive charge gains potential along a path, then the
potential along that path must have increased. The electric field produced by a very large sheet of charge is uniform
and is independent of the distance from the sheet.
(a) SET UP: No matter what the reference point, we must do work on a positive charge to move it away from the
negative sheet.
EXECUTE: Since we must do work on the positive charge, it gains potential energy, so the potential increases.
(b) SET UP: Since the electric field is uniform and is equal to = /2P0, we have BV # Ed # = 2P0 d. EXECUTE: Solving for d gives d# 2P0 BV # 2 . 8.85 " 10!12 C2 /N + m 2 / (1.00 V) = 0.00295 m = 2.95 mm
6.00 " 10!9 C/m 2
EVALUATE: Since the spacing of the equipotential surfaces (d = 2.95 mm) is independent of the distance from the
sheet, the equipotential surfaces are planes parallel to the sheet and spaced 2.95 mm apart. = Electric Potential 23.47 2317 !
Use Eq.(23.19) to calculate the components of E . IDENTIFY and SET UP:
EXECUTE: V # Axy ! Bx 2 0 Cy
(a) Ex # ! QV
# ! Ay 0 2 Bx
Qx QV
# ! Ax ! C
Qy
QV
Ez # !
#0
Qz
(b) E # 0 requires that Ex # E y # Ez # 0.
Ey # ! E z # 0 everywhere.
E y # 0 at x # !C/A. 23.48. And Ex is also equal zero for this x, any value of z, and y # 2Bx /A # (2 B/A)(!C/A) # !2 BC/A2 .
EVALUATE: V doesn’t depend on z so E z # 0 everywhere.
IDENTIFY: Apply Eq.(21.19).
!
1q
SET UP: Eq.(21.7) says E #
ˆ
r is the electric field due to a point charge q.
4, P0 r 2 &
QV
Q%
kQ
kQx
kQx
(#
#! '
# 3.
Qx
Qx ' x 2 0 y 2 0 z 2 ( ( x 2 0 y 2 0 z 2 )3 2
r
)
*
kQy
kQz
Similarly, E y # 3 and Ez # 3 .
r
r
ˆ
ˆ
% xi yˆ zk & kQ
j
kQ
(b) From part (a), E # 2 ' 0 0
ˆ
r , which agrees with Equation (21.7).
(#
'r
r)
r
r ( r2
*
!
EVALUATE: V is a scalar. E is a vector and has components.
kq
kq
IDENTIFY and SET UP: For a solid metal sphere or for a spherical shell, V #
outside the sphere and V #
at
R
r
QV
all points inside the sphere, where R is the radius of the sphere. When the electric field is radial, E # !
.
Qr
%1 1&
kq kq
EXECUTE: (a) (i) r 1 ra : This region is inside both spheres. V #
!
# kq ' ! ( .
ra rb
) ra rb *
EXECUTE: 23.49. (a) Ex # ! (ii) ra 1 r 1 rb : This region is outside the inner shell and inside the outer shell. V # %1 1 &
kq kq
!
# kq ' ! ( .
r
rb
) r rb * (iii) r 5 rb : This region is outside both spheres and V # 0 since outside a sphere the potential is the same as for point
charge. Therefore the potential is the same as for two oppositely charged point charges at the same location. These
potentials cancel.
1 %q q&
1 %1 1&
(b) Va #
q' ! ( .
' ! ( and Vb # 0 , so Vab #
4, P0 ) ra rb *
4, P0 ) ra rb *
%1 1 &
QV
q Q %1 1 &
1q
Vab
1
(c) Between the spheres ra 1 r 1 rb and V # kq ' ! ( . E # !
#!
#
.
' ! (#0
2
Qr
% 1 1 & r2
4, P0 Qr ) r rb *
4, P0 r
) r rb *
!(
'
) ra rb *
(d) From Equation (23.23): E # 0, since V is constant (zero) outside the spheres.
(e) If the outer charge is different, then outside the outer sphere the potential is no longer zero but is
1q
1Q
1 (q ! Q)
V#
!
#
. All potentials inside the outer shell are just shifted by an amount
4, P0 r 4, P0 r 4, P0
r
1Q
V #!
. Therefore relative potentials within the shells are not affected. Thus (b) and (c) do not change.
4, P0 rb
However, now that the potential does vary outside the spheres, there is an electric field there:
QV
Q % kq !kQ & kq % Q & k
E#!
#! ' 0
( # '1 ! ( # 2 ( q ! Q) .
Qr
Qr ) r
r * r2 )
q* r
EVALUATE: In part (a) the potential is greater than zero for all r 1 rb . 2318 Chapter 23 23.50. IDENTIFY: %1 1&
%1 1 &
Exercise 23.49 shows that V # kq ' ! ( for r 1 ra , V # kq ' ! ( for ra 1 r 1 rb and
ra rb *
)
) r rb *
%1 1&
Vab # kq ' ! ( .
) ra rb *
kq
SET UP: E # 2 , radially outward, for ra N r N rb
r
%1 1&
500 V
EXECUTE: (a) Vab # kq ' ! ( # 500 V gives q #
# 7.62 " 10!10 C .
%
&
ra rb *
1
1
)
!
k'
(
) 0.012 m 0.096 m *
11V
#0
. V # 400 V at
r ra kq
r # 1.45 cm , V # 300 V at r # 1.85 cm , V # 200 V at r # 2.53 cm , V # 100 V at r # 4.00 cm , V # 0 at
r # 9.60 cm . The equipotential surfaces are sketched in Figure 23.50.
EVALUATE: (c) The equipotential surfaces are concentric spheres and the electric field lines are radial, so the field
lines and equipotential surfaces are mutually perpendicular. The equipotentials are closest at smaller r, where the
electric field is largest.
(b) Vb # 0 so Va # 500 V . The inner metal sphere is an equipotential with V # 500 V . Figure 23.50
23.51. IDENTIFY: Outside the cylinder it is equivalent to a line of charge at its center.
SET UP: The difference in potential between the surface of the cylinder (a distance R from the central axis) and a general point a distance r from the central axis is given by BV # 8
ln( r / R ) .
2, P0 EXECUTE: (a) The potential difference depends only on r, and not direction. Therefore all points at the same value
of r will be at the same potential. Thus the equipotential surfaces are cylinders coaxial with the given cylinder. 23.52. 8 ln( r / R ) for r, gives r # R e 2, P0 BV/8 .
2, P0
For 10 V, the exponent is (10 V)/[(2 " 9.00 " 109 N · m2/C2)(1.50 " 10–9 C/m)] = 0.370, which gives r = (2.00 cm)
e0.370 = 2.90 cm. Likewise, the other radii are 4.20 cm (for 20 V) and 6.08 cm (for 30 V).
(c) Br1 = 2.90 cm – 2.00 cm = 0.90 cm; Br2 = 4.20 cm – 2.90 cm = 1.30 cm; Br3 = 6.08 cm – 4.20 cm = 1.88 cm
EVALUATE: As we can see, Br increases, so the surfaces get farther apart. This is very different from a sheet of
charge, where the surfaces are equally spaced planes.
IDENTIFY: The electric field is the negative gradient of the potential.
QV
SET UP: Ex # !
, so Ex is the negative slope of the graph of V as a function of x.
Qx
EXECUTE: The graph is sketched in Figure 23.52. Up to a, V is constant, so Ex = 0. From a to b, V is linear with a
positive slope, so Ex is a negative constant. Past b, the Vx graph has a decreasing positive slope which approaches
zero, so Ex is negative and approaches zero.
(b) Solving BV # Electric Potential 2319 Notice that an increasing potential does not necessarily mean that the electric field is increasing. EVALUATE: Figure 23.52
23.53. (a) IDENTIFY: Apply the workenergy theorem, Eq.(6.6).
SET UP: Points a and b are shown in Figure 23.53a. Figure 23.53a
EXECUTE: Wtot # BK # K b ! K a # K b # 4.35 " 10 !5 J The electric force FE and the additional force F both do work, so that Wtot # WFE 0 WF .
WFE # Wtot ! WF # 4.35 " 10 !5 J ! 6.50 " 10!5 J # !2.15 " 10 !5 J The forces on the charged particle are shown in Figure 23.53b. EVALUATE: Figure 23.53b The electric force is to the left (in the direction of the electric field since the particle has positive charge). The
displacement is to the right, so the electric force does negative work. The additional force F is in the direction of the
displacement, so it does positive work.
(b) IDENTIFY and SET UP: For the work done by the electric force, Wa Eb # q (Va ! Vb )
EXECUTE: Va ! Vb # Wa Eb !2.15 " 10!5 J
#
# !2.83 " 103 V.
7.60 " 10!9 C
q EVALUATE: The starting point (point a) is at 2.83 " 103 V lower potential than the ending point (point b). We
know that Vb 5 Va because the electric field always points from high potential toward low potential.
(c) IDENTIFY: Calculate E from Va ! Vb and the separation d between the two points. SET UP: Since the electric field is uniform and directed opposite to the displacement Wa Eb # ! FE d # !qEd , where
d # 8.00 cm is the displacement of the particle.
W
V ! V !2.83 " 103 V
# 3.54 " 104 V/m.
EXECUTE: E # ! a Eb # ! a b #
0.0800 m
qd
d 23.54. EVALUATE: In part (a), Wtot is the total work done by both forces. In parts (b) and (c) Wa Eb is the work done just
by the electric force.
ke 2
IDENTIFY: The electric force between the electron and proton is attractive and has magnitude F # 2 . For
r
e2
circular motion the acceleration is arad # v 2 /r . U # ! k .
r
SET UP: e # 1.60 " 10!19 C . 1 eV # 1.60 " 10!19 J .
EXECUTE: (a) mv 2 ke2
ke 2
# 2 and v #
.
r
r
mr 1
1 ke2
1
(b) K # mv 2 #
#! U
2
2r
2 2320 23.55. Chapter 23 1
1 ke2
1 k (1.60 " 10!19 C)2
(c) E # K 0 U # U # !
#!
# !2.17 " 10!18 J # !13.6 eV .
2
2r
2 5.29 " 10!11 m
EVALUATE: The total energy is negative, so the electron is bound to the proton. Work must be done on the
electron to take it far from the proton.
!
!
!
IDENTIFY and SET UP: Calculate the components of E from Eq.(23.19). Eq.(21.3) gives F from E.
EXECUTE: (a) V # Cx 4 / 3
C # V / x 4 / 3 # 240 V /(13.0 " 10!3 m)4 / 3 # 7.85 " 104 V/m4 / 3
QV
4
# ! Cx1 / 3 # !(1.05 " 105 V/m 4 / 3 ) x1/ 3
Qx
3
!
The minus sign means that Ex is in the ! x direction, which says that E points from the positive anode toward the
negative cathode.
!
!
(c) F = qE so Fx # !eEx # 4 eCx1/3
3 (b) E x # ! Halfway between the electrodes means x # 6.50 " 10!3 m.
4
Fx # 3 (1.602 " 10!19 C)(7.85 " 104 V/m 4 / 3 )(6.50 " 10!3 m)1/ 3 # 3.13 " 10!15 N Fx is positive, so the force is directed toward the positive anode.
!
EVALUATE: V depends only on x, so E y # Ez # 0. E is directed from high potential (anode) to low potential
23.56. (cathode). The electron has negative charge, so the force on it is directed opposite to the electric field.
IDENTIFY: At each point (a and b), the potential is the sum of the potentials due to both spheres. The voltmeter
reads the difference between these two potentials. The spheres behave like a pointcharges since the meter is
connected to the surface of each one.
SET UP: (a) Call a the point on the surface of one sphere and b the point on the surface of the other sphere, call r
the radius of each sphere, and call d the centertocenter distance between the spheres. The potential difference Vab
between points a and b is then
Vb – Va = Vab #
EXECUTE: 1 9 !q
q
!q & :
2q % 1
1&
%q
!(
0
!' 0
'
(=
4,P 0 ) d ! r r *
4, P0 > r d ! r ) r d ! r * ?
;
< Substituting the numbers gives
1
1
%
&
6
Vb – Va = 2(175 µC) . 9.00 " 109 N + m 2 /C2 / '
!
( = –8.40 " 10 V
0.750 m 0.250 m *
) 23.57. The meter reads 8.40 MV.
(b) Since Vb – Va is negative, Va > Vb, so point a is at the higher potential.
EVALUATE: An easy way to see that the potential at a is higher than the potential at b is that it would take positive
work to move a positive test charge from b to a since this charge would be attracted by the negative sphere and
repelled by the positive sphere.
kq q
IDENTIFY: U # 1 2
r
SET UP: Eight charges means there are 8(8 ! 1) / 2 # 28 pairs. There are 12 pairs of q and !q separated by d, 12
pairs of equal charges separated by 2d and 4 pairs of q and ! q separated by 3d . 4&
12kq %
1
1&
% 12 12
2
(a) U # kq 2 ' ! 0
!
0
( # ! d '1 !
( # !1.46q /, P0 d
d
2d
3d *
2 3 3*
)
)
EVALUATE: (b) The fact that the electric potential energy is less than zero means that it is energetically favorable
for the crystal ions to be together.
kq q
IDENTIFY: For two small spheres, U # 1 2 . For part (b) apply conservation of energy.
r
SET UP: Let q1 # 2.00 $ C and q2 # !3.50 $ C . Let ra # 0.250 m and rb E F .
2 EXECUTE: 23.58. (8.99 " 109 N + m 2 /C2 )(2.00 " 10!6 C)(!3.50 " 10!6 C)
# !0.252 J
0.250 m
2
(b) K b # 0 . U b # 0 . U a # !0.252 J . K a 0 U a # K b 0 U b gives K a # 0.252 J . K a # 1 mva , so
2
EXECUTE: va # (a) U # 2Ka
2(0.252 J)
#
# 18.3 m/s
1.50 " 10!3 kg
m Electric Potential 23.59. 2321 EVALUATE: As the sphere moves away, the attractive electrical force exerted by the other sphere does negative
work and removes all the kinetic energy it initially had. Note that it doesn’t matter which sphere is held fixed and
which is shot away; the answer to part (b) is unaffected.
(a) IDENTIFY: Use Eq.(23.10) for the electron and each proton.
SET UP: The positions of the particles are shown in Figure 23.59a. r # (1.07 " 10!10 m) / 2 # 0.535 " 10!10 m
Figure 23.59a
EXECUTE: The potential energy of interaction of the electron with each proton is
1 ( !e 2 )
, so the total potential energy is
U#
4, P0 r U #! 2e 2
2(8.988 " 109 N + m 2 /C2 )(1.60 " 10!19 C)2
#!
# !8.60 " 10!18 J
4, P0 r
0.535 " 10!10 m U # !8.60 " 10!18 J(1 eV /1.602 " 10!19 J) # !53.7 eV
EVALUATE: The electron and proton have charges of opposite signs, so the potential energy of the system is
negative.
(b) IDENTIFY and SET UP: The positions of the protons and points a and b are shown in Figure 23.59b. rb # ra2 0 d 2 ra # r # 0.535 " 10!10 m
Figure 23.59b Apply K a 0 U a 0 Wother # K b 0 U b with point a midway between the protons and point b where the electron
instantaneously has v # 0 (at its maximum displacement d from point a).
EXECUTE: Only the Coulomb force does work, so Wother # 0. U a # !8.60 " 10!18 J (from part (a)) K a # 1 mv 2 # 1 (9.109 " 10!31 kg)(1.50 " 106 m/s)2 # 1.025 " 10!18 J
2
2
Kb # 0
U b # !2ke2 /rb
Then U b # K a 0 U a ! K b # 1.025 " 10!18 J ! 8.60 " 10!18 J # !7.575 " 10!18 J. rb # ! 2ke2
2(8.988 " 109 N + m 2 /C2 )(1.60 " 10!19 C) 2
#!
# 6.075 " 10!11 m
Ub
!7.575 " 10!18 J Then d # rb2 ! ra2 # (6.075 " 10!11 m) 2 ! (5.35 " 10!11 m) 2 # 2.88 " 10!11 m. 23.60. EVALUATE: The force on the electron pulls it back toward the midpoint. The transverse distance the electron
moves is about 0.27 times the separation of the protons.
IDENTIFY: Apply 7 Fx # 0 and 7 Fy # 0 to the sphere. The electric force on the sphere is Fe # qE . The potential difference between the plates is V # Ed .
SET UP: The freebody diagram for the sphere is given in Figure 23.56.
2
EXECUTE: T cos4 # mg and T sin 4 # Fe gives Fe # mg tan4 # (1.50 " 10 !3 kg)(9.80 m s )tan(303) # 0.0085 N .
Fe # Eq # Fd (0.0085 N)(0.0500 m)
Vq
#
# 47.8 V.
and V #
d
q
8.90 " 10!6 C 2322 Chapter 23 EVALUATE: E # V/d # 956 V/m . E # = /P0 and = # EP0 # 8.46 " 10!9 C/m 2 . Figure 23.60
23.61. (a) IDENTIFY: The potential at any point is the sum of the potentials due to each of the two charged conductors.
SET UP: From Example 23.10, for a conducting cylinder with charge per unit length 8 the potential outside the
cylinder is given by V # (8 /2, P0 )ln( r0 /r ) where r is the distance from the cylinder axis and r0 is the distance from the axis for which we take V # 0. Inside the cylinder the potential has the same value as on the cylinder surface. The
electric field is the same for a solid conducting cylinder or for a hollow conducting tube so this expression for V
applies to both. This problem says to take r0 # b.
EXECUTE: For the hollow tube of radius b and charge per unit length !8 : outside V # !(8 /2, P0 )ln(b /r ); inside
V # 0 since V # 0 at r # b.
For the metal cylinder of radius a and charge per unit length 8 :
outside V # (8 /2, P0 )ln(b /r ), inside V # (8 /2, P0 )ln(b /a ), the value at r # a. (i) r 1 a; inside both V # (8 /2, P0 )ln(b /a ) (ii) a 1 r 1 b; outside cylinder, inside tube V # (8 /2, P0 )ln(b /r )
(iii) r 5 b; outside both the potentials are equal in magnitude and opposite in sign so V # 0.
(b) For r # a, Va # (8 /2, P0 )ln(b /a ).
For r # b, Vb # 0.
Thus Vab # Va ! Vb # (8 /2, P0 )ln(b /a).
(c) IDENTIFY and SET UP: Use Eq.(23.23) to calculate E.
QV
8 Q %b&
8 % r &% b &
Vab 1
#!
ln ' ( # !
.
EXECUTE: E # !
' (' ! 2 ( #
Qr
2, P0 Qr ) r *
2, P0 ) b *) r * ln(b /a) r
(d) The electric field between the cylinders is due only to the inner cylinder, so Vab is not changed, Vab # (8 /2, P0 )ln(b /a ).
EVALUATE: The electric field is not uniform between the cylinders, so Vab R E (b ! a ).
23.62. IDENTIFY: The wire and hollow cylinder form coaxial cylinders. Problem 23.61 gives E ( r ) # Vab 1
.
ln(b /a ) r a # 145 " 10!6 m , b # 0.0180 m .
Vab 1
EXECUTE: E #
and Vab # E ln (b/a)r # (2.00 " 104 N C)(ln (0.018 m 145 " 10!6 m))0.012 m # 1157 V.
ln( b a ) r
EVALUATE: The electric field at any r is directly proportional to the potential difference between the wire and the
cylinder.
!
!
!
!
IDENTIFY and SET UP: Use Eq.(21.3) to calculate F and then F = ma gives a.
!
!
!
!
!
!
EXECUTE: (a) FE # qE. Since q # !e is negative FE and E are in opposite directions; E is upward so FE is
SET UP: 23.63. downward. The magnitude of FE is
FE # q E # eE # (1.602 " 10!19 C)(1.10 " 103 N/C) # 1.76 " 10!16 N.
(b) Calculate the acceleration of the electron produced by the electric force: a# F
1.76 " 10!16 N
#
# 1.93 " 1014 m/s 2
m 9.109 " 10!31 kg !
EVALUATE: This is much larger than g # 9.80 m/s 2 , so the gravity force on the electron can be neglected. FE is
!
downward, so a is downward.
(c) IDENTIFY and SET UP: The acceleration is constant and downward, so the motion is like that of a projectile.
Use the horizontal motion to find the time and then use the time to find the vertical displacement. Electric Potential 2323 EXECUTE: xcomponent
v0 x # 6.50 " 106 m/s; a x # 0; x ! x0 # 0.060 m; t # ? x ! x0 # v0 xt 0 1 axt 2 and the a x term is zero, so
2
x ! x0
0.060 m
t#
#
# 9.231" 10!9 s
v0 x
6.50 " 106 m/s
ycomponent
v0 y # 0; a y # 1.93 " 1014 m/s2 ; t # 9.231 " 10 !9 m/s; y ! y0 # ? y ! y0 # v0 y t 0 1 a y t 2
2
y ! y0 # 1 (1.93 " 1014 m/s 2 )(9.231" 10!9 s)2 # 0.00822 m # 0.822 cm
2
(d) The velocity and its components as the electron leaves the plates are sketched in Figure 23.63.
vx # v0 x # 6.50 " 106 m/s (since ax # 0 )
v y # v0 y 0 a yt v y # 0 0 (1.93 " 1014 m/s2 )(9.231" 10!9 s)
v y # 1.782 " 106 m/s
Figure 23.63
1.782 " 10 m/s
# 0.2742 so 2 # 15.33.
vx 6.50 " 106 m/s
EVALUATE: The greater the electric field or the smaller the initial speed the greater the downward deflection.
(e) IDENTIFY and SET UP: Consider the motion of the electron after it leaves the region between the plates.
Outside the plates there is no electric field, so a # 0. (Gravity can still be neglected since the electron is traveling at
such high speed and the times are small.) Use the horizontal motion to find the time it takes the electron to travel
0.120 m horizontally to the screen. From this time find the distance downward that the electron travels.
EXECUTE: xcomponent
v0 x # 6.50 " 106 m/s; a x # 0; x ! x0 # 0.120 m; t # ?
tan 2 # vy # 6 x ! x0 # v0 xt 0 1 axt 2 and the a x term is term is zero, so
2
x ! x0
0.120 m
t#
#
# 1.846 " 10!8 s
v0 x
6.50 " 106 m/s
ycomponent
v0 y # 1.782 " 106 m/s (from part (b)); a y # 0; t # 1.846 " 10 !8 m/s; y ! y0 # ? y ! y0 # v0 yt 0 1 a yt 2 # (1.782 " 106 m/s)(1.846 " 10!8 s) # 0.0329 m # 3.29 cm
2 23.64. EVALUATE: The electron travels downward a distance 0.822 cm while it is between the plates and a distance
3.29 cm while traveling from the edge of the plates to the screen. The total downward deflection is
0.822 cm 0 3.29 cm # 4.11 cm.
The horizontal distance between the plates is half the horizontal distance the electron travels after it leaves the
plates. And the vertical velocity of the electron increases as it travels between the plates, so it makes sense for it to
have greater downward displacement during the motion after it leaves the plates.
IDENTIFY: The charge on the plates and the electric field between them depend on the potential difference across the
plates. Since we do not know the numerical potential, we shall call this potential V and find the answers in terms of V.
=
Qd
(a) SET UP: For two parallel plates, the potential difference between them is V # Ed # d #
.
P0
P0 A
EXECUTE: Solving for Q gives Q # P0 AV / d # (8.85 " 10–12 C2/N + m2)(0.030 m)2V/(0.0050 m)
Q = 1.59V " 10–12 C = 1.59V pC, when V is in volts.
(b) E = V/d = V/(0.0050 m) = 200V V/m, with V in volts.
(c) SET UP: Energy conservation gives 1 mv 2 # eV .
2
EXECUTE: Solving for v gives
v# 2 .1.60 " 10!19 C /V
2eV
#
# 5.93 " 105V 1/ 2 m/s , with V in volts
m
9.11 " 10!31 kg EVALUATE: Typical voltages in student laboratory work run up to around 25 V, so the charge on the plates is typically
about around 40 pC, the electric field is about 5000 V/m, and the electron speed would be about 3 million m/s. 2324 Chapter 23 23.65. (a) IDENTIFY and SET UP: Problem 23.61 derived that E # Vab 1
, where a is the radius of the inner cylinder
ln(b / a ) r (wire) and b is the radius of the outer hollow cylinder. The potential difference between the two cylinders is Vab .
Use this expression to calculate E at the specified r.
EXECUTE: Midway between the wire and the cylinder wall is at a radius of
r # ( a 0 b)/2 # (90.0 " 10!6 m 0 0.140 m)/2 # 0.07004 m.
E# Vab 1
50.0 " 103 V
#
# 9.71 " 104 V/m
ln(b / a ) r ln(0.140 m / 90.0 " 10!6 m)(0.07004 m) (b) IDENTIFY and SET UP: The electric force is given by Eq.(21.3). Set this equal to ten times the weight of the
particle and solve for q , the magnitude of the charge on the particle. FE # 10mg EXECUTE: 10mg 10(30.0 " 10!9 kg)(9.80 m/s 2 )
#
# 3.03 " 10!11 C
E
9.71 " 104 V/m
EVALUATE: It requires only this modest net charge for the electric force to be much larger than the weight.
(a) IDENTIFY: Calculate the potential due to each thin ring and integrate over the disk to find the potential. V is a
scalar so no components are involved.
SET UP: Consider a thin ring of radius y and width dy. The ring has area 2, y dy so the charge on the ring is
dq # = (2, y dy ).
EXECUTE: The result of Example 23.11 then says that the potential due to this thin ring at the point on the axis at a
distance x from the ring is
1
dq
2,=
y dy
dV #
#
2
2
4, P0 x 0 y
4, P0 x 2 0 y 2
q E # 10mg and q # 23.66. V # H dV #
EVALUATE: =
2P0 H R
0 y dy
x 0y
2 2 # R
=9 2
=
x 0 y2 : #
( x 2 0 R 2 ! x) 2P0 ; <0 For x # R this result should reduce to the potential of a point charge with Q # =, R 2 .
x 2 0 R 2 # x(1 0 R 2 /x 2 )1/ 2 C x(1 0 R 2 /2 x 2 ) so Then V C = R2
2P0 2 x # x 2 0 R 2 ! x C R 2 /2 x =, R 2
Q
#
, as expected.
4, P0 x 4, P0 x (b) IDENTIFY and SET UP: Use Eq.(23.19) to calculate Ex . & =x%1
&
QV
=%
x
1
#!
! 1( #
'
'! 2
(.
2
Qx
2P0 ) x 2 0 R 2
2P0 ) x
x 0R *
*
EVALUATE: Our result agrees with Eq.(21.11) in Example 21.12.
!
b!
(a) IDENTIFY: Use Va ! Vb # H E + dl.
EXECUTE: 23.67. 2P0 Ex # ! a SET UP: E (r ) # From Problem 22.48, E ( r ) # 8r
for r N R (inside the cylindrical charge distribution) and
2, P0 R 2 8r
for r O R. Let V # 0 at r # R (at the surface of the cylinder).
2, P0 r EXECUTE: r 5 R
!
!!
!
Take point a to be at R and point b to be at r, where r 5 R. Let dl = dr. E and dr are both radially outward, so
!!
r
r
E + d r # E dr. Thus VR ! Vr # H E dr. Then VR # 0 gives Vr # ! H E dr. In this interval (r 5 R ), E (r ) # 8 /2, P0 r , so
R R Vr # ! H r
R 8
8 r dr
8
%r&
dr # !
H R r # ! 2, P0 ln ' R (.
2, P0 r
2, P0
)* EVALUATE: This expression gives Vr # 0 when r # R and the potential decreases (becomes a negative number of
larger magnitude) with increasing distance from the cylinder. Electric Potential EXECUTE: 2325 r1R !!
R
Take point a at r, where r 1 R, and point b at R. E + dr # E dr as before. Thus Vr ! VR # H E dr. Then VR # 0 gives
r R Vr # H E dr. In this interval (r 1 R ), E ( r ) # 8 r / 2, P0 R , so
2 r Vr # H R
r 8
2, P0 R 2 dr # 8
2, P0 R Vr # 2 H R
r r dr # 8 % %r&
'1 ! ' (
4, P0 ' ) R *
) 2 8 % R2 r 2 &
! (.
'
2, P0 R ) 2
2*
2 &
(.
(
* EVALUATE: This expression also gives Vr # 0 when r # R. The potential is 8 / 4, P0 at r # 0 and decreases with
increasing r.
(b) EXECUTE: Graphs of V and E as functions of r are sketched in Figure 23.67. Figure 23.67
23.68. EVALUATE: E at any r is the negative of the slope of V (r ) at that r (Eq.23.23).
IDENTIFY: The alpha particles start out with kinetic energy and wind up with electrical potential energy at closest
approach to the nucleus.
SET UP: (a) The energy of the system is conserved, with U # (1/ 4, P0 )(qq0 / r ) being the electric potential energy.
With the charge of the alpha particle being 2e and that of the gold nucleus being Ze, we have
12
1 2 Ze 2
mv #
2
4, P0 R EXECUTE: Solving for v and using Z = 79 for gold gives % 1 & 4 Ze 2
v# '
=
(
) 4, P0 * mR 23.69. . 9.00 " 10 N + m /C / (4)(79) .1.60 " 10
. 6.7 "10 kg /. 5.6 " 10 m /
9 2 2 !27 !15 C/ 2 = 4.4 " 107 m/s We have neglected any relativistic effects.
(b) Outside the atom, it is neutral. Inside the atom, we can model the 79 electrons as a uniform spherical shell, which
produces no electric field inside of itself, so the only electric field is that of the nucleus.
EVALUATE: Neglecting relativistic effects was not such a good idea since the speed in part (a) is over 10% the
speed of light. Modeling 79 electrons as a uniform spherical shell is reasonable, but we would not want to do this
with small atoms.
!
b!
IDENTIFY: Va ! Vb # H E + dl .
a SET UP: From Example 21.10, we have: Ex # !!
1
Qx
. E + dl # Ex dx . Let a # F so Va # 0 .
2
2 3/ 2
4, P0 ( x 0 a )
u # x2 0 a2 EXECUTE:
23.70. !19 x
Q
xQ !1/ 2
V #!
H ( x2 0 a 2 )3/ 2 dx # 4, P0 u
4, P0 F
u #F # 1
4, P0 Q
x 0 a2
2 . EVALUATE: Our result agrees with Eq.(23.16) in Example 23.11.
IDENTIFY: Divide the rod into infinitesimal segments with charge dq. The potential dV due to the segment is
1 dq
dV #
. Integrate over the rod to find the total potential.
4, P0 r
SET UP: dq # 8 dl , with 8 # Q / , a and dl # a d4 .
, EXECUTE: dV # 1 dq
1 8 dl
1 Q dl
1 Q d4
1 Q d4
1Q
#
#
#
.V#
#
.
4, P0 r 4, P0 a
4, P0 , a a 4, P0 , a
4, P0 H , a
4, P0 a
0 2326 23.71. 23.72. Chapter 23 EVALUATE: All the charge of the ring is the same distance a from the center of curvature.
IDENTIFY: We must integrate to find the total energy because the energy to bring in more charge depends on the
charge already present.
SET UP: If G is the uniform volume charge density, the charge of a spherical shell or radius r and thickness dr is dq
= G 4!r2 dr, and G = Q/(4/3 !R3). The charge already present in a sphere of radius r is q = G(4/3 !r3). The energy to
bring the charge dq to the surface of the charge q is Vdq, where V is the potential due to q, which is q / 4, P0 r.
EXECUTE: The total energy to assemble the entire sphere of radius R and charge Q is sum (integral) of the tiny
increments of energy.
4
G , r3
2
R
&
q
3% 1
3
U # H Vdq # H
dq # H
. G 4, r 2dr / # 5 ' 4, P Q (
0 4, P r
4, P0 r
0
0 R*
) where we have substituted G = Q/(4/3 !R3) and simplified the result.
EVALUATE: For a pointcharge, R E 0 so U E F, which means that a pointcharge should have infinite selfenergy. This suggests that either pointcharges are impossible, or that our present treatment of physics is not
adequate at the extremely small scale, or both.
!
!!
b!
IDENTIFY: Va ! Vb # H E + dl . The electric field is radially outward, so E + dl = E dr .
a SET UP: Let a # F , so Va # 0 .
dr  kQ
kQ
and V # ! kQ H 2 #
.
2
r
rr
F
r EXECUTE: From Example 22.9, we have the following. For r 5 R : E # r R
! ! r ! ! kQ kQ r
kQ kQ 1 2
kQ kQ kQr 2 kQ 9
r2 :
kQr
! 3 H r dr  #
! 3 r #
0
!
#
and V # ! H E + dr  ! H E + dr  #
>3 ! 2 ? .
R3
R RR
R R 2 R R 2 R 2 R3 2 R ; R <
R
F
(b) The graphs of V and E versus r are sketched in Figure 23.72.
EVALUATE: For r 1 R the potential depends on the electric field in the region r to F . For r 1 R : E # Figure 23.72
23.73. IDENTIFY:
SET UP: Problem 23.70 shows that Vr #
V0 # Q
Q
(3 ! r 2 R 2 ) for r N R and Vr #
for r O R .
8, P0 R
4, P0 r 3Q
Q
, VR #
8, P0 R
4, P0 R Q
8, P0 R
(b) If Q 5 0 , V is higher at the center. If Q 1 0 , V is higher at the surface.
!
EVALUATE: For Q 5 0 the electric field is radially outward, E is directed toward lower potential, so V is higher at
the center. If Q 1 0 , the electric field is directed radially inward and V is higher at the surface.
EXECUTE: 23.74. (a) V0 ! VR # IDENTIFY: For r 1 c , E # 0 and the potential is constant. For r 5 c , E is the same as for a point charge and V # SET UP:
EXECUTE: VF # 0
(a) Points a, b, and c are all at the same potential, so Va ! Vb # Vb ! Vc # Va ! Vc # 0 .
2 kq (8.99 " 109 N + m 2 C )(150 " 10!6 C)
#
# 2.25 " 106 V
R
0.60 m
(b) They are all at the same potential.
(c) Only Vc ! VF would change; it would be !2.25 " 106 V.
Vc ! VF # kq
.
r Electric Potential 23.75. 23.76. 2327 EVALUATE: The voltmeter reads the potential difference between the two points to which it is connected.
IDENTIFY and SET UP: Apply Fr # ! dU / dr and Newton's third law.
EXECUTE: (a) The electrical potential energy for a spherical shell with uniform surface charge density and a point
charge q outside the shell is the same as if the shell is replaced by a point charge at its center. Since Fr # ! dU dr ,
this means the force the shell exerts on the point charge is the same as if the shell were replaced by a point charge at
its center. But by Newton’s 3rd law, the force q exerts on the shell is the same as if the shell were a point charge. But
q can be replaced by a spherical shell with uniform surface charge and the force is the same, so the force between the
shells is the same as if they were both replaced by point charges at their centers. And since the force is the same as for
point charges, the electrical potential energy for the pair of spheres is the same as for a pair of point charges.
(b) The potential for solid insulating spheres with uniform charge density is the same outside of the sphere as for a
spherical shell, so the same result holds.
(c) The result doesn’t hold for conducting spheres or shells because when two charged conductors are brought close
together, the forces between them causes the charges to redistribute and the charges are no longer distributed
uniformly over the surfaces.
qq
kq q
EVALUATE: For the insulating shells or spheres, F # k 1 2 2 and U # 1 2 , where q1 and q2 are the charges of
r
r
the objects and r is the distance between their centers.
IDENTIFY: Apply Newton's second law to calculate the acceleration. Apply conservation of energy and
conservation of momentum to the motions of the spheres.
qq
kq q
SET UP: Problem 23.75 shows that F # k 1 2 2 and U # 1 2 , where q1 and q2 are the charges of the objects and
r
r
r is the distance between their centers.
EXECUTE: Maximum speed occurs when the spheres are very far apart. Energy conservation gives
kq1q2 1
1
2
2
# m50v50 0 m150v150 . Momentum conservation gives m50v50 # m150v150 and v50 # 3v150 . r # 0.50 m. Solve for v50
r
2
2
and v150 : v50 # 12.7 m s, v150 # 4.24 m s . Maximum acceleration occurs just after spheres are released. 7 F # ma gives (9 " 109 N + m 2 C 2 )(10!5 C)(3 " 10!5 C)
kq1q2
2
# (0.15 kg)a150 . a150 # 72.0 m s and
# m150 a150 .
2
(0.50 m) 2
r
2 23.77. 23.78. a50 # 3a150 # 216 m s .
EVALUATE: The more massive sphere has a smaller acceleration and a smaller final speed.
IDENTIFY: Use Eq.(23.17) to calculate Vab .
SET UP: From Problem 22.43, for R N r N 2 R (between the sphere and the shell) E # Q / 4, P0 r 2
Take a at R and b at 2R.
2R
2R
Q 2 R dr
Q 9 1:
Q %1 1 &
#
!? #
EXECUTE: Vab # Va ! Vb # H E dr #
'!
(
>
R
4, P0 H R r 2 4, P0 ; r < R
4, P0 ) R 2 R *
Q
Vab #
8, P0 R
EVALUATE: The electric field is radially outward and points in the direction of decreasing potential, so the sphere
is at higher potential than the shell.
!
b!
IDENTIFY: Va ! Vb # H E + dl
a
!
!!
SET UP: E is radially outward, so E + dl = E dr . Problem 22.42 shows that E ( r ) # 0 for r N a , E (r ) # kq / r 2 for
a 1 r 1 b , E ( r ) # 0 for b 1 r 1 c and E (r ) # kq / r 2 for r 5 c .
c EXECUTE: kq
kq
dr # .
2
c
Fr (a) At r # c : Vc # ! H c
! ! b ! ! kq
kq
!0#
(b) At r # b : Vb # ! H E + dr ! H E + dr #
.
c
c
c
F
c
a
! ! b ! ! a ! ! kq
dr
91 1 1 :
! kq H 2 # kq > ! 0 ?
(c) At r # a : Va # ! H E + dr ! H E + dr ! H E + dr #
c
r
;c b a <
F
c
b
b 91 1 1 :
(d) At r # 0 : V0 # kq > ! 0 ? since it is inside a metal sphere, and thus at the same potential as its surface.
;c b a <
91 1:
EVALUATE: The potential difference between the two conductors is Va ! Vb # kq > ! ? .
;a b< 2328 Chapter 23 23.79. IDENTIFY: Slice the rod into thin slices and use Eq.(23.14) to calculate the potential due to each slice. Integrate
over the length of the rod to find the total potential at each point.
(a) SET UP: An infinitesimal slice of the rod and its distance from point P are shown in Figure 23.79a. Figure 23.79a Use coordinates with the origin at the lefthand end of the rod and one axis along the rod. Call the axes x and y so
as not to confuse them with the distance x given in the problem.
EXECUTE: Slice the charged rod up into thin slices of width dx. Each slice has charge dQ # Q (dx/a ) and a
distance r # x 0 a ! x from point P. The potential at P due to the small slice dQ is
dV # 1 % dQ &
1 Q % dx &
'
(#
'
(.
4, P0 ) r * 4, P0 a ) x 0 a ! x * Compute the total V at P due to the entire rod by integrating dV over the length of the rod ( x # 0 to x # a ) :
a
Q
dxQ
Q
% x0a&
a
'
(
H 0 ( x 0 a ! x) # 4, P0a [! ln( x 0 a ! x)]0 # 4, P0a ln ) x *.
4, P0 a V # H dV #
EVALUATE:
(b) SET UP: Q
% x&
ln ' ( # 0.
4, P0 a ) x *
An infinitesimal slice of the rod and its distance from point R are shown in Figure 23.79b.
As x E F, V E Figure 23.79b dQ # (Q / a )dx as in part (a) Each slice dQ is a distance r # y 2 0 (a ! x) 2 from point R.
EXECUTE: The potential dV at R due to the small slice dQ is
dV # dx 1 % dQ &
1Q
'
(#
4, P0 ) r * 4, P0 a V # H dV # Q
4, P0 a H y 0 (a ! x) 2
2 dx a
0 y 0 (a ! x) 2
2 . . In the integral make the change of variable u # a ! x; du # ! dxV #! V #! 0
0
Q
du
Q9
#!
ln(u 0 y 2 0 u 2 ) :
<a
4, P0 a H a y 2 0 u 2
4, P0 a ; Q
Q 9 % a 0 a2 0 y2
> ln '
[ln y ! ln(a 0 y 2 0 a 2 )] #
4, P0 a
4, P0 a > '
y
;) (The expression for the integral was found in appendix B.)
% y&
Q
EVALUATE: As y E F, V E
ln ' ( # 0.
4, P0 a ) y * &:
(? .
(?
*< Electric Potential part (a): V # (c) SET UP: 2329 Q
Q
% x0a&
% a&
ln '
ln ' 1 0 ( .
(#
4, P0 a ) x * 4, P0 a )
x* From Appendix B, ln(1 0 u ) # u ! u 2 / 2 . . . , so ln(1 0 a / x) # a / x ! a 2 / 2 x 2 and this becomes a / x when x is large.
Thus V E EXECUTE: part (b): V # Q %a&
Q
. For large x, V becomes the potential of a point charge.
' (#
4, P0 a ) x * 4, P0 a Q 9 % a 0 a2 0 y2
>ln '
y
4, P0 a > '
;) From Appendix B, &:
%a
Q
a2 &
(? #
ln ' 0 1 0 2 ( .
( ? 4, P0 a ' y
y(
)
*
*< 1 0 a 2 / y 2 # (1 0 a 2 / y 2 )1/ 2 # 1 0 a 2 / 2 y 2 0 $ Thus a / y 0 1 0 a 2 / y 2 E 1 0 a / y 0 a 2 / 2 y 2 0 $ E 1 0 a / y. And then using ln(1 0 u ) C u gives
VE
EVALUATE:
23.80. IDENTIFY: Q
Q %a&
Q
ln(1 0 a / y ) E
.
' (#
4, P0 a
4, P0 a ) y * 4, P0 y For large y, V becomes the potential of a point charge.
The potential at the surface of a uniformly charged sphere is V # kQ
.
R 4
For a sphere, V # , R 3 . When the raindrops merge, the total charge and volume is conserved.
3
kQ k (!1.20 " 10!12 C)
#
# !16.6 V .
EXECUTE: (a) V #
R
6.50 " 10!4 m
SET UP: (b) The volume doubles, so the radius increases by the cube root of two: Rnew # 3 2 R # 8.19 " 10!4 m and the new kQnew k (!2.40 " 10!12 C)
#
# !26.4 V .
8.19 " 10!4 m
Rnew
The charge doubles but the radius also increases and the potential at the surface increases by only a charge is Qnew # 2Q # ! 2.40 " 10!12 C. The new potential is Vnew # 23.81. EVALUATE:
2
factor of 1/ 3 # 22 / 3 .
2
(a) IDENTIFY and SET UP: The potential at the surface of a charged conducting sphere is given by Example 23.8:
1q
. For spheres A and B this gives
V#
4, P0 R VA #
EXECUTE: QA
QB
and VB #
.
4, P0 RA
4, P0 RB VA # VB gives QA / 4, P0 RA # QB / 4, P0 RB and QB / QA # RB / RA . And then RA # 3RB implies QB / QA # 1/ 3.
(b) IDENTIFY and SET UP:
Example 22.5: The electric field at the surface of a charged conducting sphere is given in
E# EXECUTE: 1q
.
4, P0 R 2 For spheres A and B this gives
EA # QA
QB
and EB #
.
2
2
4, P0 RA
4, P0 RB 2
EB % QB & % 4, P0 RA &
2
2
#'
( # QB /QA ( RA /RB ) # (1/3)(3) # 3.
2 ('
E A ) 4, P0 R B * ' QA (
)
* 23.82. EVALUATE: The sphere with the larger radius needs more net charge to produce the same potential. We can write
E # V / R for a sphere, so with equal potentials the sphere with the smaller R has the larger V.
IDENTIFY: Apply conservation of energy, K a 0 U a # K b 0 U b .
SET UP: Assume the particles initially are far apart, so U a # 0 , The alpha particle has zero speed at the distance of closest approach, so K b # 0 . 1 eV # 1.60 " 10!19 J . The alpha particle has charge 02e and the lead nucleus has
charge 082e . 2330 Chapter 23 Set the alpha particle’s kinetic energy equal to its potential energy: K a # U b gives EXECUTE: 23.83. k (164)(1.60 " 10!19 C)2
k (2e)(82e)
11.0 MeV #
and r #
# 2.15 " 10!14 m .
r
(11.0 " 106 eV)(1.60 " 10!19 J eV)
EVALUATE: The calculation assumes that at the distance of closest approach the alpha particle is outside the radius
of the lead nucleus.
IDENTIFY and SET UP: The potential at the surface is given by Example 23.8 and the electric field at the surface is
given by Example 22.5. The charge initially on sphere 1 spreads between the two spheres such as to bring them to
the same potential.
1 Q1
1 Q1
EXECUTE: (a) E1 #
# R1E1
, V1 #
2
4, P0 R1
4, P0 R1
(b) Two conditions must be met:
1) Let q1and q2 be the final potentials of each sphere. Then q1 0 q2 # Q1 (charge conservation)
2) Let V1 and V2 be the final potentials of each sphere. All points of a conductor are at the same potential, so V1 # V2 .
V1 # V2 requires that 1 q1
1 q2
#
and then q1 / R1 # q2 / R2
4, P0 R1 4, P0 R2
q1R2 # q2 R1 # (Q1 ! q1 ) R1 This gives q1 # ( R1 /[ R1 0 R2 ])Q1 and q2 # Q1 ! q1 # Q1 (1 ! R1 /[ R1 0 R2 ]) # Q1 ( R2 /[ R1 0 R2 ])
(c) V1 # 1 q1
Q1
1 q2
Q1
#
#
, which equals V1 as it should.
and V2 #
4, P0 R1 4, P0 ( R1 0 R2 )
4, P0 R2 4, P0 ( R1 0 R2 ) (d) E1 # V1
Q1
V
Q1
#
. E2 # 2 #
.
R1 4, P0 R1 ( R1 0 R2 )
R2 4, P0 R2 ( R1 0 R2 ) EVALUATE: 23.84. Part (a) says q2 # q1 ( R2 / R1 ). The sphere with the larger radius needs more charge to produce the same potential at its surface. When R1 # R2 , q1 # q2 # Q1 / 2. The sphere with the larger radius has the smaller
electric field at its surface.
!
b!
IDENTIFY: Apply Va ! Vb # H E + dl
a From Problem 22.57, for r O R , E # SET UP: kQ 9 r 3
r4 :
kQ
. For r N R , E # 2 > 4 3 ! 3 4 ? .
2
r
r;R
R< r EXECUTE: (a) r O R : E # kQ
kQ
kQ
I V # ! H 2 dr  #
, which is the potential of a point charge.
2
r
rr
F R
r
:
kQ 9 r 3
r4 :
kQ 9
r2
R 2 r 3 R 3 : kQ 9 r 3
r2
4 3 ! 3 4 ? and V # ! H Edr ! H Edr #
>1 ! 2 2 0 2 2 0 3 ! 3 ? #
> 3 ! 2 2 0 2? .
2>
r;R
R<
R;
R
R
R R < R ;R
R
<
F
R
kQ
2kQ
At r # R , V #
. At r # 0 , V #
. The electric field is radially outward and V increases as r
R
R (b) r N R : E #
EVALUATE:
23.85. decreases.
IDENTIFY: Apply conservation of energy: Ei # Ef .
SET UP: In the collision the initial kinetic energy of the two particles is converted into potential energy at the
distance of closest approach.
EXECUTE: (a) The two protons must approach to a distance of 2rp , where rp is the radius of a proton.
2
k (1.60 " 10!19 C)2
91
: ke
Ei # Ef gives 2 > mpv 2 ? #
and v #
# 7.58 " 106 m s .
2(1.2 " 10!15 m)(1.67 " 10!27 kg)
;2
< 2rp
(b) For a heliumhelium collision, the charges and masses change from (a) and v# k (2(1.60 " 10!19 C)) 2
# 7.26 " 106 m s.
(3.5 " 10!15 m)(2.99)(1.67 " 10!27 kg) (c) K # m v 2 (1.67 " 10!27 kg)(7.58 " 106 m s)2
3kT mv 2
#
. Tp # p #
# 2.3 " 109 K .
2
2
3k
3(1.38 " 10!23 J K)
THe # mHev 2 (2.99)(1.67 " 10!27 kg)(7.26 " 106 m s)2
#
# 6.4 " 109 K .
3k
3(1.38 " 10!23 J K) Electric Potential 23.86. 2331 (d) These calculations were based on the particles’ average speed. The distribution of speeds ensures that there are
always a certain percentage with a speed greater than the average speed, and these particles can undergo the
necessary reactions in the sun’s core.
EVALUATE: The kinetic energies required for fusion correspond to very high temperatures.
!
b!
W
IDENTIFY and SET UP: Apply Eq.(23.20). a Eb # Va ! Vb and Va ! Vb # H E + dl .
a
q0
!
QV ˆ QV ˆ QV ˆ
ˆ
ˆ
EXECUTE: (a) E = !
i!
j!
k = !2 Axi 0 6 Ayˆ ! 2 Azk
j
Qx
Qy
Qz
0
0
!
2
ˆ
(b) A charge is moved in along the z axis. The work done is given by W # q H E + kdz # q H (!2 Az ) dz # 0 ( Aq ) z0 . z0 z0 !5 Wa Eb
6.00 " 10 J
#
# 640 V m 2 .
2
qz0
(1.5 " 10!6 C)(0.250 m) 2
!
ˆ
ˆ
(c) E (0,0,0.250) = !2(640 V m 2 )(0.250 m)k = !(320 V m)k . Therefore, A # (d) In every plane parallel to the xz plane, y is constant, so V ( x, y, z ) # Ax 2 0 Az 2 ! C , where C # 3Ay 2 . V 0C
# R 2 , which is the equation for a circle since R is constant as long as we have constant potential on
A
those planes.
2
1280 V 0 3(640 V m )(2.00 m)2
(e) V # 1280 V and y # 2.00 m , so x 2 0 z 2 #
# 14.0 m 2 and the radius of the circle
2
640 V m
is 3.74 m.
!
EVALUATE: In any plane parallel to the xzplane, E projected onto the plane is radial and hence perpendicular to
the equipotential circles.
IDENTIFY: Apply conservation of energy to the motion of the daughter nuclei.
SET UP: Problem 23.73 shows that the electrical potential energy of the two nuclei is the same as if all their charge
was concentrated at their centers.
EXECUTE: (a) The two daughter nuclei have half the volume of the original uranium nucleus, so their radii are
7.4 " 10!15 m
smaller by a factor of the cube root of 2: r #
# 5.9 " 10!15 m.
3
2
k (46e) 2 k (46) 2 (1.60 " 10!19 C)2
(b) U #
#
# 4.14 " 10!11 J . U # 2 K , where K is the final kinetic energy of each
2r
1.18 " 10!14 m
x2 0 z 2 # 23.87. nucleus. K # U 2 # (4.14 " 10!11 J) 2 # 2.07 " 10!11 J .
(c) If we have 10.0 kg of uranium, then the number of nuclei is n # 23.88. 10.0 kg
# 2.55 " 1025 nuclei .
(236 u)(1.66 " 10!27 kg u) And each releases energy U, so E # nU # (2.55 " 1025 )(4.14 " 10!11 J) # 1.06 " 1015 J # 253 kilotons of TNT .
(d) We could call an atomic bomb an “electric” bomb since the electric potential energy provides the kinetic energy
of the particles.
EVALUATE: This simple model considers only the electrical force between the daughter nuclei and neglects the
nuclear force.
QV
IDENTIFY and SET UP: In part (a) apply E # !
. In part (b) apply Gauss's law.
Qr
!
QV
G a2 9 r
QV
r2 : G a 9 r r2 :
EXECUTE: (a) For r N a , E # !
# 0 . E has
# ! 0 > !6 2 0 6 3 ? # 0 > ! 2 ? . For r O a , E # !
Qr
Qr
18P0 ; a
a < 3P0 ; a a <
only a radial component because V depends only on r.
Q
G a 9 r r2 :
(b) For r N a , Gauss's law gives Er 4, r 2 # r # 0 > ! 2 ? 4, r 2 and
P0
3P0 ; a a <
Er 0 dr 4, (r 2 0 2rdr ) # Qr 0 dr G 0 a 9 r 0 dr (r 2 0 2rdr ) :
2
#
!
>
? 4, (r 0 2rdr ) . Therefore,
P0
a2
3P0 ; a
< Qr 0 dr ! Qr G (r )4, r 2 dr G0 a 4, r 2 dr 9 2r 2 2r 1 :
G0 9 4r :
9 4r :
#
C
> ! a 2 0 a ! a 2 0 a ? and G (r ) # 3 >3 ! a ? # G0 >1 ! 3a ? .
3P0
P0
P0
;
<
;
<
;
< 2332 Chapter 23 (c) For r O a , G (r ) # 0 , so the total charge enclosed will be given by
a 23.89. a
a9
91 3 r4 :
4r 3 :
Q # 4, H G (r )r 2 dr # 4,G 0 H > r 2 !
? dr # 4,G 0 > r ! ? # 0 .
0
3a <
3a < 0
;
;3
0
EVALUATE: Apply Gauss's law to a sphere of radius r 5 R . The result of part (c) says that Qencl # 0 , so E # 0 .
This agrees with the result we calculated in part (a).
IDENTIFY: Angular momentum and energy must be conserved.
SET UP: At the distance of closest approach the speed is not zero. E # K 0 U . q1 # 2e , q2 # 82e . EXECUTE: 1
kq q
mv1b # mv2 r2 . E1 # E2 gives E1 # mv2 2 0 1 2 . E1 # 11 MeV # 1.76 " 10!12 J . r2 is the distance of
2
r2 %b&
b 2 kq q
closest approach. Substituting in for v2 # v1 ' ( we find E1 # E1 2 0 1 2 .
r2
r2
) r2 *
2
2
!12
!12
( E1 )r2 ! (kq1q2 )r2 ! E1b # 0 . For b # 10 m , r2 # 1.01 " 10 m . For b # 10!13 m , r2 # 1.11 " 10!13 m . And for 23.90. b # 10!14 m , r2 # 2.54 " 10!14 m .
EVALUATE: As b decreases the collision is closer to being headon and the distance of closest approach decreases.
Problem 23.82 shows that the distance of closest approach is 2.15 " 10!14 m when b # 0 .
IDENTIFY: Consider the potential due to an infinitesimal slice of the cylinder and integrate over the length of the
QV
.
cylinder to find the total potential. The electric field is along the axis of the tube and is given by E # !
Qx
SET UP: Use the expression from Example 23.11 for the potential due to each infinitesimal slice. Let the slice be at
coordinate z along the xaxis, relative to the center of the tube.
EXECUTE: (a) For an infinitesimal slice of the finite cylinder, we have the potential
k dQ
kQ
dz
dV #
#
. Integrating gives
L ( x ! z )2 0 R 2
( x ! z )2 0 R 2
L 2! x L2 V# kQ
dz
kQ
du
#
where u # x ! z . Therefore,
H
L ! L 2 ( x ! z )2 0 R 2
L ! LH ! x u 2 0 R2
2 V# kQ 9 ( L 2 ! x) 2 0 R 2 0 ( L 2 ! x ) :
? on the cylinder axis.
ln >
L > ( L 2 0 x)2 0 R 2 ! L 2 ! x ?
;
< (b) For L 11 R , V C VC kQ 9 ( L 2 ! x ) 2 0 R 2 0 L 2 ! x : kQ 9 x 2 ! xL 0 R 2 0 L 2 ! x :
?C
ln >
ln >
?.
L > ( L 2 0 x) 2 0 R 2 ! L 2 ! x ? L > x 2 0 xL 0 R 2 ! L 2 ! x ?
;
<
;
< kQ 9 1 ! xL (R 2 0 x 2 ) 0 ( L 2 ! x ) R 2 0 x 2 : kQ 9 1 ! xL 2( R 2 0 x 2 ) 0 ( L 2 ! x) R 2 0 x 2 :
?#
ln >
ln >
?.
L > 1 0 xL ( R 2 0 x 2 ) 0 (! L 2 ! x ) R 2 0 x 2 ? L >1 0 xL 2( R 2 0 x 2 ) 0 (! L 2 ! x) R 2 0 x 2 ?
;
<
;
< VC
VC kQ
2L
#
L 2 x2 0 R2 :
9
:&
kQ 91 0 L 2 R 2 0 x 2 : kQ % 9
L
L
ln >
?#
' ln >1 0
? ! ln >1 !
?( .
'
2
2
2
2
2
2(
L >1 ! L 2 R 0 x ? L ) ; 2 R 0 x <
; 2 R 0 x <*
;
<
kQ
x 0 R2
2 . , which is the same as for a ring. 2
2
2
2
QV 2kQ ( L ! 2 x) 0 4 R ! ( L 0 2 x) 0 4 R
(c) Ex # !
#
Qx
( L ! 2 x)2 0 4 R 2 ( L 0 2 x)2 0 4 R 2 EVALUATE:
23.91. IDENTIFY: / For L 11 R the expression for Ex reduces to that for a ring of charge, as given in Example 23.14.
When the oil drop is at rest, the upward force q E from the electric field equals the downward weight of the drop. When the drop is falling at its terminal speed, the upward viscous force equals the downward weight of
the drop.
4
SET UP: The volume of the drop is related to its radius r by V # , r 3 .
3
4, r 3
4, G r 3 gd
EXECUTE: (a) Fg # mg #
G g . Fe # q E # q VAB d . Fe # Fg gives q #
.
3 VAB
3 Electric Potential (b) 2333 4, r 3
9Svt
G g # 6,S rvt gives r #
. Using this result to replace r in the expression in part (a) gives
3
2G g
3 4, G gd 9 9S vt :
d S 3vt3
.
q#
>
? # 18,
3 VAB > 2 G g ?
VAB 2 G g
;
<
(c) q # 18, 10!3 m (1.81 " 10!5 N + s m 2 )3 (1.00 " 10!3 m 39.3 s)3
# 4.80 " 10!19 C # 3e . The drop has acquired three
9.16 V
2(824 kg m 3 )(9.80 m s 2 ) excess electrons.
r# EVALUATE: 9(1.81 " 10!5 N + s m 2 )(1.00 " 10!3 m 39.3 s)
# 5.07 " 10!7 m # 0.507 $ m .
2(824 kg m 3 )(9.80 m s 2 ) % 4, r 3 &
!15
The weight of the drop is '
( G g # 4.4 " 10 N . The density of air at room temperature is
)3* 1.2 kg/m3 , so the buoyancy force is G airVg # 6.4 " 10!18 N and can be neglected.
23.92. IDENTIFY:
SET UP:
EXECUTE: vcm # m1v1 0 m2v2
m1 0 m2
kq1q2
.
r
(6 " 10!5 kg)(400m s) 0 (3 " 10!5 kg)(1300 m s)
#
# 700 m s
6.0 " 10!5 kg 0 3.0 " 10!5 kg E # K1 0 K 2 0 U , where U #
(a) vcm 1
1
kq q 1
2
(b) Erel # m1v12 0 m2v2 2 0 1 2 ! ( m1 0 m2 )vcm . After expanding the center of mass velocity and collecting like
2
2
2
r
1 m1m2
1
kq q
kq q
terms Erel #
[v12 0 v2 2 ! 2v1v2 ] 0 1 2 # $ (v1 ! v2 ) 2 0 1 2 .
2 m1 0 m2
2
r
r
1
k (2.0 " 10!6 C)( ! 5.0 " 10!6 C)
(c) Erel # (2.0 " 10!5 kg)(900 m s)2 0
# !1.9 J
2
0.0090 m
(d) Since the energy is less than zero, the system is “bound.”
kq q
(e) The maximum separation is when the velocity is zero: !1.9 J # 1 2 gives
r
k (2.0 " 10!6 C)(!5.0 " 10!6 C)
r#
# 0.047 m .
!1.9 J
(f) Now using v1 # 400 m s and v2 # 1800 m s , we find Erel # 09.6 J . The particles do escape, and the final relative velocity is v1 ! v2 # 2 Erel $ # 2(9.6 J)
# 980 m s .
2.0 " 10!5 kg EVALUATE: For an isolated system the velocity of the center of mass is constant and the system must retain the
kinetic energy associated with the motion of the center of mass. 24 CAPACITANCE AND DIELECTRICS 24.1. 24.2. 24.3. Q
Vab
SET UP: 1 $ F # 10 !6 F
EXECUTE: Q # CVab # (7.28 " 10 !6 F)(25.0 V) # 1.82 " 10 !4 C # 182 $ C
EVALUATE: One plate has charge 0 Q and the other has charge !Q .
PA
Q
and V # Ed .
IDENTIFY and SET UP: C # 0 , C #
d
V
A
0.00122 m 2
(a) C # P0 # P0
# 3.29 pF
d
0.00328 m
!8
Q 4.35 " 10 C
(b) V # #
# 13.2 kV
C 3.29 " 10 !12 F
V 13.2 " 103 V
(c) E # #
# 4.02 " 106 V/m
d
0.00328 m
EVALUATE:! The electric field is uniform between the plates, at points that aren't close to the edges.
IDENTIFY and SET UP:! It is a parallelplate air capacitor, so we can apply the equations of Sections 24.1.
Q
Q 0.148 " 10 !6 C
EXECUTE:! (a) C #
so Vab # #
# 604 V
Vab
C
245 " 10 !12 F IDENTIFY:! C # !12
!3
P0 A
Cd . 245 " 10 F / . 0.328 " 10 m /
so A #
#
# 9.08 " 10!3 m 2 # 90.8 cm 2
d
P0
8.854 " 10 !12 C 2 / N + m 2
V
604 V
(c) Vab # Ed so E # ab #
# 1.84 " 106 V/m
d
0.328 " 10!3 m (b) C # (d) E # = P0 so = # EP0 # .1.84 " 106 V/m / . 8.854 " 10!12 C2 / N + m 2 / # 1.63 " 10!5 C/m 2 EVALUATE:! We could also calculate = directly as Q/A. = #
24.4. Q 0.148 " 10 !6 C
#
# 1.63 " 10 !5 C/m 2 , which checks.
A 9.08 " 10!3 m 2 A
when there is air between the plates.
d
A # (3.0 " 10!2 m) 2 is the area of each plate. IDENTIFY:! C # P0
SET UP:! (8.854 " 10 !12 F/m)(3.0 " 10 !2 m) 2
# 1.59 " 10 !12 F # 1.59 pF
5.0 " 10!3 m
EVALUATE:! C increases when A increases and C increases when d decreases.
PA
Q
IDENTIFY:! C #
. C# 0 .
Vab
d
SET UP:! When the capacitor is connected to the battery, Vab # 12.0 V . EXECUTE:! C # 24.5. EXECUTE:! (a) Q # CVab # (10.0 " 10!6 F)(12.0 V) # 1.20 " 10!4 C # 120 $ C
(b) When d is doubled C is halved, so Q is halved. Q # 60 $ C .
(c) If r is doubled, A increases by a factor of 4. C increases by a factor of 4 and Q increases by a factor of 4.
Q # 480 $ C.
EVALUATE:! When the plates are moved apart, less charge on the plates is required to produce the same potential
difference. With the separation of the plates constant, the electric field must remain constant to produce the same
potential difference. The electric field depends on the surface charge density, = . To produce the same = , more
charge is required when the area increases.
241 242 24.6. 24.7. Chapter 24 IDENTIFY:! C # Q
PA
. C# 0 .
Vab
d SET UP:! When the capacitor is connected to the battery, enough charge flows onto the plates to make Vab # 12.0 V.
EXECUTE:! (a) 12.0 V
Q
(b) (i) When d is doubled, C is halved. Vab #
and Q is constant, so V doubles. V # 24.0 V .
C
(ii) When r is doubled, A increases by a factor of 4. V decreases by a factor of 4 and V # 3.0 V .
EVALUATE:! The electric field between the plates is E # Q / P0 A . Vab # Ed . When d is doubled E is unchanged and
V doubles. When A is increased by a factor of 4, E decreases by a factor of 4 so V decreases by a factor of 4.
PA
IDENTIFY:! C # 0 . Solve for d.
d
SET UP:! Estimate r # 1.0 cm . A # , r 2 . P , r 2 P0, (0.010 m) 2
P0 A
#
# 2.8 mm .
so d # 0
C
d
1.00 " 10!12 F
EVALUATE:! The separation between the pennies is nearly a factor of 10 smaller than the diameter of a penny, so it
is a reasonable approximation to treat them as infinite sheets.
Q
PA
INCREASE:! C #
. Vab # Ed . C # 0 .
Vab
d
EXECUTE:! C # 24.8. SET UP:! We want E # 1.00 " 104 N/C when V # 100 V .
V
1.00 " 102 V
EXECUTE:! (a) d # ab #
# 1.00 " 10!2 m # 1.00 cm .
E 1.00 " 104 N/C
A# Cd (5.00 " 10!12 F)(1.00 " 10!2 m)
A
# 4.24 " 10!2 m # 4.24 cm .
#
# 5.65 " 10!3 m 2 . A # , r 2 so r #
!12
2
2
P0
8.854 " 10 C /(N + m )
, (b) Q # CVab # (5.00 " 10!12 F)(1.00 " 102 V) # 5.00 " 10!10 C # 500 pC
P0 A
. We could have a larger d, along with a larger A, and still achieve the required C without
d
exceeding the maximum allowed E.
IDENTIFY:! Apply the results of Example 24.4. C # Q / V . EVALUATE:! C # 24.9. SET UP:! ra # 0.50 mm , rb # 5.00 mm
EXECUTE:! (a) C # L 2, P0
(0.180 m)2, P0
#
# 4.35 " 10!12 F .
ln(rb ra ) ln(5.00 0.50) (b) V # Q / C # (10.0 " 10!12 C) /(4.35 " 10!12 F) # 2.30 V C
# 24.2 pF . This value is similar to those in Example 24.4. The capacitance is determined entirely by
L
the dimensions of the cylinders.
IDENTIFY:! Capacitance depends on the geometry of the object.
2, P0 L
(a) SET UP:! The capacitance of a cylindrical capacitor is C #
. Solving for rb gives rb # ra e 2, P0 L / C .
ln . rb / ra /
EVALUATE:! 24.10. EXECUTE:! Substituting in the numbers for the exponent gives 2, . 8.85 " 10!12 C2 /N + m 2 / (0.120 m) # 0.182
3.67 " 10!11 F
Now use this value to calculate rb: rb = ra e0.182 = (0.250 cm)e0.182 = 0.300 cm
(b) SET UP:! For any capacitor, C = Q/V and 8 = Q/L. Combining these equations and substituting the numbers
gives 8 = Q/L = CV/L.
EXECUTE:! Numerically we get
!11
CV . 3.67 " 10 F / .125 V /
#
# 3.82 " 10!8 C/m = 38.2 nC/m
0.120 m
L
EVALUATE:! The distance between the surfaces of the two cylinders would be only 0.050 cm, which is just
0.50 mm. These cylinders would have to be carefully constructed. 8# Capacitance and Dielectrics 24.11. 243 IDENTIFY and SET UP:! Use the expression for C/L derived in Example 24.4. Then use Eq.(24.1) to calculate Q.
C
2, P0
EXECUTE:! (a) From Example 24.4,
#
L ln . rb / ra /
!12
2
2
C 2, . 8.854 " 10 C / N + m /
#
# 6.57 " 10!11 F/m # 66 pF/m
ln . 3.5 mm/1.5 mm /
L (b) C # . 6.57 " 10!11 F/m / . 2.8 m / # 1.84 " 10!10 F. Q # CV # .1.84 " 10!10 F / . 350 " 10!3 V / # 6.4 " 10!11 C # 64 pC 24.12. The conductor at higher potential has the positive charge, so there is +64 pC on the inner conductor and !64 pC on
the outer conductor.
EVALUATE:! C depends only on the dimensions of the capacitor. Q and V are proportional.
IDENTIFY:! Apply the results of Example 24.3. C # Q / V .
SET UP:! ra # 15.0 cm . Solve for rb . 1% r r &
EXECUTE:! (a) For two concentric spherical shells, the capacitance is C # ' a b ( . kCrb ! kCra # ra rb and
k ) rb ! ra *
!12
kCra
k (116 " 10 F)(0.150 m)
rb #
#
# 0.175 m .
kC ! ra k (116 " 10!12 F) ! 0.150 m
(b) V # 220 V and Q # CV # (116 " 10!12 F)(220 V) # 2.55 " 10!8 C .
EVALUATE:! A parallelplate capacitor with A # 4, ra rb # 0.33 m 2 and d # rb ! ra # 2.5 " 10!2 m has P0 A
# 117 pF , in excellent agreement with the value of C for the spherical capacitor.
d
IDENTIFY:! We can use the definition of capacitance to find the capacitance of the capacitor, and then relate the
capacitance to geometry to find the inner radius.
(a) SET UP:! By the definition of capacitance, C = Q/V. C# 24.13. EXECUTE:! C # Q 3.30 " 10!9 C
#
# 1.50 " 10!11 F = 15.0 pF
V 2.20 " 102 V (b) SET UP:! The capacitance of a spherical capacitor is C # 4, P0 ra rb
.
rb ! ra EXECUTE:! Solve for ra and evaluate using C = 15.0 pF and rb = 4.00 cm, giving ra = 3.09 cm.
(c) SET UP:! We can treat the inner sphere as a pointcharge located at its center and use Coulomb’s law,
1q
E#
.
4, P0 r 2 . 9.00 "10 9 EXECUTE:! E # 24.14. N + m 2 /C2 / . 3.30 " 10!9 C / . 0.0309 m / 2 # 3.12 " 104 N/C EVALUATE:! Outside the capacitor, the electric field is zero because the charges on the spheres are equal in
magnitude but opposite in sign.
IDENTIFY:! The capacitors between b and c are in parallel. This combination is in series with the 15 pF capacitor.
SET UP:! Let C1 # 15 pF , C2 # 9.0 pF and C3 # 11 pF .
EXECUTE:! (a) For capacitors in parallel, Ceq # C1 0 C2 0 % so C23 # C2 0 C3 # 20 pF
(b) C1 # 15 pF is in series with C23 # 20 pF . For capacitors in series, C123 # 1
1
1
1
1
1
#0
and
#0
0 % so
Ceq C1 C2
C123 C1 C23 C1C23
(15 pF)(20 pF)
#
# 8.6 pF .
C1 0 C23 15 pF 0 20 pF EVALUATE:! For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors. For
capacitors in series the equivalent capacitance is smaller than any of the individual capacitors. 244 Chapter 24 24.15. IDENTIFY:! Replace series and parallel combinations of capacitors by their equivalents. In each equivalent network
apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work back to the
original circuit.
SET UP:! Do parts (a) and (b) together. The capacitor network is drawn in Figure 24.15a. C1 # C2 # C3 # C4 # 400 $ F
Vab # 28.0 V Figure 24.15a
EXECUTE:! Simplify the circuit by replacing the capacitor combinations by their equivalents: C1 and C2 are in
series and are equivalent to C12 (Figure 24.15b). 1
1
1
#0
C12 C1 C2
Figure 24.15b . 4.00 " 10 F/. 4.00 " 10!6 F/ # 2.00 " 10!6 F
C1C2
#
C1 0 C2
4.00 " 10!6 F 0 4.00 " 10!6 F
C12 and C3 are in parallel and are equivalent to C123 (Figure 24.15c).
!6 C12 # C123 # C12 0 C3
C123 # 2.00 " 10!6 F 0 4.00 " 10!6 F C123 # 6.00 " 10!6 F
Figure 24.15c C123 and C4 are in series and are equivalent to C1234 (Figure 24.15d).
1
1
1
#
0
C1234 C123 C4
Figure 24.15d . 6.00 "10!6 F /. 4.00 " 10!6 F/ # 2.40 " 10!6 F
C123C4
#
C123 0 C4
6.00 " 10!6 F 0 4.00 " 10!6 F
The circuit is equivalent to the circuit shown in Figure 24.15e.
C1234 # V1234 # V # 28.0 V Q1234 # C1234V # . 2.40 " 10!6 F / . 28.0 V / # 67.2 $ C Figure 24.15e Now build back up the original circuit, step by step. C1234 represents C123 and C4 in series (Figure 24.15f).
Q123 # Q4 ! Q1234 # 67.2 $ C
(charge same for capacitors in series)
Figure 24.15f Q123 67.2 $ C
#
# 11.2 V
C123 6.00 $ F
Q 67.2 $ C
# 16.8 V
V4 # 4 #
C4 4.00 $ F
Note that V4 0 V123 # 16.8 V 0 11.2 V # 28.0 V, as it should. Then V123 # Capacitance and Dielectrics 245 Next consider the circuit as written in Figure 24.15g.
V3 # V12 # 28.0 V ! V4
V3 # 11.2 V Q3 # C3V3 # . 4.00 $ F / .11.2 V /
Q3 # 44.8 $ C Q12 ! C12V12 # . 2.00 $ F /.11.2 V /
Q12 # 22.4 $ C
Figure 24.15g Finally, consider the original circuit, as shown in Figure 24.15h.
Q1 # Q2 # Q12 # 22.4 $ C
(charge same for capacitors in series)
Q 22.4 $ C
V1 # 1 #
# 5.6 V
C1 4.00 $ F
V2 # Q2 22.4 $ C
#
# 5.6 V
C2 4.00 $ F Figure 24.15h Note that V1 0 V2 # 11.2 V, which equals V3 as it should.
Summary: Q1 # 22.4 $ C, V1 # 5.6 V
Q2 # 22.4 $ C, V2 # 5.6 V
Q3 # 44.8 $ C, V3 # 11.2 V
Q4 # 67.2 $ C, V4 # 16.8 V
(c) Vad # V3 # 11.2 V
EVALUATE:! V1 0 V2 0 V4 # V , or V3 0 V4 # V . Q1 # Q2 , Q1 0 Q3 # Q4 and Q4 # Q1234 .
24.16. IDENTIFY:! The two capacitors are in series. The equivalent capacitance is given by 1
1
1
#0
.
Ceq C1 C2 SET UP:! For capacitors in series the charges are the same and the potentials add to give the potential across the
network.
1
1
1
1
1
EXECUTE:! (a)
#0
#
0
# 5.33 " 105 F!1 . Ceq # 1.88 " 10!6 F . Then
!6
Ceq C1 C2 (3.0 " 10 F) (5.0 " 10!6 F) Q # VCeq # (52.0 V)(1.88 " 10!6 F) # 9.75 " 10!5 C . Each capacitor has charge 9.75 " 10!5 C .
(b) V1 # Q / C1 # 9.75 " 10!5 C 3.0 " 10!6 F # 32.5 V . V2 # Q / C2 # 9.75 " 10!5 C 5.0 " 10!6 F # 19.5 V . 24.17. EVALUATE:! V1 0 V2 # 52.0 V , which is equal to the applied potential Vab . The capacitor with the smaller C has the
larger V.
IDENTIFY:! The two capacitors are in parallel so the voltage is the same on each, and equal to the applied voltage Vab .
SET UP:! Do parts (a) and (b) together. The network is sketched in Figure 24.17.
EXECUTE:! V1 # V2 # V V1 # 52.0 V
V2 # 52.0 V
Figure 24.17 C # Q / V so Q # CV
Q1 # C1V1 # . 3.00 $ F /. 52.0 V / # 156 $ C. Q2 # C2V2 # . 5.00 $ F / . 52.0 V / # 260 $ C.
EVALUATE:! To produce the same potential difference, the capacitor with the larger C has the larger Q. 246 Chapter 24 24.18. IDENTIFY:! For capacitors in parallel the voltages are the same and the charges add. For capacitors in series, the
charges are the same and the voltages add. C # Q / V .
SET UP:! C1 and C2 are in parallel and C3 is in series with the parallel combination of C1 and C2 .
EXECUTE:! (a) C1 and C2 are in parallel and so have the same potential across them: Q2 40.0 " 10!6 C
#
# 13.33 V . Therefore, Q1 # V1C1 # (13.33 V)(3.00 " 10!6 F) # 80.0 " 10!6 C . Since C3 is
C2 3.00 " 10!6 F
in series with the parallel combination of C1 and C2 , its charge must be equal to their combined charge:
V1 # V2 # 24.19. 24.20. C3 # 40.0 " 10!6 C 0 80.0 " 10!6 C # 120.0 " 10!6 C .
1
1
1
1
1
(b) The total capacitance is found from
and Ctot # 3.21 $ F .
#
0
#
0
!6
Ctot C12 C3 9.00 " 10 F 5.00 " 10!6 F
Q
120.0 " 10!6 C
Vab # tot #
# 37.4 V .
Ctot
3.21 " 10!6 F
Q 120.0 " 10!6 C
EVALUATE:! V3 # 3 #
# 24.0 V . Vab # V1 0 V3 .
C3 5.00 " 10!6 F
IDENTIFY and SET UP:! Use the rules for V for capacitors in series and parallel: for capacitors in parallel the voltages
are the same and for capacitors in series the voltages add.
EXECUTE:! V1 # Q1 / C1 # .150 $ C / / . 3.00 $ F / # 50 V
C1 and C2 are in parallel, so V2 # 50 V
V3 # 120 V ! V1 # 70 V
EVALUATE:! Now that we know the voltages, we could also calculate Q for the other two capacitors.
1
1
1
PA
IDENTIFY and SET UP:! C # 0 . For two capacitors in series,
#0
.
Ceq C1 C2
d
!1 24.21. 24.22. !1 %d
PA
d&
%
&
EXECUTE:! Ceq # ' 1 0 1 ( # ' 1 0 2 ( # 0
. This shows that the combined capacitance for two
C1 C2 *
d1 0 d 2
P0 A P0 A *
)
)
capacitors in series is the same as that for a capacitor of area A and separation (d1 0 d 2 ) .
EVALUATE:! Ceq is smaller than either C1 or C2 .
PA
IDENTIFY and SET UP:! C # 0 . For two capacitors in parallel, Ceq # C1 0 C2 .
d
P0 A1 P0 A2 P0 ( A1 0 A2 )
EXECUTE:! Ceq # C1 0 C2 #
. So the combined capacitance for two capacitors in parallel is
0
#
d
d
d
that of a single capacitor of their combined area ( A1 0 A2 ) and common plate separation d.
EVALUATE:! Ceq is larger than either C1 or C2 . IDENTIFY:! Simplify the network by replacing series and parallel combinations of capacitors by their equivalents.
1
1
1
SET UP:! For capacitors in series the voltages add and the charges are the same;
#0
0 % For capacitors
Ceq C1 C2
Q
in parallel the voltages are the same and the charges add; Ceq # C1 0 C2 0 % C # .
V
EXECUTE:! (a) The equivalent capacitance of the 5.0 $ F and 8.0 $ F capacitors in parallel is 13.0 $ F. When these
two capacitors are replaced by their equivalent we get the network sketched in Figure 24.22. The equivalent
capacitance of these three capacitors in series is 3.47 $ F.
(b) Qtot # CtotV # (3.47 $ F)(50.0 V) # 174 $ C
(c) Qtot is the same as Q for each of the capacitors in the series combination shown in Figure 24.22, so Q for each of
the capacitors is 174 $ C.
Q
Q
EVALUATE:! The voltages across each capacitor in Figure 24.22 are V10 # tot # 17.4 V , V13 # tot # 13.4 V and
C10
C13
Qtot
V9 #
# 19.3 V . V10 0 V13 0 V9 # 17.4 V 0 13.4 V 0 19.3 V # 50.1 V . The sum of the voltages equals the applied
C9
voltage, apart from a small difference due to rounding. Figure 24.22 Capacitance and Dielectrics 24.23. 247 IDENTIFY:! Refer to Figure 24.10b in the textbook. For capacitors in parallel, Ceq # C1 0 C2 0 %. For capacitors in series, 1
1
1
#0
0 %.
Ceq C1 C2 SET UP:! The 11 $ F , 4 $ F and replacement capacitor are in parallel and this combination is in series with the
9.0 $ F capacitor.
EXECUTE:! 24.24. %
1
1
1
1&
#
#'
0
( . (15 0 x ) $ F # 72 $ F and x # 57 $ F .
Ceq 8.0 $ F ) (11 0 4.0 0 x ) $ F 9.0 $ F * EVALUATE:! Increasing the capacitance of the one capacitor by a large amount makes a small increase in the
equivalent capacitance of the network.
PA
IDENTIFY:! Apply C # Q / V . C # 0 . The work done to double the separation equals the change in the stored
d
energy.
1
Q2
SET UP:! U # CV 2 #
.
2
2C
EXECUTE:! (a) V # Q / C # (2.55 $ C) (920 " 10!12 F) # 2770 V
P0 A
says that since the charge is kept constant while the separation doubles, that means that the capacitance
d
halves and the voltage doubles to 5540 V.
Q 2 (2.55 " 10!6 C) 2
(c) U #
#
# 3.53 " 10!3 J . When if the separation is doubled while Q stays the same, the
2C 2(920 " 10!12 F)
capacitance halves, and the energy stored doubles. So the amount of work done to move the plates equals the
difference in energy stored in the capacitor, which is 3.53 " 10!3 J.
EVALUATE:! The oppositely charged plates attract each other and positive work must be done by an external force to
pull them farther apart.
IDENTIFY and SET UP:! The energy density is given by Eq.(24.11): u # 1 P0 E 2 . Use V # Ed to solve for E.
2 (b) C # 24.25. EXECUTE:! Calculate E : E # Then u # 1 P0 E 2 #
2 24.26. 1
2 .8.854 " 10 V
400 V
#
# 8.00 " 104 V/m.
d 5.00 " 10!3 m !12 C2 / N + m 2 /. 8.00 " 104 V/m / # 0.0283 J/m3
2 EVALUATE:! E is smaller than the value in Example 24.8 by about a factor of 6 so u is smaller by about a factor of
62 # 36.
Q
PA
IDENTIFY:! C #
. C # 0 . Vab # Ed . The stored energy is 1 QV .
2
d
Vab
SET UP:! d # 1.50 " 10!3 m . 1 $ C # 10!6 C 0.0180 " 10!6 C
# 9.00 " 10!11 F # 90.0 pF
200 V
PA
Cd (9.00 " 10!11 F)(1.50 " 10!3 m)
(b) C # 0 so A #
#
# 0.0152 m 2 .
d
8.854 " 10!12 C2 /(N + m 2 )
P0
EXECUTE:! (a) C # (c) V # Ed # (3.0 " 106 V/m)(1.50 " 10!3 m) # 4.5 " 103 V
(d) Energy # 1 QV # 1 (0.0180 " 10!6 C)(200 V) # 1.80 " 10!6 J # 1.80 $ J
2
2
EVALUATE:! We could also calculate the stored energy as
24.27. Q 2 (0.0180 " 10!6 C) 2
#
# 1.80 $ J .
2C
2(9.00 " 10!11 F) IDENTIFY:! The energy stored in a charged capacitor is 1 CV 2 .
2
SET UP:! 1 $ F # 10!6 F
EXECUTE:! 1
2 CV 2 # 1 (450 " 10!6 F)(295 V)2 # 19.6 J
2 EVALUATE:! Thermal energy is generated in the wire at the rate I 2 R , where I is the current in the wire. When the
capacitor discharges there is a flow of charge that corresponds to current in the wire. 248 Chapter 24 24.28. IDENTIFY:! After the two capacitors are connected they must have equal potential difference, and their combined
charge must add up to the original charge.
Q2 1
SET UP:! C # Q / V . The stored energy is U #
# CV 2
2C 2
EXECUTE:! (a) Q # CV0 .
Q
Q2
QQ
C
Q
3
(b) V # 1 # 2 and also Q1 0 Q2 # Q # CV0 . C1 # C and C2 #
so 1 #
and Q2 # 1 . Q # Q1 .
C (C 2)
C1 C2
2
2
2
2
Q1 2 Q 2
#
# V0 .
Q1 # Q and V #
3
C 3C 3
1 % Q 2 Q 2 & 1 9 ( 2 Q) 2 2( 1 Q) 2 : 1 Q 2 1
03
# CV0 2
(c) U # ' 1 0 2 ( # > 3
?#
C < 3C 3
2 ) C1 C2 * 2 ; C
1
(d) The original U was U # 1 CV0 2 , so BU # ! CV0 2 .
2
6
(e) Thermal energy of capacitor, wires, etc., and electromagnetic radiation.
EVALUATE:! The original charge of the charged capacitor must distribute between the two capacitors to make the
potential the same across each capacitor. The voltage V for each after they are connected is less than the original
voltage V0 of the charged capacitor.
IDENTIFY and SET UP:! Combine Eqs. (24.9) and (24.2) to write the stored energy in terms of the separation
between the plates.
Q2
PA
xQ 2
EXECUTE:! (a) U #
; C # 0 so U #
2C
x
2P0 A 24.29. (b) x E x 0 dx gives U # . x 0 dx / Q 2
2P0 A . x 0 dx / Q 2 ! xQ 2 # % Q 2 & dx
dU #
'
(
2P0 A
2P0 A ) 2P0 A *
Q2
2P0 A
(d) EVALUATE:! The capacitor plates and the field between the plates are shown in Figure 24.29a.
=
Q
E# #
P0 P0 A
F # 1 QE , not QE
2
Figure 24.29a
(c) dW # F dx # dU , so F # The reason for the difference is that E is the field due to both plates. If we consider the positive plate only and
calculate its electric field using Gauss’s law (Figure 24.29b):
!!Q
úE + dA # Pe0ncl
=A
2 EA #
P0
=
Q
#
E#
2P0 2P0 A
Figure 24.29b
The force this field exerts on the other plate, that has charge !Q, is F #
24.30. IDENTIFY:! C # P0 A . The stored energy can be expressed either as Q2
.
2P0 A Q2
CV 2
or as
, whichever is more convenient
2C
2 d
for the calculation.
SET UP:! Since d is halved, C doubles.
EXECUTE:! (a) If the separation distance is halved while the charge is kept fixed, then the capacitance increases and
the stored energy, which was 8.38 J, decreases since U # Q 2 2C. Therefore the new energy is 4.19 J.
(b) If the voltage is kept fixed while the separation is decreased by one half, then the doubling of the capacitance
leads to a doubling of the stored energy to 16.8 J, using U # CV 2 2 , when V is held constant throughout. Capacitance and Dielectrics 24.31. 249 EVALUATE:! When the capacitor is disconnected, the stored energy decreases because of the positive work done by
the attractive force between the plates. When the capacitor remains connected to the battery, Q # CV tells us that the
charge on the plates increases. The increased stored energy comes from the battery when it puts more charge onto the
plates.
Q
IDENTIFY and SET UP:! C # . U # 1 CV 2 .
2
V
EXECUTE:! (a) Q # CV # (5.0 $ F)(1.5 V) # 7.5 $ C . U # 1 CV 2 # 1 (5.0 $ F)(1.5 V)2 # 5.62 $ J
2
2
(b) U # 1 CV 2 # 1 C (Q / C ) 2 # Q 2 / 2C . Q # 2CU # 2(5.0 " 10!6 F)(1.0 J) # 3.2 " 10!3 C .
2
2 Q 3.2 " 10!3 C
#
# 640 V .
C 5.0 " 10!6 F
EVALUATE:! The stored energy is proportional to Q 2 and to V 2 .
V# 24.32. IDENTIFY:! The two capacitors are in series. 1
1
1
Q
#0
0 % C # . U # 1 CV 2 .
2
V
Ceq C1 C2 SET UP:! For capacitors in series the voltages add and the charges are the same.
CC
(150 nF)(120 nF)
1
1
1
# 66.7 nF .
EXECUTE:! (a)
so Ceq # 1 2 #
#0
Ceq C1 C2
C1 0 C2 150 nF 0 120 nF Q # CV # (66.7 nF)(36 V) # 2.4 " 10!6 C # 2.4 $ C
(b) Q # 2.4 $ C for each capacitor.
(c) U # 1 CeqV 2 # 1 (66.7 " 10!9 F)(36 V)2 # 43.2 $ J
2
2
(d) We know C and Q for each capacitor so rewrite U in terms of these quantities. U # 1 CV 2 # 1 C (Q / C ) 2 # Q 2 / 2C
2
2 (2.4 " 10!6 C) 2
(2.4 " 10!6 C) 2
# 19.2 $ J ; 120 nF: U #
# 24.0 $ J
!9
2(150 " 10 F)
2(120 " 10!9 F)
Note that 19.2 $ J 0 24.0 $ J # 43.2 $ J , the total stored energy calculated in part (c).
150 nF: U # Q 2.4 " 10!6 C
Q 2.4 " 10!6 C
#
# 16 V ; 120 nF: V # #
# 20 V
!9
C 150 " 10 F
C 120 " 10!9 F
Note that these two voltages sum to 36 V, the voltage applied across the network.
EVALUATE:! Since Q is the same the capacitor with smaller C stores more energy ( U # Q 2 / 2C ) and has a larger
voltage ( V # Q / C ).
(e) 150 nF: V # 24.33. Q
. U # 1 CV 2 .
2
V
SET UP:! For capacitors in parallel, the voltages are the same and the charges add.
EXECUTE:! (a) Ceq # C1 0 C2 # 35 nF 0 75 nF # 110 nF . Qtot # CeqV # (110 " 10!9 F)(220 V) # 24.2 $ C IDENTIFY:! The two capacitors are in parallel. Ceq # C1 0 C2 . C # (b) V # 220 V for each capacitor.
35 nF: Q35 # C35V # (35 " 10!9 F)(220 V) # 7.7 $ C ; 75 nF: Q75 # C75V # (75 " 10!9 F)(220 V) # 16.5 $ C . Note that Q35 0 Q75 # Qtot .
(c) U tot # 1 CeqV 2 # 1 (110 " 10!9 F)(220 V) 2 # 2.66 mJ
2
2
(d) 35 nF: U 35 # 1 C35V 2 # 1 (35 " 10!9 F)(220 V)2 # 0.85 mJ ;
2
2 24.34. 75 nF: U 75 # 1 C75V 2 # 1 (75 " 10!9 F)(220 V) 2 # 1.81 mJ . Since V is the same the capacitor with larger C stores more
2
2
energy.
(e) 220 V for each capacitor.
EVALUATE:! The capacitor with the larger C has the larger Q.
IDENTIFY:! Capacitance depends on the geometry of the object.
(a) SET UP:! The potential difference between the core and tube is V # density gives 8 # 8
ln . rb / ra / . Solving for the linear charge
2, P0 2, P0V
4, P0V
.
#
ln . rb / ra / 2 ln . rb / ra / EXECUTE:! Using the given values gives 8 # 6.00 V
% 2.00 &
2 . 9.00 " 10 N + m /C / ln '
(
) 1.20 *
9 2 2 # 6.53 " 10!10 C/m 2410 Chapter 24 (b) SET UP:! Q # 8 L
EXECUTE:! Q # (6.53 " 10!10 C/m)(0.350 m) = 2.29 " 10!10 C
(c) SET UP:! The definition of capacitance is C # Q / V . 2.29 " 1010 C
# 3.81 " 10!11 F
6.00 V
(d) SET UP:! The energy stored in a capacitor is U # 1 CV 2 .
2
EXECUTE:! C # 24.35. EXECUTE:! U # 1 (3.81" 10!11 F)(6.00 V)2 # 6.85 " 10!10 J
2
EVALUATE:! The stored energy could be converted to heat or other forms of energy.
IDENTIFY:! U # 1 QV . Solve for Q. C # Q / V .
2
SET UP:! Example 24.4 shows that for a cylindrical capacitor,
EXECUTE:! (a) U # 1 QV gives Q #
2 C
2, P0
#
.
L ln( rb /ra ) 2U 2(3.20 " 10!9 J)
#
# 1.60 " 10!9 C.
V
4.00 V C
2, P0
r
#
. b # exp(2, P0 L/C ) # exp(2, P0 LV/Q) # exp(2, P0 (15.0 m)(4.00 V) (1.60 " 10!9 C)) # 8.05.
L ln( rb ra ) ra
The radius of the outer conductor is 8.05 times the radius of the inner conductor.
EVALUATE:! When the ratio rb / ra increases, C / L decreases and less charge is stored for a given potential
difference.
IDENTIFY:! Apply Eq.(24.11).
% rr &
Q
Q
# ' a b (Vab .
SET UP:! Example 24.3 shows that E #
between the conducting shells and that
2
4, P0 r
4, P0 ) rb ! ra *
(b) 24.36. % r r &V
% [0.125 m][0.148 m] & 120 V 96.5 V + m
EXECUTE:! E # ' a b ( ab # '
(2#
2
r2
) 0.148 m ! 0.125 m * r
) rb ! ra * r
(a) For r # 0.126 m , E # 6.08 " 103 V/m . u # 1 P0 E 2 # 1.64 " 10!4 J/m3 .
2 24.37. (b) For r # 0.147 m , E # 4.47 " 103 V/m . u # 1 P0 E 2 # 8.85 " 10!5 J/m3 .
2
EVALUATE:! (c) No, the results of parts (a) and (b) show that the energy density is not uniform in the region between
the plates. E decreases as r increases, so u decreases also.
IDENTIFY:! Use the rules for series and for parallel capacitors to express the voltage for each capacitor in terms of
the applied voltage. Express U, Q, and E in terms of the capacitor voltage.
SET UP:! Le the applied voltage be V. Let each capacitor have capacitance C. U # 1 CV 2 for a single capacitor with
2
voltage V.
EXECUTE:! (a) series
Voltage across each capacitor is V /2. The total energy stored is Us # 2 . 1 C[V/2]2 / # 1 CV 2
2
4
parallel
Voltage across each capacitor is V. The total energy stored is Up # 2 . 1 CV 2 / # CV 2
2 Up # 4Us
(b) Q # CV for a single capacitor with voltage V. Qs # 2 . C[V/2]/ # CV ; Qp # 2(CV ) # 2CV ; Q p # 2Qs
(c) E # V/d for a capacitor with voltage V. Es # V/2d ; Ep # V/d ; Ep # 2 Es 24.38. EVALUATE:! The parallel combination stores more energy and more charge since the voltage for each capacitor is
larger for parallel. More energy stored and larger voltage for parallel means larger electric field in the parallel case.
IDENTIFY:! V # Ed and C # Q / V . With the dielectric present, C # KC0 .
SET UP:! V # Ed holds both with and without the dielectric.
EXECUTE:! (a) V # Ed # (3.00 " 104 V/m)(1.50 " 10!3 m) # 45.0 V . Q # C0V # (5.00 " 10!12 F)(45.0 V) # 2.25 " 10!10 C .
(b) With the dielectric, C # KC0 # (2.70)(5.00 pF) # 13.5 pF . V is still 45.0 V, so Q # CV # (13.5 " 10!12 F)(45.0 V) # 6.08 " 10!10 C .
EVALUATE:! The presence of the dielectric increases the amount of charge that can be stored for a given potential
difference and electric field between the plates. Q increases by a factor of K. Capacitance and Dielectrics 24.39. 2411 IDENTIFY and SET UP:! Q is constant so we can apply Eq.(24.14). The charge density on each surface of the
dielectric is given by Eq.(24.16).
E
E
3.20 " 105 V/m
EXECUTE:! E # 0 so K # 0 #
# 1.28
K
E 2.50 " 105 V/m
(a) = i # = (1 ! 1/ K ) = # P0 E0 # (8.854 " 10!12 C 2 /N + m 2 )(3.20 " 105 N/C) # 2.833 " 10!6 C/m 2 24.40. = i # (2.833 " 10!6 C/m 2 )(1 ! 1/1.28) # 6.20 " 10!7 C/m 2
(b) As calculated above, K # 1.28.
EVALUATE:! The surface charges on the dielectric produce an electric field that partially cancels the electric field
produced by the charges on the capacitor plates.
IDENTIFY:! Capacitance depends on geometry, and the introduction of a dielectric increases the capacitance.
SET UP:! For a parallelplate capacitor, C # K P0 A/d .
EXECUTE:! (a) Solving for d gives
d# K P0 A (3.0)(8.85 " 10 !12 C 2 /N + m 2 )(0.22 m)(0.28 m)
#
# 1.64 " 10!3 m = 1.64 mm .
C
1.0 " 10!9 F 1.64 mm
C 8 sheets .
0.20 mm/sheet
Cd
(1.0 " 10!9 F)(0.012 m)
(b) Solving for the area of the plates gives A #
#
# 0.45 m 2 .
K P0 (3.0)(8.85 " 10!12 C2 /N + m 2 )
(c) Teflon has a smaller dielectric constant (2.1) than the posterboard, so she will need more area to achieve the same
capacitance.
EVALUATE:! The use of dielectric makes it possible to construct reasonablesized capacitors since the dielectric
increases the capacitance by a factor of K.
IDENTIFY and SET UP:! For a parallelplate capacitor with a dielectric we can use the equation C # K P0 A / d . Dividing this result by the thickness of a sheet of paper gives 24.41. 24.42. 24.43. Minimum A means smallest possible d. d is limited by the requirement that E be less than 1.60 " 107 V/m when V is
as large as 5500 V.
5500 V
V
EXECUTE:! V # Ed so d # #
# 3.44 " 10!4 m
E 1.60 " 107 V/m
Cd
(1.25 " 10!9 F)(3.44 " 10!4 m)
Then A #
#
# 0.0135 m 2 .
K P0 (3.60)(8.854 " 10!12 C2 / N + m2 )
EVALUATE:! The relation V = Ed applies with or without a dielectric present. A would have to be larger if there were
no dielectric.
IDENTIFY and SET UP:! Adapt the derivation of Eq.(24.1) to the situation where a dielectric is present.
EXECUTE:! Placing a dielectric between the plates just results in the replacement of P for P0 in the derivation of
Equation (24.20). One can follow exactly the procedure as shown for Equation (24.11).
EVALUATE:! The presence of the dielectric increases the energy density for a given electric field.
IDENTIFY:! The permittivity P of a material is related to its dielectric constant by P # K P0 . The maximum voltage is
E0
=
#
, where
K K P0
= is the surface charge density on each plate. The induced surface charge density on the surface of the dielectric is
given by = i # = (1 ! 1/ K ) .
SET UP:! From Table 24.2, for polystyrene K # 2.6 and the dielectric strength (maximum allowed electric field) is
2 " 107 V/m .
EXECUTE:! (a)! P # K P0 # (2.6)P0 # 2.3 " 10!11 C2 N + m 2
related to the maximum possible electric field before dielectric breakdown by Vmax # Emax d . E # (b) Vmax # Emax d # (2.0 " 107 V m)(2.0 " 10!3 m) # 4.0 " 104 V
(c) E # %
) =
K P0 and = # PE # (2.3 " 10!11 C2 /N + m 2 )(2.0 " 107 V/m) # 0.46 " 10!3 C/m2 .
1&
* = i # = '1 ! ( # (0.46 " 10!3 C/m 2 )(1 ! 1/ 2.6) # 2.8 " 10!4 C/m 2 .
K
EVALUATE:! The net surface charge density is = net # = ! = i # 1.8 " 10!4 C/m 2 and the electric field between the plates is E # = net / P0 . 2412 Chapter 24 24.44. IDENTIFY:! C # Q / V . C # KC0 . V # Ed .
SET UP:! Table 24.1 gives K # 3.1 for mylar.
EXECUTE:! (a) BQ # Q ! Q0 # ( K ! 1)Q0 # ( K ! 1)C0V0 # (2.1)(2.5 " 10!7 F)(12 V) # 6.3 " 10!6 C . 24.45. (b) = i # = (1 ! 1/ K ) so Qi # Q(1 ! 1/K ) # (9.3 " 10!6 C)(1 ! 1/ 3.1) # 6.3 " 10!6 C .
(c) The addition of the mylar doesn’t affect the electric field since the induced charge cancels the additional charge
drawn to the plates.
EVALUATE:! E # V / d and V is constant so E doesn't change when the dielectric is inserted.
(a) IDENTIFY and SET UP:! Since the capacitor remains connected to the power supply the potential difference
doesn’t change when the dielectric is inserted. Use Eq.(24.9) to calculate V and combine it with Eq.(24.12) to obtain a
relation between the stored energies and the dielectric constant and use this to calculate K.
EXECUTE:! Before the dielectric is inserted U 0 # 1 C0V 2 so V #
2 2 .1.85 " 10!5 J /
2U 0
#
# 10.1 V
C0
360 " 10!9 F (b) K # C/C0 U 0 # 1 C0V 2 , U # 1 CV 2 so C/C0 # U / U 0
2
2
K# 24.46. U 1.85 " 10!5 J 0 2.32 " 10!5 J
#
# 2.25
1.85 " 10!5 J
U0 EVALUATE:! K increases the capacitance and then from U # 1 CV 2 , with V constant an increase in C gives an
2
increase in U.
IDENTIFY:! C # KC0 . C # Q / V . V # Ed .
SET UP:! Since the capacitor remains connected to the battery the potential between the plates of the capacitor
doesn't change.
EXECUTE:! (a) The capacitance changes by a factor of K when the dielectric is inserted. Since V is unchanged (the
C
Q
45.0 pC
# K # 1.80 .
battery is still connected), after # after #
Cbefore Qbefore 25.0 pC
(b) The area of the plates is , r 2 # , (0.0300 m) 2 # 2.827 " 10!3 m 2 and the separation between them is thus P0 A (8.85 " 10!12 C2 N + m 2 )(2.827 " 10!3 m 2 )
PA Q
#
# 2.00 " 10!3 m . Before the dielectric is inserted, C # 0 #
12.5 " 10!12 F
C
d
V
Qd
(25.0 " 10!12 C)(2.00 " 10!3 m)
and V #
#
# 2.00 V . The battery remains connected, so the potential
P0 A (8.85 " 10!12 C 2 /N + m 2 )(2.827 " 10!3 m 2 )
difference is unchanged after the dielectric is inserted.
Q
25.0 " 10!12 C
(c) Before the dielectric is inserted, E #
#
# 1000 N/C
P0 A (8.85 " 10!12 C 2 /N + m 2 )(2.827 " 10!3 m 2 )
Again, since the voltage is unchanged after the dielectric is inserted, the electric field is also unchanged.
2.00 V
V
EVALUATE:! E # #
# 1000 N/C , whether or not the dielectric is present. This agrees with the result
d 2.00 " 10!3 m
in part (c). The electric field has this value at any point between the plates. We need d to calculate E because V is the
potential difference between points separated by distance d.
IDENTIFY:! C # KC0 . U # 1 CV 2 .
2
d# 24.47. SET UP:! C0 # 12.5 $ F is the value of the capacitance without the dielectric present.
EXECUTE:! (a) With the dielectric, C # (3.75)(12.5 $ F) # 46.9 $ F . before: U # 1 C0V 2 # 1 (12.5 " 10!6 F)(24.0 V)2 # 3.60 mJ
2
2 24.48. after: U # 1 CV 2 # 1 (46.9 " 10!6 F)(24.0 V) 2 # 13.5 mJ
2
2
(b) BU # 13.5 mJ ! 3.6 mJ # 9.9 mJ . The energy increased.
EVALUATE:! The power supply must put additional charge on the plates to maintain the same potential difference
when the dielectric is inserted. U # 1 QV , so the stored energy increases.
2
IDENTIFY:! Gauss’s law in dielectrics has the same form as in vacuum except that the electric field is multiplied by a
factor of K and the charge enclosed by the Gaussian surface is the free charge. The capacitance of an object depends
on its geometry.
(a) SET UP:! The capacitance of a parallelplate capacitor is C # K P0 A/d and the charge on its plates is Q = CV. Capacitance and Dielectrics 2413 EXECUTE:! First find the capacitance: C# K P0 A (2.1)(8.85 " 10!12 C2 /N + m 2 )(0.0225 m2 )
#
# 4.18 " 10!10 F .
d
1.00 " 10!3 m Now find the charge on the plates: Q # CV # (4.18 " 10!10 F)(12.0 V) = 5.02 " 10!9 C .
(b) SET UP:! Gauss’s law within the dielectric gives KEA # Qfree /P0 .
EXECUTE:! Solving for E gives
Q
5.02 " 10!9 C
E # free #
# 1.20 " 104 N/C
KAP0 (2.1)(0.0225 m 2 )(8.85 " 10 !12 C 2 /N + m 2 )
(c) SET UP:! Without the Teflon and the voltage source, the charge is unchanged but the potential increases, so
C # P0 A/d and Gauss’s law now gives EA # Q/P0 .
EXECUTE:! First find the capacitance: P0 A (8.85 " 10!12 C2 /N + m 2 )(0.0225 m 2 )
#
# 1.99 " 10!10 F.
d
1.00 " 10!3 m C#
The potential difference is V #
E# 24.49. Q 5.02 " 10!9 C
#
# 25.2 V. From Gauss’s law, the electric field is
C 1.99 " 10!10 F
Q
5.02 " 10!9 C
#
# 2.52 " 10 4 N/C.
!12
P0 A (8.85 " 10 C 2 /N + m 2 )(0.0225 m 2 ) EVALUATE:! The dielectric reduces the electric field inside the capacitor because the electric field due to the dipoles
of the dielectric is opposite to the external field due to the free charge on the plates.
IDENTIFY:! Apply Eq.(24.23) to calculate E. V = Ed and C = Q/V apply whether there is a dielectric between the
plates or not.
(a) SET UP:! Apply Eq.(24.23) to the dashed surface in Figure 24.49:
!!Q
EXECUTE:! úKE + dA # enclfree
P0
!!
úKE + dA # KEA since E = 0 outside the plates
Qenclfree # = A # . Q / A / AFigure 24.49 Thus KEA # . Q/A/ A and E # Q
P0 AK P0
Qd
(b) V # Ed #
P0 AK
PA
Q
Q
(c) C # #
# K 0 # KC0 .
V (Qd/P0 AK )
d 24.50. EVALUATE:! Our result shows that K # C/C0 , which is Eq.(24.12).
PA
IDENTIFY:! C # 0 . C # Q / V . V # Ed . U # 1 CV 2 .
2
d
SET UP:! With the battery disconnected, Q is constant. When the separation d is doubled, C is halved.
P A P (0.16 m) 2
EXECUTE:! (a) C # 0 # 0
# 4.8 " 10!11 F
d
4.7 " 10!3 m
(b) Q # CV # (4.8 " 10!11 F)(12 V) # 0.58 " 10!9 C
(c) E # V/d # (12 V) /(4.7 " 10!3 m) # 2550 V/m
(d) U # 1 CV 2 # 1 (4.8 " 10!11 F)(12 V) 2 # 3.46 " 10!9 J
2
2
(e) If the battery is disconnected, so the charge remains constant, and the plates are pulled further apart to 0.0094 m, then
the calculations above can be carried out just as before, and we find:! (a) C # 2.41"10!11 F ! ! (b) Q # 0.58 "10!9 C ! Q 2 (0.58 " 10!9 C)2
#
# 6.91 " 10!9 J
2C 2(2.41 " 10!11 F)
Q
EVALUATE:! Q is unchanged. E #
so E is unchanged. U doubles because C is halved. The additional stored
P0 A
energy comes from the work done by the force that pulled the plates apart.
(c) E # 2550 V m ! ! (d) U # 2414 Chapter 24 24.51. IDENTIFY and SET UP:! If the capacitor remains connected to the battery, the battery keeps the potential difference
between the plates constant by changing the charge on the plates.
PA
EXECUTE:! (a) C # 0
d
!12
2
(8.854 " 10 C / N + m 2 )(0.16 m)2
C#
# 2.4 " 10!11 F # 24 pF
9.4 " 10!3 m
(b) Remains connected to the battery says that V stays 12 V. Q # CV # (2.4 " 10!11 F)(12 V) # 2.9 " 10!10 C
V
12 V
#
# 1.3 " 103 V/m
d 9.4 " 10!3 m
(d) U # 1 QV # 1 (2.9 " 10!10 C)(12.0 V) # 1.7 " 10!9 J
2
2
EVALUATE:! Increasing the separation decreases C. With V constant, this means that Q decreases and U decreases.
Q decreases and E # Q / P0 A so E decreases. We come to the same conclusion from E # V / d . (c) E # 24.52. 24.53. 24.54. 24.55. IDENTIFY:! C # KC0 # K P0 A
. V # Ed for a parallel plate capacitor; this equation applies whether or not a dielectric
d is present.
SET UP:! A # 1.0 cm 2 # 1.0 " 10!4 m 2 .
(8.85 " 10!12 F/m)(1.0 " 10!4 m 2 )
EXECUTE:! (a) C # (10)
# 1.18 $ F per cm2.
7.5 " 10!9 m
V
85 mV
(b) E # #
# 1.13 " 107 V/m .
K 7.5 " 10!9 m
EVALUATE:! The dielectric material increases the capacitance. If the dielectric were not present, the same charge
density on the faces of the membrane would produce a larger potential difference across the membrane.
IDENTIFY:! P # E/t , where E is the total light energy output. The energy stored in the capacitor is U # 1 CV 2 .
2
SET UP:! E # 0.95U
EXECUTE:! (a) The power output is 600 W, and 95% of the original energy is converted, so
E # Pt # (2.70 " 105 W)(1.48 " 10!3 s) # 400 J . E0 # 400 J # 421 J .
0.95
2U 2(421 J)
2
1
# 0.054 F .
(b) U # 2 CV so C # 2 #
V
(125 V) 2
EVALUATE:! For a given V, the stored energy increases linearly with C.
PA
IDENTIFY:! C # 0
d
SET UP:! A # 4.2 " 10!5 m 2 . The original separation between the plates is d # 0.700 " 10 !3 m . d  is the separation
between the plates at the new value of C.
AP (4.20 " 10!5 m 2 )P0
# 5.31" 10!13 F . The new value of C is C # C0 0 0.25 pF # 7.81 " 10!13 F .
EXECUTE:! C0 # 0 #
d
7.00 " 10!4 m
AP (4.20 " 10!5 m 2 )P0
AP
But C # 0 , so d  # 0 #
# 4.76 " 10!4 m . Therefore the key must be depressed by a distance of
C
7.81 " 10!13 F
d7.00 " 10!4 m ! 4.76 " 10!4 m # 0.224 mm .
EVALUATE:! When the key is depressed, d decreases and C increases.
2, P0 L
IDENTIFY:! Example 24.4 shows that C #
for a cylindrical capacitor.
ln(rb / ra )
SET UP:! ln(1 0 x ) C x when x is small. The area of each conductor is approximately A # 2, ra L . 2, P0 L
2, P0 L
2, P0 L
2, ra LP0 P0 A
#
#
C
#
ln( rb ra ) ln((d 0 ra ) ra ) ln(1 0 d ra )
d
d
EVALUATE:! (b) At the scale of part (a) the cylinders appear to be flat, and so the capacitance should appear like that
of flat plates.
IDENTIFY:! Initially the capacitors are connected in parallel to the source and we can calculate the charges Q1 and
EXECUTE:! (a) d 11 ra : C # 24.56. Q2
.
2C
SET UP:! After they are reconnected, the charges add and the voltages are the same, so Ceq # C1 0 C2 , as for
Q2 on each. After they are reconnected to each other the total charge is Q # Q2 ! Q1 . U # 1 CV 2 #
2 capacitors in parallel. Capacitance and Dielectrics 2415 EXECUTE:! Originally Q1 # C1V1 # (9.0 $ F) (28 V) # 2.52 " 10!4 C and Q2 # C2V2 # (4.0 $ F)(28 V) # 1.12 " 10!4 C .
Ceq # C1 0 C2 # 13.0 $ F . The original energy stored is U # 1 CeqV 2 # 1 (13.0 " 10!6 F)(28 V)2 # 5.10 " 10!3 J .
2
2 Disconnect and flip the capacitors, so now the total charge is Q # Q2 ! Q1 # 1.4 " 10!4 C and the equivalent capacitance
is still the same, Ceq # 13.0 $ F . The new energy stored is U # 24.57. Q2
(1.4 " 10!4 C) 2
#
# 7.54 " 10!4 J . The change in
2Ceq 2(13.0 " 10!6 F) stored energy is BU # 7.45 " 10!4 J ! 5.10 " 10!3 J # !4.35 " 10!3 J .
EVALUATE:! When they are reconnected, charge flows and thermal energy is generated and energy is radiated as
electromagnetic waves.
IDENTIFY:! Simplify the network by replacing series and parallel combinations by their equivalent. The stored
energy in a capacitor is U # 1 CV 2 .
2
SET UP:! For capacitors in series the voltages add and the charges are the same; 1
1
1
#0
0 %. For capacitors
Ceq C1 C2 Q
. U # 1 CV 2 .
2
V
EXECUTE:! (a) Find Ceq for the network by replacing each series or parallel combination by its equivalent. The
successive simplified circuits are shown in Figure 24.57a–c.
U tot # 1 CeqV 2 # 1 (2.19 " 10!6 F)(12.0 V)2 # 1.58 " 10!4 J # 158 $ J
2
2 in parallel the voltages are the same and the charges add; Ceq # C1 0 C2 0 % C # (b) From Figure 24.57c, Qtot # CeqV # (2.19 "10!6 F)(12.0 V) # 2.63 "10!5 C. From Figure 24.57b, Qtot # 2.63 "10!5 C.
Q4.8 2.63 " 10!5 C
#
# 5.48 V . U 4.8 # 1 CV 2 # 1 (4.80 " 10!6 F)(5.48 V)2 # 7.21" 10!5 J # 72.1 $ J
2
2
C4.8 4.80 " 10!6 F
This one capacitor stores nearly half the total stored energy.
Q2
EVALUATE:! U #
. For capacitors in series the capacitor with the smallest C stores the greatest amount of
2C
energy.
V4.8 # Figure 24.57
24.58. IDENTIFY:! Apply the rules for combining capacitors in series and parallel. For capacitors in series the voltages add
and in parallel the voltages are the same.
SET UP:! When a capacitor is a moderately good conductor it can be replaced by a wire and the potential across it is zero.
EXECUTE:! (a) A network that has the desired properties is sketched in Figure 24.58a. Ceq # C 0 C # C . The total
22
capacitance is the same as each individual capacitor, and the voltage is spilt over each so that V # 480 V.
(b) If one capacitor is a moderately good conductor, then it can be treated as a “short” and thus removed from the
circuit, and one capacitor will have greater than 600 V across it.
EVALUATE:! An alternative solution is two in parallel in series with two in parallel, as sketched in Figure 24.58b. Figure 24.58 2416 Chapter 24 24.59. (a) IDENTIFY:! Replace series and parallel combinations of capacitors by their equivalents.
SET UP:! The network is sketched in Figure 24.59a. C1 # C5 # 8.4 $ F
C2 # C3 # C4 # 4.2 $ F
Figure 24.59a
EXECUTE:! Simplify the circuit by replacing the capacitor combinations by their equivalents: C3 and C4 are in series and can be replaced by C34 (Figure 24.59b):
1
1
1
#
0
C34 C3 C4
1
C 0 C4
#3
C34
C3C4
Figure 24.59b
C34 # C3C4
. 4.2 $ F /. 4.2 $ F / # 2.1 $ F
#
4.2 $ F 0 4.2 $ F
C3 0 C4 C2 and C34 are in parallel and can be replaced by their equivalent (Figure 24.59c):
C234 # C2 0 C34
C234 # 4.2 $ F 0 2.1 $ F
C234 # 6.3 $ F
Figure 24.59c C1 , C5 and C234 are in series and can be replaced by Ceq (Figure 24.59d):
1
1
1
1
#0
0
Ceq C1 C5 C234
1
2
1
#
0
Ceq 8.4 $ F 6.3 $ F
Ceq # 2.5 $ F
Figure 24.59d
EVALUATE:! For capacitors in series the equivalent capacitor is smaller than any of those in series. For capacitors in
parallel the equivalent capacitance is larger than any of those in parallel.
(b) IDENTIFY and SET UP:! In each equivalent network apply the rules for Q and V for capacitors in series and
parallel; start with the simplest network and work back to the original circuit.
EXECUTE:! The equivalent circuit is drawn in Figure 24.59e. Qeq # CeqV
Qeq # . 2.5 $ F / . 220 V / # 550 $ C
Figure 24.59e Q1 # Q5 # Q234 # 550 $ C (capacitors in series have same charge)
V1 # Q1 550 $ C
#
# 65 V
C1 8.4 $ F V5 # Q5 550 $ C
#
# 65 V
C5 8.4 $ F V234 # Q234 550 $ C
#
# 87 V
C234 6.3 $ F Capacitance and Dielectrics 2417 Now draw the network as in Figure 24.59f.
V2 # V34 # V234 # 87 V
capacitors in parallel have the same potential
Figure 24.59f Q2 # C2V2 # . 4.2 $ F /. 87 V / # 370 $ C
Q34 # C34V34 # . 2.1 $ F /. 87 V / # 180 $ C
Finally, consider the original circuit (Figure 24.59g).
Q3 # Q4 # Q34 # 180 $ C
capacitors in series have the same charge
Figure 24.59g V3 # Q3 180 $ C
#
# 43 V
C3 4.2 $ F V4 # Q4 180 $ C
#
# 43 V
C4 4.2 $ F Summary: Q1 # 550 $ C, V1 # 65 V
Q2 # 370 $ C, V2 # 87 V
Q3 # 180 $ C, V3 # 43 V
Q4 # 180 $ C, V4 # 43 V
Q5 # 550 $ C, V5 # 65 V
EVALUATE:! V3 0 V4 # V2 and V1 0 V2 0 V5 # 220 V (apart from some small rounding error)
24.60. Q1 # Q2 0 Q3 and Q5 # Q2 0 Q4
IDENTIFY:! Apply the rules for combining capacitors in series and in parallel.
SET UP:! With the switch open each pair of 3.00 $ F and 6.00 $ F capacitors are in series with each other and each
pair is in parallel with the other pair. When the switch is closed each pair of 3.00 $ F and 6.00 $ F capacitors are in
parallel with each other and the two pairs are in series.
!1
!1
%%
&
%
&&
EXECUTE:! (a) With the switch open Ceq # ' ' 1 0 1 ( 0 ' 1 0 1 ( ( # 4.00 $ F .
' ) 3 $F 6 $F *
) 3 $F 6 $F * (
)
*
Qtotal # CeqV # (4.00 $ F) (210 V) # 8.40 " 10!4 C . By symmetry, each capacitor carries 4.20 " 10!4 C. The
voltages are then calculated via V # Q / C . This gives Vad # Q / C3 # 140 V and Vac # Q / C6 # 70 V .
Vcd # Vad ! Vac # 70 V .
(b) When the switch is closed, the points c and d must be at the same potential, so the equivalent capacitance is
!1 %
&
1
1
!4
0
Ceq # '
( # 4.5 $ F . Qtotal # CeqV # (4.50 $ F)(210 V) # 9.5 " 10 C , and each
) (3.00 0 6.00) $ F (3.00 0 6.00) $ F *
capacitor has the same potential difference of 105 V (again, by symmetry).
(c) The only way for the sum of the positive charge on one plate of C2 and the negative charge on one plate of
C1 to change is for charge to flow through the switch. That is, the quantity of charge that flows through the
switch is equal to the change in Q2 ! Q1 . With the switch open, Q1 # Q2 and Q2 ! Q1 # 0. After the switch is
closed, Q2 ! Q1 # 315 $ C , so 315 $ C of charge flowed through the switch.
EVALUATE:! When the switch is closed the charge must redistribute to make points c and d be at the same
potential. 2418 Chapter 24 24.61. (a) IDENTIFY:! Replace the three capacitors in series by their equivalent. The charge on the equivalent capacitor
equals the charge on each of the original capacitors.
SET UP:! The three capacitors can be replaced by their equivalent as shown in Figure 24.61a. Figure 24.61a
EXECUTE:! C3 # C1 / 2 so 1
1
1
1
4
#0
0
#
and Ceq # 8.4 $ F/4 # 2.1 $ F
Ceq C1 C2 C3 8.4 $ F Q # CeqV # . 2.1 $ F /. 36 V / # 76 $ C
The three capacitors are in series so they each have the same charge: Q1 # Q2 # Q3 # 76 $ C
EVALUATE: The equivalent capacitance for capacitors in series is smaller than each of the original capacitors.
(b) IDENTIFY and SET UP:! Use U # 1 QV . We know each Q and we know that V1 0 V2 0 V3 # 36 V.
2
EXECUTE:! U # 1 QV1 0 1 Q2V2 0 1 Q3V3
21
2
2 But Q1 # Q2 # Q3 # Q so U # 1 Q (V1 0 V2 0 V3 )
2
But also V1 0 V2 0 V3 # V # 36 V, so U # 1 QV # 1 (76 $ C)(36 V) # 1.4 " 10!3 J.
2
2
EVALUATE:! We could also use U # Q 2 / 2C and calculate U for each capacitor.
(c) IDENTIFY:! The charges on the plates redistribute to make the potentials across each capacitor the same.
SET UP:! The capacitors before and after they are connected are sketched in Figure 24.61b. Figure 24.61b
EXECUTE:! The total positive charge that is available to be distributed on the upper plates of the three capacitors is
Q0 # Q01 0 Q02 0 Q03 # 3 . 76 $ C / # 228 $ C. Thus Q1 0 Q2 0 Q3 # 228 $ C. After the circuit is completed the charge distributes to make V1 # V2 # V3 . V # Q / C and V1 # V2 so Q1 / C1 # Q2 / C2 and then C1 # C2 says Q1 # Q2 . V1 # V3 says
Q1 / C1 # Q3 / C3 and Q1 # Q3 . C1 / C3 / # Q3 . 8.4 $ F/4.2 $ F / # 2Q3 Using Q2 # Q1 and Q1 # 2Q3 in the above equation gives 2Q3 0 2Q3 0 Q3 # 228 $ C.
5Q3 # 228 $ C and Q3 # 45.6 $ C, Q1 # Q2 # 91.2 $ C
Q1 91.2 $ C
Q 91.2 $ C
Q 45.6 $ C
#
# 11 V, V2 # 2 #
# 11 V, and V3 # 3 #
# 11 V.
C1 8.4 $ F
C2
8.4 $ F
C3
4.2 $ F
The voltage across each capacitor in the parallel combination is 11 V.
(d) U # 1 QV1 0 1 Q2V2 0 1 Q3V3 .
21
2
2
Then V1 # But V1 # V2 # V3 so U # 1 V1 . Q1 0 Q2 0 Q3 / #
2 24.62. 1
2 .11 V / . 228 $ C / # 1.3 " 10!3 J. EVALUATE:! This is less than the original energy of 1.4 " 10!3 J. The stored energy has decreased, as in
Example 24.7.
PA
Q
IDENTIFY:! C # 0 . C # . V # Ed . U # 1 QV .
2
d
V
3
2
SET UP:! d # 3.0 " 10 m . A # , r , with r # 1.0 " 103 m .
P A (8.854 " 10 !12 C 2 /N + m 2 ), (1.0 " 103 m) 2
EXECUTE:! (a) C # 0 #
# 9.3 " 10!9 F .
d
3.0 " 103 m
Q
20 C
(b) V # #
# 2.2 " 109 V
C 9.3 " 10!9 F
V 2.2 " 109 V
(c) E # #
# 7.3 " 105 V/m
d 3.0 " 103 m
(d) U # 1 QV # 1 (20 C)(2.2 " 109 V) # 2.2 " 1010 J
2
2
EVALUATE:! Thunderclouds involve very large potential differences and large amounts of stored energy. Capacitance and Dielectrics 24.63. 2419 IDENTIFY:! Replace series and parallel combinations of capacitors by their equivalents. In each equivalent network
apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work back to the
original circuit.
(a) SET UP:! The network is sketched in Figure 24.63a. C1 # 6.9 $ F
C2 # 4.6 $ F
Figure 24.63a
EXECUTE:! Simplify the network by replacing the capacitor combinations by their equivalents. Make the
replacement shown in Figure 24.63b. 1
3
#
Ceq C1
Ceq # C1 6.9 $ F
#
# 2.3 $ F
3
3 Figure 24.63b Next make the replacement shown in Figure 24.63c. Ceq # 2.3 $ F 0 C2
Ceq # 2.3 $ F 0 4.6 $ F # 6.9 $ F
Figure 24.63c Make the replacement shown in Figure 24.63d.
1
2
1
3
#0
#
Ceq C1 6.9 $ F 6.9 $ F
Ceq # 2.3 $ F
Figure 24.63d Make the replacement shown in Figure 24.63e.
Ceq # C2 0 2.3 $ F # 4.6 $ F 0 2.3 $ F Ceq # 6.9 $ F
Figure 24.63e Make the replacement shown in Figure 24.63f.
1
2
1
3
#0
#
Ceq C1 6.9 $ F 6.9 $ F
Ceq # 2.3 $ F
Figure 24.63f
(b) Consider the network as drawn in Figure 24.63g. From part (a) 2.3 $ F is the equivalent
capacitance of the rest of the network.
Figure 24.63g 2420 Chapter 24 The equivalent network is shown in Figure 24.63h.
The capacitors are in series,
so all three capacitors have
the same Q.
Figure 24.63h But here all three have the same C, so by V = Q/C all three must have the same V. The three voltages must add
to 420 V, so each capacitor has V = 140 V. The 6.9 $ F to the right is the equivalent of C2 and the 2.3 $ F
capacitor in parallel, so V2 # 140 V. (Capacitors in parallel have the same potential difference.) Hence Q1 # C1V1 # (6.9 $ F)(140 V) # 9.7 " 10!4 C and Q2 # C2V2 # (4.6 $ F)(140 V) # 6.4 " 10!4 C.
(c) From the potentials deduced in part (b) we have the situation shown in Figure 24.63i.
From part (a) 6.9 $ F is the
equivalent capacitance of the
rest of the network. Figure 24.63i 24.64. The three rightmost capacitors are in series and therefore have the same charge. But their capacitances are also equal,
so by V = Q/C they each have the same potential difference. Their potentials must sum to 140 V, so the potential
across each is 47 V and Vcd # 47 V.
EVALUATE:! In each capacitor network the rules for combining V for capacitors in series and parallel are obeyed.
Note that Vcd 1 V , in fact V ! 2(140 V) ! 2(47 V) # Vcd .
IDENTIFY:! Find the total charge on the capacitor network when it is connected to the battery. This is the amount of
charge that flows through the signal device when the switch is closed.
SET UP:! For capacitors in parallel, Ceq # C1 0 C2 0 C3 0 %
EXECUTE:! Cequiv # C1 0 C2 0 C3 # 60.0 $ F . Q # CV # (60.0 $ F)(120 V) # 7200 $ C . 24.65. EVALUATE:! More charge is stored by the three capacitors in parallel than would be stored in each capacitor used
alone.
(a) IDENTIFY and SET UP:! Q is constant. C # KC0 ; use Eq.(24.1) to relate the dielectric constant K to the ratio of
the voltages without and with the dielectric.
EXECUTE:! With the dielectric: V # Q / C # Q / . KC0 / without the dielectric: V0 # Q / C0
V0 / V # K , so K # . 45.0 V / / .11.5 V / # 3.91
EVALUATE:! Our analysis agrees with Eq.(24.13).
(b) IDENTIFY:! The capacitor can be treated as equivalent to two capacitors C1 and C2 in parallel, one with
area 2A/3 and air between the plates and one with area A/3 and dielectric between the plates.
SET UP:! The equivalent network is shown in Figure 24.65. Figure 24.65
EXECUTE:! Let C0 # P0 A / d be the capacitance with only air between the plates. C1 # KC0 / 3, C2 # 2C0 / 3; Ceq # C1 0 C2 # (C0 / 3)( K 0 2)
V# Q
Q% 3 &
%3&
%3&
#'
( # V0 '
( # . 45.0 V / '
( # 22.8 V
Ceq C0 ) K 0 2 *
K 02*
)
) 5.91 * EVALUATE:! The voltage is reduced by the dielectric. The voltage reduction is less when the dielectric doesn’t
completely fill the volume between the plates. Capacitance and Dielectrics 24.66. IDENTIFY:! This situation is analogous to having two capacitors C1 in series, each with separation 1
2 2421 (d ! a ). 1
1
1
#0
.
SET UP:! For capacitors in series,
Ceq C1 C2
!1 24.67. P0 A
PA
%
&
EXECUTE:! (a) C # ' 1 0 1 ( # 1 C1 # 1
#0
2
2
C1 C1 *
(d ! a) 2 d ! a
)
P0 A P0 A d
d
#
# C0
(b) C #
d !a
d d !a
d !a
(c) As a E 0 , C E C0 . The metal slab has no effect if it is very thin. And as a E d , C E F . V # Q / C . V # Ey is
the potential difference between two points separated by a distance y parallel to a uniform electric field. When the
distance is very small, it takes a very large field and hence a large Q on the plates for a given potential difference.
Since Q # CV this corresponds to a very large C.
(a) IDENTIFY:! The conductor can be at some potential V, where V = 0 far from the conductor. This potential
depends on the charge Q on the conductor so we can define C = Q/V where C will not depend on V or Q.
(b) SET UP:! Use the expression for the potential at the surface of the sphere in the analysis in part (a).
EXECUTE:! For any point on a solid conducting sphere V # Q / 4, P0 R if V # 0 at r E F.
C# % 4, P0 R &
Q
# Q'
( # 4, P0 R
V
)Q* (c) C # 4, P0 R # 4, . 8.854 " 10!12 F/m / . 6.38 " 106 m / # 7.10 " 10!4 F # 710 $ F. 24.68. EVALUATE:! The capacitance of the earth is about seven times larger than the largest capacitances in this range. The
capacitance of the earth is quite small, in view of its large size.
Q2
IDENTIFY:! The electric field energy density is 1 P0 E 2 . For a capacitor U #
.
2
2C
Q
SET UP:! For a solid conducting sphere of radius R, E # 0 for r 1 R and E #
for r 5 R .
4, P0 r 2
EXECUTE:! (a) r 1 R : u # 1 P0 E 2 # 0.
2
2 %Q&
Q2
(b) r 5 R : u # P E # P '
.
#
2(
2
4
) 4, P0 r * 32, P0 r
1
20 2 F 1
20 (c) U # H udV # 4, H r 2udr #
R 24.69. F Q 2 dr
Q2
H r 2 # 8, P0 R .
8, P0 R Q2
(d) This energy is equal to 1
which is just the energy required to assemble all the charge into a spherical
2 4, P0 R
distribution. (Note that being aware of double counting gives the factor of 1/2 in front of the familiar potential energy
formula for a charge Q a distance R from another charge Q.)
Q2
Q2
EVALUATE:! (e) From Equation (24.9), U #
.U#
from part (c) , C # 4, P0 R , as in Problem (24.67).
2C
8, P0 R
IDENTIFY:! We model the earth as a spherical capacitor.
rr
SET UP:! The capacitance of the earth is C # 4, P0 a b and, the charge on it is Q = CV, and its stored energy is
rb ! ra U # 1 CV 2 .
2
EXECUTE:! (a) C # . 6.38 "106 m / . 6.45 "106 m / # 6.5 "10!2 F
1
9.00 " 109 N + m 2 /C 2 6.45 " 106 m ! 6.38 " 106 m (b) Q # CV # . 6.54 " 10!2 F / (350,000 V) = 2.3 " 104 C (c) U # 1 CV 2 # 1 . 6.54 " 10!2 F / (350,000 V)2 # 4.0 " 109 J
2
2
EVALUATE:! While the capacitance of the earth is larger than ordinary laboratory capacitors, capacitors much larger
than this, such as 1 F, are readily available. 2422 Chapter 24 24.70. IDENTIFY:! The electric field energy density is u # 1 P0 E 2 . U #
2
SET UP:! For this charge distribution, E # 0 for r 1 ra , E # Example 24.4 shows that Q2
.
2C 8
for ra 1 r 1 rb and E # 0 for r 5 rb .
2, P0 r U
2, P0
for a cylindrical capacitor.
#
L ln( rb / ra )
2 %8&
82
EXECUTE:! (a) u # 1 P0 E 2 # 1 P0 '
(# 2 2
2
2
) 2, P0 r * 8, P0 r
r (b) U # H udV # 2, L H urdr # U
82
L8 2 b dr
and
#
ln(rb / ra ) .
L 4, P0
4, P0 rH r
a Q2
Q2
8 2L
#
ln(rb / ra ) #
ln(rb / ra ) . This agrees with the result of part (b).
2C 4, P0 L
4, P0
Q2
EVALUATE:! We could have used the results of part (b) and U #
to calculate U / L and would obtain the same
2C
result as in Example 24.4.
IDENTIFY:! C # Q / V , so we need to calculate the effect of the dielectrics on the potential difference between the
plates.
SET UP:! Let the potential of the positive plate be Va , the potential of the negative plate be Vc , and the potential
(c) Using Equation (24.9), U # 24.71. midway between the plates where the dielectrics meet be Vb , as shown in Figure 24.71. C# Q
Q
#
.
Va ! Vc Vac Vac # Vab 0 Vbc .
Figure 24.71
EXECUTE:! The electric field in the absence of any dielectric is E0 # reduced to E1 #
E2 # Q
. In the first dielectric the electric field is
P0 A E0
Q
Qd
%d &
#
and Vab # E1 ' ( #
. In the second dielectric the electric field is reduced to
K1 K1P0 A
) 2 * K1 2P0 A E0
Q
Qd
Qd
Qd
Qd % 1
1&
%d &
#
0
#
and Vbc # E2 ' ( #
. Thus Vac # Vab 0 Vbc #
'0
(.
K 2 K 2P0 A
K1 2P0 A K 2 2P0 A 2P0 A ) K1 K 2 *
) 2 * K 2 2P0 A % 2P A & % K K & 2P A % K K &
Qd % K1 0 K 2 &
Q
# Q ' 0 ( ' 1 2 ( # 0 ' 1 2 (.
'
( . This gives C #
Vac
d ) K1 0 K 2 *
2P0 A ) K1K 2 *
) Qd * ) K1 0 K 2 *
EVALUATE:! An equivalent way to calculate C is to consider the capacitor to be two in series, one with dielectric
constant K1 and the other with dielectric constant K 2 and both with plate separation d/2. (Can imagine inserting a
thin conducting plate between the dielectric slabs.)
PA
PA
C1 # K1 0 # 2 K1 0
d /2
d
P0 A
P0 A
C2 # K 2
# 2K2
d /2
d
11
1
CC
2P A % K K &
Since they are in series the total capacitance C is given by
so C # 1 2 # 0 ' 1 2 (
#0
C C1 C2
C1 0 C2
d ) K1 0 K 2 *
Vac # 24.72. IDENTIFY:! This situation is analogous to having two capacitors in parallel, each with an area A/2. SET UP:! For capacitors in parallel, Ceq # C1 0 C2 . For a parallelplate capacitor with plates of area A/ 2, C #
P0 A / 2 P0 A / 2 P0 A
0
#
( K1 0 K 2 )
d
d
2d
PA
EVALUATE:! If K1 # K 2 , Ceq # K 0 , which is Eq.(24.19).
d EXECUTE:! Ceq # C1 0 C2 # P0 ( A/2)
.
d Capacitance and Dielectrics 24.73. 2423 IDENTIFY and SET UP:! Show the transformation from one circuit to the other: Figure 24.73a
EXECUTE:! (a) Consider the two networks shown in Figure 24.73a. From Circuit 1: Vac # q1 ! q3
q 0 q3
and Vbc # 2
.
Cy
Cx %q
CxC yCz % q1 q2 &
q3 q1 ! q3 q2 ! q3
q&
!
T K' 1 ! 2 (.
#
#
. This gives q3 #
'
' C y Cx (
(
' C y Cx (
Cz
Cy
Cx
Cx 0 C y 0 Cz )
*
)
*
%1
%1
q1 q1 0 q2
1&
1
q2 q1 0 q2
1
1&
# q1 ' 0 ( 0 q2
and Vbc #
0
# q1 0 q2 '
0 ( . Setting the
From Circuit 2: Vac # 0
C1
C3
C3
C2
C3
C3
) C1 C3 *
) C2 C3 *
q3 is derived from Vab : Vab # coefficients of the charges equal to each other in matching potential equations from the two circuits results in three
1
1%
1
1&
independent equations relating the two sets of capacitances. The set of equations are
#
!
'1 !
(,
C1 C y ' KC y KCx (
)
*
1
1%
1
1&
1
1
# '1 !
!
and
#
. From these, subbing in the expression for K , we get
' KC y KC x (
(
C2 C x )
C3 KC yC x
*
C1 # (C xC y 0 C yCz 0 Cz Cx ) Cx , C2 # (CxC y 0 C yC z 0 C zC x ) C y and C3 # (CxC y 0 C yC z 0 C zC x ) C z .
(b) Using the transformation of part (a) we have the equivalent networks shown in Figure 24.73b: Figure 24.73b C1 # 126 $ F , C2 # 28 $ F , C3 # 42 $ F , C4 # 42 $ F , C5 # 147 $ F and C6 # 32 $ F . The total equivalent capacitance
%1
1
1
1
1&
0
0
0
0
is Ceq # '
(
72 $ F 126 $ F 34.8 $ F 147 $ F 72 $ F *
)
!1
!1
%% 1
%1
1&
1 &&
0
0'
0
34.8 $ F # ' '
( (.
' ) 42 $ F 32 $ F (
*
) 28 $ F 42 $ F * (
)
* !1 # 14.0 $ F, where the 34.8 $ F comes from 2424 Chapter 24 (c) The circuit diagram can be redrawn as shown in Figure 25.73c. The overall charge is given by
Q # CeqV # (14.0 $ F)(36 V) # 5.04 " 10!4 C . And this is also the charge on the 72 $ F capacitors, so V72 # 5.04 " 10!4 C
# 7.0 V .
72 " 10!6 F Figure 24.73c Next we will find the voltage over the numbered capacitors, and their associated voltages. Then those voltages
will be changed back into voltage of the original capacitors, and then their charges. QC1 # QC5 # Q72 # 5.04 " 10 !4 C .
VC5 # 5.04 " 10!4 C
5.04 " 10!4 C
# 3.43 V and VC1 #
# 4.00 V . Therefore,
!6
147 " 10 F
126 " 10!6 F
!1 %1
1&
VC2 C4 # VC3C8 # (36.0 ! 7.00 ! 7.00 ! 4.00 ! 3.43) V # 14.6 V . But Ceq (C2C4 ) # ' 0 ( # 16.8 $ F and
C2 C4 *
)
!1 %1
1&
Ceq (C3C6 ) # ' 0 ( # 18.2 $ F , so QC2 # QC4 # VC2 C4 Ceq (C2 C4 ) # 2.45 " 10!4 C and
) C3 C6 *
QC
QC
QC
QC3 # QC6 # VC3C6 Ceq( C3C6 ) # 2.64 " 10!4 C . Then VC2 # 2 # 8.8 V , VC3 # 3 # 6.3 V , VC4 # 4 # 5.8 V and
C3
C2
C4
QC
VC6 # 6 # 8.3 V . Vac # VC1 0 VC2 # V18 # 13 V and Q18 # C18V18 # 2.3 " 10!4 C . Vab # VC1 0 VC3 # V27 # 10 V and
C6 Q27 # C27V27 # 2.8 " 10!4 C . Vcd # VC4 0 VC5 # V28 # 9 V and Q28 # C28V28 # 2.6 " 10!4 C . Vbd # VC5 0 VC6 # V21 # 12 V
and Q21 # C21V21 # 2.5 " 10!4 C . Vbc # VC3 ! VC2 # V6 # 2.5 V and Q6 # C6V6 # 1.5 " 10!5 C .
24.74. EVALUATE:! Note that 2V72 0 V18 0 V28 # 2(7.0 V) 0 13 V 0 9 V # 36 V, as it should.
IDENTIFY:! The force on one plate is due to the electric field of the other plate. The electrostatic force must be
balanced by the forces from the springs.
SET UP:! The electric field due to one plate is E # = 2P0 . The force exerted by a spring compressed a distance z0 ! z from equilibrium is k ( z0 ! z ) .
EXECUTE:! (a) The force between the two parallel plates is F # qE # 2
q=
q2
(CV ) 2 P0 A2 V 2
P AV 2
#
#
#2
#0 2 .
2P0 2P0 A 2P0 A
z 2P0 A
2z (b) When V # 0, the separation is just z0 . When V R 0 , the total force from the four springs must equal the P0 AV 2
P AV 2
and 2 z 3 ! 2 z 3 z0 0 0
#0.
2
2z
4k
(c) For A # 0.300 m 2 , z0 # 1.2 " 10!3 m , k # 25 N/m and V # 120 V , so 2z 3 ! (2.4 " 10!3 m)z 2 0 3.82 " 10!10 m3 # 0 .
The physical solutions to this equation are z # 0.537 mm and 1.014 mm.
EVALUATE:! (d) Stable equilibrium occurs if a slight displacement from equilibrium yields a force back toward the
equilibrium point. If one evaluates the forces at small displacements from the equilibrium positions above, the
1.014 mm separation is seen to be stable, but not the 0.537 mm separation. electrostatic force calculated in part (a). F4 springs # 4k ( z0 ! z ) # Capacitance and Dielectrics 24.75. 2425 IDENTIFY:! The system can be considered to be two capacitors in parallel, one with plate area L( L ! x ) and air
between the plates and one with area Lx and dielectric filling the space between the plates.
KP A
SET UP:! C # 0 for a parallelplate capacitor with plate area A.
d
P
PL
EXECUTE:! (a) C # 0 (( L ! x) L 0 xKL) # 0 ( L 0 ( K ! 1) x )
D
D
P0 L
PL
2
1
(b) dU # 2 ( dC )V , where C # C0 0
(! dx 0 dxK ) , with C0 # 0 ( L 0 ( K ! 1) x) . This gives
D
D
2
% P0 Ldx
& 2 ( K ! 1)P0V L
dU # 1 '
dx .
( K ! 1) (V #
2
2D
)D
*
(c) If the charge is kept constant on the plates, then Q # UC %C&
P0 LV
( L 0 ( K ! 1) x ) and U # 1 CV 2 # 1 C0V 2 ' ( .
2
2
D
) C0 * &
P0 L
C0V 2 %
( K ! 1)P0V 2 L
( K ! 1) dx ( and BU # U ! U 0 # !
dx .
'1 !
2 ) DC0
2D
* ( K ! 1)P0V 2 L
dx , the force is in the opposite direction to the motion dx , meaning that the
2D
slab feels a force pushing it out.
EVALUATE:! (e) When the plates are connected to the battery, the plates plus slab are not an isolated system. In
addition to the work done on the slab by the charges on the plates, energy is also transferred between the battery and
( K ! 1)P0V 2 L
.
the plates. Comparing the results for dU in part (c) to dU # ! Fdx gives F #
2D
IDENTIFY:! C # Q / V . Apply Gauss's law and the relation between potential difference and electric field.
r!
! r!
!
SET UP:! Each conductor is an equipotential surface. Va ! Vb # H b E U + d r # H b E L + d r , so EU # EL , where these (d) Since dU # ! Fdx # ! 24.76. r0 ra are the fields between the upper and lower hemispheres. The electric field is the same in the air space as in the
dielectric.
% rr &
EXECUTE:! (a) For a normal spherical capacitor with air between the plates, C0 # 4, P0 ' a b ( . The capacitor in
) rb ! ra *
this problem is equivalent to two parallel capacitors, CL and CU , each with half the plate area of the normal
capacitor. CL # % rr &
% rr &
% rr &
KC0
C
# 2, K P0 ' a b ( and CU # 0 # 2, P0 ' a b ( . C # CU 0 CL # 2, P0 (1 0 K ) ' a b ( .
2
2
) rb ! ra *
) rb ! ra *
) rb ! ra * (b) Using a hemispherical Gaussian surface for each respective half, EL EU QL
4, r 2 QL
, so EL #
, and
#
K P0
2, K P0 r 2
2 QU
4, r 2 QU
VC K
#
, so EU #
. But QL # VCL and QU # VCU . Also, QL 0 QU # Q . Therefore, QL # 0 # KQU
2
2
P0
2, P0 r
2 and QU #
EU # KQ
1
2
Q
Q
KQ
, QL #
. This gives EL #
#
and
2
10 K
10 K
1 0 K 2, K P0 r
1 0 K 4, P0 r 2 Q
1
2
Q
#
. We do find that EU # EL .
2
1 0 K 2, K P0 r
1 0 K 4, K P0 r 2 (c) The free charge density on upper and lower hemispheres are: (= f,ra ) U # (= f , rb ) U # QU
Q
#
and
2
2
2, ra
2, ra (1 0 K ) QU
Q
Q
KQ
Q
KQ
#
; (= f,ra ) L # L 2 #
and (= f,rb ) L # L 2 #
.
2
2
2
2, rb
2, rb (1 0 K )
2, ra
2, ra (1 0 K )
2, rb
2, rb 2 (1 0 K ) % ( K ! 1) & Q % K & % K ! 1 & Q
(d) = i,ra # = f,ra (1 ! 1 K ) # '
(
(#'
(
2'
2
) K * 2, ra ) K 0 1 * ) K 0 1 * 2, ra
% ( K ! 1) & Q % K & % K ! 1 & Q
(
(#'
(
2'
2
) K * 2, rb ) K 0 1 * ) K 0 1 * 2, rb
(e) There is zero bound charge on the flat surface of the dielectricair interface, or else that would imply a
circumferential electric field, or that the electric field changed as we went around the sphere.
EVALUATE:! The charge is not equally distributed over the surface of each conductor. There must be more charge on
the lower half, by a factor of K, because the polarization of the dielectric means more free charge is needed on the
lower half to produce the same electric field. = i,rb # = f,rb (1 ! 1 K ) # ' 2426 Chapter 24 24.77. IDENTIFY:! The object is equivalent to two identical capacitors in parallel, where each has the same area A, plate
separation d and dielectric with dielectric constant K.
PA
SET UP:! For each capacitor in the parallel combination, C # 0 .
d
EXECUTE:! (a) The charge distribution on the plates is shown in Figure 24.77.
2
% P A & 2(4.2)P0 (0.120 m)
(b) C # 2 ' 0 ( #
# 2.38 " 10!9 F .
4.5 " 10!4 m
)d*
EVALUATE:! If two of the plates are separated by both sheets of paper to form a capacitor, C # P0 A 2.38 " 10!9 F
,
#
2d
4 smaller by a factor of 4 compared to the capacitor in the problem. Figure 24.77
24.78. IDENTIFY:! As in Problem 24.72, the system is equivalent to two capacitors in parallel. One of the capacitors has
plate separation d, plate area w( L ! h) and air between the plates. The other has the same plate separation d, plate area
wh and dielectric constant K.
K PA
SET UP:! Define K eff by Ceq # eff 0 , where A # wL . For two capacitors in parallel, Ceq # C1 0 C2 .
d
P w( L ! h) K P0 wh P0 wL % Kh h &
EXECUTE:! (a) The capacitors are in parallel, so C # 0
0
#
! ( . This gives
'1 0
d
d
d)
L L* % Kh h &
! (.
K eff # '1 0
L L*
)
(b) For gasoline, with K # 1.95 : L&
L&
1
1
%
%
full: K eff ' h # ( # 1.24 ; full: K eff ' h # ( # 1.48 ;
4
2
4*
2*
)
) 3L &
3
%
full: K eff ' h #
( # 1.71.
4*
4
)
1
L&
1
L&
3
3L &
%
%
%
full: K eff ' h # ( # 9 ;
full: K eff ' h # ( # 17 ; full: K eff ' h #
( # 25.
4
2
4
4*
2*
4*
)
)
)
(d) This kind of fuel tank sensor will work best for methanol since it has the greater range of K eff values. (c) For methanol, with K # 33 : EVALUATE:! When h # 0 , K eff # 1 . When h # L , K eff # K . 25 CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE 25.1. 25.2. IDENTIFY:! I # Q / t .
SET UP:! 1.0 h # 3600 s
EXECUTE:! Q # It # (3.6 A)(3.0)(3600 s) # 3.89 " 104 C.
EVALUATE:! Compared to typical charges of objects in electrostatics, this is a huge amount of charge.
IDENTIFY:! I # Q / t . Use I # n q vd A to calculate the drift velocity vd .
SET UP:! n # 5.8 " 1028 m !3 .
EXECUTE:! (a) I # q # 1.60 "10!19 C . 420 C
Q
#
# 8.75 " 10!2 A.
t
80(60 s) (b) I # n q vd A. This gives vd # I
8.75 "10!2 A
#
# 1.78 "10!6 m s.
28
nqA (5.8 " 10 )(1.60 " 10!19 C)(, (1.3 " 10!3 m)2 ) EVALUATE:! vd is smaller than in Example 25.1, because I is smaller in this problem.
25.3. IDENTIFY:! I # Q / t . J # I / A . J # n q vd
SET UP:! A # (, / 4) D 2 , with D # 2.05 " 10!3 m . The charge of an electron has magnitude 0 e # 1.60 "10!19 C. EXECUTE:! (a) Q # It # (5.00 A)(1.00 s) # 5.00 C. The number of electrons is
(b) J # 25.4. 25.5. I
5.00 A
#
# 1.51" 106 A/m 2 .
(, / 4) D 2 (, / 4)(2.05 "10!3 m) 2 (c) vd # Q
# 3.12 "1019.
e J
1.51"106 A/m 2
#
# 1.11" 10!4 m/s # 0.111 mm/s.
n q (8.5 "1028 m !3 )(1.60 " 10!19 C) EVALUATE:! (a) If I is the same, J # I / A would decrease and vd would decrease. The number of electrons
passing through the light bulb in 1.00 s would not change.
(a) IDENTIFY:! By definition, J = I/A and radius is onehalf the diameter.
SET UP:! Solve for the current: I = JA = J!(D/2)2
EXECUTE:! I = (1.50 " 106 A/m2)(!)[(0.00102 m)/2]2 = 1.23 A
EVALUATE:! This is a realistic current.
(b) IDENTIFY:! The current density is J = nqvd
SET UP:! Solve for the drift velocity: vd = J/nq
EXECUTE:! Since most laboratory wire is copper, we use the value of n for copper, giving
vd # (1.50 "106 A/m 2 ) /[(8.5 " 1028 el/m3)(1.60 " 10!19 C) = 1.1 " 10!4 m/s = 0.11 mm/s
EVALUATE:! This is a typical drift velocity for ordinary currents and wires.
IDENTIFY and SET UP:! Use Eq. (25.3) to calculate the drift speed and then use that to find the time to travel the
length of the wire.
EXECUTE:! (a) Calculate the drift speed vd : J# I
I
4.85 A
# 2#
# 1.469 " 106 A/m 2
2
A ,r
, .1.025 "10!3 m / vd # J
1.469 " 106 A/m 2
#
# 1.079 " 10!4 m/s
n q . 8.5 " 1028 / m3 /.1.602 " 10!19 C / t# L
0.710 m
#
# 6.58 " 103 s # 110 min.
vd 1.079 "10!4 m/s
251 252 Chapter 25 (b) vd # t# I , r 2n q 2
L ,r n q L
#
vd
I t is proportional to r 2 and hence to d 2 where d # 2r is the wire diameter.
2 25.6. % 4.12 mm &
4
t # . 6.58 " 103 s / '
( # 2.66 " 10 s # 440 min.
) 2.05 mm *
(c) EVALUATE:! The drift speed is proportional to the current density and therefore it is inversely proportional to
the square of the diameter of the wire. Increasing the diameter by some factor decreases the drift speed by the
square of that factor.
IDENTIFY:! The number of moles of copper atoms is the mass of 1.00 m3 divided by the atomic mass of copper.
There are N A # 6.023 " 1023 atoms per mole.
SET UP:! The atomic mass of copper is 63.55 g mole, and its density is 8.96 g cm3 . Example 25.1 says there are
8.5 "1028 free electrons per m3 .
EXECUTE:! The number of copper atoms in 1.00 m3 is
(8.96 g cm3 )(1.00 " 106 cm3 m3 )(6.023 " 1023 atoms mole)
# 8.49 "1028 atoms m3 .
63.55 g mole 25.7. EVALUATE:! Since there are the same number of free electrons m3 as there are atoms of copper m3 , the number
of free electrons per copper atom is one.
IDENTIFY and SET UP:! Apply Eq. (25.1) to find the charge dQ in time dt. Integrate to find the total charge in the
whole time interval.
EXECUTE:! (a) dQ # I dt
Q#H 8.0 s
0 .55 A ! . 0.65 A / s / t / dt # 9.55 A / t ! . 0.217 A / s / t :
;
<
2 2 Q # . 55 A /. 8.0 s / ! . 0.217 A / s 2 2 / . 8.0 s / 3 3 8.0 s
0 # 330 C Q 330 C
= 41 A
#
8.0 s
t
EVALUATE:! The current decreases from 55 A to 13.4 A during the interval. The decrease is not linear and the
average current is not equal to (55A + 13.4 A) / 2.
IDENTIFY:! I # Q / t . Positive charge flowing in one direction is equivalent to negative charge flowing in the (b) I # 25.8. opposite direction, so the two currents due to Cl ! and Na + are in the same direction and add.
SET UP:! Na + and Cl! each have magnitude of charge q # 0e
EXECUTE:! (a) Qtotal # (nCl 0 nNa )e # (3.92 " 1016 0 2.68 " 1016 )(1.60 " 10!19 C) # 0.0106 C. Then I# 25.9. Qtotal 0.0106 C
#
# 0.0106 A # 10.6 mA.
t
1.00 s (b) Current flows, by convention, in the direction of positive charge. Thus, current flows with Na 0 toward the
negative electrode.
EVALUATE:! The Cl ! ions have negative charge and move in the direction opposite to the conventional current
direction.
IDENTIFY:! The number of moles of silver atoms is the mass of 1.00 m3 divided by the atomic mass of silver.
There are N A # 6.023 "1023 atoms per mole.
SET UP:! For silver, density # 10.5 " 103 kg/m3 and the atomic mass is M # 107.868 " 10!3 kg/mol.
EXECUTE:! Consider 1 m3 of silver. m # (density)V # 10.5 " 103 kg . n # m M # 9.734 "104 mol and the number of atoms is N # nN A # 5.86 "1028 atoms . If there is one free electron per atom, there are 25.10. 5.86 " 1028 free electrons m3 . This agrees with the value given in Exercise 25.2.
EVALUATE:! Our result verifies that for silver there is approximately one free electron per atom. Exercise 25.6
showed that for copper there is also one free electron per atom.
(a) IDENTIFY:! Start with the definition of resisitivity and solve for E.
SET UP:! E = G J = GI/!r2
EXECUTE:! E = (1.72 " 10!8 & + m)(2.75 A)/[!(0.001025 m)2] = 1.43 " 10!2 V/m Current, Resistance, and Electromotive Force! ! 253 25.11. 25.12. EVALUATE:! The field is quite weak, since the potential would drop only a volt in 70 m of wire.
(b) IDENTIFY:! Take the ratio of the field in silver to the field in copper.
SET UP:! Take the ratio and solve for the field in silver: ES = EC(GS/GC)
EXECUTE:! ES = (0.0143 V/m)[(1.47)/(1.72)] = 1.22 " 10!2 V/m
EVALUATE:! Since silver is a better conductor than copper, the field in silver is smaller than the field in copper.
IDENTIFY:! First use Ohm’s law to find the resistance at 20.0°C; then calculate the resistivity from the resistance.
Finally use the dependence of resistance on temperature to calculate the temperature coefficient of resistance.
SET UP:! Ohm’s law is R = V/I, R = GL/A, R = R0[1 + 2(T – T0)], and the radius is onehalf the diameter.
EXECUTE:! (a) At 20.0°C, R = V/I = (15.0 V)/(18.5 A) = 0.811 &. Using R = GL/A and solving for G gives G =
RA/L = R!(D/2)2/L = (0.811 &)![(0.00500 m)/2]2/(1.50 m) = 1.06 " 10!6 & + m.
(b) At 92.0°C, R = V/I = (15.0 V)/(17.2 A) = 0.872 &. Using R = R0[1 + 2(T – T0)] with T0 taken as 20.0°C, we
have 0.872 & = (0.811 &)[1 + 2 ( 92.0°C – 20.0°C)]. This gives 2 = 0.00105 (C3) !1
EVALUATE:! The results are typical of ordinary metals.
IDENTIFY:! E # G J , where J # I / A . The drift velocity is given by I # n q vd A.
SET UP:! For copper, G # 1.72 "10!8 U + m . n # 8.5 "1028 / m3 . 3.6 A
EXECUTE:! (a) J # I #
# 6.81"105 A/m 2 .
A (2.3 " 10!3 m)2 25.13. (b) E # G J # (1.72 " 10!8 U + m)(6.81 " 105 A/m 2 ) # 0.012 V m.
(c) The time to travel the wire’s length l is
l ln q A (4.0 m)(8.5 " 1028 m3 )(1.6 " 10!19 C)(2.3 " 10!3 m)2
t# #
#
# 8.0 " 104 s.
vd
I
3.6 A
t # 1333 min C 22 hrs!
EVALUATE:! The currents propagate very quickly along the wire but the individual electrons travel very slowly.
IDENTIFY:! E # G J , where J # I / A.
SET UP:! For tungsten G # 5.25 "10!8 U + m and for aluminum G # 2.75 "10!8 U + m.
EXECUTE:! (a) tungsten: E # G J # GI
A # (5.25 "10!8 U + m)(0.820 A)
# 5.16 "10!3 V m.
(, 4)(3.26 " 10!3 m)2 GI 25.14. (2.75 " 10!8 U + m)(0.820 A)
(b) aluminum: E # G J #
#
# 2.70 "10!3 V m.
A
(, 4)(3.26 " 10!3 m) 2
EVALUATE:! A larger electric field is required for tungsten, because it has a larger resistivity.
IDENTIFY:! The resistivity of the wire should identify what the material is.
SET UP:! R = GL/A and the radius of the wire is half its diameter.
EXECUTE:! Solve for G and substitute the numerical values. , .[0.00205 m]/2 / (0.0290 U)
2 G # AR / L # , ( D / 2) 2 R / L # 25.15. = 1.47 " 10!8 & + m
6.50 m
EVALUATE:! This result is the same as the resistivity of silver, which implies that the material is silver.
(a) IDENTIFY:! Start with the definition of resistivity and use its dependence on temperature to find the electric
field.
I
SET UP:! E # G J # G 20 [1 0 2 (T ! T0 )] 2
,r
EXECUTE:! E = (5.25 " 10!8 & + m)[1 + (0.0045/ C! )(120°C – 20°C)](12.5 A)/[!(0.000500 m)2] = 1.21 V/m.
(Note that the resistivity at 120°C turns out to be 7.61 " 10!8 & + m.)
EVALUATE:! This result is fairly large because tungsten has a larger resisitivity than copper.
(b) IDENTIFY:! Relate resistance and resistivity.
SET UP:! R = GL/A = GL/!r2
EXECUTE:! R = (7.61 " 10!8 & + m)(0.150 m)/[!(0.000500 m)2] = 0.0145 &
EVALUATE:! Most metals have very low resistance.
(c) IDENTIFY:! The potential difference is proportional to the length of wire.
SET UP:! V = EL
EXECUTE:! V = (1.21 V/m)(0.150 m) = 0.182 V
EVALUATE:! We could also calculate V # IR # (12.5 A)(0.0145 U) # 0.181 V , in agreement with part (c). 254 Chapter 25 25.16. IDENTIFY:! Apply R # 25.17. GL and solve for L.
A
SET UP:! A # , D 2 / 4 , where D # 0.462 mm .
RA (1.00 U)(, 4)(0.462 " 10!3 m)2
#
# 9.75 m.
EXECUTE:! L #
G
1.72 " 10!8 U + m
EVALUATE:! The resistance is proportional to the length of the wire.
GL
IDENTIFY:! R #
.
A
SET UP:! For copper, G # 1.72 "10!8 U + m . A # , r 2 .
(1.72 "10!8 U + m)(24.0 m)
# 0.125 U
, (1.025 " 10!3 m)2
EVALUATE:! The resistance is proportional to the length of the piece of wire.
GL
GL
IDENTIFY:! R #
#
.
A ,d2 /4
SET UP:! For aluminum, G al # 2.63 " 10!8 U + m . For copper, G c # 1.72 "10!8 U + m.
EXECUTE:! R # 25.18. EXECUTE:! 25.19. G
d 2 # G
Gc
Gc
1.72 "10!8 U + m
R,
# constant , so al # 2 . d c # d al
# (3.26 mm)
# 2.64 mm.
2
d al d c
G al
2.63 "10!8 U + m
4L EVALUATE:! Copper has a smaller resistivity, so the copper wire has a smaller diameter in order to have the same
resistance as the aluminum wire.
IDENTIFY and SET UP:! Use Eq. (25.10) to calculate A. Find the volume of the wire and use the density to
calculate the mass.
EXECUTE:! Find the volume of one of the wires:
GL
GL
R#
so A #
and
A
R volume # AL # G L2
R # .1.72 "10 !8 U + m / . 3.50 m / 0.125 U 2 # 1.686 " 10!6 m3 m # . density /V # .8.9 "103 kg/m3 / .1.686 " 10!6 m3 / # 15 g
25.20. EVALUATE:! The mass we calculated is reasonable for a wire.
GL
IDENTIFY:! R #
.
A
SET UP:! The length of the wire in the spring is the circumference , d of each coil times the number of coils.
EXECUTE:! L # (75), d # (75), (3.50 " 10!2 m) # 8.25 m. A # , r 2 # , d 2 / 4 # , (3.25 "10!3 m)2 / 4 # 8.30 " 10!6 m 2 .
RA (1.74 U)(8.30 " 10!6 m 2 )
#
# 1.75 "10!6 U + m.
L
8.25 m
EVALUATE:! The value of G we calculated is about a factor of 100 times larger than G for copper. The metal of
the spring is not a very good conductor.
GL
IDENTIFY:! R #
.
A
SET UP:! L # 1.80 m, the length of one side of the cube. A # L2 . G# 25.21. 25.22. GL GL G 2.75 "10!8 U + m
# 1.53 " 10!8 U
A
L2 L
1.80 m
EVALUATE:! The resistance is very small because A is very much larger than the typical value for a wire.
IDENTIFY:! Apply RT # R0 (1 0 2 (T ! T0 )).
EXECUTE:! R # # # # SET UP:! Since V # IR and V is the same, RT I 20
#
. For tungsten, 2 # 4.5 "10!3 (C!)!1.
R20 IT Current, Resistance, and Electromotive Force! ! 255 EXECUTE:! The ratio of the current at 203C to that at the higher temperature is (0.860 A) (0.220 A) # 3.909. RT
# 1 0 2 (T ! T0 ) # 3.909 , where T0 # 20!C.
R20
3.909 ! 1
# 6663C.
2
4.5 "10!3 (C!)!1
EVALUATE:! As the temperature increases, the resistance increases and for constant applied voltage the current
decreases. The resistance increases by nearly a factor of four.
IDENTIFY:! Relate resistance to resistivity.
SET UP:! R = GL/A
EXECUTE:! (a) R = GL/A = (0.60 & + m)(0.25 m)/(0.12 m)2 = 10.4 &
(b) R = GL/A = (0.60 & + m)(0.12 m)/(0.12 m)(0.25 m) = 2.4 &
EVALUATE:! The resistance is greater for the faces that are farther apart.
GL
IDENTIFY:! Apply R #
and V # IR .
A
SET UP:! A # , r 2
RA VA (4.50 V), (6.54 " 10!4 m)2
#
#
# 1.37 " 10!7 U + m.
EXECUTE:! G #
L
IL
(17.6 A)(2.50 m)
EVALUATE:! Our result for G shows that the wire is made of a metal with resistivity greater than that of good
metallic conductors such as copper and aluminum.
IDENTIFY and SET UP:! Eq. (25.5) relates the electric field that is given to the current density. V = EL gives the
potential difference across a length L of wire and Eq. (25.11) allows us to calculate R.
EXECUTE:! (a) Eq. (25.5): G # E / J so J # E / G
T # T0 0 25.23. 25.24. 25.25. RT / R20 ! 1 # 203C 0 From Table 25.1 the resistivity for gold is 2.44 " 10!8 U + m.
E
0.49 V/m
# 2.008 " 107 A/m 2
J# #
G 2.44 " 10!8 U + m
I # JA # J, r 2 # . 2.008 " 107 A/m 2 /, . 0.41" 10!3 m / # 11 A
2 (b) V # EL # . 0.49 V/m /. 6.4 m / # 3.1 V
(c) We can use Ohm’s law (Eq. (25.11)): V # IR.
V 3.1 V
R# #
# 0.28 U
I 11 A
EVALUATE:! We can also calculate R from the resistivity and the dimensions of the wire (Eq. 25.10):
!8
G L G L . 2.44 " 10 U + m / . 6.4 m /
R#
# 2#
# 0.28 U, which checks.
2
A ,r
, . 0.42 " 10!3 m /
25.26. IDENTIFY and SET UP:! Use V = EL to calculate E and then G # E / J to calculate G .
V 0.938 V
#
# 1.25 V/m
L 0.750 m
E
1.25 V/m
(b) E # G J so G # #
# 2.84 " 10!8 U + m
J 4.40 " 107 A/m 2
EVALUATE:! This value of G is similar to that for the good metallic conductors in Table 25.1. EXECUTE:! (a) E # 25.27.
IDENTIFY:! Apply R # R0 91 0 2 .T ! T0 / : to calculate the resistance at the second temperature.
;
<
(a) SET UP:! 2 # 0.0004 . C3 /
EXECUTE:! R0 # !1 (Table 25.1). Let T0 be 0.03C and T be 11.53C. R
100.0 U
#
# 99.54 U
1 0 2 .T ! T0 / 1+ 0.0004 . C3 /!1 .11.5 C3 / (b) SET UP:! 2 # !0.0005 . C3 / . !1 / (Table 25.2). Let T0 # 0.03C and T # 25.83C. . / EXECUTE:! R # R0 91 0 2 .T ! T0 / : # 0.0160 U 91+ !0.0005 . C3 / . 25.8 C3 / : # 0.0158 U
;
<
>
?
;
<
EVALUATE:! Nichrome, like most metallic conductors, has a positive 2 and its resistance increases with
temperature. For carbon, 2 is negative and its resistance decreases as T increases.
!1 256 Chapter 25 25.28. IDENTIFY:! RT # R0 [1 0 2 (T ! T0 )]
SET UP:! R0 # 217.3 U . RT # 215.8 U . For carbon, 2 # !0.00050 (C!) !1. (215.8 U / 217.3 U) ! 1
# 13.8 C! . T # 13.8 C! 0 4.0!C # 17.8!C.
!0.00050 (C!) !1
2
EVALUATE:! For carbon, 2 is negative so R decreases as T increases.
GL
IDENTIFY and SET UP:! Apply R #
to determine the effect of increasing A and L.
A
EXECUTE:! (a) If 120 strands of wire are placed side by side, we are effectively increasing the area of the current
carrier by 120. So the resistance is smaller by that factor: R # (5.60 "10!6 U) /120 # 4.67 " 10!8 U.
(b) If 120 strands of wire are placed end to end, we are effectively increasing the length of the wire by 120, and so
R # (5.60 " 10!6 U)120 # 6.72 " 10!4 U.
EVALUATE:! Placing the strands side by side decreases the resistance and placing them end to end increases the
resistance.
IDENTIFY:! When the ohmmeter is connected between the opposite faces, the current flows along its length, but
when the meter is connected between the inner and outer surfaces, the current flows radially outward.
(a) SET UP:! For a hollow cylinder, R = GL/A, where A = !(b2 – a2).
EXECUTE:! T ! T0 # 25.29. 25.30. ( RT / R0 ) ! 1 EXECUTE:! R # G L / A # 25.31. 25.33. 25.34. , .b2 ! a2 / # . 2.75 "10 !8 U + m / (2.50 m) , 9(0.0460 m) 2 ! (0.0320 m) 2 :
;
< = 2.00 " 10!5 & (b) SET UP:! For radial current flow from r = a to r = b, R = (G/2!L) ln(b/a) (Example 25.4)
G
2.75 " 10!8 U + m % 4.60 cm &
!10
EXECUTE:! R #
ln(b / a ) #
ln '
( = 6.35 " 10 &
2, L
2, (2.50 m)
3.20 cm *
)
EVALUATE:! The resistance is much smaller for the radial flow because the current flows through a much smaller
distance and the area through which it flows is much larger.
GL
IDENTIFY:! Use R #
to calculate R and then apply V # IR . P # VI and energy # Pt
A
SET UP:! For copper, G # 1.72 "10!8 U + m . A # , r 2 , where r # 0.050 m.
EXECUTE:! (a) R # 25.32. GL # GL
A # (1.72 " 10!8 U + m)(100 " 103 m)
# 0.219 U . V # IR # (125 A)(0.219 U) # 27.4 V.
, (0.050 m) 2 (b) P # VI # (27.4 V)(125 A) # 3422 W # 3422 J/s and energy # Pt # (3422 J/s)(3600 s) # 1.23 " 107 J.
EVALUATE:! The rate of electrical energy loss in the cable is large, over 3 kW.
IDENTIFY:! When current passes through a battery in the direction from the ! terminal toward the + terminal, the
terminal voltage Vab of the battery is Vab # E ! Ir . Also, Vab # IR, the potential across the circuit resistor.
SET UP:! E # 24.0 V . I # 4.00 A.
E ! Vab 24.0 V ! 21.2 V
EXECUTE:! (a) Vab # E ! Ir gives r #
#
# 0.700 U.
I
4.00 A
V
21.2 V
# 5.30 U.
(b) Vab ! IR # 0 so R # ab #
I
4.00 A
EVALUATE:! The voltage drop across the internal resistance of the battery causes the terminal voltage of the
24.0 V
battery to be less than its emf. The total resistance in the circuit is R 0 r # 6.00 U . I #
# 4.00 A, which
6.00 U
agrees with the value specified in the problem.
IDENTIFY:! V # E ! Ir .
SET UP:! The graph gives V # 9.0 V when I # 0 and I # 2.0 A when V # 0.
EXECUTE:! (a) E is equal to the terminal voltage when the current is zero. From the graph, this is 9.0 V.
(b) When the terminal voltage is zero, the potential drop across the internal resistance is just equal in magnitude to
the internal emf, so rI = E , which gives r = E /I = (9.0 V)/(2.0 A) = 4.5 &.
EVALUATE:! The terminal voltage decreases as the current through the battery increases.
(a) IDENTIFY:! The idealized ammeter has no resistance so there is no potential drop across it. Therefore it acts
like a short circuit across the terminals of the battery and removes the 4.00& resistor from the circuit. Thus the
only resistance in the circuit is the 2.00& internal resistance of the battery.
SET UP:! Use Ohm’s law: I = E /r.
EXECUTE:! I = (10.0 V)/(2.00 &) = 5.00 A. Current, Resistance, and Electromotive Force! ! 257 25.35. 25.36. 25.37. (b) The zeroresistance ammeter is in parallel with the 4.00& resistor, so all the current goes through the ammeter.
If no current goes through the 4.00& resistor, the potential drop across it must be zero.
(c) The terminal voltage is zero since there is no potential drop across the ammeter.
EVALUATE:! An ammeter should never be connected this way because it would seriously alter the circuit!
IDENTIFY:! The terminal voltage of the battery is Vab # E ! Ir. The voltmeter reads the potential difference
between its terminals.
SET UP:! An ideal voltmeter has infinite resistance.
EXECUTE:! (a) Since an ideal voltmeter has infinite resistance, so there would be NO current through the
2.0 U resistor.
(b) Vab # E # 5.0 V; since there is no current there is no voltage lost over the internal resistance.
(c) The voltmeter reading is therefore 5.0 V since with no current flowing there is no voltage drop across either
resistor.
EVALUATE:! This not the proper way to connect a voltmeter. If we wish to measure the terminal voltage of the
battery in a circuit that does not include the voltmeter, then connect the voltmeter across the terminals of the battery.
IDENTIFY:! The sum of the potential changes around the circuit loop is zero. Potential decreases by IR when
going through a resistor in the direction of the current and increases by E when passing through an emf in the
direction from the ! to + terminal.
SET UP:! The current is counterclockwise, because the 16 V battery determines the direction of current flow.
EXECUTE:! 016.0 V ! 8.0 V ! I (1.6 U 0 5.0 U 0 1.4 U 0 9.0 U ) # 0
16.0 V ! 8.0 V
# 0.47 A
I#
1.6 U 0 5.0 U 0 1.4 U 0 9.0 U
(b) Vb 0 16.0 V ! I (1.6 U) # Va , so Va ! Vb # Vab # 16.0 V ! (1.6 U)(0.47 A) # 15.2 V.
(c) Vc 0 8.0 V 0 I (1.4 U 0 5.0 U) # Va so Vac # (5.0 U)(0.47 A) 0 (1.4 U)(0.47 A) 0 8.0 V # 11.0 V.
(d) The graph is sketched in Figure 25.36.
EVALUATE:! Vcb # (0.47 A)(9.0 U) # 4.2 V. The potential at point b is 15.2 V below the potential at point a and
the potential at point c is 11.0 V below the potential at point a, so the potential of point c is
15.2 V ! 11.0 V # 4.2 V above the potential of point b. Figure 25.36
IDENTIFY:! The voltmeter reads the potential difference Vab between the terminals of the battery.
SET UP:! open circuit I # 0. The circuit is sketched in Figure 25.37a. EXECUTE:! Vab # E # 3.08 V Figure 25.37a
SET UP:! switch closed The circuit is sketched in Figure 35.37b.
EXECUTE:!
Vab # E ! Ir # 2.97 V
E ! 2.97 V
r#
I
3.08 V ! 2.97 V
# 0.067 U
r#
1.65 A
Figure 25.37b Vab 2.97 V
#
# 1.80 U.
1.65 A
I
EVALUATE:! When current flows through the battery there is a voltage drop across its internal resistance and its
terminal voltage V is less than its emf. And Vab # IR so R # 258 Chapter 25 25.38. IDENTIFY:! The sum of the potential changes around the loop is zero.
SET UP:! The voltmeter reads the IR voltage across the 9.0 U resistor. The current in the circuit is
counterclockwise because the 16 V battery determines the direction of the current flow.
EXECUTE:! (a) Vbc # 1.9 V gives I # Vbc Rbc # 1.9 V 9.0 U # 0.21 A.
(b) 16.0 V ! 8.0 V # (1.6 U 0 9.0 U 0 1.4 U 0 R)(0.21 A) and R # 5.48 V
# 26.1 U.
0.21 A (c) The graph is sketched in Figure 25.38.
EVALUATE:! In Exercise 25.36 the current is 0.47 A. When the 5.0 U resistor is replaced by the 26.1 U resistor
the current decreases to 0.21 A. Figure 25.38
25.39. (a) IDENTIFY and SET UP:! Assume that the current is clockwise. The circuit is sketched in Figure 25.39a. Figure 25.39a Add up the potential rises and drops as travel clockwise around the circuit.
EXECUTE:! 16.0 V ! I .1.6 U / ! I . 9.0 U / 0 8.0 V ! I .1.4 U / ! I . 5.0 U / # 0
16.0 V 0 8.0 V
24.0 V
#
# 1.41 A, clockwise
9.0 U 0 1.4 U 0 5.0 U 0 1.6 U 17.0 U
EVALUATE:! The 16.0 V battery drives the current clockwise more strongly than the 8.0 V battery does in the
opposite direction.
(b) IDENTIFY and SET UP:! Start at point a and travel through the battery to point b, keeping track of the potential
changes. At point b the potential is Vb .
I# EXECUTE:! Va 0 16.0 V ! I .1.6 U / # Vb Va ! Vb # !16.0 V 0 .1.41 A /.1.6 U /
Vab # !16.0 V 0 2.3 V # !13.7 V (point a is at lower potential; it is the negative terminal)
EVALUATE:! Could also go counterclockwise from a to b:
Va 0 .1.41 A /. 5.0 U / 0 .1.41 A /.1.4 U / ! 8.0 V 0 .1.41 A / . 9.0 U / # Vb
Vab # !13.7 V, which checks.
(c) IDENTIFY and SET UP:! State at point a and travel through the battery to point c, keeping track of the potential
changes.
EXECUTE:! Va 0 16.0 V ! I .1.6 U / ! I . 9.0 U / # Vc
Va ! Vc # !16.0 V 0 .1.41 A /.1.6 U 0 9.0 U / Va c # !16.0 V 0 15.0 V # !1.0 V (point a is at lower potential than point c)
EVALUATE:! Could also go counterclockwise from a to c:
Va 0 .1.41 A /. 5.0 U / 0 .1.41 A /.1.4 U / ! 8.0 V # Vc Vac # !1.0 V, which checks. Current, Resistance, and Electromotive Force! ! 259 (d) Call the potential zero at point a. Travel clockwise around the circuit. The graph is sketched in Figure 25.39b. Figure 25.39b
25.40. Vab
is a constant.
I
SET UP:! (a) The graph is given in Figure 25.40a.
EXECUTE:! (b) No. The graph of Vab versus I is not a straight line so Thyrite does not obey Ohm’s law.
(c) The graph of R versus I is given in Figure 25.40b. R is not constant; it decreases as I increases.
EVALUATE:! Not all materials obey Ohm’s law. IDENTIFY:! Ohm’s law says R # Figure 25.40
25.41. Vab
is a constant.
I
SET UP:! (a) The graph is given in Figure 25.41.
EXECUTE:! (b) The graph of Vab versus I is a straight line so Nichrome obeys Ohm’s law.
15.52 V ! 1.94 V
(c) R is the slope of the graph in part (a). R #
# 3.88 U.
4.00 A ! 0.50 A
EVALUATE:! Vab / I for every I gives the same result for R, R # 3.88 U. IDENTIFY:! Ohm’s law says R # Figure 25.41
25.42. IDENTIFY and SET UP:! For a resistor, P # VI # V 2 / R and V # IR.
V 2 (15.0 V)2
#
# 0.688 U
EXECUTE:! (a) R #
327 W
P
V
15.0 V
(b) I # #
# 21.8 A
R 0.688 U
P 327 W
EVALUATE:! We could also write P # VI to calculate I # #
# 21.8 A.
V 15.0 V 2510 Chapter 25 25.43. IDENTIFY:! The bulbs are each connected across a 120V potential difference.
SET UP:! Use P = V2/R to solve for R and Ohm’s law (I = V/R) to find the current.
EXECUTE:! (a) R = V2/P = (120 V)2/(100 W) = 144 &.
(b) R = V2/P = (120 V)2/(60 W) = 240 &
(c) For the 100W bulb: I = V/R = (120 V)/(144 &) = 0.833 A
For the 60W bulb: I = (120 V)/(240 &) = 0.500 A
EVALUATE:! The 60W bulb has more resistance than the 100W bulb, so it draws less current.
IDENTIFY:! Across 120 V, a 75W bulb dissipates 75 W. Use this fact to find its resistance, and then find the
power the bulb dissipates across 220 V.
SET UP:! P = V2/R, so R = V2/P
EXECUTE:! Across 120 V: R = (120 V)2/(75 W) = 192 &. Across a 220V line, its power will be P = V2/R =
(220 V)2/(192 &) = 252 W.
EVALUATE:! The bulb dissipates much more power across 220 V, so it would likely blow out at the higher
voltage. An alternative solution to the problem is to take the ratio of the powers.
2
2
2
2
P220 V220 / R % V220 & % 220 &
% 220 &
#2
#'
. This gives P220 # (75 W) '
( #'
(
( = 252 W.
P
V120 / R ) V120 * ) 120 *
) 120 *
120 25.44. 25.45. 25.46. 25.47. 25.48. IDENTIFY:! A “100W” European bulb dissipates 100 W when used across 220 V.
(a) SET UP:! Take the ratio of the power in the US to the power in Europe, as in the alternative method for
problem 25.44, using P = V2/R.
2
2
2
2
PUS VUS / R % VUS & % 120 V &
% 120 V &
#2
#'
EXECUTE:!
. This gives PUS # (100 W) '
( #'
( = 29.8 W.
(
PE VE / R ) VE * ) 220 V *
) 220 V * (b) SET UP:! Use P = IV to find the current.
EXECUTE:! I = P/V = (29.8 W)/(120 V) = 0.248 A
EVALUATE:! The bulb draws considerably less power in the U.S., so it would be much dimmer than in Europe.
IDENTIFY:! P # VI . Energy # Pt.
SET UP:! P # (9.0 V)(0.13 A) # 1.17 W
EXECUTE:! Energy # (1.17 W)(1.5 h)(3600 s/h) # 6320 J
EVALUATE:! The energy consumed is proportional to the voltage, to the current and to the time.
P
IDENTIFY and SET UP:! By definition p #
. Use P # VI , E # VL and I # JA to rewrite this expression in terms
LA
of the specified variables.
VI
EXECUTE:! (a) E is related to V and J is related to I, so use P = VI. This gives p #
LA
V
I
# E and # J so p # EJ
L
A
I 2R
(b) J is related to I and G is related to R, so use P # IR 2 . This gives p #
.
LA
GL
J 2 A2 G L 2
I # JA and R #
so p #
GJ
A
LA2
V2
(c) E is related to V and G is related to R, so use P # V 2 / R. This gives p #
.
RLA
GL
E 2 L2 % A & E 2
so p #
.
V # EL and R #
'
(#
A
LA ) G L * G
EVALUATE:! For a given material ( G constant), p is proportional to J 2 or to E 2 .
IDENTIFY:! Calculate the current in the circuit. The power output of a battery is its terminal voltage times the
current through it. The power dissipated in a resistor is I 2 R .
SET UP:! The sum of the potential changes around the circuit is zero.
8.0 V
EXECUTE:! (a) I #
# 0.47 A . Then P9U # I 2 R # (0.47 A)2 (5.0 U) # 1.1 W and
17 U
P9U # I 2 R # (0.47 A)2 (9.0 U) # 2.0 W.
(b) P # EI ! I 2r # (16 V)(0.47 A) ! (0.47 A) 2 (1.6 U) # 7.2 W.
16V
(c) P8V # EI 0 Ir 2 # (8.0 V)(0.47 A) 0 (0.47 A) 2 (1.4 U) # 4.1 W.
EVALUATE:! (d) (b) # (a) 0 (c) . The rate at which the 16.0 V battery delivers electrical energy to the circuit
equals the rate at which it is consumed in the 8.0 V battery and the 5.0 U and 9.0 U resistors. Current, Resistance, and Electromotive Force! ! 2511 25.49. (a) IDENTIFY and SET UP:! P # VI and energy = (power) " (time).
EXECUTE:! P # VI # .12 V /. 60 A / # 720 W The battery can provide this for 1.0 h, so the energy the battery has stored is
U # Pt # . 720 W /. 3600 s / # 2.6 " 106 J
(b) IDENTIFY and SET UP:! For gasoline the heat of combustion is Lc # 46 " 106 J/kg. Solve for the mass m
required to supply the energy calculated in part (a) and use density G # m / V to calculate V.
EXECUTE:! The mass of gasoline that supplies 2.6 "106 J is m # 25.50. 25.51. 25.52. 25.53. 2.6 "106 J
# 0.0565 kg.
46 "106 J/kg The volume of this mass of gasoline is
m 0.0565 kg
% 1000 L &
# 6.3 "10!5 m3 '
# 0.063 L
V# #
3(
G 900 kg/m3
) 1m *
(c) IDENTIFY and SET UP:! Energy = (power) " (time); the energy is that calculated in part (a).
U 2.6 "106 J
# 5800 s # 97 min # 1.6 h.
EXECUTE:! U # Pt , t # #
450 W
P
EVALUATE:! The battery discharges at a rate of 720 W (for 60 A) and is charged at a rate of 450 W, so it takes
longer to charge than to discharge.
IDENTIFY:! The rate of conversion of chemical to electrical energy in an emf is EI . The rate of dissipation of
electrical energy in a resistor R is I 2 R .
SET UP:! Example 25.10 finds that I # 1.2 A for this circuit. In Example 25.9, EI # 24 W and I 2 r # 8 W . In
Example 25.10, I 2 R # 12 W , or 11.5 W if expressed to three significant figures.
EXECUTE:! (a) ! P # EI # (12 V)(1.2 A) # 14.4 W . This is less than the previous value of 24 W.
(b) The energy dissipated in the battery is P # I 2 r # (1.2 A)2 (2.0 U) # 2.9 W. This is less than 8 W, the amount
found in Example (25.9).
(c) The net power output of the battery is 14.4 W ! 2.9 W # 11.5 W . This is the same as the power dissipated in
the 8.0 U resistor.
EVALUATE:! With the larger circuit resistance the current is less and the power input and power consumption are
less.
IDENTIFY:! Some of the power generated by the internal emf of the battery is dissipated across the battery’s
internal resistance, so it is not available to the bulb.
SET UP:! Use P = I2R and take the ratio of the power dissipated in the internal resistance r to the total power.
Pr
I 2r
r
3.5 U
#2
#
#
# 0.123 # 12.3%
EXECUTE:!
PTotal I (r 0 R) r 0 R 28.5 U
EVALUATE:! About 88% of the power of the battery goes to the bulb. The rest appears as heat in the internal
resistance.
IDENTIFY:! The voltmeter reads the terminal voltage of the battery, which is the potential difference across the
appliance. The terminal voltage is less than 15.0 V because some potential is lost across the internal resistance of
the battery.
(a) SET UP:! P = V2/R gives the power dissipated by the appliance.
EXECUTE:! P = (11.3 V)2/(75.0 &) = 1.70 W
(b) SET UP:! The drop in terminal voltage ( E – Vab) is due to the potential drop across the internal resistance r.
Use Ir = E – Vab to find the internal resistance r, but first find the current using P = IV.
EXECUTE:! I = P/V = (1.70 W)/(11.3 V) = 0.151 A. Then Ir = E – Vab gives
(0.151 A)r = 15.0 V – 11.3 V and r = 24.6 &.
EVALUATE:! The full 15.0 V of the battery would be available only when no current (or a very small current) is
flowing in the circuit. This would be the base if the appliance had a resistance much greater than 24.6 &.
IDENTIFY:! Solve for the current I in the circuit. Apply Eq. (25.17) to the specified circuit elements to find the
rates of energy conversion.
SET UP:! The circuit is sketched in Figure 25.53.
EXECUTE:! Compute I:
E ! Ir ! IR # 0
E
12.0 V
I#
#
# 2.00 A
r 0 R 1.0 U 0 5.0 U
Figure 25.53 2512 Chapter 25 (a) The rate of conversion of chemical energy to electrical energy in the emf of the battery is
P # EI # .12.0 V /. 2.00 A / # 24.0 W.
(b) The rate of dissipation of electrical energy in the internal resistance of the battery is
2
P # I 2 r # . 2.00 A / .1.0 U / # 4.0 W.
(c) The rate of dissipation of electrical energy in the external resistor R is P # I 2 R # . 2.00 A / . 5.0 U / # 20.0 W.
2 25.54. EVALUATE:! The rate of production of electrical energy in the circuit is 24.0 W. The total rate of consumption of
electrical energy in the circuit is 4.00 W + 20.0 W = 24.0 W. Equal rate of production and consumption of
electrical energy are required by energy conservation.
IDENTIFY:! The power delivered to the bulb is I 2 R . Energy # Pt .
SET UP: The circuit is sketched in Figure 25.54. rtotal is the combined internal resistance of both batteries. EXECUTE:! (a) rtotal # 0 . The sum of the potential changes around the circuit is zero, so 1.5 V 0 1.5 V ! I (17 U) # 0 . I # 0.1765 A . P # I 2 R # (0.1765 A)2 (17 U) # 0.530 W . This is also
(3.0 V)(0.1765 A) .
(b) Energy # (0.530 W)(5.0 h)(3600 s/h) # 9540 J
0.265 W
P
0.530 W
#
# 0.125 A .
# 0.265 W . P # I 2 R so I #
2
17 U
R
The sum of the potential changes around the circuit is zero, so 1.5 V 0 1.5 V ! IR ! Irtotal # 0 . (c) P # 3.0 V ! (0.125 A)(17 U)
# 7.0 U.
0.125 A
EVALUATE:! When the power to the bulb has decreased to half its initial value, the total internal resistance of the
two batteries is nearly half the resistance of the bulb. Compared to a single battery, using two identical batteries in
series doubles the emf but also doubles the total internal resistance.
rtotal # Figure 25.54
25.55. 2 V
# VI . V # IR .
R
SET UP:! The heater consumes 540 W when V # 120 V . Energy # Pt. IDENTIFY:! P # I 2 R # V2
V 2 (120 V)2
#
# 26.7 U
so R #
540 W
R
P
P 540 W
(b) P # VI so I # #
# 4.50 A
V 120 V
V 2 (110 V) 2
#
# 453 W . P is smaller by a factor of (110 /120)2 .
(c) Assuming that R remains 26.7 U , P #
26.7 U
R
EVALUATE:! (d) With the lower line voltage the current will decrease and the operating temperature will
decrease. R will be less than 26.7 U and the power consumed will be greater than the value calculated in part (c).
m
IDENTIFY:! From Eq. (25.24), G # 2 .
ne @
SET UP:! For silicon, G # 2300 U + m.
EXECUTE:! (a) P # 25.56. EXECUTE:! (a) @ # m
9.11 " 10!31 kg
#
# 1.55 " 10!12 s.
2
16
!3
ne G (1.0 " 10 m )(1.60 " 10!19 C)2 (2300 U + m) EVALUATE:! (b) The number of free electrons in copper (8.5 "1028 m !3 ) is much larger than in pure silicon 25.57. (1.0 "1016 m !3 ). A smaller density of current carriers means a higher resistivity.
GL
(a) IDENTIFY and SET UP:! Use R #
.
A
!3
RA . 0.104 U /, .1.25 "10 m /
#
# 3.65 "10!8 U + m
EXECUTE:! G #
L
14.0 m
2 Current, Resistance, and Electromotive Force! ! 2513 EVALUATE:! This value is similar to that for good metallic conductors in Table 25.1.
(b) IDENTIFY and SET UP:! Use V = EL to calculate E and then Ohm's law gives I.
EXECUTE:! V # EL # .1.28 V/m /.14.0 m / # 17.9 V V 17.9 V
#
# 172 A
R 0.104 U
EVALUATE:! We could do the calculation another way:
E
1.28 V/m
E # G J so J # #
# 3.51"107 A/m 2
G 3.65 "10!8 U + m
I# I # JA # . 3.51" 107 A/m 2 /, .1.25 " 10!3 m / # 172 A, which checks
2 (c) IDENTIFY and SET UP:! Calculate J # I / A or J # E / G and then use Eq. (25.3) for the target variable vd .
EXECUTE:! J # n q vd # nevd
vd # 25.58. J
3.51 " 107 A/m 2
#
# 2.58 " 10!3 m/s # 2.58 mm/s
ne . 8.5 " 10 28 m !3 /.1.602 " 10 !19 C / EVALUATE:! Even for this very large current the drift speed is small.
GL
IDENTIFY:! Use R #
to calculate the resistance of the silver tube. Then I # V / R.
A
SET UP:! For silver, G # 1.47 "10!8 U + m. The silver tube is sketched in Figure 25.58. Since the thickness
T # 0.100 mm is much smaller than the radius, r # 2.00 cm , the cross section area of the silver is 2, rT . The
length of the tube is l # 25.0 m.
V
V
VA V (2, rT ) (12 V)(2, )(2.00 " 10!2 m)(0.100 " 10!3 m)
EXECUTE:! I # #
#
#
#
# 410 A
R Gl A Gl
(1.47 " 10!8 U + m)(25.0 m)
Gl
EVALUATE:! The resistance is small, R # 0.0292 U , so 12.0 V produces a large current. Figure 25.58
25.59. IDENTIFY and SET UP:! With the voltmeter connected across the terminals of the battery there is no current
through the battery and the voltmeter reading is the battery emf; E # 12.6 V.
With a wire of resistance R connected to the battery current I flows and E ! Ir ! IR # 0, where r is the internal
resistance of the battery. Apply this equation to each piece of wire to get two equations in the two unknowns.
EXECUTE:! Call the resistance of the 20.0m piece R1; then the resistance of the 40.0m piece is R2 # 2 R1. E ! I1r ! I1R1 # 0; E ! I 2 r ! I 2 . 2 R1 / # 0; 12.6 V ! . 7.00 A / r ! . 7.00 A / R1 # 0 12.6 V ! . 4.20 A / r ! . 4.20 A / . 2R1 / # 0 Solving these two equations in two unknowns gives R1 # 1.20 U. This is the resistance of 20.0 m, so the resistance 25.60. of one meter is 91.20 U / . 20.0 m / : .1.00 m / # 0.060 U
;
<
EVALUATE:! We can also solve for r and we get r # 0.600 U. When measuring small resistances, the internal
resistance of the battery has a large effect.
IDENTIFY:! Conservation of charge requires that the current is the same in both sections. The voltage drops
across each section add, so R # RCu 0 RAg . The total resistance is the sum of the resistances of each section. GI IR
, so E # , where R is the resistance of a section and L is its length.
A
L
SET UP:! For copper, G Cu # 1.72 "10!8 U + m . For silver, G Ag # 1.47 "10!8 U + m.
E # GJ # 2514 Chapter 25 EXECUTE:! (a) I # RAg # G Ag LAg
AAg # V
V
GL
(1.72 "10!8 U + m)(0.8 m)
#
. RCu # Cu Cu #
# 0.049 U and
R RCu 0 RAg
ACu
(, /4)(6.0 " 10!4 m)2 (1.47 "10!8 U + m)(1.2 m)
5.0 V
# 0.062 U. This gives I #
# 45 A.
0.049 U 0 0.062 U
(, /4)(6.0 "10!4 m)2 The current in the copper wire is 45 A.
(b) The current in the silver wire is 45 A, the same as that in the copper wire or else charge would build up at their
interface.
(45 A)(0.049 U)
IR
(c) ECu # J GCu # Cu #
# 2.76 V m.
0.8 m
LCu
(d) EAg # J G Ag # IRAg
LAg # (45 A)(0.062 U)
# 2.33 V m.
1.2 m (e) VAg # IRAg # (45 A)(0.062 U) # 2.79 V.
EVALUATE:! For the copper section, VCu # IRCu # 2.21 V. Note that VCu 0 VAg # 5.0 V, the voltage applied across
25.61. the ends of the composite wire.
IDENTIFY:! Conservation of charge requires that the current be the same in both sections of the wire.
% EA & % G L &
GI
E # GJ #
. For each section, V # IR # JAR # '
('
( # EL. The voltages across each section add.
A
) G *) A *
SET UP:! A # (, / 4) D 2 , where D is the diameter.
EXECUTE:! (a) The current must be the same in both sections of the wire, so the current in the thin end is 2.5 mA.
G I (1.72 "10!8 U + m)(2.5 " 10!3 A)
(b) E1.6mm # G J #
#
# 2.14 " 10!5 V/m.
A
(, 4)(1.6 " 10!3 m) 2
(c) E0.8mm # G J # GI
A # (1.72 " 10!8 U + m)(2.5 " 10!3 A)
# 8.55 "10!5 V/m . This is 4E1.6mm .
(, 4)(0.80 "10!3 m) 2 (d) V # E1.6mm L1.6 mm 0 E0.8 mm L0.8 mm . V # (2.14 " 10!5 V/m)(1.20 m) 0 (8.55 " 10!5 V/m)(1.80 m) # 1.80 " 10!4 V. 25.62. EVALUATE:! The currents are the same but the current density is larger in the thinner section and the electric field
is larger there.
IDENTIFY:! I # JA.
SET UP:! From Example 25.1, an 18gauge wire has A # 8.17 " 10!3 cm 2 .
EXECUTE:! (a) I # JA # (1.0 "105 A/cm 2 )(8.17 " 10!3 cm 2 ) # 820 A
(b) A # I / J # (1000 A) (1.0 " 106 A/cm2 ) # 1.0 " 10!3 cm 2 . A # , r 2 so 25.63. r # A , # (1.0 "10!3 cm 2 ) , # 0.0178 cm and d # 2r # 0.36 mm .
EVALUATE:! These wires can carry very large currents.
(a) IDENTIFY:! Apply Eq. (25.10) to calculate the resistance of each thin disk and then integrate over the
truncated cone to find the total resistance.
SET UP:
EXECUTE:! The radius of a
truncated cone a distance y
above the bottom is given by
r # r2 0 . y / h / . r1 ! r2 / # r2 0 y V
with V # . r1 ! r2 / / h
Figure 25.63 Consider a thin slice a distance y above the bottom. The slice has thickness dy and radius r. The resistance of the
slice is
G dy G dy
G dy
# 2#
dR #
2
,r
A
, . r2 0 V y /
The total resistance of the cone if obtained by integrating over these thin slices:
h R # H dR # Gh
dy
G9 1
G9 1
1:
!1 :
#
! . r2 0 y V / ? # !
!?
>
, H0 . r2 0 V y /2 , > V
,V ; r2 0 hV r2 <
;
<0 Current, Resistance, and Electromotive Force! ! 2515 But r2 0 hV # r1
9 1 1 : G % h &% r1 ! r2 & G h
('
(#
> ! ?# '
; r2 r1 < , ) r1 ! r2 *) r1r2 * , r1r2
(b) EVALUATE:! Let r1 # r2 # r. Then R # G h / , r 2 # G L / A where A # , r 2 and L # h. This agrees with
Eq. (25.10).
IDENTIFY:! Divide the region into thin spherical shells of radius r and thickness dr. The total resistance is the
sum of the resistances of the thin shells and can be obtained by integration.
SET UP:! I # V / R and J # I / 4, r 2 , where 4, r 2 is the surface area of a shell of radius r.
R# 25.64. G
,V b EXECUTE:! (a) dR #
(b) I # G dr
G b dr
G1
G %1 1& G %b!a&
IR#
2
H r 2 # ! 4, r a # 4, ' a ! b ( # 4, ' ab (.
4, r
4, a
)
*
)
* Vab Vab 4, ab
I
Vab 4, ab
Vab ab
#
and J # #
#
.
2
A G (b ! a )4, r
G (b ! a)r 2
R
G (b ! a) (c) If the thickness of the shells is small, then 4, ab C 4, a 2 is the surface area of the conducting material.
G % 1 1 & G (b ! a)
GL
GL
R#
C
#
, where L # b ! a.
' ! (#
4, ) a b *
4, ab
4, a 2
A
25.65. EVALUATE:! The current density in the material is proportional to 1/ r 2 .
IDENTIFY and SET UP:! Use E # G J to calculate the current density between the plates. Let A be the area of each
plate; then I = JA.
E
=
Q
EXECUTE:! J # and E #
#
G
K P0 KAP0 Thus J # 25.66. Q
Q
and I # JA #
, as was to be shown.
KAP0 G
K P0 G EVALUATE:! C # K P0 A / d and V # Q / C # Qd / K P0 A so the result can also be written as I # VA / d G . The
resistance of the dielectric is R # V / I # d G / A, which agrees with Eq. (25.10).
IDENTIFY:! As the resistance R varies, the current in the circuit also varies, which causes the potential drop across
the internal resistance of the battery to vary.
SET UP:! The largest current will occur when R = 0, and the smallest current will occur when R E #. The largest
terminal voltage will occur when the current is zero (R E #) and the smallest terminal voltage will be when the
current is a maximum (R = 0).
EXECUTE:! (a) As R E #, I E 0, so Vab E E = 15.0 V, which is the largest reading of the voltmeter. When R =
0, the current is largest at (15.0 V)/(4.00 &) = 3.75 A, so the smallest terminal voltage is Vab = E – rI = 15.0 V –
(4.00 &)(3.75 A) = 0.
(b) From part (a), the maximum current is 3.75 A when R = 0, and the minimum current is 0.00 A when R E #.
(c) The graphs are sketched in Figure 25.66.
EVALUATE:! Increasing the resistance R increases the terminal voltage, but at the same time it decreases the
current in the circuit. Figure 25.66
25.67. IDENTIFY:! Apply R # GL .
A
SET UP:! For mercury at 20!C , G # 9.5 "10!7 U + m , 2 # 0.00088 (C!)!1 and V # 18 "10!5 (C!) !1.
EXECUTE:! (a) R # GL
A # (9.5 " 10!7 U + m)(0.12 m)
# 0.057 U.
(, 4)(0.0016 m) 2 2516 Chapter 25 (b) G (T ) # G 0 (1 0 2BT ) gives G (603 C) # (9.5 "10!7 U + m)(1 0 (0.00088 (C3)!1 )(40 C!) # 9.83 " 10!7 U + m, so BG # 3.34 " 10!8 U + m.
(c) BV # V V0 BT gives ABL # A ( V L0 BT ) . Therefore 25.68. BL # V L0 BT # (18 " 10!5 (C3) !1 )(0.12 m)(40 C3) # 8.64 " 10!4 m # 0.86 mm. The cross sectional area of the
mercury remains constant because the diameter of the glass tube doesn't change. All of the change in volume of the
mercury must be accommodated by a change in length of the mercury column.
GL
LBG GBL
0
(d) R #
gives BR #
.
A
A
A
(3.34 "10!8 U + m)(0.12 m) (95 " 10!8 U + m)(0.86 " 10!3 m)
BR #
0
# 2.40 " 10!3 U.
(, /4)(0.0016 m)2
(, /4)(0.0016 m)2
EVALUATE:! (e) From Equation (25.12),
% (0.057 U 0 2.40 " 10!3 U) &
%
&
2 # 1 ' R ! 1( # 1 '
! 1( # 1.1" 10!3 (C3)!1.
0.057 U
BT ) R0
* 40 C! )
*
This value is 25% greater than the temperature coefficient of resistivity and the length increase is important.
IDENTIFY:! Consider the potential changes around the circuit. For a complete loop the sum of the potential
changes is zero.
SET UP:! There is a potential drop of IR when you pass through a resistor in the direction of the current.
8.0 V ! 4.0 V
EXECUTE:! (a) I #
# 0.167 A . Vd 0 8.00 V ! I (0.50 U 0 8.00 U) # Va , so
24.0 U
Vad # 8.00 V ! (0.167 A) (8.50 U ) # 6.58 V.
(b) The terminal voltage is Vbc # Vb ! Vc . Vc 0 4.00 V 0 I (0.50 U) # Vb and Vbc # 0 4.00 V 0 (0.167 A) (0.50 U) # 0 4.08 V.
(c) Adding another battery at point d in the opposite sense to the 8.0 V battery produces a counterclockwise current
10.3 V ! 8.0 V 0 4.0 V
# 0.257 A . Then Vc 0 4.00 V ! I (0.50 U) # Vb and
with magnitude I #
24.5 U 25.69. Vbc # 4.00 V ! (0.257 A) (0.50 U) # 3.87 V.
EVALUATE:! When current enters the battery at its negative terminal, as in part (c), the terminal voltage is less
than its emf. When current enters the battery at the positive terminal, as in part (b), the terminal voltage is greater
than its emf.
IDENTIFY:! In each case write the terminal voltage in terms of E , I, and r. Since I is known, this gives two
equations in the two unknowns E and r.
SET UP:! The battery with the 1.50 A current is sketched in Figure 25.69a.
Vab # 8.4 V
Vab # E ! Ir E ! .1.50 A / r # 8.4 V Figure 25.69a The battery with the 3.50 A current is sketched in Figure 25.69b.
Vab # 9.4 V
Vab # E 0 Ir E 0 . 3.5 A / r # 9.4 V Figure 25.69b
EXECUTE:! (a) Solve the first equation for E and use that result in the second equation:
E # 8.4 V 0 .1.50 A / r 8.4 V 0 .1.50 A / r 0 . 3.50 A / r # 9.4 V . 5.00 A / r # 1.0 V so r # 1.0 V
# 0.20 U
5.00 A Current, Resistance, and Electromotive Force! ! 2517 (b) Then E # 8.4 V 0 .1.50 A / r # 8.4 V 0 .1.50 A / . 0.20 U / # 8.7 V
EVALUATE:! When the current passes through the emf in the direction from ! to 0 , the terminal voltage is less
than the emf and when it passes through from 0 to !, the terminal voltage is greater than the emf.
25.70. IDENTIFY:! V # IR . P # I 2 R.
SET UP:! The total resistance is the resistance of the person plus the internal resistance of the power supply.
14 "103 V
V
EXECUTE:! (a) I #
#
# 1.17 A
Rtot 10 "103 U 0 2000 U
(b) P # I 2 R # (1.17 A)2 (10 " 103 U) # 1.37 " 104 J # 13.7 kJ 14 "103 V
V
#
# 14 " 106 U. The resistance of the power supply would need to be
I 1.00 "10!3 A
14 " 106 U ! 10 " 103 U # 14 " 106 U # 14 MU.
EVALUATE:! The current through the body in part (a) is large enough to be fatal.
GL
IDENTIFY:! R #
. V # IR . P # I 2 R.
A
SET UP:! The area of the end of a cylinder of radius r is , r 2 .
(5.0 U + m)(1.6 m)
EXECUTE:! (a) R #
# 1.0 "103 U
, (0.050 m) 2
(c) Rtot # 25.71. (b) V # IR # (100 "10!3 A)(1.0 " 103 U) # 100 V 25.72. (c) P # I 2 R # (100 " 10!3 A) 2 (1.0 " 103 U) # 10 W
EVALUATE:! The resistance between the hands when the skin is wet is about a factor of ten less than when the
skin is dry (Problem 25.70).
IDENTIFY:! The cost of operating an appliance is proportional to the amount of energy consumed. The energy
depends on the power the item consumes and the length of time for which it is operated.
SET UP:! At a constant power, the energy is equal to Pt, and the total cost is the cost per kilowatthour (kWh)
times the time the energy (in kWh).
EXECUTE:! (a) Use the fact that 1.00 kWh = (1000 J/s)(3600 s) = 3.60 " 106 J, and one year contains
3.156 "107 s. % 3.156 "107 s & % $0.120 &
('
( = $78.90
6
1 yr
)
* ) 3.60 "10 J * . 75 J/s / ' (b) At 8 h/day, the refrigerator runs for 1/3 of a year. Using the same procedure as above gives 1 & % 3.156 "107 s & % $0.120
('
('
6
1 yr
) 3 *)
* ) 3.60 "10 . 400 J/s / %
' 25.73. &
( = $140.27
J* EVALUATE:! Electric lights can be a substantial part of the cost of electricity in the home if they are left on for a
long time!
IDENTIFY:! Set the sum of the potential rises and drops around the circuit equal to zero and solve for I.
SET UP:! The circuit is sketched in Figure 25.73.
EXECUTE:!
E ! IR ! V # 0
E ! IR ! 2 I ! V I 2 # 0 V I 2 0 .R 02 /I ! E # 0
Figure 25.73
2
The quadratic formula gives I # .1/ 2 V / 9 ! . R 0 2 / 6 . R 0 2 / 0 4 V E :
>
?
;
<
I must be positive, so take the + sign
2
I # .1/ 2 V / 9 ! . R 0 2 / 0 . R 0 2 / 0 4 V E :
>
?
;
<
I # !2.692 A 0 4.116 A # 1.42 A
EVALUATE:! For this I the voltage across the thermistor is 8.0 V. The voltage across the resistor must then be
12.6 V ! 8.0 V # 4.6 V, and this agrees with Ohm's law for the resistor. 2518 Chapter 25 25.74. (a) IDENTIFY:! The rate of heating (power) in the cable depends on the potential difference across the cable and
the resistance of the cable.
SET UP:! The power is P = V2/R and the resistance is R = GL/A. The diameter D of the cable is twice its radius.
V2
V2
AV 2 , r 2V 2
#
#
#
. The electric field in the cable is equal to the potential difference across its
P#
GL
R . G L / A/ G L ends divided by the length of the cable: E = V/L.
EXECUTE:! Solving for r and using the resistivity of copper gives
r# 25.75. (50.0 W) .1.72 " 10!8 U + m / (1500 m)
PG L
#
= 9.21 " 10–5 m. D = 2r = 0.184 mm
,V 2
, (220.0 V) 2 (b) SET UP:! E = V/L
EXECUTE:! E = (220 V)/(1500 m) = 0.147 V/m
EVALUATE:! This would be an extremely thin (and hence fragile) cable.
IDENTIFY:! The ammeter acts as a resistance in the circuit loop. Set the sum of the potential rises and drops
around the circuit equal to zero.
(a) SET UP:! The circuit with the ammeter is sketched in Figure 25.75a.
EXECUTE:!
E
IA #
r 0 R 0 RA E # I A . r 0 R 0 RA / Figure 25.75a
SET UP:! The circuit with the ammeter removed is sketched in Figure 25.75b.
EXECUTE:!
E
I#
R0r Figure 25.75b Combining the two equations gives
RA &
%1&
%
I #'
( I A . r 0 R 0 RA / # I A ' 1 0
(
) R0r*
) r0R*
(b) Want I A # 0.990 I . Use this in the result for part (a).
R&
%
I # 0.990 I '1 0 A (
) r0R*
%R &
0.010 # 0.990 ' A (
)r0R*
RA # . r 0 R /. 0.010 / 0.990 / # . 0.45 U 0 3.80 U / . 0.010 / 0.990/ # 0.0429 U
(c) I ! I A # E
E
!
r 0 R r 0 R 0 RA % r 0 R 0 RA ! r ! R &
ERA
I ! IA # E '
' . r 0 R /. r 0 R 0 R / ( # . r 0 R /. r 0 R 0 R / .
(
A*
A
)
EVALUATE:! The difference between I and I A increases as RA increases. If RA is larger than the value
25.76. calculated in part (b) then I A differs from I by more than 1.0%.
IDENTIFY:! Since the resistivity is a function of the position along the length of the cylinder, we must integrate to
find the resistance.
(a) SET UP:! The resistance of a crosssection of thickness dx is dR = Gdx/A.
EXECUTE:! Using the given function for the resistivity and integrating gives
R#H G dx
A #H L 0 . a 0 bx / dx # aL 0 bL / 3.
2 ,r 2 3 ,r 2 Current, Resistance, and Electromotive Force! ! 2519 Now get the constants a and b: G ( 0) = a = 2.25 " 10!8 & + m and
G ( L) = a + bL2 gives 8.50 " 10!8 & + m = 2.25 " 10!8 & + m + b(1.50 m)2
which gives b = 2.78 " 10!8 &/m. Now use the above result to find R.
R# . 2.25 "10 !8 U + m / (1.50 m) 0 . 2.78 "10!8 U / m / (1.50 m)3 / 3 , (0.0110 m) 2 = 1.71 " 10–4 & = 171 µ& (b) IDENTIFY:! Use the definition of resistivity to find the electric field at the midpoint of the cylinder, where x =
L/2.
SET UP:! E = GJ. Evaluate the resistivity, using the given formula, for x = L/2.
2
9
:
G I ;a 0 b . L / 2/ < I
EXECUTE:! At the midpoint, x = L/2, giving E =
.
#
2
2
,r
,r 9 2.25 "10!8 U + m 0 . 2.78 " 10!8 U / m / (0.750 m)2 : (1.75 A)
<
E= ;
= 1.76 " 10!4 V/m
, (0.0110 m) 2
(c) IDENTIFY:! For the first segment, the result is the same as in part (a) except that the upper limit of the integral
is L/2 instead of L.
SET UP:! Integrating using the upper limit of L/2 gives R1 # 25.77. a . L / 2 / 0 (b / 3) . L3 /8 / , r2 . EXECUTE:! Substituting the numbers gives
. 2.25 " 10!8 U + m / (0.750 m) 0 (2.78 " 10!8 U / m)/3. (1.50 m)3 /8/ = 5.47 " 10!5 U
R1 =
, (0.0110 m) 2
The resistance R2 of the second half is equal to the total resistance minus the resistance of the first half.
R2 = R – R1 = 1.71 " 10!4 & – 5.47 " 10!5 & = 1.16 " 10!4 &
EVALUATE:! The second half has a greater resistance than the first half because the resistance increases with
distance along the cylinder.
IDENTIFY:! The power supplied to the house is P # VI . The rate at which electrical energy is dissipated in the
GL
wires is I 2 R, where R #
.
A
SET UP:! For copper, G # 1.72 " 10!8 U + m.
EXECUTE:! (a) The line voltage, current to be drawn, and wire diameter are what must be considered in
household wiring.
P 4200 W
# 35 A, so the 8gauge wire is necessary, since it can carry up to 40 A.
(b) P # VI gives I # #
V
120 V
(c) P # I 2 R # I 2 G L (35 A)2 (1.72 " 10!8 U + m) (42.0 m)
#
# 106 W.
A
(, 4) (0.00326 m) 2 I 2 G L (35 A)2 (1.72 " 10!8 U + m) (42 m)
#
# 66 W . The decrease in energy
A
(, 4) ) (0.00412 m)2
consumption is BE # BPt # ( 40 W) (365 days/yr) (12 h/day) # 175 kWh/yr and the savings is (d) If 6gauge wire is used, P # 25.78. (175 kWh/yr) ($0.11 kWh) # $19.25 per year.
EVALUATE:! The cost of the 4200 W used by the appliances is $2020. The savings is about 1%.
V
IDENTIFY:! RT # R0 (1 0 2 [T ! T0 ]) . R # . P # VI .
I
SET UP:! When the temperature increases the resistance increases and the current decreases.
VV
EXECUTE:! (a)
# (1 0 2 [T ! T0 ]) . I 0 # IT (1 0 2 [T ! T0 ]) .
IT I 0
I 0 ! IT
1.35 A ! 1.23 A
#
# 217 C! . T # 20!C 0 2173C # 237!C
(1.23 A)(4.5 " 10!4 (C!)!1 )
2 IT
(b) (i) P # VI # (120 V)(1.35 A) # 162 W (ii) P # (120 V)(1.23 A) # 148 W
T ! T0 # 25.79. EVALUATE:! P # V 2 / R shows that the power dissipated decreases when the resistance increases.
(a) IDENTIFY:! Set the sum of the potential rises and drops around the circuit equal to zero and solve for the
resulting equation for the current I. Apply Eq. (25.17) to each circuit element to find the power associated with it. 2520 Chapter 25 SET UP:! The circuit is sketched in Figure 25.79. EXECUTE:!
E1 ! E2 ! I . r1 0 r2 0 R / # 0 I# E1 ! E2
r1 0 r2 0 R 12.0 V ! 8.0 V
1.0 U 0 1.0 U 0 8.0 U
I # 0.40 A I# Figure 25.79
(b) P # I 2 R 0 I 2 r1 0 I 2 r2 # I 2 . R 0 r1 0 r2 / # . 0.40 A / . 8.0 U 0 1.0 U 0 1.0 U /
2 P # 1.6 W
(c) Chemical energy is converted to electrical energy in a battery when the current goes through the battery from
the negative to the positive terminal, so the electrical energy of the charges increases as the current passes through.
This happens in the 12.0 V battery, and the rate of production of electrical energy is
P # E1I # .12.0 V /. 0.40 A / # 4.8 W.
(d) Electrical energy is converted to chemical energy in a battery when the current goes through the battery from
the positive to the negative terminal, so the electrical energy of the charges decreases as the current passes through.
This happens in the 8.0 V battery, and the rate of consumption of electrical energy is
P # E2 I # . 8.0 V /. 0.40 V / # 3.2 W. 25.80. (e) EVALUATE:! Total rate of production of electrical energy = 4.8 W. Total rate of consumption of electrical
energy = 1.6 W + 3.2 W = 4.8 W, which equals the rate of production, as it must.
GL
IDENTIFY:! Apply R #
for each material. The total resistance is the sum of the resistances of the rod and the
A
wire. The rate at which energy is dissipated is I 2 R.
SET UP:! For steel, G # 2.0 "10!7 U + m . For copper, G # 1.72 " 10!8 U + m.
EXECUTE:! (a) Rsteel # RCu # GL
A # GL
A # (2.0 " 10!7 U + m) (2.0 m)
# 1.57 " 10!3 U and
(, 4) (0.018 m) 2 !8 (1.72 " 10 U + m) (35 m)
# 0.012 U . This gives
(, 4) (0.008 m)2 V # IR # I ( Rsteel 0 RCu ) # (15000 A) (1.57 " 10!3 U 0 0.012 U) # 204 V.
(b) E # Pt # I 2 Rt # (15000 A) 2 (0.0136 U) (65 " 10!6 s) # 199 J. 25.81. EVALUATE:! I 2 R is large but t is very small, so the energy deposited is small. The wire and rod each have a
mass of about 1 kg, so their temperature rise due to the deposited energy will be small.
IDENTIFY and SET UP:! The terminal voltage is Vab # P ! Ir # IR , where R is the resistance connected to the battery. During the charging the terminal voltage is Vab # P 0 Ir . P # VI and energy is E # Pt . I 2 r is the rate
at which energy is dissipated in the internal resistance of the battery.
EXECUTE:! (a) Vab # P 0 Ir # 12.0 V 0 (10.0 A) (0.24 U) # 14.4 V.
(b) E # Pt # IVt # (10 A) (14.4 V) (5) (3600 s) # 2.59 " 106 J.
(c) Ediss # Pdisst # I 2 rt # (10 A)2 (0.24 U) (5) (3600 s) # 4.32 " 105 J.
(d) Discharged at 10 A: I # E
r0R IR# E ! Ir 12.0 V ! (10 A) (0.24 U)
#
# 0.96 U.
I
10 A (e) E # Pt # IVt # (10 A) (9.6 V) (5) (3600 s) # 1.73 " 106 J.
(f) Since the current through the internal resistance is the same as before, there is the same energy dissipated as in
(c): Ediss # 4.32 " 105 J.
(g) Part of the energy originally supplied was stored in the battery and part was lost in the internal resistance. So
the stored energy was less than what was supplied during charging. Then when discharging, even more energy is
lost in the internal resistance, and only what is left is dissipated by the external resistor. Current, Resistance, and Electromotive Force! ! 2521 25.82. IDENTIFY and SET UP:! The terminal voltage is Vab # P ! Ir # IR, where R is the resistance connected to the battery. During the charging the terminal voltage is Vab # P 0 Ir . P # VI and energy is E # Pt . I 2 r is the rate at
which energy is dissipated in the internal resistance of the battery.
EXECUTE:! (a) Vab # E 0 Ir # 12.0 V 0 ( 30 A) (0.24 U) # 19.2 V.
(b) E # Pt # IVt # (30 A) (19.2 V) (1.7) (3600 s) # 3.53 " 106 J.
(c) Ediss # Pdisst # I 2 Rt # (30 A) 2 (0.24 U) (1.7) (3600 s) # 1.32 " 106 J.
(d) Discharged at 30 A: I # 25.83. 25.84. E
E ! Ir 12.0 V ! (30 A) (0.24 U)
#
# 0.16 U.
gives R #
r0R
I
30 A (e) E # Pt # I 2 Rt # (30 A)2 (0.16 U) (1.7) (3600 s) # 8.81 " 105 J.
(f) Since the current through the internal resistance is the same as before, there is the same energy dissipated as in
(c): Ediss # 1.32 " 106 J.
(g) Again, part of the energy originally supplied was stored in the battery and part was lost in the internal
resistance. So the stored energy was less than what was supplied during charging. Then when discharging, even
more energy is lost in the internal resistance, and what is left is dissipated over the external resistor. This time, at a
higher current, much more energy is lost in the internal resistance. Slow charging and discharging is more energy
efficient.
IDENTIFY and SET UP:! Follow the steps specified in the problem.
q a
#.
EXECUTE:! (a) W F # ma # q E gives
mE
 q  aL
#
(b) If the electric field is constant, Vbc # EL and
m Vbc
(c) The free charges are “left behind” so the left end of the rod is negatively charged, while the right end is
positively charged. Thus the right end, point c, is at the higher potential.
V  q  (1.0 " 10!3 V) (1.6 " 10!19 C)
#
# 3.5 " 108 m/s 2 .
(d) a # bc
mL
(9.11 " 10!31 kg) (0.50 m)
EVALUATE:! (e) Performing the experiment in a rotational way enables one to keep the experimental apparatus in
a localized area—whereas an acceleration like that obtained in (d), if linear, would quickly have the apparatus
moving at high speeds and large distances. Also, the rotating spool of thin wire can have many turns of wire and
the total potential is the sum of the potentials in each turn, the potential in each turn times the number of turns.
IDENTIFY:! E ! IR ! V # 0
e
SET UP:! With T # 293 K ,
# 39.6 V !1.
kT
EXECUTE:! (a) E # IR 0 V gives 2.00 V # I (1.0 U) 0 V . Dropping units and using the expression given in the problem for I, this becomes 2.00 # I S[exp(eV kT ) ! 1] 0 V.
(b) For I S # 1.50 " 10!3 A and T # 293 K, 1333 # exp [39.6 V ] ! 1 0 667V . Trial and error shows that the righthand
side (rhs) above, for specific V values, equals 1333 V when V # 0.179 V. The current then is just I # I S[exp(39.6 V ) ! 1] # (1.5 " 10!3 A) . exp([39.6][0.179]) ! 1]/ # 1.80 A.
25.85. EVALUATE:! The voltage across the resistor R is 1.80 V. The diode does not obey Ohm’s law.
GL
IDENTIFY:! Apply R #
to find the resistance of a thin slice of the rod and integrate to find the total R.
A
V # IR . Also find R ( x ), the resistance of a length x of the rod.
SET UP:! E ( x) # G ( x) J
EXECUTE:! (a) dR # R# G0
A G dx
A L H exp [! x L] dx # 0 # G0
A G 0 exp[! x L] dx
A L
[! L exp[! x L]]0 # rather than L in the integration, R ( x) #
(b) E ( x) # G ( x) J # G0 L
A I G 0e ! x L
V e! x L
# 0 !1 .
A
L .1 ! e / so G0 L
A (1 ! e !1 ) and I # .1 ! e / .
!x/ L V0
V0 A
#
. With an upper limit of x
R G 0 L(1 ! e !1 ) 2522 Chapter 25
!x/ L
%
& % G0 L &
! !1
V0 A
!x/ L
(c) V # V0 ! IR ( x) . V # V0 ! '
/ # V0 (e(1 ! e!e) )
( .1 ! e
1
!1 ( '
) G0 L[1 ! e ] * ) A *
(d) Graphs of resistivity, electric field and potential from x # 0 to L are given in Figure 25.85. Each quantity is
given in terms of the indicated unit.
EVALUATE:! The current is the same at all points in the rod. Where the resistivity is larger the electric field must
be larger, in order to produce the same current density. Figure 25.85
25.86. IDENTIFY:! The power output of the source is VI # (E ! Ir ) I .
SET UP:! The shortcircuit current is I short circuit # E / r. 1E 1
dP
# E ! 2 Ir # 0 for maximum power output and I P max #
# I short circuit .
2r 2
dI
E
1E
(b) For the maximum power output of part (a),! I #
. r 0 R # 2r and R # r.
#
r0R 2r
EXECUTE:! (a) P # EI ! I 2 r , so 2 E2
%E&
Then, P # I 2 R # ' ( r # .
4r
) 2r * 25.87. EVALUATE:! When R is smaller than r, I is large and the I 2 r losses in the battery are large. When R is larger than
r, I is small and the power output EI of the battery emf is small.
1 dG
IDENTIFY:! Use 2 # !n / T in 2 #
to get a separable differential equation that can be integrated.
G dT
SET UP:! For carbon, G # 3.5 "10!5 U + m and 2 # !5 " 10!4 (K) !1.
EXECUTE:! (a) 2 # 1 % dG &
n
ndT d G
a
#
I ln (T ! n ) # ln( G ) I G # n .
'
(#! I
G ) dT *
T
T
G
T (b) n # !2T # ! (! 5 " 10!4 (K) !1 ) (293 K) # 0.15. a
I a # GT n # (3.5 " 10!5 U + m) (293 K)0.15 # 8.0 " 10!5 U + m + K 0.15 .
Tn
8.0 " 10!5
(c) T # ! 1963C # 77 K : G #
# 4.3 " 10!5 U + m.
(77 K)0.15 G# 8.0 " 10!5
# 3.2 " 10!5 U + m.
(573 K)0.15
EVALUATE:! 2 is negative and decreases as T decreases, so G changes more rapidly with temperature at lower
temperatures.
T # ! 3003C # 573 K : G # 26 DIRECTCURRENT CIRCUITS 26.1. 26.2. 26.3. IDENTIFY: The newlyformed wire is a combination of series and parallel resistors.
SET UP:! Each of the three linear segments has resistance R/3. The circle is two R/6 resistors in parallel.
EXECUTE:! The resistance of the circle is R/12 since it consists of two R/6 resistors in parallel. The equivalent
resistance is two R/3 resistors in series with an R/6 resistor, giving Requiv = R/3 + R/3 + R/12 = 3R/4.
EVALUATE:! The equivalent resistance of the original wire has been reduced because the circle’s resistance is less
than it was as a linear wire.
IDENTIFY:! It may appear that the meter measures X directly. But note that X is in parallel with three other
resistors, so the meter measures the equivalent parallel resistance between ab.
SET UP:! We use the formula for resistors in parallel.
EXECUTE:! 1/(2.00 &) = 1/X + 1/(15.0 &) + 1/(5.0 &) + 1/(10.0 &), so X = 7.5 &.
EVALUATE:! X is greater than the equivalent parallel resistance of 2.00 &.
(a) IDENTIFY:! Suppose we have two resistors in parallel, with R1 1 R2 .
SET UP:! The equivalent resistance is
EXECUTE:! It is always true that 1
1
1
#0
Req R1 R2 1
1
1
1
1
0
5 . Therefore
5
and Req 1 R1 .
R1 R2 R1
Req R1 EVALUATE:! The equivalent resistance is always less than that of the smallest resistor.
(b) IDENTIFY:! Suppose we have N resistors in parallel, with R1 1 R2 1 % 1 RN .
SET UP:! The equivalent resistance is
EXECUTE:! It is always true that
26.4. 1
1
1
1
#0
0% 0
Req R1 R2
RN 1
1
1
1
1
1
0
0% 0
5 . Therefore
5
and Req 1 R1 .
Req R1
R1 R2
RN R1 EVALUATE:! The equivalent resistance is always less than that of the smallest resistor.
IDENTIFY:! For resistors in parallel the voltages are the same and equal to the voltage across the equivalent
resistance.
1
1
1
SET UP:! V # IR .
#0
.
Req R1 R2
!1 1&
%1
EXECUTE:! (a) Req # '
0
( # 12.3 U.
32 U 20 U *
)
V
240 V
(b) I #
#
# 19.5 A.
Req 12.3 U
V 240 V
V 240 V
#
# 7.5 A; I 20 U # #
# 12 A.
R 32 U
R 20 U
EVALUATE:! More current flows through the resistor that has the smaller R.
IDENTIFY:! The equivalent resistance will vary for the different connections because the seriesparallel
combinations vary, and hence the current will vary.
SET UP:! First calculate the equivalent resistance using the seriesparallel formulas, then use Ohm’s law (V = RI)
to find the current.
EXECUTE:! (a) 1/R = 1/(15.0 &) + 1/(30.0 &) gives R = 10.0 &. I = V/R = (35.0 V)/(10.0 &) = 3.50 A.
(b) 1/R = 1/(10.0 &) + 1/(35.0 &) gives R = 7.78 &. I = (35.0 V)/(7.78 &) = 4.50 A
(c) 1/R = 1/(20.0 &) + 1/(25.0 &) gives R = 11.11 &, so I = (35.0 V)/(11.11 &) = 3.15 A.
(c) I 32 U # 26.5. 261 262 26.6. 26.7. 26.8. Chapter 26 (d) From part (b), the resistance of the triangle alone is 7.78 &. Adding the 3.00& internal resistance of the battery
gives an equivalent resistance for the circuit of 10.78 &. Therefore the current is I = (35.0 V)/(10.78 &) = 3.25 A
EVALUATE:! It makes a big difference how the triangle is connected to the battery.
IDENTIFY:! The potential drop is the same across the resistors in parallel, and the current into the parallel
combination is the same as the current through the 45.0& resistor.
(a) SET UP:! Apply Ohm’s law in the parallel branch to find the current through the 45.0& resistor. Then apply
Ohm’s law to the 45.0& resistor to find the potential drop across it.
EXECUTE:! The potential drop across the 25.0& resistor is V25 = (25.0 &)(1.25 A) = 31.25 V. The potential drop
across each of the parallel branches is 31.25 V. For the 15.0& resistor: I15 = (31.25 V)/(15.0 &) = 2.083 A. The
resistance of the 10.0& + 15.0 & combination is 25.0 &, so the current through it must be the same as the current
through the upper 25.0 & resistor: I10+15 = 1.25 A. The sum of currents in the parallel branch will be the current
through the 45.0& resistor.
ITotal = 1.25 A + 2.083 A + 1.25 A = 4.58 A
Apply Ohm’s law to the 45.0 & resistor: V45 = (4.58 A)(45.0 &) = 206 V
(b) SET UP:! First find the equivalent resistance of the circuit and then apply Ohm’s law to it.
EXECUTE:! The resistance of the parallel branch is 1/R = 1/(25.0 &) + 1/(15.0 &) + 1/(25.0 &), so R = 6.82 &.
The equivalent resistance of the circuit is 6.82 & + 45.0 & + 35.00 & = 86.82 &. Ohm’s law gives VBat =
(86.62 &)(4.58 A) = 398 V.
EVALUATE:! The emf of the battery is the sum of the potential drops across each of the three segments (parallel
branch and two series resistors).
IDENTIFY:! First do as much seriesparallel reduction as possible.
SET UP:! The 45.0& and 15.0& resistors are in parallel, so first reduce them to a single equivalent resistance.
Then find the equivalent series resistance of the circuit.
EXECUTE:! 1/Rp = 1/(45.0 &) + 1/(15.0 &) and Rp = 11.25 &. The total equivalent resistance is
18.0 & + 11.25 & + 3.26 & = 32.5 &. Ohm’s law gives I = (25.0 V)/(32.5 &) = 0.769 A.
EVALUATE:! The circuit appears complicated until we realize that the 45.0& and 15.0& resistors are in parallel.
IDENTIFY:! Eq.(26.2) gives the equivalent resistance of the three resistors in parallel. For resistors in parallel, the
voltages are the same and the currents add.
(a) SET UP:! The circuit is sketched in Figure 26.8a.
EXECUTE:! parallel
1
1
1
1
#0
0
Req R1 R2 R3 1
1
1
1
#
0
0
Req 1.60 U 2.40 U 4.80 U
Req # 0.800 U
Figure 26.8a
(b) For resistors in parallel the voltage is the same across each and equal to the applied voltage;
V1 # V2 # V3 # E # 28.0 V V # IR so I1 # V1 28.0 V
#
# 17.5 A
R1 1.60 U V2 28.0 V
V
28.0 V
#
# 11.7 A and I 3 # 3 #
# 5.8 A
R2 2.40 U
R3 4.8 U
(c) The currents through the resistors add to give the current through the battery:
I # I1 0 I 2 0 I 3 # 17.5 A 0 11.7 A 0 5.8 A # 35.0 A
I2 # EVALUATE:! Alternatively, we can use the equivalent resistance Req as shown in Figure 26.8b. E ! IReq # 0
I# E
28.0 V
#
# 35.0 A,
Req 0.800 U which checks
Figure 26.8b
(d) As shown in part (b), the voltage across each resistor is 28.0 V.
(e) IDENTIFY and SET UP:! We can use any of the three expressions for P : P # VI # I 2 R # V 2 / R. They will all
give the same results, if we keep enough significant figures in intermediate calculations. DirectCurrent Circuits! ! 263 EXECUTE:! Using P # V 2 / R, P # V12 / R1 #
1
P3 # V32 / R3 # . 28.0 V / . 28.0 V / 2 1.60 U # 490 W, P2 # V22 / R2 # . 28.0 V /
2.40 U 2 # 327 W, and 2 # 163 W
4.80 U
EVALUATE:! The total power dissipated is Pout # P 0 P2 0 P3 # 980 W. This is the same as the power
1
Pin # EI # . 2.80 V /. 35.0 A / # 980 W delivered by the battery. 26.9. (f ) P # V 2 / R. The resistors in parallel each have the same voltage, so the power P is largest for the one with the
least resistance.
IDENTIFY:! For a series network, the current is the same in each resistor and the sum of voltages for each resistor
equals the battery voltage. The equivalent resistance is Req # R1 0 R2 0 R3 . P # I 2 R .
SET UP:! Let R1 # 1.60 U , R2 # 2.40 U , R3 # 4.80 U .
EXECUTE:! (a) Req # 1.60 U 0 2.40 U 0 4.80 U # 8.80 U
(b) I # V
28.0 V
#
# 3.18 A
Req 8.80 U (c) I # 3.18 A , the same as for each resistor.
(d) V1 # IR1 # (3.18 A)(1.60 U) # 5.09 V . V2 # IR2 # (3.18 A)(2.40 U) # 7.63 V . V3 # IR3 # (3.18 A)(4.80 U) # 15.3 V . Note that V1 0 V2 0 V3 # 28.0 V .
(e) P # I 2 R1 # (3.18 A) 2 (1.60 U) # 16.2 W . P2 # I 2 R2 # (3.18 A) 2 (2.40 U) # 24.3 W .
1
P3 # I 2 R3 # (3.18 A) 2 (4.80 U ) # 48.5 W . 26.10. (f ) Since P # I 2 R and the current is the same for each resistor, the resistor with the greatest R dissipates the
greatest power.
EVALUATE:! When resistors are connected in parallel, the resistor with the smallest R dissipates the greatest power.
(a) IDENTIFY:! The current, and hence the power, depends on the potential difference across the resistor.
SET UP:! P # V D /R
EXECUTE:! (a) V # PR # (5.0 W)(15,000 U) # 274 V 26.11. (b) P # V D /R # (120 V) D /(9,000 U ) # 1.6 W
SET UP:! (c) If the larger resistor generates 2.00 W, the smaller one will generate less and hence will be safe.
Therefore the maximum power in the larger resistor must be 2.00 W. Use P # I 2 R to find the maximum current
through the series combination and use Ohm’s law to find the potential difference across the combination.
EXECUTE:! P # I D R gives I = P/R = (2.00 W)/(150 &) = 0.0133 A. The same current flows through both
resistors, and their equivalent resistance is 250 &. Ohm’s law gives V = IR = (0.0133 A)(250 &) = 3.33 V.
Therefore P150 = 2.00 W and P 00 # I D R = (0.0133 A)2(100 &) = 0.0177 W.
1
EVALUATE:! If the resistors in a series combination all have the same power rating, it is the largest resistance that
limits the amount of current.
1
11
RR
#0
IDENTIFY:! For resistors in parallel, the voltages are the same and the currents add.
so Req # 1 2 ,
Req R1 R2
R1 0 R2 For resistors in series, the currents are the same and the voltages add. Req # R1 0 R2 .
SET UP:! The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits
shown in Figure 26.11.
60.0 V
EXECUTE:! Req # 5.00 U . In Figure 26.11c, I #
# 12.0 A . This is the current through each of the
5.00 U
resistors in Figure 26.11b. V12 # IR12 # (12.0 A)(2.00 U) # 24.0 V . V34 # IR34 # (12.0 A)(3.00 U) # 36.0 V . Note that V12 0 V34 # 60.0 V . V12 is the voltage across R1 and across R2 , so I1 # V12 24.0 V
#
# 8.00 A and
R1 3.00 U I2 # 36.0 V
V12 24.0 V
V
#
# 4.00 A . V34 is the voltage across R3 and across R4 , so I 3 # 34 #
# 3.00 A and
R2 6.00 U
R3 12.0 U I4 # V34 36.0 V
#
# 9.00 A .
R4 4.00 U 264 Chapter 26 EVALUATE:! Note that I1 0 I 2 # I 3 0 I 4 . Figure 26.11
26.12. IDENTIFY:! Replace the series combinations of resistors by their equivalents. In the resulting parallel network the
battery voltage is the voltage across each resistor.
SET UP:! The circuit is sketched in Figure 26.12a.
EXECUTE:! R1 and R2 in series have an equivalent resistance of R12 # R1 0 R2 # 4.00 U
R3 and R4 in series have an equivalent resistance of
R34 # R3 0 R4 # 12.0 U
Figure 26.12a The circuit is equivalent to the circuit sketched in Figure 26.12b.
R12 and R34 in parallel are equivalent to
Req given by
Req #
Req # 1
1
1
R 0 R34
#
0
# 12
Req R12 R34
R12 R34 R12 R34
R12 0 R34 . 4.00 U / .12.0 U / # 3.00 U
4.00 U 0 12.0 U Figure 26.12b The voltage across each branch of the parallel combination is E , so E ! I12 R12 # 0.
E
48.0 V
I12 #
#
# 12.0 A
R12 4.00 U
E
48.0 V
E ! I 34 R34 # 0 so I 34 #
#
# 4.0 A
R34 12.0 U
The current is 12.0 A through the 1.00 U and 3.00 U resistors, and it is 4.0 A through the 7.00 U and 5.00 U resistors.
EVALUATE:! The current through the battery is I # I12 0 I 34 # 12.0 A 0 4.0 A # 16.0 A, and this is equal to
E / Req ! 48.0 V / 3.00 U # 16.0 A.
26.13. IDENTIFY:! In both circuits, with and without R4 , replace series and parallel combinations of resistors by their
equivalents. Calculate the currents and voltages in the equivalent circuit and infer from this the currents and
voltages in the original circuit. Use P # I 2 R to calculate the power dissipated in each bulb.
(a) SET UP:! The circuit is sketched in Figure 26.13a.
EXECUTE:! R2 , R3 , and R4 are in
parallel, so their equivalent resistance
1
1
1
1
#
00
Req is given by
Req R2 R3 R4
Figure 26.13a 1
3
#
and Req # 1.50 U.
Req 4.50 U DirectCurrent Circuits! ! 265 The equivalent circuit is drawn in Figure 26.13b.
E ! I . R1 0 Req / # 0
I# E
R1 0 Req Figure 26.13b
9.00 V
# 1.50 A and I1 # 1.50 A
4.50 U 0 1.50 U
Then V1 # I1R1 # .1.50 A /. 4.50 U / # 6.75 V I# I eq # 1.50 A, Veq # I eq Req # .1.50 A /.1.50 U / # 2.25 V
For resistors in parallel the voltages are equal and are the same as the voltage across the equivalent resistor, so
V2 # V3 # V4 # 2.25 V.
I2 # V2 2.25 V
V
V
#
# 0.500 A, I 3 # 3 # 0.500 A, I 4 # 4 # 0.500 A
R2 4.50 U
R3
R4 EVALUATE:! Note that I 2 0 I 3 0 I 4 # 1.50 A, which is I eq . For resistors in parallel the currents add and their sum is the current through the equivalent resistor.
(b) SET UP:! P # I 2 R
2
EXECUTE:! P # .1.50 A / . 4.50 U / # 10.1 W
1
P2 # P3 # P4 # . 0.500 A / . 4.50 U / # 1.125 W, which rounds to 1.12 W. R1 glows brightest.
2 2
EVALUATE:! Note that P2 0 P3 0 P4 # 3.37 W. This equals Peq # I eq Req # (1.50 A)2 (1.50 U) # 3.37 W, the power dissipated in the equivalent resistor.
(c) SET UP:! With R4 removed the circuit becomes the circuit in Figure 26.13c.
EXECUTE:! R2 and R3 are in parallel and their equivalent resistance Req is given by
1
1
1
2
and Req # 2.25 U
#
0
#
Req R2 R3 4.50 U
Figure 26.13c The equivalent circuit is shown in Figure 26.13d. E ! I . R1 0 Req / # 0
I# E
R1 0 Req I# 9.00 V
# 1.333 A
4.50 U 0 2.25 U Figure 26.13d I1 # 1.33 A, V1 # I1R1 # .1.333 A /. 4.50 U / # 6.00 V
I eq # 1.33 A, Veq # I eq Req # .1.333 A /. 2.25 U / # 3.00 V and V2 # V3 # 3.00 V.
I2 # V2 3.00 V
V
#
# 0.667 A, I 3 # 3 # 0.667 A
R2 4.50 U
R3 (d) SET UP:! P # I 2 R
EXECUTE:! P # (1.333 A) D (4.50 U/ # XYZZ W !
1 P2 # P3 # (0.667 A) D (4.50 U) # 2.00 W.
(e) EVALUATE:! When R4 is removed, P1 decreases and P2 and P3 increase. Bulb R1 glows less brightly and bulbs
R2 and R3 glow more brightly. When R4 is removed the equivalent resistance of the circuit increases and the current
through R1 decreases. But in the parallel combination this current divides into two equal currents rather than three,
so the currents through R2 and R3 increase. Can also see this by noting that with R4 removed and less current
through R1 the voltage drop across R1 is less so the voltage drop across R2 and across R3 must become larger. 266 Chapter 26 26.14. IDENTIFY:! Apply Ohm's law to each resistor.
SET UP:! For resistors in parallel the voltages are the same and the currents add. For resistors in series the currents
are the same and the voltages add.
EXECUTE:! From Ohm’s law, the voltage drop across the 6.00 U resistor is V = IR = (4.00 A)(6.00 U ) = 24.0 V.
The voltage drop across the 8.00 U resistor is the same, since these two resistors are wired in parallel. The current
through the 8.00 U resistor is then I = V/R = 24.0 V/8.00 U = 3.00 A. The current through the 25.0 U resistor is the
sum of these two currents: 7.00 A. The voltage drop across the 25.0 U resistor is V = IR = (7.00 A)(25.0 U ) = 175 V,
and total voltage drop across the top branch of the circuit is 175 V + 24.0 V = 199 V, which is also the voltage
drop across the 20.0 U resistor. The current through the 20.0 U resistor is then I # V/R # 199 V/20 U # 9.95 A.
EVALUATE:! The total current through the battery is 7.00 A 0 9.95 A # 16.95 A . Note that we did not need to
calculate the emf of the battery.
IDENTIFY:! Apply Ohm's law to each resistor.
SET UP:! For resistors in parallel the voltages are the same and the currents add. For resistors in series the currents
are the same and the voltages add.
EXECUTE:! The current through 2.00 U resistor is 6.00 A. Current through 1.00U resistor also is 6.00 A and the
voltage is 6.00 V. Voltage across the 6.00U resistor is 12.0 V + 6.0 V = 18.0 V. Current through the 6.00U
resistor is (18.0 V)/(6.00 U ) = 3.00 A. The battery emf is 18.0 V.
EVALUATE:! The current through the battery is 6.00 A + 3.00 A = 9.00 A. The equivalent resistor of the resistor
network is 2.00 U , and this equals (18.0 V)/(9.00 A).
IDENTIFY:! The filaments must be connected such that the current can flow through each separately, and also
through both in parallel, yielding three possible current flows. The parallel situation always has less resistance than
any of the individual members, so it will give the highest power output of 180 W, while the other two must give
power outputs of 60 W and 120 W.
SET UP:! P = V 2/R, where R is the equivalent resistance.
V2
V2
(120 V)2
(120 V)2
# 240 U . 120 W #
# 120 U. For these
EXECUTE:! (a) 60 W #
gives R1 #
gives R2 #
60 W
120 W
R1
R2
V 2 (120 V)2
RR
#
# 180 W , which is the desired value.
two resistors in parallel, Req # 1 2 # 80 U and P #
R1 0 R2
80 U
Req 26.15. 26.16. 26.17. (b) If R1 burns out, the 120 W setting stays the same, the 60 W setting does not work and the 180 W setting goes to
120 W: brightnesses of zero, medium and medium.
(c) If R2 burns out, the 60 W setting stays the same, the 120 W setting does not work, and the 180 W setting is now
60 W: brightnesses of low, zero and low.
EVALUATE:! Since in each case 120 V is supplied to each filament network, the lowest resistance dissipates the
greatest power.
IDENTIFY:! For resistors in series, the voltages add and the current is the same. For resistors in parallel, the
voltages are the same and the currents add. P = I 2R.
(a) SET UP:! The circuit is sketched in Figure 26.17a. For resistors in series the current
is the same through each.
Figure 26.17a
EXECUTE:! Req # R1 0 R2 # 1200 U. I # 120 V
V
#
# 0.100 A. This is the current drawn from the line.
Req 1200 U (b) P # I12 R1 # (0.100 A)2 (400 U) # 4.0 W
1
2
P2 # I 2 R2 # (0.100 A)2 (800 U) # 8.0 W
(c) Pout # P 0 P2 # 12.0 W, the total power dissipated in both bulbs. Note that
1
Pin # Vab I # (120 V)(0.100 A) # 12.0 W, the power delivered by the potential source, equals Pout.
(d) SET UP:! The circuit is sketched in Figure 26.17b. For resistors in parallel the voltage
across each resistor is the same. Figure 26.17b DirectCurrent Circuits! ! 267 V1 120 V
V 120 V
#
# 0.300 A, I 2 # 2 #
# 0.150 A
R1 400 U
R2 800 U
EVALUATE:! Note that each current is larger than the current when the resistors are connected in series.
(e) EXECUTE:! P # I12 R1 # (0.300 A)2 (400 U) # 36.0 W
1
EXECUTE:! I1 # 26.18. 2
P2 # I 2 R2 # (0.150 A) 2 (800 U) # 18.0 W
(f ) Pout # P 0 P2 # 54.0 W
1
EVALUATE:! Note that the total current drawn from the line is I # I1 0 I 2 # 0.450 A. The power input from the
line is Pin # Vab I # (120 V)(0.450 A) # 54.0 W, which equals the total power dissipated by the bulbs.
(g) The bulb that is dissipating the most power glows most brightly. For the series connection the currents are the
same and by P # I 2 R the bulb with the larger R has the larger P; the 800 U bulb glows more brightly. For the
parallel combination the voltages are the same and by P # V 2 / R the bulb with the smaller R has the larger P; the
400 U bulb glows more brightly.
(h) The total power output Pout equals Pin # Vab I , so Pout is larger for the parallel connection where the current
drawn from the line is larger (because the equivalent resistance is smaller.)
IDENTIFY:! Use P # V 2 / R with V # 120 V and the wattage for each bulb to calculate the resistance of each bulb.
When connected in series the voltage across each bulb will not be 120 V and the power for each bulb will be
different.
SET UP:! For resistors in series the currents are the same and Req # R1 0 R2 . V 2 (120 V)2
V 2 (120 V) 2
#
# 240 U ; R200W #
#
# 72 U.
P
60 W
P
200 W
E
240 V
##
# 0.769 A.
R (240 U 0 72 U) EXECUTE:! (a) R60W # Therefore, I 60W # I 200W 26.19. (b) P60W # I 2 R # (0.769 A)2 (240 U) # 142 W; P200W # I 2 R # (0.769 A) 2 (72 U) # 42.6 W.
(c) The 60 W bulb burns out quickly because the power it delivers (142 W) is 2.4 times its rated value.
EVALUATE:! In series the largest resistance dissipates the greatest power.
IDENTIFY and SET UP:! Replace series and parallel combinations of resistors by their equivalents until the circuit
is reduced to a single loop. Use the loop equation to find the current through the 20.0 U resistor. Set P # I 2 R for
the 20.0 U resistor equal to the rate Q /t at which heat goes into the water and set Q # mcBT .
EXECUTE:! Replace the network by the equivalent resistor, as shown in Figure 26.19. Figure 26.19 30.0 V ! I . 20.0 U 0 5.0 U 0 5.0 U / # 0; I # 1.00 A
For the 20.0U resistor thermal energy is generated at the rate P # I 2 R # 20.0 W. Q # Pt and Q # mcBT gives
mcBT . 0.100 kg /. 4190 J/kg + K / . 48.0 C3 /
#
# 1.01 " 103 s
P
20.0 W
EVALUATE:! The battery is supplying heat at the rate P # EI # 30.0 W. In the series circuit, more energy is
dissipated in the larger resistor (20.0 U) than in the smaller ones (5.00 U).
t# 26.20. IDENTIFY:! P # I 2 R determines R1 . R1 , R2 and the 10.0 U resistor are all in parallel so have the same voltage. Apply the junction rule to find the current through R2 .
SET UP:! P # I 2 R for a resistor and P # EI for an emf. The emf inputs electrical energy into the circuit and
electrical energy is removed in the resistors.
EXECUTE:! (a) P # I12 R1 . 20 W # (2 A)2 R1 and R1 # 5.00 U . R1 and 10 U are in parallel, so
1 (10 U) I10 # (5 U)(2 A) and I10 # 1 A . So I 2 # 3.50 A ! I1 ! I10 # 0.50 A . R1 and R2 are in parallel, so
(0.50 A) R2 # (2 A)(5 U) and R2 # 20.0 U .
(b) E # V1 # (2.00 A)(5.00 U) # 10.0 V
(c) From part (a),! I 2 # 0.500 A, I10 # 1.00 A 268 Chapter 26
2
2
(d) P # 20.0 W (given). P2 # I 2 R2 # (0.50 A) 2 (20 U) # 5.00 W . P # I10 R10 # (1.0 A) 2 (10 U) # 10.0 W . The total
10
1 rate at which the resistors remove electrical energy is PResist # 20 W 0 5 W 0 10 W # 35.0 W . The total rate at which the battery inputs electrical energy is PBattery # I P # (3.50 A)(10.0 V) # 35.0 W . PResist # PBattery , which agrees with 26.21. conservation of energy.
EVALUATE:! The three resistors are in parallel, so the voltage for each is the battery voltage, 10.0 V. The currents
in the three resistors add to give the current in the battery.
IDENTIFY:! Apply Kirchhoff's point rule at point a to find the current through R. Apply Kirchhoff's loop rule to
loops (1) and (2) shown in Figure 26.21a to calculate R and E . Travel around each loop in the direction shown.
(a) SET UP:! ! Figure 26.21a
EXECUTE:! Apply Kirchhoff's point rule to point a: 7 I # 0 so I 0 4.00 A ! 6.00 A # 0 I = 2.00 A (in the direction shown in the diagram).
(b) Apply Kirchhoff's loop rule to loop (1): ! . 6.00 A / . 3.00 U / ! . 2.00 A / R 0 28.0 V # 0
!18.0 V ! . 2.00 U / R 0 28.0 V # 0
28.0 V ! 18.0 V
# 5.00 U
2.00 A
(c) Apply Kirchhoff's loop rule to loop (2): ! . 6.00 A / . 3.00 U / ! . 4.00 A / . 6.00 U / 0 E # 0
R# E # 18.0 V 0 24.0 V # 42.0 V
EVALUATE:! Can check that the loop rule is satisfied for loop (3), as a check of our work:
28.0 V ! E 0 . 4.00 A /. 6.00 U / ! . 2.00 A / R # 0 28.0 V ! 42.0 V 0 24.0 V ! . 2.00 A / . 5.00 U / # 0
52.0 V # 42.0 V 0 10.0 V
52.0 V # 52.0 V, so the loop rule is satisfied for this loop.
(d) IDENTIFY:! If the circuit is broken at point x there can be no current in the 6.00 U resistor. There is now only
a single current path and we can apply the loop rule to this path.
SET UP:! The circuit is sketched in Figure 26.21b. Figure 26.21b
EXECUTE:! 028.0 V ! . 3.00 U / I ! . 5.00 U / I # 0 28.0 V
# 3.50 A
8.00 U
EVALUATE:! Breaking the circuit at x removes the 42.0 V emf from the circuit and the current through the
3.00 U resistor is reduced.
I# DirectCurrent Circuits! ! 269 26.22. IDENTIFY:! Apply the loop rule and junction rule.
SET UP:! The circuit diagram is given in Figure 26.22. The junction rule has been used to find the magnitude and
direction of the current in the middle branch of the circuit. There are no remaining unknown currents.
EXECUTE:! The loop rule applied to loop (1) gives:
020.0 V ! (1.00 A)(1.00 U) 0 (1.00 A)(4.00 U) 0 (1.00 A)(1.00 U) ! E1 ! (1.00 A)(6.00 U) # 0 E1 # 20.0 V ! 1.00 V 0 4.00 V 0 1.00 V ! 6.00 V # 18.0 V . The loop rule applied to loop (2) gives:
020.0 V ! (1.00 A)(1.00 U) ! (2.00 A)(1.00 U) ! E2 ! (2.00 A)(2.00 U ) ! (1.00 A)(6.00 U ) # 0 E2 # 20.0 V ! 1.00 V ! 2.00 V ! 4.00 V ! 6.00 V # 7.0 V . Going from b to a along the lower branch,
Vb 0 (2.00 A)(2.00 U) 0 7.0 V 0 (2.00 A)(1.00 U) # Va . Vb ! Va # !13.0 V ; point b is at 13.0 V lower potential
than point a.
EVALUATE:! We can also calculate Vb ! Va by going from b to a along the upper branch of the circuit.
Vb ! (1.00 A)(6.00 U) 0 20.0 V ! (1.00 A)(1.00 U ) # Va and Vb ! Va # !13.0 V . This agrees with Vb ! Va calculated
along a different path between b and a. Figure 26.22
26.23. IDENTIFY:! Apply the junction rule at points a, b, c and d to calculate the unknown currents. Then apply the loop
rule to three loops to calculate E1 , E2 and R.
(a) SET UP:! The circuit is sketched in Figure 26.23. Figure 26.23
EXECUTE:! Apply the junction rule to point a: 3.00 A 0 5.00 A ! I 3 # 0
I 3 # 8.00 A Apply the junction rule to point b: 2.00 A 0 I 4 ! 3.00 A # 0
I 4 # 1.00 A Apply the junction rule to point c: I 3 ! I 4 ! I 5 # 0
I 5 # I 3 ! I 4 # 8.00 A ! 1.00 A # 7.00 A EVALUATE:! As a check, apply the junction rule to point d: I 5 ! 2.00 A ! 5.00 A # 0
I 5 # 7.00 A 2610 Chapter 26 (b) EXECUTE:! Apply the loop rule to loop (1): E1 ! . 3.00 A / . 4.00 U / ! I 3 . 3.00 U / # 0 E1 # 12.0 V 0 . 8.00 A /. 3.00 U / # 36.0 V
Apply the loop rule to loop (2): E2 ! . 5.00 A / . 6.00 U / ! I 3 . 3.00 U / # 0 E2 # 30.0 V 0 . 8.00 A /. 3.00 U / # 54.0 V
(c) Apply the loop rule to loop (3): ! . 2.00 A / R ! E1 0 E2 # 0
E2 ! E1 54.0 V ! 36.0 V
#
# 9.00 U
2.00 A
2.00 A
EVALUATE:! Apply the loop rule to loop (4) as a check of our calculations:
! . 2.00 A / R ! . 3.00 A /. 4.00 U / 0 . 5.00 A / . 6.00 U / # 0 R# ! . 2.00 A /. 9.00 U / ! 12.0 V 0 30.0 V # 0
26.24. !18.0 V 0 18.0 V # 0
IDENTIFY:! Use Kirchhoff’s Rules to find the currents.
SET UP:! Since the 1.0 V battery has the larger voltage, assume I1 is to the left through the 10 V battery, I 2 is to
the right through the 5 V battery, and I 3 is to the right through the 10 U resistor. Go around each loop in the
counterclockwise direction.
EXECUTE:! Upper loop: 10.0 V ! . 2.00 U 0 3.00 U / I1 ! .1.00 U 0 4.00 U / I 2 ! 5.00 V # 0 . This gives
5.0 V ! . 5.00 U / I1 ! . 5.00 U / I 2 # 0 , and I I1 0 I 2 # 1.00 A .
Lower loop: 5.00 V 0 .1.00 U 0 4.00 U / I 2 ! .10.0 U / I 3 # 0 . This gives 5.00 V 0 . 5.00 U / I 2 ! .10.0 U / I 3 # 0 , and
I 2 ! 2 I 3 # !1.00 A
Along with I1 # I 2 0 I 3 , we can solve for the three currents and find:
I1 # 0.800 A, I 2 # 0.200 A, I 3 # 0.600 A.
(b) Vab # ! . 0.200 A /. 4.00 U / ! . 0.800 A / . 3.00 U / # !3.20 V. 26.25. EVALUATE:! Traveling from b to a through the 4.00 U and 3.00 U resistors you pass through the resistors in the
direction of the current and the potential decreases; point b is at higher potential than point a.
IDENTIFY:! Apply the junction rule to reduce the number of unknown currents. Apply the loop rule to two loops
to obtain two equations for the unknown currents I1 and I 2
(a) SET UP:! The circuit is sketched in Figure 26.25. Figure 26.25 Let I1 be the current in the 3.00 U resistor and I2 be the current in the 4.00 U resistor and assume that these
currents are in the directions shown. Then the current in the 10.0 U resistor is I 3 # I1 ! I 2 , in the direction shown,
where we have used Kirchhoff's point rule to relate I3 to I1 and I2. If we get a negative answer for any of these
currents we know the current is actually in the opposite direction to what we have assumed. Three loops and
directions to travel around the loops are shown in the circiut diagram. Apply Kirchhoff's loop rule to each loop.
EXECUTE:! loop (1)
010.0 V ! I1 . 3.00 U / ! I 2 . 4.00 U / 0 5.00 V ! I 2 .1.00 U / ! I1 . 2.00 U / # 0 15.00 V ! . 5.00 U / I1 ! . 5.00 U / I 2 # 0
3.00 A ! I1 ! I 2 # 0 DirectCurrent Circuits! ! 2611 loop (2)
05.00 V ! I 2 (1.00 U) 0 ( I1 ! I 2 )10.0 U ! I 2 (4.00 U) # 0
5.00 V 0 (10.0 U) I1 ! (15.0 U) I 2 # 0
1.00 A 0 2.00 I1 ! 3.00 I 2 # 0
The first equation says I 2 # 3.00 A ! I1.
Use this in the second equation: 1.00 A 0 2.00 I1 ! 9.00 A 0 3.00 I1 # 0
5.00 I1 # 8.00 A, I1 # 1.60 A
Then I 2 # 3.00 A ! I1 # 3.00 A ! 1.60 A # 1.40 A.
I 3 # I1 ! I 2 # 1.60 A ! 1.40 A # 0.20 A
EVALUATE:! Loop (3) can be used as a check.
010.0 V ! (1.60 A)(3.00 U) ! (0.20 A)(10.00 U) ! (1.60 A)(2.00 U ) # 0
10.0 V # 4.8 V 0 2.0 V 0 3.2 V
10.0 V # 10.0 V
We find that with our calculated currents the loop rule is satisfied for loop (3). Also, all the currents came out to be
positive, so the current directions in the circuit diagram are correct.
(b) IDENTIFY and SET UP:! To find Vab # Va ! Vb start at point b and travel to point a. Many different routes can
be taken from b to a and all must yield the same result for Vab .
EXECUTE:! Travel through the 4.00 U resistor and then through the 3.00 U resistor:
Vb 0 I 2 (4.00 U) 0 I1 (3.00 U) # Va
Va ! Vb # (1.40 A)(4.00 U) 0 (1.60 A)(3.00 U ) # 5.60 V 0 4.8 V # 10.4 V (point a is at higher potential than point b)
EVALUATE:! Alternatively, travel through the 5.00 V emf, the 1.00 U resistor, the 2.00 U resistor, and the 10.0 V emf.
Vb 0 5.00 V ! I 2 .1.00 U / ! I1 . 2.00 U / 0 10.0 V # Va Va ! Vb # 15.0 V ! .1.40 A /.1.00 U / ! .1.60 A / . 2.00 U / # 15.0 V ! 1.40 V ! 3.20 V # 10.4 V, the same as before.
26.26. IDENTIFY:! Use Kirchhoff’s rules to find the currents
SET UP:! Since the 20.0 V battery has the largest voltage, assume I1 is to the right through the 10.0 V battery, I 2 is to the left through the 20.0 V battery, and I 3 is to the right through the 10 U resistor. Go around each loop in
the counterclockwise direction.
EXECUTE:! Upper loop: 10.0 V+ . 2.00 U 0 3.00 U / I1 0 .1.00 U 0 4.00 U / I 2 ! 20.00 V # 0 .
!10.0 V+ . 5.00 U / I1 0 . 5.00 U / I 2 # 0 , so I1 0 I 2 # 02.00 A.
Lower loop: 20.00 V ! .1.00 U 0 4.00 U / I 2 ! .10.0 U / I 3 # 0 . 20.00 V ! . 5.00 U / I 2 ! .10.0 U / I 3 # 0 , so I 2 0 2 I 3 # 4.00 A.
Along with I 2 # I1 0 I 3 , we can solve for the three currents and find I1 # 00.4 A, I 2 # 01.6 A, I 3 # 01.2 A. 26.27. (b) Vab # I 2 (4 U) 0 I1 (3 U) # (1.6 A)(4 U) 0 (0.4 A)(3 U) # 7.6 V
EVALUATE:! Traveling from b to a through the 4.00 U and 3.00 U resistors you pass through each resistor
opposite to the direction of the current and the potential increases; point a is at higher potential than point b.
(a) IDENTIFY:! With the switch open, the circuit can be solved using seriesparallel reduction.
SET UP:! Find the current through the unknown battery using Ohm’s law. Then use the equivalent resistance of
the circuit to find the emf of the battery.
EXECUTE:! The 30.0& and 50.0& resistors are in series, and hence have the same current. Using Ohm’s law
I50 = (15.0 V)/(50.0 &) = 0.300 A = I30. The potential drop across the 75.0& resistor is the same as the potential
drop across the 80.0& series combination. We can use this fact to find the current through the 75.0& resistor
using Ohm’s law: V75 = V80 = (0.300 A)(80.0 &) = 24.0 V and I75 = (24.0 V)/(75.0 &) = 0.320 A.
The current through the unknown battery is the sum of the two currents we just found:
ITotal = 0.300 A + 0.320 A = 0.620 A
The equivalent resistance of the resistors in parallel is 1/Rp = 1/(75.0 &) + 1/(80.0 &). This gives Rp = 38.7 &. The
equivalent resistance “seen” by the battery is Requiv = 20.0 & + 38.7 & = 58.7 &.
Applying Ohm’s law to the battery gives P = RequivITotal = (58.7 &)(0.620 A) = 36.4 V
(b) IDENTIFY:! With the switch closed, the 25.0V battery is connected across the 50.0& resistor.
SET UP:! Taking a loop around the right part of the circuit.
EXECUTE:! Ohm’s law gives I = (25.0 V)/(50.0 &) = 0.500 A
EVALUATE:! The current through the 50.0& resistor, and the rest of the circuit, depends on whether or not the
switch is open. 2612 Chapter 26 26.28. IDENTIFY:! We need to use Kirchhoff’s rules.
SET UP:! Take a loop around the outside of the circuit, use the current at the upper junction, and then take a loop
around the right side of the circuit.
EXECUTE:! The outside loop gives 75.0 V – (12.0 &)(1.50 A) – (48.0 &)I48 = 0 , so I48 = 1.188 A. At a
junction we have 1.50 A = IP + 1.188 A , and IP = 0.313 A. A loop around the right part of the circuit gives
E ! (48 &)(1.188 A) + (15.0 &)(0.313 A). E = 52.3 V, with the polarity shown in the figure in the problem.
EVALUATE:! The unknown battery has a smaller emf than the known one, so the current through it goes against
its polarity.
(a) IDENTIFY:! With the switch open, we have a series circuit with two batteries.
SET UP:! Take a loop to find the current, then use Ohm’s law to find the potential difference between a and b.
EXECUTE:! Taking the loop: I = (40.0 V)/(175 &) = 0.229 A. The potential difference between a and b is
Vb – Va = +15.0 V – (75.0 &)(0.229 A) = ! 2.14 V.
EVALUATE:! The minus sign means that a is at a higher potential than b.
(b) IDENTIFY:! With the switch closed, the ammeter part of the circuit divides the original circuit into two
circuits. We can apply Kirchhoff’s rules to both parts.
SET UP:! Take loops around the left and right parts of the circuit, and then look at the current at the junction.
EXECUTE:! The lefthand loop gives I100 = (25.0 V)/(100.0 &) = 0.250 A. The righthand loop gives
I75 = (15.0 V)/(75.0 &) = 0.200 A. At the junction just above the switch we have I100 = 0.250 A (in) and
I75 = 0.200 A (out) , so IA = 0.250 A – 0.200 A = 0.050 A, downward. The voltmeter reads zero because the
potential difference across it is zero with the switch closed.
EVALUATE:! The ideal ammeter acts like a short circuit, making a and b at the same potential. Hence the
voltmeter reads zero.
IDENTIFY:! The circuit is sketched in Figure 26.30a. Since all the external resistors are equal, the current must be
symmetrical through them. That is, there can be no current through the resistor R for that would imply an
imbalance in currents through the other resistors. With no current going through R, the circuit is like that shown in
Figure 26.30b.
SET UP:! For resistors in series, the equivalent resistance is Rs # R1 0 R2 . For resistors in parallel, the equivalent 26.29. 26.30. resistance is 1
1
1
#0
Rp R1 R2
!1 1&
13 V
%1
EXECUTE:! The equivalent resistance of the circuit is Req # '
0
# 13 A. The two
( # 1 U and I total #
1U
)2U 2U*
1
parallel branches have the same resistance, so I each branch # I total # 6.5 A. The current through each 1 U resistor is 6.5 A
2
and no current passes through R.
(b) As worked out above, Req # 1 U .
(c) Vab # 0, since no current flows through R.
EVALUATE:! (d) R plays no role since no current flows through it and the voltage across it is zero. Figure 26.30
26.31. IDENTIFY:! To construct an ammeter, add a shunt resistor in parallel with the galvanometer coil. To construct a
voltmeter, add a resistor in series with the galvanometer coil.
SET UP:! The fullscale deflection current is 500 $ A and the coil resistance is 25.0 U .
EXECUTE:! (a) For a 20mA ammeter, the two resistances are in parallel and the voltages across each are the
same. Vc # Vs gives I c Rc # I s Rs . . 500 " 10!6 A / . 25.0 U / # . 20 " 10!3 A ! 500 " 10!6 A / Rs and Rs # 0.641 U . DirectCurrent Circuits! ! 2613 (b) For a 500mV voltmeter, the resistances are in series and the current is the same through each: Vab # I . Rc 0 Rs / Vab
500 " 10!3 V
! Rc #
! 25.0 U # 975 U.
I
500 " 10!6 A
EVALUATE:! The equivalent resistance of the voltmeter is Req # Rs 0 Rc # 1000 U . The equivalent resistance of
and Rs # the ammeter is given by
26.32. 1
1
1
#
0
and Req # 0.625 U . The voltmeter is a highresistance device and the
Req Rsh Rc ammeter is a lowresistance device.
IDENTIFY:! The galvanometer is represented in the circuit as a resistance Rc. Use the junction rule to relate the
current through the galvanometer and the current through the shunt resistor. The voltage drop across each parallel
path is the same; use this to write an equation for the resistance R.
SET UP:! The circuit is sketched in Figure 26.32. Figure 26.32 We want that I a # 20.0 A in the external circuit to produce I fs # 0.0224 A through the galvanometer coil.
EXECUTE:! Applying the junction rule to point a gives I a ! I fs ! I sh # 0
I sh # I a ! I fs # 20.0 A ! 0.0224 A # 19.98 A
The potential difference Vab between points a and b must be the same for both paths between these two points:
I fs . R 0 Rc / # I sh Rsh I sh Rsh
.19.98 A /. 0.0250 U / ! 9.36 U # 22.30 U ! 9.36 U # 12.9 U
! Rc #
I fs
0.0224 A
EVALUATE:! Rsh 11 R 0 Rc ; most of the current goes through the shunt. Adding R decreases the fraction of the
current that goes through Rc .
IDENTIFY:! The meter introduces resistance into the circuit, which affects the current through the 5.00k& resistor
and hence the potential drop across it.
SET UP:! Use Ohm’s law to find the current through the 5.00k& resistor and then the potential drop across it.
EXECUTE:! (a) The parallel resistance with the voltmeter is 3.33 k&, so the total equivalent resistance across the
battery is 9.33 k&, giving I = (50.0 V)/(9.33 k&) = 5.36 mA. Ohm’s law gives the potential drop across the
5.00k& resistor: V5 k& = (3.33 k&)(5.36 mA) = 17.9 V
(b) The current in the circuit is now I = (50.0 V)/(11.0 k&) = 4.55 mA. V5 k& = (5.00 k&)(4.55 mA) = 22.7 V.
(c) % error = (22.7 V – 17.9 V)/(22.7 V) = 0.214 = 21.4%. (We carried extra decimal places for accuracy since we
had to subtract our answers.)
EVALUATE:! The presence of the meter made a very large percent error in the reading of the “true” potential
across the resistor.
IDENTIFY:! The resistance of the galvanometer can alter the resistance in a circuit.
SET UP:! The shunt is in parallel with the galvanometer, so we find the parallel resistance of the ammeter. Then
use Ohm’s law to find the current in the circuit.
EXECUTE:! (a) The resistance of the ammeter is given by 1/RA = 1/(1.00 &) + 1/(25.0 &), so RA = 0.962 &. The
current through the ammeter, and hence the current it measures, is I = V/R = (25.0 V)/(15.96 &) = 1.57 A.
(b) Now there is no meter in the circuit, so the total resistance is only 15.0 &. I = (25.0 V)/(15.0 &) = 1.67 A
(c) (1.67 A – 1.57 A)/(1.67 A) = 0.060 = 6.0%
EVALUATE:! A 1& shunt can introduce noticeable error in the measurement of an ammeter.
IDENTIFY:! When the galvanometer reading is zero E2 # IRcb and E1 # IRab .
SET UP:! Rcb is proportional to x and Rab is proportional to l.
R
x
EXECUTE:! (a) E2 # E1 cb # E1 .
Rab
l
(b) The value of the galvanometer’s resistance is unimportant since no current flows through it.
x
0.365 m
(c) E2 # E1 # . 9.15 V /
# 3.34 V
l
1.000 m
EVALUATE:! The potentiometer measures the emf E2 of the source directly, unaffected by the internal resistance
of the source, since the measurement is made with no current through E2 .
R# 26.33. 26.34. 26.35. 2614 Chapter 26 26.36. IDENTIFY:! A halfscale reading occurs with R # 600 U , so the current through the galvanometer is half the fullscale current.
SET UP:! The resistors Rs , Rc and R are in series, so the total resistance of the circuit is Rtotal # Rs 0 Rc 0 R . 26.37. % 3.60 " 10!3 A &
EXECUTE:! E # IRtotal . 1.50 V # '
( .15.0 U 0 600 U 0 Rs / and Rs # 218 U .
2
)
*
EVALUATE:! We have assumed that the device is linear, in the sense that the deflection is proportional to the
current through the meter.
IDENTIFY:! Apply E # IRtotal to relate the resistance Rx to the current in the circuit.
SET UP:! R, Rx and the meter are in series, so Rtotal # R 0 Rx 0 RM , where RM # 65.0 U is the resistance of the meter. I fsd # 2.50 mA is the current required for fullscale deflection.
EXECUTE:! (a) When the wires are shorted, the fullscale deflection current is obtained: E # IRtotal . 1.52 V # . 2.50 " 10!3 A / . 65.0 U 0 R / and R # 543 U . (b) If the resistance Rx # 200 U : I # V
1.52 V
#
# 1.88 mA.
Rtotal 65.0 U 0 543 U 0 Rx (c) I x # For I x # 26.40. 1.52 V
1
! 608 U # 608 U.
I fsd # 1.25 " 10!3 A , Rx #
1.25 " 10!3 A
2 For I x # 26.39. 1.52 V
1
I fsd # 6.25 " 10!4 A , Rx #
! 608 U # 1824 U .
6.25 " 10!4 A
4 For I x # 26.38. E
1.52 V
1.52 V
#
and Rx #
! 608 U . For each value of I x we have:
Rtotal 65.0 U 0 543 U 0 Rx
Ix 3
1.52 V
I fsd # 1.875 " 10!3 A , Rx #
! 608 U # 203 U .
4
1.875 " 10!3 A EVALUATE:! The deflection of the meter increases when the resistance Rx decreases.
IDENTIFY:! An uncharged capacitor is placed into a circuit. Apply the loop rule at each time.
SET UP:! The voltage across a capacitor is VC # q / C .
EXECUTE:! (a) At the instant the circuit is completed, there is no voltage over the capacitor, since it has no charge
stored.
(b) Since VC # 0 , the full battery voltage appears across the resistor VR # E # 125 V.
(c) There is no charge on the capacitor.
E
125 V
(d) The current through the resistor is i #
#
# 0.0167 A.
Rtotal 7500 U
(e) After a long time has passed the full battery voltage is across the capacitor and i # 0 . The voltage across the
capacitor balances the emf: VC #125 V. The voltage across the resister is zero. The capacitor’s charge is q # CVC # (4.60 " 10!6 F) (125 V) # 5.75 " 10!4 C. The current in the circuit is zero.
EVALUATE:! The current in the circuit starts at 0.0167 A and decays to zero. The charge on the capacitor starts at
zero and rises to q # 5.75 " 10!4 C .
IDENTIFY:! The capacitor discharges exponentially through the voltmeter. Since the potential difference across
the capacitor is directly proportional to the charge on the plates, the voltage across the plates decreases
exponentially with the same time constant as the charge.
SET UP:! The reading of the voltmeter obeys the equation V = V0e–t/RC, where RC is the time constant.
EXECUTE:! (a) Solving for C and evaluating the result when t = 4.00 s gives
t
4.00 s
C#
#
= 8.49 " 10–7 F
R ln .V / V0 /
% 12.0 V &
6
(3.40 "10 U)ln '
(
) 3.00 V *
6
–7
(b) @ = RC = (3.40 " 10 &)(8.49 " 10 F) = 2.89 s
EVALUATE:! In most laboratory circuits, time constants are much shorter than this one.
E
IDENTIFY:! For a charging capacitor q (t ) # CE (1 ! e !t / @ ) and i (t ) # e ! t / @ .
R
SET UP:! The time constant is RC # (0.895 " 106 U) (12.4 " 10!6 F) # 11.1 s.
EXECUTE:! (a) At t # 0 s: q # CE (1 ! e ! t / RC ) # 0. DirectCurrent Circuits! ! 2615 At t # 5 s: q # C E (1 ! e ! t / RC ) # (12.4 " 10!6 F)(60.0 V)(1 ! e! (5.0 s) /(11.1 s) ) # 2.70 " 10!4 C.
At t # 10 s: q # CE (1 ! e! t / RC ) # (12.4 " 10!6 F)(60.0 V)(1 ! e! (10.0 s) /(11.1 s) ) # 4.42 " 10!4 C.
At t # 20 s: q # CE (1 ! e! t / RC ) # (12.4 " 10!6 F)(60.0 V)(1 ! e ! (20.0 s) /(11.1 s) ) # 6.21" 10!4 C.
At t # 100 s: q # C E (1 ! e ! t / RC ) # (12.4 " 10!6 F)(60.0 V)(1 ! e! (100 s) /(11.1 s) ) # 7.44 " 10!4 C.
(b) The current at time t is given by: i # E ! t / RC
e
.
R 60.0 V
e!0 /11.1 # 6.70 " 10!5 A.
8.95 " 105 U
60.0 V
At t # 5 s: i #
e !5 /11.1 # 4.27 " 10!5 A.
8.95 " 105 U
60.0 V
At t # 10 s: i #
e!10 /11.1 # 2.27 " 10!5 A.
8.95 " 105 U
60.0 V
At t # 20 s: i #
e!20 /11.1 # 1.11 " 10!5 A.
8.95 " 105 U
60.0 V
At t # 100 s: i #
e !100 /11.1 # 8.20 " 10!9 A.
8.95 " 105 U
(c) The graphs of q (t ) and i (t ) are given in Figure 26.40a and 26.40b
EVALUATE:! The charge on the capacitor increases in time as the current decreases.
At t # 0 s: i # Figure 26.40
26.41. 26.42. IDENTIFY:! The capacitors, which are in parallel, will discharge exponentially through the resistors.
SET UP:! Since V is proportional to Q, V must obey the same exponential equation as Q,
V = V0 e–t/RC. The current is I = (V0 /R) e–t/RC.
EXECUTE:! (a) Solve for time when the potential across each capacitor is 10.0 V:
t = ! RC ln(V/V0) = –(80.0 &)(35.0 µF) ln(10/45) = 4210 µs = 4.21 ms
(b) I = (V0 /R) e–t/RC. Using the above values, with V0 = 45.0 V, gives I = 0.125 A.
EVALUATE:! Since the current and the potential both obey the same exponential equation, they are both reduced
by the same factor (0.222) in 4.21 ms.
IDENTIFY:! In @ # RC use the equivalent capacitance of the two capacitors.
1
1
1
SET UP:! For capacitors in series,
#0
. For capacitors in parallel, Ceq # C1 0 C2 .! Originally,
Ceq C1 C2 @ # RC # 0.870 s.
EXECUTE:! (a) The combined capacitance of the two identical capacitors in series is given by 1
112
#0#
,
Ceq C C C C
0.870 s
# 0.435 s.
. The new time constant is thus R (C / 2) #
2
2
(b) With the two capacitors in parallel the new total capacitance is simply 2C. Thus the time constant is
R (2C ) # 2(0.870 s) # 1.74 s.
so Ceq # EVALUATE:! The time constant is proportional to Ceq . For capacitors in series the capacitance is decreased and for capacitors in parallel the capacitance is increased. 2616 Chapter 26 26.43. IDENTIFY and SET UP:! Apply the loop rule. The voltage across the resistor depends on the current through it and
the voltage across the capacitor depends on the charge on its plates.
EXECUTE:! E ! VR ! VC # 0 E # 120 V, VR # IR # . 0.900 A /. 80.0 U / # 72 V, so VC # 48 V Q # CV # . 4.00 " 10!6 F / . 48 V / # 192 $ C 26.44. 26.45. EVALUATE:! The initial charge is zero and the final charge is CE # 480 $ C. Since current is flowing at the
instant considered in the problem the capacitor is still being charged and its charge has not reached its final value.
IDENTIFY:! The charge is increasing while the current is decreasing. Both obey exponential equations, but they
are not the same equation.
SET UP:! The charge obeys the equation Q = Qmax(1 – e–t/RC), but the equation for the current is I = Imaxe–t/RC.
EXECUTE:! When the charge has reached 1 of its maximum value, we have Qmax/4 = Qmax(1 – e–t/RC) , which
4 says that the exponential term has the value e–t/RC = 3 . The current at this time is I = Imaxe–t/RC = Imax(3/4) =
4
(3/4)[(10.0 V)/(12.0 &)] = 0.625 A
EVALUATE:! Notice that the current will be 3 , not 1 , of its maximum value when the charge is 1 of its
4
4
4
maximum. Although current and charge both obey exponential equations, the equations have different forms for a
charging capacitor.
IDENTIFY:! The stored energy is proportional to the square of the charge on the capacitor, so it will obey an
exponential equation, but not the same equation as the charge.
SET UP:! The energy stored in the capacitor is U = Q2/2C and the charge on the plates is Q0 e–t/RC. The current is
I = I0 e–t/RC.
EXECUTE:! U = Q2/2C = (Q0 e–t/RC)2/2C = U0 e–2t/RC
When the capacitor has lost 80% of its stored energy, the energy is 20% of the initial energy, which is U0/5. U0/5 =
U0 e–2t/RC gives t = (RC/2) ln 5 = (25.0 &)(4.62 pF)(ln 5)/2 = 92.9 ps.
At this time, the current is I = I0 e–t/RC = (Q0/RC) e–t/RC, so
I = (3.5 nC)/[(25.0 &)(4.62 pF)] e–(92.9 ps)/[(25.0 &)(4.62 pF)] = 13.6 A. 26.46. EVALUATE:! When the energy reduced by 80%, neither the current nor the charge are reduced by that percent.
IDENTIFY:! Both the charge and energy decay exponentially, but not with the same time constant since the energy
is proportional to the square of the charge.
SET UP:! The charge obeys the equation Q = Q0 e–t/RC but the energy obeys the equation
U = Q2/2C = (Q0 e–t/RC)/2C = U0 e–2t/RC.
EXECUTE:! (a) The charge is reduced by half: Q0/2 = Q0 e–t/RC. This gives t = RC ln 2 = (175 &)(12.0 µF)(ln 2) = 1.456 ms = 1.46 ms.
(b) The energy is reduced by half: U0/2 = U0 e–2t/RC. This gives t = (RC ln 2)/2 = (1.456 ms)/2 = 0.728 ms.
26.47. 26.48. EVALUATE:! The energy decreases faster than the charge because it is proportional to the square of the charge.
IDENTIFY:! In both cases, simplify the complicated circuit by eliminating the appropriate circuit elements. The
potential across an uncharged capacitor is initially zero, so it behaves like a short circuit. A fully charged capacitor
allows no current to flow through it.
(a) SET UP:! Just after closing the switch, the uncharged capacitors all behave like short circuits, so any resistors
in parallel with them are eliminated from the circuit.
EXECUTE:! The equivalent circuit consists of 50 & and 25 & in parallel, with this combination in series with
75 &, 15 &, and the 100V battery. The equivalent resistance is 90 & + 16.7 & = 106.7 &, which gives
I = (100 V)/(106.7 &) = 0.937 A.
(b) SET UP:! Long after closing the switch, the capacitors are essentially charged up and behave like open circuits
since no charge can flow through them. They effectively eliminate any resistors in series with them since no
current can flow through these resistors.
EXECUTE:! The equivalent circuit consists of resistances of 75 &, 15 &, and three 25& resistors, all in series with
the 100V battery, for a total resistance of 165 &. Therefore I = (100 V)/(165 &) = 0.606 A
EVALUATE:! The initial and final behavior of the circuit can be calculated quite easily using simple seriesparallel
circuit analysis. Intermediate times would require much more difficult calculations!
IDENTIFY:! When the capacitor is fully charged the voltage V across the capacitor equals the battery emf and
Q # CV . For a charging capacitor, q # Q .1 ! e !t / RC / .
SET UP:! ln e x # x DirectCurrent Circuits! ! 2617 EXECUTE:! (a) Q # CV # (5.90 " 10!6 F)(28.0 V) # 1.65 " 10!4 C.
(b) q # Q(1 ! e! t / RC ) , so e ! t / RC # 1 !
t # 3 " 10!3 s: R # q
!t
and R #
. After
Q
C ln(1 ! q/Q ) !3 " 10!3 s
# 463 U.
(5.90 " 10 F)(ln(1 ! 110 /165))
!6 (c) If the charge is to be 99% of final value: 26.49. q
# (1 ! e ! t / RC ) gives
Q t # ! RC ln(1 ! q / Q ) # ! (463 U) (5.90 " 10!6 F) ln(0.01) # 0.0126 s.
EVALUATE:! The time constant is @ # RC # 2.73 ms . The time in part (b) is a bit more than one time constant and
the time in part (c) is about 4.6 time constants.
IDENTIFY:! For each circuit apply the loop rule to relate the voltages across the circuit elements.
(a) SET UP:! With the switch in position 2 the circuit is the charging circuit shown in Figure 26.49a. At t = 0, q = 0. Figure 26.49a
EXECUTE:! The charge q on the capacitor is given as a function of time by Eq.(26.12):
q # C E .1 ! e! t / RC / Qf # C E # .1.50 " 10!5 F / .18.0 V / # 2.70 " 10!4 C.
RC # . 980 U / .1.50 " 10!5 F / # 0.0147 s . / Thus, at t # 0.0100 s, q # . 2.70 " 10!4 C / 1 ! e ! . 0.0100 s / / . 0.0147 s / # 133 $ C.
q
133 $ C
#
# 8.87 V
C 1.50 " 10!5 F
The loop rule says E ! vC ! vR # 0
(b) vC # vR # E ! vC # 18.0 V ! 8.87 V # 9.13 V
(c) SET UP:! Throwing the switch back to position 1 produces the discharging circuit shown in Figure 26.49b.
The initial charge Q0 is the
charge calculated in part (b),
Q0 # 133 $ C.
Figure 26.49b
q
133 $ C
#
# 8.87 V, the same as just before the switch is thrown. But now
C 1.50 " 10!5 F
vC ! vR # 0, so vR # vC # 8.87 V.
(d) SET UP:! In the discharging circuit the charge on the capacitor as a function of time is given by Eq.(26.16):
q # Q0e ! t / RC . EXECUTE:! vC # EXECUTE:! RC # 0.0147 s, the same as in part (a). Thus at t # 0.0100 s, q # (133 $ C)e !. (0.0100 s) /(0.0147 s) / # 67.4 $ C.
EVALUATE:! t # 10.0 ms is less than one time constant, so at the instant described in part (a) the capacitor is not
fully charged; its voltage (8.87 V) is less than the emf. There is a charging current and a voltage drop across the
resistor. In the discharging circuit the voltage across the capacitor starts at 8.87 V and decreases. After t # 10.0 ms
it has decreased to vC # q/C # 4.49 V.
26.50. IDENTIFY:! P # VI # I 2 R
SET UP:! Problem 25.77 says that for 12gauge wire the maximum safe current is 2.5 A.
P 4100 W
EXECUTE:! (a) I # #
# 17.1 A. So we need at least 14gauge wire (good up to 18 A). 12 gauge is also
240 V
V
ok (good up to 25 A). 2618 Chapter 26 V 2 (240 V) 2
V2
#
# 14 U .
and R #
P
4100 W
R
(c) At 11c per kWH, for 1 hour the cost is (11c/kWh)(1 h)(4.1 kW) # 45c .
[
[
[
EVALUATE:! The cost to operate the device is proportional to its power consumption.
IDENTIFY and SET UP:! The heater and hair dryer are in parallel so the voltage across each is 120 V and the
current through the fuse is the sum of the currents through each appliance. As the power consumed by the dryer
increases the current through it increases. The maximum power setting is the highest one for which the current
through the fuse is less than 20 A.
EXECUTE:! Find the current through the heater. P # VI so I # P / V # 1500 W /120 V # 12.5 A. The maximum
total current allowed is 20 A, so the current through the dryer must be less than 20 A ! 12.5 A # 7.5 A. The power
dissipated by the dryer if the current has this value is P # VI # .120 V / . 7.5 A / # 900 W. For P at this value or
(b) P # 26.51. 26.52. larger the circuit breaker trips.
EVALUATE:! P # V 2 / R and for the dryer V is a constant 120 V. The higher power settings correspond to a
smaller resistance R and larger current through the device.
IDENTIFY:! The current gets split evenly between all the parallel bulbs.
P 90 W
# 0.75 A .
SET UP:! A single bulb will draw I # #
V 120 V
20 A
# 26.7. So you can attach 26 bulbs safely.
0.75 A
EVALUATE:! In parallel the voltage across each bulb is the circuit voltage.
IDENTIFY and SET UP:! Ohm's law and Eq.(25.18) can be used to calculate I and P given V and R. Use Eq.(25.12)
to calculate the resistance at the higher temperature.
(a) EXECUTE:! When the heater element is first turned on it is at room temperature and has resistance R # 20 U .
V 120 V
I# #
# 6.0 A
R 20 U
EXECUTE:! Number of bulbs N 26.53. V 2 .120 V /
#
# 720 W
R
20 U
(b) Find the resistance R(T) of the element at the operating temperature of 2803C.
Take T0 # 23.03C and R0 # 20 U. Eq.(25.12) gives
2 P# .. R .T / # R0 .1 0 2 .T ! T0 / / # 20 U 1 0 2.8 " 10!3 . C3 / I# !1 / . 2803C ! 23.03C// # 34.4 U. V 120 V
#
# 3.5 A
R 34.4 U V 2 .120 V /
#
# 420 W
34.4 U
R
EVALUATE:! When the temperature increases, R increases and I and P decrease. The changes are substantial.
(a) IDENTIFY:! Two of the resistors in series would each dissipate onehalf the total, or 1.2 W, which is ok. But
the series combination would have an equivalent resistance of 800 U, not the 400 U that is required. Resistors in
parallel have an equivalent resistance that is less than that of the individual resistors, so a solution is two in series
in parallel with another two in series.
SET UP:! The network can be simplified as shown in Figure 26.54a.
2 P# 26.54. Figure 26.54a
EXECUTE:! Rs is the resistance equivalent to two of the 400 U resistors in series. Rs # R 0 R # 800 U . Req is the resistance equivalent to the two Rs # 800 U resistors in parallel: 1
1
1
2
800 U
# 0 # ; Req #
# 400 U .
Req Rs Rs Rs
2 EVALUATE:! This combination does have the required 400 U equivalent resistance. It will be shown in part (b)
that a total of 2.4 W can be dissipated without exceeding the power rating of each individual resistor. DirectCurrent Circuits! ! 2619 IDENTIFY:! Another solution is two resistors in parallel in series with two more in parallel.
SET UP:! The network can be simplified as shown in Figure 26.54b. EXECUTE:! Figure 26.54b
111
2
#0#
; Rp # 200 U ; Req # Rp 0 Rp # 400 U
Rp R R 400 U EVALUATE:! This combination has the required 400 U equivalent resistance It will be shown in part (b) that a
total of 2.4 W can be dissipated without exceeding the power rating of each individual resistor.
(b) IDENTIFY and SET UP:! Find the applied voltage Vab such that a total of 2.4 W is dissipated and then for this
Vab find the power dissipated by each resistor.
EXECUTE:! For a combination with equivalent resistance Req # 400 U to dissipate 2.4 W the voltage Vab applied
2
to the network must be given by P # Vab / Req so Vab # PReq # . 2.4 W / . 400 U / # 31.0 V and the current through the equivalent resistance is I # Vab / R # 31.0 V / 400 U # 0.0775 A. For the first combination this means 31.0 V
across each parallel branch and 1 . 31.0 V / # 15.5 V across each 400 U resistor. The power dissipated by each
2
individual resistor is then P # V 2 / R # .15.5 V / / 400 U # 0.60 W, which is less than the maximum allowed value
2 of 1.20 W. For the second combination this means a voltage of IRp # . 0.0775 A / . 200 U / # 15.5 V across each
parallel combination and hence across each separate resistor. The power dissipated by each resistor is again
2
P # V 2 / R # .15.5 V / / 400 U # 0.60 W, which is less than the maximum allowed value of 1.20 W. 26.55. EVALUATE:! The symmetry of each network says that each resistor in the network dissipates the same power. So,
for a total of 2.4 W dissipated by the network, each resistor dissipates (2.4 W) / 4 # 0.60 W, which agrees with the
above analysis.
GL
IDENTIFY:! The Cu and Ni cables are in parallel. For each, R #
.
A
SET UP:! The composite cable is sketched in Figure 26.55. The cross sectional area of the nickel segment is , a 2
and the area of the copper portion is , (b 2 ! a 2 ). For nickel G # 7.8 "10!8 U + m and for copper G # 1.72 "10!8 U + m. L
1
1
L
. Therefore,
0
. RNi # "Ni L / A # "Ni 2 and RCu # "Cu L/A # "Cu
2
! (b ! a 2 )
!a
RCable RNi RCu
!a 2 ! (b 2 ! a 2 )
.
#
0
"Ni L
"Cu L
! % a2 b2 ! a2 &
! 9 (0.050 m) 2
(0.100 m) 2 ! (0.050 m) 2 :
!6
#'
0
0
(#
>
? and RCable # 13.6 "10 U # 13.6 $U.
L ) "Ni
"Cu * 20 m ; 7.8 " 10!8 U + m
1.72 " 10!8 U + m
< EXECUTE:!
1
Rcable 1
Rcable 1 # L
L
!b 2 R ! (0.10 m) 2 (13.6 " 10!6 U)
# G eff 2 . This gives "eff #
#
# 2.14 " 10!8 U + m
A
!b
L
20 m
EVALUATE:! The effective resistivity of the cable is about 25% larger than the resistivity of copper. If nickel had
infinite resitivity and only the copper portion conducted, the resistance of the cable would be 14.6 $U , which is
not much larger than the resistance calculated in part (a).
(b) R # "eff Figure 26.55 2620 Chapter 26 26.56. IDENTIFY and SET UP:! Let R # 1.00 U, the resistance of one wire. Each half of the wire has
Rh # R / 2 # 0.500 U . The combined wires are the same as a resistor network. Use the rules for equivalent
resistance for resistors in series and parallel to find the resistance of the network, as shown in Figure 26.56.
EXECUTE:! ! Figure 26.56 The equivalent resistance is Rh 0 Rh / 2 0 Rh # 5Rh / 2 # 26.57. 5
2 . 0.500 U / # 1.25 U EVALUATE:! If the two wires were connected endtoend, the total resistance would be 2.00 U . If they were
joined sidebyside, the total resistance would be 0.500 U . Our answer is between these two limiting values.
IDENTIFY:! The terminal voltage of the battery depends on the current through it and therefore on the equivalent
resistance connected to it. The power delivered to each bulb is P # I 2 R , where I is the current through it.
SET UP:! The terminal voltage of the source is E ! Ir .
EXECUTE:! (a) The equivalent resistance of the two bulbs is1.0 U. This equivalent resistance is in series with the V
8.0 V
#
# 4.4 A and the
Rtotal 1.0 U 0 0.80 U
current through each bulb is 2.2 A. The voltage applied to each bulb is E ! Ir # 8.0 V ! (4.4 A)(0.80 U ) # 4.4 V .
internal resistance of the source, so the current through the battery is I # Therefore, Pbulb # I 2 R # (2.2 A) 2 (2.0 U) # 9.7 W .
(b) If one bulb burns out, then I # V
8.0 V
#
# 2.9 A. . The current through the remaining bulb is
Rtotal 2.0 U 0 0.80 U 26.58. 2.9 A, and P # I 2 R # (2.9 A) 2 (2.0 U) # 16.3 W . The remaining bulb is brighter than before, because it is
consuming more power.
EVALUATE:! In Example 26.2 the internal resistance of the source is negligible and the brightness of the
remaining bulb doesn’t change when one burns out.
IDENTIFY:! Half the current flows through each parallel resistor and the full current flows through the third resistor,
that is in series with the parallel combination. Therefore, only the series resistor will be at its maximum power.
SET UP:! P # I 2 R
EXECUTE:! The maximum allowed power is when the total current is the maximum allowed value of
I # P / R # 36 W / 2.4 U # 3.9 A. Then half the current flows through the parallel resistors and the maximum 26.59. power is Pmax # ( I / 2) 2 R 0 ( I / 2) 2 R 0 I 2 R # 3 I 2 R # 3 (3.9 A) 2 (2.4 U) # 54 W.
2
2
EVALUATE:! If all three resistors were in series or all three were in parallel, then the maximum power would be
3(36 W) # 108 W . For the network in this problem, the maximum power is half this value.
IDENTIFY:! The ohmmeter reads the equivalent resistance between points a and b. Replace series and parallel
combinations by their equivalent.
1
1
1
SET UP:! For resistors in parallel,
#0
. For resistors in series, Req # R1 0 R2
Req R1 R2
EXECUTE:! Circuit (a): The 75.0 U and 40.0 U resistors are in parallel and have equivalent resistance 26.09 U .
The 25.0 U and 50.0 U resistors are in parallel and have an equivalent resistance of 16.67 U . The equivalent
1
1
1
#
0
network is given in Figure 26.59a.
, so Req # 18.7 U .
Req 100.0 U 23.05 U Figure 26.59a DirectCurrent Circuits! ! 2621 Circuit (b): The 30.0 U and 45.0 U resistors are in parallel and have equivalent resistance 18.0 U . The
equivalent network is given in Figure 26.59b. 1
1
1
#
0
, so Req # 7.5 U .
Req 10.0 U 30.3 U Figure 26.59b 26.60. EVALUATE:! In circuit (a) the resistance along one path between a and b is 100.0 U , but that is not the equivalent
resistance between these points. A similar comment can be made about circuit (b).
IDENTIFY:! Heat, which is generated in the resistor, melts the ice.
SET UP:! Find the rate at which heat is generated in the 20.0& resistor using P # V 2 /R. Then use the heat of
fusion of ice to find the rate at which the ice melts. The heat dH to melt a mass of ice dm is dH = LF dm, where LF
is the latent heat of fusion. The rate at which heat enters the ice, dH /dt, is the power P in the resistor, so P = LF
dm/dt. Therefore the rate of melting of the ice is dm/dt = P/LF.
EXECUTE:! The equivalent resistance of the parallel branch is 5.00 &, so the total resistance in the circuit is 35.0 &.
Therefore the total current in the circuit is ITotal = (45.0 V)/(35.0 &) = 1.286 A. The potential difference across
the 20.0& resistor in the ice is the same as the potential difference across the parallel branch: Vice = ITotalRp =
(1.286 A)(5.00 &) = 6.429 V. The rate of heating of the ice is Pice = Vice2/R = (6.429 V)2/(20.0 &) = 2.066 W. This
power goes into to heat to melt the ice, so dm/dt = P/LF = (2.066 W)/(3.34 " 105 J/kg) = 6.19 " 10–6 kg/s = 6.19 " 10–3 g/s
26.61. EVALUATE:! The melt rate is about 6 mg/s, which is not much. It would take 1000 s to melt just 6 g of ice.
IDENTIFY:! Apply the junction rule to express the currents through the 5.00 U and 8.00 U resistors in terms of
I1 , I 2 and I 3 . Apply the loop rule to three loops to get three equations in the three unknown currents.
SET UP:! The circuit is sketched in Figure 26.61. Figure 26.61 The current in each branch has been written in terms of I1 , I 2 and I 3 such that the junction rule is satisfied at each
junction point.
EXECUTE:! Apply the loop rule to loop (1).
!12.0 V 0 I 2 .1.00 U / 0 . I 2 ! I 3 /. 5.00 U / # 0
I 2 . 6.00 U / ! I 3 . 5.00 U / # 12.0 V eq.(1) 2622 Chapter 26 Apply the loop rule to loop (2).
! I1 .1.00 U / 0 9.00 V ! . I1 0 I 3 /. 8.00 U / # 0
I1 . 9.00 U / 0 I 3 . 8.00 U / # 9.00 V eq.(2) Apply the loop rule to loop (3).
! I 3 .10.0 U / ! 9.00 V 0 I1 .1.00 U / ! I 2 .1.00 U / 0 12.0 V # 0
! I1 .1.00 U / 0 I 2 .1.00 U / 0 I 3 .10.0 U / # 3.00 V eq.(3) Eq.(1) gives I 2 # 2.00 A 0 I 3 ; eq.(2) gives I1 # 1.00 A ! I 3
5
6 8
9 5
Using these results in eq.(3) gives ! .1.00 A ! 8 I 3 / .1.00 U / 0 . 2.00 A 0 6 I 3 / .1.00 U / 0 I 3 .10.0 U / # 3.00 V
9
18
. 160150180 / I3 # 2.00 A; I3 # 211 . 2.00 A / # 0.171 A
18 Then I 2 # 2.00 A 0 5 I 3 # 2.00 A 0 5 . 0.171 A / # 2.14 A and I1 # 1.00 A ! 8 I 3 # 1.00 A ! 8 . 0.171 A / # 0.848 A.
6
6
9
9
EVALUATE:! We could check that the loop rule is satisfied for a loop that goes through the 5.00 U, 8.00 U and
10.0 U resistors. Going around the loop clockwise: ! . I 2 ! I 3 / . 5.00 U / 0 . I1 0 I 3 / . 8.00 U / 0 I 3 .10.0 U / #
26.62. !9.85 V 0 8.15 V 0 1.71 V, which does equal zero, apart from rounding.
IDENTIFY:! Apply the junction rule and the loop rule to the circuit.
SET UP:! Because of the polarity of each emf, the current in the 7.00 U resistor must be in the direction shown in
Figure 26.62a. Let I be the current in the 24.0 V battery.
EXECUTE:! The loop rule applied to loop (1) gives: 024.0 V ! (1.80 A)(7.00 U) ! I (3.00 U) # 0 . I # 3.80 A . The
junction rule then says that the current in the middle branch is 2.00 A, as shown in Figure 26.62b. The loop rule
applied to loop (2) gives: 0 E ! (1.80 A)(7.00 U) 0 (2.00 A)(2.00 U) # 0 and E # 8.6 V .
EVALUATE:! We can check our results by applying the loop rule to loop (3) in Figure 26.62b:
024.0 V ! E ! (2.00 A)(2.00 U) ! (3.80 A)(3.00 U) # 0 and P # 24.0 V ! 4.0 V ! 11.4 V # 8.6 V , which agrees
with our result from loop (2). Figure 26.62
26.63. IDENTIFY and SET UP:! The circuit is sketched in Figure 26.63. Two unknown currents I1 (through the 2.00 U
resistor) and I 2 (through the 5.00 U resistor)
are labeled on the circuit diagram. The current
through the 4.00 U resistor has been written as
I 2 ! I1 using the junction rule. Figure 26.63 Apply the loop rule to loops (1) and (2) to get two equations for the unknown currents, I1 and I 2 . Loop (3) can
then be used to check the results. DirectCurrent Circuits! ! 2623 EXECUTE:! loop (1): 020.0 V ! I1 . 2.00 U / ! 14.0 V 0 . I 2 ! I1 / . 4.00 U / # 0 6.00 I1 ! 4.00 I 2 # 6.00 A
3.00 I1 ! 2.00 I 2 # 3.00 A eq.(1) loop (2): 036.0 V ! I 2 . 5.00 U / ! . I 2 ! I1 / . 4.00 U / # 0
!4.00 I1 0 9.00 I 2 # 36.0 A eq.(2) 2
Solving eq. (1) for I1 gives I1 # 1.00 A 0 3 I 2 Using this in eq.(2) gives !4.00 .1.00 A 0 2 I 2 / 0 9.00 I 2 # 36.0 A
3 . ! 83 0 9.00 / I 2 # 40.0 A and I 2 # 6.32 A.
Then I1 # 1.00 A 0 2 I 2 # 1.00 A 0 2 . 6.32 A / # 5.21 A.
3
3
In summary then
Current through the 2.00 U resistor: I1 # 5.21 A.
Current through the 5.00 U resistor: I 2 # 6.32 A.
Current through the 4.00 U resistor: I 2 ! I1 # 6.32 A ! 5.21 A # 1.11 A. EVALUATE:! Use loop (3) to check. 020.0 V ! I1 . 2.00 U / ! 14.0 V 0 36.0 V ! I 2 . 5.00 U / # 0 . 5.21 A /. 2.00 U / 0 . 6.32 A /. 5.00 U / # 42.0 V
26.64. 10.4 V 0 31.6 V # 42.0 V, so the loop rule is satisfied for this loop.
IDENTIFY:! Apply the loop and junction rules.
SET UP:! Use the currents as defined on the circuit diagram in Figure 26.64 and obtain three equations to solve for
the currents.
EXECUTE:! Left loop: 14 ! I1 ! 2( I1 ! I 2 ) # 0 and 3I1 ! 2 I 2 # 14 .
Top loop : ! 2( I ! I1 ) 0I 2 0 I1 # 0 and !2 I 0 3I1 0 I 2 # 0 .
Bottom loop : ! ( I ! I1 0 I 2 ) 0 2( I1 ! I 2 ) ! I 2 # 0 and ! I 0 3I1 ! 4 I 2 # 0.
Solving these equations for the currents we find: I # I battery # 10.0 A; I1 # I R1 # 6.0 A; I 2 # I R3 # 2.0 A.
So the other currents are: I R2 # I ! I1 # 4.0 A; I R4 # I1 ! I 2 # 4.0 A; I R5 # I ! I1 0 I 2 # 6.0 A.
14.0 V
(b) Req # V #
# 1.40 U.
I 10.0 A
EVALUATE:! It isn’t possible to simplify the resistor network using the rules for resistors in series and parallel.
But the equivalent resistance is still defined by V # IReq . Figure 26.64
26.65. (a) IDENTIFY:! Break the circuit between points a and b means no current in the middle branch that contains the
3.00 U resistor and the 10.0 V battery. The circuit therefore has a single current path. Find the current, so that
potential drops across the resistors can be calculated. Calculate Vab by traveling from a to b, keeping track of the
potential changes along the path taken. 2624 Chapter 26 SET UP:! The circuit is sketched in Figure 26.65a. Figure 26.65a
EXECUTE:! Apply the loop rule to loop (1).
012.0 V ! I .1.00 U 0 2.00 U 0 2.00 U 0 1.00 U / ! 8.0 V ! I . 2.00 U 0 1.00 U / # 0 12.0 V ! 8.0 V
# 0.4444 A.
9.00 U
To find Vab start at point b and travel to a, adding up the potential rises and drops. Travel on path (2) shown on
the diagram. The 1.00 U and 3.00 U resistors in the middle branch have no current through them and hence no
voltage across them. Therefore, Vb ! 10.0 V 0 12.0 V ! I .1.00 U 0 1.00 U 0 2.00 U / # Va ; thus
I# Va ! Vb # 2.0 V ! . 0.4444 A /. 4.00 U / # 00.22 V (point a is at higher potential)
EVALUATE:! As a check on this calculation we also compute Vab by traveling from b to a on path (3). Vb ! 10.0 V 0 8.0 V 0 I . 2.00 U 0 1.00 U 0 2.00 U / # Va Vab # !2.00 V 0 . 0.4444 A /. 5.00 U / # 00.22 V, which checks.
(b) IDENTIFY and SET UP:! With points a and b connected by a wire there are three current branches, as shown in
Figure 26.65b. Figure 26.65b The junction rule has been used to write the third current (in the 8.0 V battery) in terms of the other currents. Apply
the loop rule to loops (1) and (2) to obtain two equations for the two unknowns I1 and I 2 .
EXECUTE:! Apply the loop rule to loop (1).
12.0 V ! I1 .1.00 U / ! I1 . 2.00 U / ! I 2 .1.00 U / ! 10.0 V ! I 2 . 3.00 U / ! I1 .1.00 U / # 0 2.0 V ! I1 . 4.00 U / ! I 2 . 4.00 U / # 0 . 2.00 U / I1 0 . 2.00 U / I 2 # 1.0 V eq.(1) Apply the loop rule to loop (2).
! . I1 ! I 2 /. 2.00 U / ! . I1 ! I 2 /.1.00 U / ! 8.0 V ! . I1 ! I 2 / . 2.00 U / 0 I 2 . 3.00 U / 0 10.0 V 0 I 2 .1.00 U / # 0
2.0 V ! . 5.00 U / I1 0 . 9.00 U / I 2 # 0 eq.(2) DirectCurrent Circuits! ! 2625 Solve eq.(1) for I 2 and use this to replace I 2 in eq.(2).
I 2 # 0.50 A ! I1 2.0 V ! . 5.00 U / I1 0 . 9.00 U /. 0.50 A ! I1 / # 0 .14.0 U / I1 # 6.50 V so I1 # . 6.50 V / / .14.0 U / # 0.464 A
I 2 # 0.500 A ! 0.464 A # 0.036 A. 26.66. The current in the 12.0 V battery is I1 # 0.464 A.
EVALUATE:! We can apply the loop rule to loop (3) as a check.
012.0 V ! I1 (1.00 U 0 2.00 U 0 1.00 U) ! ( I1 ! I 2 )(2.00 U 0 1.00 U 0 2.00 U ) ! 8.0 V # 4.0 V ! 1.86 V ! 2.14 V # 0,
as it should.
IDENTIFY:! Simplify the resistor networks as much as possible using the rule for series and parallel combinations
of resistors. Then apply Kirchhoff’s laws.
SET UP:! First do the series/parallel reduction. This gives the circuit in Figure 26.66. The rate at which the
10.0 U resistor generates thermal energy is P # I 2 R .
EXECUTE:! Apply Kirchhoff’s laws and solve for E . BVadefa # 0 : ! (20 U)(2 A) ! 5 V ! (20 U) I 2 # 0 .
This gives I 2 # ! 2.25 A . Then I1 0 I 2 # 2 A gives I1 # 2 A ! (!2.25 A) # 4.25 A .
BVabcdefa # 0: (15 U)(4.25 A) 0 E ! (20 U)(!2.25 A) # 0 . This gives E # !109 V . Since E is calculated to be
negative, its polarity should be reversed.
(b) The parallel network that contains the 10.0 U resistor in one branch has an equivalent resistance of 10 U . The
voltage across each branch of the parallel network is Vpar # RI # (10 U)(2A) # 20 V . The current in the upper
20 V 2
# A . Pt # E , so I 2 Rt # E , where E # 60.0 J .
branch is I # V #
R 30 U 3 . 32 A / 2 (10 U)t # 60 J , and t # 13.5 s . EVALUATE:! For the 10.0 U resistor, P # I 2 R # 4.44 W . The total rate at which electrical energy is inputed to
the circuit in the emf is (5.0 V)(2.0 A) 0 (109 V)(4.25 A) # 473 J . Only a small fraction of the energy is dissipated
in the 10.0 U resistor. Figure 26.66
26.67. IDENTIFY:! In Figure 26.67, points a and c are at the same potential and points d and b are at the same potential,
so we can calculate Vab by calculating Vcd . We know the current through the resistor that is between points c and
d. We thus can calculate the terminal voltage of the 24.0 V battery without calculating the current through it.
SET UP:! ! Figure 26.67
EXECUTE:! Vd 0 I1 .10.0 U / 0 12.0 V # Vc Vc ! Vd # 12.7 V; Va ! Vb # Vc ! Vd # 12.7 V
EVALUATE:! The voltage across each parallel branch must be the same. The current through the 24.0 V battery
must be . 24.0 V ! 12.7 V / / .10.0 U / # 1.13 A in the direction b to a. 2626 Chapter 26 26.68. IDENTIFY:! The current through the 40.0 U resistor equals the current through the emf, and the current through
each of the other resistors is less than or equal to this current. So, set P40 # 1.00 W and use this to solve for the 26.69. current I through the emf. If P40 # 1.00 W , then P for each of the other resistors is less than 1.00 W.
SET UP:! Use the equivalent resistance for series and parallel combinations to simplify the circuit.
EXECUTE:! I 2 R # P gives I 2 (40 U) # 1 W , and I # 0.158 A . Now use series / parallel reduction to simplify the
circuit. The upper parallel branch is 6.38 U and the lower one is 25 U . The series sum is now 126 U . Ohm’s law
gives E # (126 U)(0.158 A) # 19.9 V .
EVALUATE:! The power input from the emf is EI # 3.14 W , so nearly onethird of the total power is dissipated in
the 40.0 U resistor.
IDENTIFY and SET UP:! Simplify the circuit by replacing the parallel networks of resistors by their equivalents. In
this simplified circuit apply the loop and junction rules to find the current in each branch.
EXECUTE:! The 20.0U and 30.0U resistors are in parallel and have equivalent resistance 12.0 U . The two
resistors R are in parallel and have equivalent resistance R/2. The circuit is equivalent to the circuit sketched in
Figure 26.69. Figure 26.69
(a) Calculate Vca by traveling along the branch that contains the 20.0 V battery, since we know the current in that
branch.
Va ! . 5.00 A /.12.0 U / ! . 5.00 A /.18.0 U / ! 20.0 V # Vc Va ! Vc # 20.0 V 0 90.0 V 0 60.0 V # 170.0 V
Vb ! Va # Vab # 16.0 V
X ! Vba # 170.0 V so X # 186.0 V, with the upper terminal 0
(b) I1 # .16.0 V / / . 8.0 U / # 2.00 A The junction rule applied to point a gives I 2 0 I1 # 5.00 A, so I 2 # 3.00 A. The current through the 200.0 V battery
is in the direction from the ! to the 0 terminal, as shown in the diagram.
(c) 200.0 V ! I 2 . R / 2/ # 170.0 V . 3.00 A / . R / 2/ # 30.0 V so R # 20.0 U
EVALUATE:! We can check the loop rule by going clockwise around the outer circuit loop. This gives
020.0 V 0 . 5.00 A /.18.0 U 0 12.0 U / 0 . 3.00 A / .10.0 U / ! 200.0 V # 20.0 V 0 150.0 V 0 30.0 V ! 200.0 V,
26.70. which does equal zero.
V2
IDENTIFY:! Ptot #
.
Req
SET UP:! Let R be the resistance of each resistor.
EXECUTE:! When the resistors are in series, Req # 3R and Ps # V2
. When the resistors are in parallel, Req # R / 3 .
3R V2
V2
#3
# 9 Ps # 9(27 W) # 243 W .
R/3
R
EVALUATE:! In parallel, the voltage across each resistor is the full applied voltage V. In series, the voltage across
each resistor is V / 3 and each resistor dissipates less power.
Pp # DirectCurrent Circuits! ! 2627 26.71. IDENTIFY and SET UP:! For part (a) use that the full emf is across each resistor. In part (b), calculate the power
dissipated by the equivalent resistance, and in this expression express R1 and R2 in terms of P , P2 and E .
1
EXECUTE:! P # E 2 / R1 so R1 # E 2 / P
1
1
P2 # E 2 / R2 so R2 # E 2 / P2
(a) When the resistors are connected in parallel to the emf, the voltage across each resistor is E and the power
dissipated by each resistor is the same as if only the one resistor were connected. Ptot # P 0 P2
1 (b) When the resistors are connected in series the equivalent resistance is Req # R1 0 R2
Ptot # E2
E2
PP
#2
# 12
R1 0 R2 E / P 0 E 2 / P2 P 0 P2
1
1 1
11
# 0 . Our results are that for parallel the powers add
Ptot P P2
1
and that for series the reciprocals of the power add. This is opposite the result for combining resistance. Since
P # E 2 / R tells us that P is proportional to 1/R, this makes sense.
IDENTIFY and SET UP:! Just after the switch is closed the charge on the capacitor is zero, the voltage across the
capacitor is zero and the capacitor can be replaced by a wire in analyzing the circuit. After a long time the current
to the capacitor is zero, so the current through R3 is zero. After a long time the capacitor can be replaced by a break
in the circuit.
EXECUTE:! (a) Ignoring the capacitor for the moment, the equivalent resistance of the two parallel resistors is
1
1
1
3
#
0
#
; Req # 2.00 U . In the absence of the capacitor, the total current in the circuit (the
Req 6.00 U 3.00 U 6.00 U
EVALUATE:! The result in part (b) can be written as 26.72. current through the 8.00 U resistor) would be i # E
42.0 V
#
# 4.20 A , of which 2 3 , or 2.80 A, would
R 8.00 U 0 2.00 U go through the 3.00 U resistor and 1 3 , or 1.40 A, would go through the 6.00 U resistor. Since the current
V ! t RC
, at the instant t # 0 the circuit behaves as through the capacitor were
e
R
not present, so the currents through the various resistors are as calculated above.
(b) Once the capacitor is fully charged, no current flows through that part of the circuit. The 8.00 U and the through the capacitor is given by i # 6.00 U resistors are now in series, and the current through them is i # E R # ( 42.0 V) /(8.00 U 0 6.00 U) # 3.00 A.
The voltage drop across both the 6.00 U resistor and the capacitor is thus V # iR # (3.00 A)(6.00 U ) # 18.0 V.
(There is no current through the 3.00 U resistor and so no voltage drop across it.) The charge on the capacitor is Q # CV # (4.00 " 10!6 F)(18.0 V) # 7.2 " 10!5 C .
EVALUATE:! The equivalent resistance of R2 and R3 in parallel is less than R3 , so initially the current through
26.73. R1 is larger than its value after a long time has elapsed.
(a) IDENTIFY and SET UP:! The circuit is sketched in Figure 26.73a. With the switch open there is no
current through it and there are
only the two currents I1 and I 2
indicated in the sketch. Figure 26.73a The potential drop across each parallel branch is 36.0 V. Use this fact to calculate I1 and I 2 . Then travel from
point a to point b and keep track of the potential rises and drops in order to calculate Vab . 2628 Chapter 26 EXECUTE:! ! I1 (6.00 U 0 3.00 U ) 0 36.0 V # 0
36.0 V
I1 #
# 4.00 A
6.00 U 0 3.00 U
! I 2 (3.00 U 0 6.00 U ) 0 36.0 V # 0
36.0 V
I2 #
# 4.00 A
3.00 U 0 6.00 U
To calculate Vab # Va ! Vb start at point b and travel to point a, adding up all the potential rises and drops along the
way. We can do this by going from b up through the 3.00 U resistor:
Vb 0 I 2 (3.00 U ) ! I1 (6.00 U ) # Va
Va ! Vb # ( 4.00 A)(3.00 U ) ! (4.00 A)(6.00 U ) # 12.0 V ! 24.0 V # !12.0 V
Vab # !12.0 V (point a is 12.0 V lower in potential than point b)
EVALUATE:! Alternatively, we can go from point b down through the 6.00 U resistor.
Vb ! I 2 (6.00 U ) 0 I1 (3.00 U ) # Va
Va ! Vb # !( 4.00 A)(6.00 U ) 0 (4.00 A)(3.00 U ) # !24.0 V 0 12.0 V # !12.0 V, which checks.
(b) IDENTIFY:! Now there are multiple current paths, as shown in Figure 26.73b. Use junction rule to write the
current in each branch in terms of three unknown currents I1, I2, and I3. Apply the loop rule to three loops to get
three equations for the three unknowns. The target variable is I3, the current through the switch. Req is calculated
from V # IReq , where I is the total current that passes through the network.
SET UP:! The three unknown currents I1 , I 2 , and I 3
are labeled on Figure 26.73b. Figure 26.73b
EXECUTE:! Apply the loop rule to loops (1), (2), and (3).
loop (1): ! I1 (6.00 U) 0 I 3 (3.00 U) 0 I 2 (3.00 U) # 0 I 2 # 2 I1 ! I 3 eq.(1) loop (2): !( I1 0 I 3 )(3.00 U) 0 ( I 2 ! I 3 )(6.00 U) ! I 3 (3.00 U) # 0
6 I 2 ! 12 I 3 ! 3I1 # 0 so 2 I 2 ! 4 I 3 ! I1 # 0
Use eq(1) to replace I 2 :
4 I1 ! 2 I 3 ! 4 I 3 ! I1 # 0
3I1 # 6 I 3 and I1 # 2 I 3
eq.(2)
loop (3) (This loop is completed through the battery [not shown], in the direction from the ! to the 0 terminal.):
! I1 (6.00 U) ! ( I1 0 I 3 )(3.00 U) 0 36.0 V # 0
9 I1 0 3I 3 # 36.0 A and 3I1 0 I 3 # 12.0 A
Use eq.(2) in eq.(3) to replace I1:
3(2 I 3 ) 0 I 3 # 12.0 A
I 3 # 12.0 A / 7 # 1.71 A
I1 # 2 I 3 # 3.42 A
I 2 # 2 I1 ! I 3 # 2(3.42 A) ! 1.71 A # 5.13 A
The current through the switch is I 3 # 1.71 A. eq.(3) DirectCurrent Circuits! ! 2629 (c) From the results in part (a) the current through the battery is I # I1 0 I 2 # 3.42 A 0 5.13 A # 8.55 A. The
equivalent circuit is a single resistor that produces the same current through the 36.0 V battery, as shown in
Figure 26.73c. ! IR 0 36.0 V # 0
36.0 V 36.0 V
R#
#
# 4.21 U
I
8.55 A
Figure 26.73c 26.74. EVALUATE:! With the switch open (part a), point b is at higher potential than point a, so when the switch is closed
the current flows in the direction from b to a. With the switch closed the circuit cannot be simplified using series
and parallel combinations but there is still an equivalent resistance that represents the network.
IDENTIFY:! With S open and after equilibrium has been reached, no current flows and the voltage across each
capacitor is 18.0 V. When S is closed, current I flows through the 6.00 U and 3.00 U resistors.
SET UP:! With the switch closed, a and b are at the same potential and the voltage across the 6.00 U resistor
equals the voltage across the 6.00 $ F capacitor and the voltage is the same across the 3.00 $ F capacitor and
3.00 U resistor.
EXECUTE:! (a) With an open switch: Vab # E # 18.0 V .
(b) Point a is at a higher potential since it is directly connected to the positive terminal of the battery.
(c) When the switch is closed 18.0 V # I (6.00 U 0 3.00 U) . I # 2.00 A and Vb # (2.00 A)(3.00 U) # 6.00 V.
(d) Initially the capacitor’s charges were Q3 # CV # (3.00 " 10!6 F)(18.0 V) # 5.40 " 10!5 C and Q6 # CV # (6.00 " 10!6 F)(18.0 V) # 1.08 " 10!4 C. After the switch is closed Q3 # CV # (3.00 " 10!6 F)(18.0 V ! 12.0 V) # 1.80 " 10!5 C and
Q6 # CV # (6.00 " 10!6 F)(18.0 V ! 6.0 V) # 7.20 " 10!5 C. Both capacitors lose 3.60 " 10!5 C. 26.75. EVALUATE:! The voltage across each capacitor decreases when the switch is closed, because there is then current
through each resistor and therefore a potential drop across each resistor.
IDENTIFY:! The current through the galvanometer for fullscale deflection is 0.0200 A. For each connection, there
are two parallel branches and the voltage across each is the same.
SET UP:! The sum of the two currents in the parallel branches for each connection equals the current into the
meter for that connection.
EXECUTE:! From the circuit we can derive three equations:
(i) ( R1 0 R2 0 R3 )(0.100 A ! 0.0200 A) # (48.0 U)(0.0200 A) and R1 0 R2 0 R3 # 12.0 U. (ii) ( R1 0 R2 )(1.00 A ! 0.0200 A) # (48.0 U 0 R3 )(0.0200 A) and R1 0 R2 ! 0.0204 R3 # 0.980 U.
(iii) R1 (10.0 A ! 0.0200 A) # (48.0 U 0 R2 0 R3 )(0.0200 A) and R1 ! 0.002 R2 ! 0.002R3 # 0.096 U. 26.76. From (i)!and (ii), R3 # 10.8 U. From (ii)!and (iii), R2 # 1.08 U . Therefore, R1 # 0.12 U .
EVALUATE:! For the 0.100 A setting the circuit consists of 48.0 U and R1 0 R2 0 R3 # 12.0 U in parallel and the
equivalent resistance of the meter is 9.6 U . For each of the other two settings the equivalent resistance of the
meter is less than 9.6 U .
IDENTIFY:! In each case the sum of the voltage drops across the resistors in the circuit must equal the fullscale
voltage reading. The resistors are in series so the total resistance is the sum of the resistances in the circuit.
SET UP:! For each range setting the circuit has the form shown in Figure 26.76. Figure 26.76
EXECUTE:! 3.00 V For V # 3.00 V, R # R1 and the total meter resistance Rm is Rm # RG 0 R1.
V # I fs Rm so Rm # V
3.00 V
#
# 3.00 " 103 U .
I fs 1.00 " 10!3 A Rm # RG 0 R1 so R1 # Rm ! RG # 3.00 " 103 U ! 40.0 U # 2960 U 2630 Chapter 26 15.0 V
For V # 15.0 V, R # R1 0 R2 and the total meter resistance is Rm # RG 0 R1 0 R2 .
V
15.0 V
V # I fs Rm so Rm #
#
# 1.50 " 104 U .
I fs 1.00 " 10!3 A R2 # Rm ! RG ! R1 # 1.50 " 104 U ! 40.0 U ! 2960 U # 1.20 " 104 U
150 V
For V # 150 V, R # R1 0 R2 0 R3 and the total meter resistance is Rm # RG 0 R1 0 R2 0 R3 .
V
150 V
V # I fs Rm so Rm #
#
# 1.50 " 105 U .
I fs 1.00 " 10!3 A 26.77. R3 # Rm ! RG ! R1 ! R2 # 1.50 " 105 U ! 40.0 U ! 2960 U ! 1.20 " 104 U # 1.35 " 105 U .
EVALUATE:! The greater the total resistance in series inside the meter the greater the potential difference between
the two connections to the meter when the same 1.00 mA current flows through it.
IDENTIFY:! Connecting the voltmeter between point b and ground gives a resistor network and we can solve for
the current through each resistor. The voltmeter reading equals the potential drop across the 200 kU resistor.
1
1
1
#0
. For resistors in series, Req # R1 0 R2 .
SET UP:! For resistors in parallel,
Req R1 R2
!1 %1
0.400 kV
1&
EXECUTE:! (a) Req # 100 kU 0 '
# 2.86 " 10!3 A .
0
( # 140 kU. The total current is I #
140 kU
) 200 kU 50 kU *
!1 The voltage across the 200 kU resistor is V200kU %1
1&
# IR # (2.86 " 10 A) '
0
( # 114.4 V.
200 kU 50 kU *
)
!3 (b) If VR # 5.00 " 106 U, then we carry out the same calculations as above to find Req # 292 kU ,
I # 1.37 " 10!3 A and V200kU # 263 V.
(c) If VR # F , then we find Req # 300 kU , I # 1.33 " 10!3 A and V200kU # 266 V. 26.78. EVALUATE:! When a voltmeter of finite resistance is connected to a circuit, current flows through the voltmeter
and the presence of the voltmeter alters the currents and voltages in the original circuit. The effect of the voltmeter
on the circuit decreases as the resistance of the voltmeter increases.
IDENTIFY:! The circuit consists of two resistors in series with 110 V applied across the series combination.
SET UP:! The circuit resistance is 30 kU 0 R . The voltmeter reading of 68 V is the potential across the voltmeter
terminals, equal to I (30 kU) . 110 V
. I (30 kU) # 68 V gives (68 V)(30 kU 0 R ) # (110 V)30 kU and R # 18.5 kU. .
(30 kU 0 R )
EVALUATE:! This is a method for measuring large resistances.
IDENTIFY and SET UP:! Zero current through the galvanometer means the current I1 through N is also the current
EXECUTE:! I # 26.79. through M and the current I 2 through P is the same as the current through X. And it means that points b and c are at
the same potential, so I1 N # I 2 P .
EXECUTE:! (a) The voltage between points a and d is E , so I1 # E
E
and I 2 #
. Using these
P0 X
N 0M E
E
N#
P . N ( P 0 X ) # P ( N 0 M ) . NX # PM and X # MP / N .
N 0M
P0 X
MP (850.0 U)(33.48 U)
#
# 1897 U
(b) X #
N
15.00 U
EVALUATE:! The measurement of X does not require that we know the value of the emf.
IDENTIFY:! Add resistors in series and parallel with the second galvanometer, so that the equivalent resistance is
65.0 U and so that for a current of 1.50 mA into the device the current through the galvanometer is 3.60 $ A .
SET UP:! In order for the second galvanometer to give the same fullscale deflection and to have the same
resistance as the first, we need two additional resistances as shown in Figure 26.80.
EXECUTE:! For 3.60 $ A through R the current through R1 is 1.496 mA . R and R1 are in parallel so have equal
expressions in I1 N # I 2 P gives 26.80. voltages: (3.6 $ A)(38.0 U) # (1.496 mA) R1 and R1 # 91.4 mU . And for the total resistance to be 65.0 U :
%1
&
1
65.0 U # R2 0 '
0
(
38.0 U 0.0914 U *
) !1 and R2 # 64.9 U . DirectCurrent Circuits! ! 2631 EVALUATE:! Adding R1 in parallel lowers the equivalent resistance so R2 must be added in series to raise the
equivalent resistance to 65.0 U . Figure 26.80
26.81. IDENTIFY and SET UP:! Without the meter, the circuit consists of the two resistors in series. When the meter is
connected, its resistance is added to the circuit in parallel with the resistor it is connected across.
(a) EXECUTE:! I # I1 # I 2
90.0 V
90.0 V
I#
#
# 0.1107 A
R1 0 R2 224 U 0 589 U
V1 # I1R1 # . 0.1107 A /. 224 U / # 24.8 V; V2 # I 2 R2 # . 0.1107 A / . 589 U / # 65.2 V
(b) SET UP:! The resistor network is sketched in Figure 26.81a. The voltmeter reads the potential
difference across its terminals,
which is 23.8 V. If we can find
the current I1 through the
voltmeter then we can use Ohm's
law to find its resistance.
Figure 26.81c
EXECUTE:! The voltage drop across the 589 U resistor is 90.0 V ! 23.8 V # 66.2 V, so
V 66.2 V
V 23.8 V
I# #
# 0.1124 A. The voltage drop across the 224 U resistor is 23.8 V, so I 2 # #
# 0.1062 A.
R 589 U
R 224 U
V
23.8 V
# 3840 U
Then I # I1 0 I 2 gives I1 # I ! I 2 # 0.1124 A ! 0.1062 A # 0.0062 A. RV # #
I1 0.0062 A
(c) SET UP:! The circuit with the voltmeter connected is sketched in Figure 26.81b. Figure 26.81b
EXECUTE:! Replace the two resistors in parallel by their equivalent, as shown in Figure 26.81c.
1
1
1
#
0
;
Req 3840 U 589 U Req # . 3840 U / . 589 U / # 510.7 U
3840 U 0 589 U Figure 26.81c 90.0 V
# 0.1225 A
224 U 0 510.7 U
The potential drop across the 224 U resistor then is IR # . 0.1225 A / . 224 U / # 27.4 V, so the potential drop I# across the 589 U resistor and across the voltmeter (what the voltmeter reads) is 90.0 V ! 27.4 V # 62.6 V. 2632 Chapter 26 26.82. (d) EVALUATE:! No, any real voltmeter will draw some current and thereby reduce the current through the
resistance whose voltage is being measured. Thus the presence of the voltmeter connected in parallel with the
resistance lowers the voltage drop across that resistance. The resistance of the voltmeter is only about a factor of
ten larger than the resistances in the circuit, so the voltmeter has a noticeable effect on the circuit.
IDENTIFY:! Just after the connection is made, q # 0 and the voltage across the capacitor is zero. After a long time i # 0.
SET UP:! The rate at which the resistor dissipates electrical energy is PR # V 2 / R , where V is the voltage across the resistor. The energy stored in the capacitor is q 2 /2C . The power output of the source is PP # Ei . V 2 (120 V) 2
dU
1 d (q 2 ) iq
#
# 3380 W . (ii)! PC #
#
# # 0. !
dt 2C dt
C
R
4.26 U
120 V
# 3380 W .
(iii)! PP # I # (120 V)
4.26 U
(b) After a long time, i # 0 , so PR # 0, PC # 0, PP # 0.
EVALUATE:! Initially all the power output of the source is dissipated in the resistor. After a long time energy is
stored in the capacitor but the amount stored isn't changing.
IDENTIFY:! Apply the loop rule to the circuit. The initial current determines R. We can then use the time constant
to calculate C.
SET UP:! The circuit is sketched in Figure 26.83.
EXECUTE:! (a) (i) PR # P 26.83. Initially, the charge of the
capacitor is zero, so by
v # q / C the voltage across
the capacitor is zero.
Figure 26.83 E
110 V
#
# 1.7 " 106 U
i 6.5 " 10!5 A
@
6.2 s
# 3.6 $ F.
The time constant is given by @ # RC (Eq.26.14), so C # #
R 1.7 " 106 U
EVALUATE:! The resistance is large so the initial current is small and the time constant is large.
IDENTIFY:! The energy stored in a capacitor is U # q 2 /2C. The electrical power dissipated in the resistor is P # i 2 R.
q
SET UP:! For a discharging capacitor, i # !
.
RC
Q2
(0.0081 C) 2
EXECUTE:! (a) U 0 # 0 #
# 7.10 J.
2C 2(4.62 " 10!6 F)
EXECUTE:! The loop rule therefore gives E ! iR # 0 and R # 26.84. 2 (0.0081 C) 2
%Q &
# 3616 W
(b) P0 # I 0 2 R # ' 0 ( R #
(850 U)(4.62 " 10!6 F)2
) RC *
2 26.85. 26.86. 2 1 % Q0 &
1
Q
1
1 Q0 2
%Q&
(c) When U # U 0 #
, Q # 0 . This gives P # '
( R# '
( R # P0 # 1808 W.
RC *
2 ) RC *
2
2
2 2C
2
)
EVALUATE:! All the energy originally stored in the capacitor is dissipated as current flow through the resistor.
IDENTIFY:! q # Q0e! t / RC . The time constant is @ # RC .
SET UP:! The charge of one electron has magnitude e # 1.60 " 10!19 C .
EXECUTE:! (a) We will say that a capacitor is discharged if its charge is less than that of one electron. The time this
takes is then given by q # Q0e ! t/RC, so t # RC ln(Q0 /e) # (6.7 "105 U)(9.2 "10!7 F)ln(7.0 "10!6 C/1.6 "10!19 C) # 19.36 s,
or 31.4 time constants.
EVALUATE:! (b) As shown in part (a), t # @ ln(Q0 q ) and so the number of time constants required to discharge
the capacitor is independent of R and C , and depends only on the initial charge.
IDENTIFY:! The energy changes exponentially, but it does not obey exactly the same equation as the charge since
it is proportional to the square of the charge.
(a) SET UP:! For charging, U = Q2/2C = (Q0 e–t/RC)2/2C = U0 e–2t/RC.
EXECUTE:! To reduce the energy to 1/e of its initial value:
U 0 /e # U 0e !2t/RC
t # RC/ 2 DirectCurrent Circuits! ! 2633 (b) SET UP:! For discharging, U = Q /2C = [Q0(1 – e
2 26.87. –t/RC 2 –t/RC 2 )] /2C = Umax (1 – e ) 1&
%
EXECUTE:! To reach 1/e of the maximum energy, Umax/e = Umax (1 – e–t/RC)2 and t = ! RC ln '1 !
(.
e*
)
EVALUATE:! The time to reach 1/e of the maximum energy is not the same as the time to discharge to 1/e of the
maximum energy.
IDENTIFY and SET UP:! For parts (a) and (b) evaluate the integrals as specified in the problem. The current as a
E
function of time is given by Eq.(26.13) i # e! t/RC . The energy stored in the capacitor is given by Q 2 / 2C.
R
EXECUTE:! (a) P # Ei
The total energy supplied by the battery is H F
0 Pdt # H Eidt # . E 2 /R / H e !t / RC dt # . E 2 / R / 9 ! RCe! t / RC : # CE 2 .
;
<
F F F 0 0 0 (b) P # i 2 R
The total energy dissipated in the resistor is H F
0 Pdt # H i 2 Rdt # . E 2 / R / H e!2t / RC dt # . E 2 / R / 9 ! . RC / 2 / e !2t / RC : # 1 CE 2 .
2
;
<
F F F 0 0 0 (c) The final charge on the capacitor is Q # C E . The energy stored is U # Q 2 / . 2C / # 1 CE 2 . The final energy stored
2 in the capacitor . (d) EVALUATE:! 1
2 C E 2 / # total energy supplied by the battery . CE 2 / – energy dissipated in the resistor
1
2 . 1
2 CE 2 / of the energy supplied by the battery is stored in the capacitor. This fraction is independent of R. The other of the energy supplied by the battery is dissipated in the resistor. When R is small the current initially
is large but dies away quickly. When R is large the current initially is small but lasts longer.
1
2 26.88. F IDENTIFY:! E # H Pdt . The energy stored in a capacitor is U # q 2 / 2C .
0 Q0 !t / RC
e
RC
Q2
Q2 F
Q 2 RC Q0 2
Q
EXECUTE:! i # ! 0 e !t RC gives P # i 2 R # 0 2 e!2t / RC and E # 0 2 H e!2t RC dt # 0 2
#
# U0.
0
RC
RC
RC
RC 2
2C
EVALUATE:! Increasing the energy stored in the capacitor increases current through the resistor as the capacitor
discharges.
IDENTIFY and SET UP:!
EXECUTE:! (a) Using Kirchhoff’s Rules on the circuit we find:
Left loop: 92 ! 140 I1 ! 210 I 2 0 55 # 0 I 147 ! 140 I1 ! 210 I 2 # 0.
SET UP:! i # ! 26.89. Right loop: 57 ! 35I 3 ! 210 I 2 0 55 # 0 I 112 ! 210 I 2 ! 35I 3 # 0.
Junction rule: I1 ! I 2 0 I 3 # 0.
Solving for the three currents we have: I1 # 0.300 A, I 2 # 0.500 A, I 3 # 0.200 A.
(b) Leaving only the 92V battery in the circuit:
Left loop: 92 ! 140 I1 ! 210 I 2 # 0. Right loop: !35I 3 ! 210 I 2 # 0.
Junction rule: I1 ! I 2 0 I 3 # 0. Solving for the three currents:
I1 # 0.541 A, I 2 # 0.077 A, I 3 # !0.464 A. (c)! Leaving only the 57V battery in the circuit:
Left loop: 140 I1 0 210 I 2 # 0. Right loop: 57 ! 35I 3 ! 210 I 2 # 0. Junction rule: I1 ! I 2 0 I 3 # 0. Solving for the three currents:
I1 # !0.287 A, I 2 # 0.192 A, I 3 # 0.480 A. (d) Leaving only the 55V battery in the circuit:
Left loop: 55 ! 140 I1 ! 210 I 2 # 0. Right loop: 55 ! 35I 3 ! 210 I 2 # 0. Junction rule: I1 ! I 2 0 I 3 # 0. Solving for the three currents:
I1 # 0.046 A, I 2 # 0.231 A, I 3 # 0.185 A. (e) If we sum the currents from the previous three parts we find: I1 # 0.300 A, I 2 # 0.500 A, I 3 # 0.200 A, just as in part (a). 2634 Chapter 26 (f ) ! Changing the 57V battery for an 80V battery just affects the calculation in part (c). It changes to: Left loop:
140 I1 0 210 I 2 # 0. Right loop: 80 ! 35I 3 ! 210 I 2 # 0.
Junction rule: I1 ! I 2 0 I 3 # 0. Solving for the three currents: I1 # !0.403 A, I 2 # 0.269 A, The total current for the full circuit is the sum of (b), (d) and (f ) above:
I1 # 0.184 A,
I 2 # 0.576 A,
26.90. I 3 # 0.672 A.
I 3 # 0.392 A. EVALUATE:! This problem presents an alternative means of solving for currents in multiloop circuits.
IDENTIFY and SET UP:! When C changes after the capacitor is charged, the voltage across the capacitor changes.
Current flows through the resistor until the voltage across the capacitor again equals the emf.
EXECUTE:! (a) Fully charged: Q # CV # (10.0 " 10!12 F)(1000 V) # 1.00 " 10!8 C.
(b) The initial current just after the capacitor is charged is I 0 # q
E ! VC  E
q
%E
#!
. This gives i (t ) # ' !
R
R RC ) R RC  & !t RC ,
(e
* where C  # 1.1C .
(c) We need a resistance such that the current will be greater than 1 $ A for longer than 200 $s. This requires that
at t # 200 $s , i # 1.0 " 10!6 A # 1%
1.0 " 10!8 C & ! (2.0"10!4 s)/R (11"10!12 F)
. This says
'1000 V !
(e
R)
1.1(1.0 " 10!11 F) * 7
1
(90.9)e ! (1.8"10 U ) R and 18.3R ! RlnR ! 1.8 " 107 # 0. Solving for R numerically we find
R
7.15 " 106 U N R N 7.01 " 107 U.
EVALUATE:! If the resistance is too small, then the capacitor discharges too quickly, and if the resistance is too
large, the current is not large enough.
IDENTIFY:! Consider one segment of the network attached to the rest of the network.
SET UP:! We can redraw the circuit as shown in Figure 26.91.
!1
1&
R2 RT
%1
2
2
EXECUTE:! RT # 2 R1 0 ' 0
( # 2 R1 0 R 0 R . RT ! 2 R1RT ! 2 R1R2 # 0 . RT # R1 6 R1 0 2 R1R2 . RT 5 0 ,
R2 RT *
2
T
) 1.0 " 10!6 A # 26.91. so RT # R1 0 R12 0 2 R1R2 .
EVALUATE:! Even though there are an infinite number of resistors, the equivalent resistance of the network is finite. Figure 26.91
26.92. IDENTIFY:! Assume a voltage V applied between points a and b and consider the currents that flow along each
path between a and b.
SET UP:! The currents are shown in Figure 26.92.
EXECUTE:! Let current I enter at a and exit at b. At a there are three equivalent branches, so current is I / 3 in each.
At the next junction point there are two equivalent branches so each gets current I / 6. Then at b there are three
equivalent branches with current I / 3 in each. The voltage drop from a to b then is
%I&
%I&
%I&
V # ' ( R 0 ' ( R 0 ' ( R # 5 IR. This must be the same as V # IReq , so R eq # 5 R.
6
6
) 3*
) 6*
) 3*
EVALUATE:! The equivalent resistance is less than R, even though there are 12 resistors in the network. Figure 26.92 DirectCurrent Circuits! ! 2635 26.93. IDENTIFY:! The network is the same as the one in Challenge Problem 26.91, and that problem shows that the equivalent resistance of the network is RT # R12 0 2 R1R2 .
SET UP:! The circuit can be redrawn as shown in Figure 26.93.
Req
1
RR
2 R ( R 0 R2 ) 2 R1
EXECUTE:! (a) Vcd # Vab
# Vab
and Req # 2 T . But V # 1 T
#
, so
2 R1 0 Req
2 R1 / Req 0 1
R2 0 RT
RT R2
Req Vcd # Vab 1
.
10 V (b) V1 # Vn ! 1
V0
V1
V0
V0
I V2 #
#
I Vn #
#
.
2
(1 0 V )
(1 0 V ) (1 0 V )
(1 0 V ) (1 0 V ) n If R1 # R2 , then RT # R1 0 R12 0 2 R1R1 # R1 (1 0 3) and V #
of the original voltage, we need: 2(2 0 3)
# 2.73 . So, for the nth segment to have 1%
10 3 1
1
#
N 0.01 . This says n # 4 , and then V4 # 0.005V0 .
n
(1 0 V )
(1 0 2.73)n (c) RT # R1 0 R12 0 2 R1R2 gives RT # 6400 U 0 (6400 U) 2 0 2(6400 U)(8.0 " 108 U) # 3.2 " 106 U and 2(6400 U)(3.2 " 106 U 0 8.0 " 108 U)
# 4.0 " 10!3 .
(3.2 " 106 U)(8.0 " 108 U)
(d) Along a length of 2.0 mm of axon, there are 2000 segments each 1.0 $ m long. The voltage therefore V# attenuates by V2000 # V0
V
1
# 3.4 " 10!4 .
, so 2000 #
V0
(1 0 V ) 2000
(1 0 4.0 " 10!3 )2000 (e) If R2 # 3.3 " 1012 U , then RT # 2.1 " 108 U and V # 6.2 " 10!5. This gives V2000
1
#
# 0.88.
V0
(1 0 6.2 " 10!5 )2000
EVALUATE:! As R2 increases, V decreases and the potential difference decrease from one section to the next is
less. Figure 26.93 36 DIFFRACTION 36.1. IDENTIFY: Use y = x tan θ to calculate the angular position θ of the first minimum. The minima are located by
mλ
, m = ±1, ± 2,… First minimum means m = 1 and sin θ1 = λ / a and λ = a sin θ1. Use this
Eq.(36.2): sin θ =
a
equation to calculate λ .
SET UP: The central maximum is sketched in Figure 36.1.
EXECUTE: y1 = x tan θ1
y
tan θ1 = 1 =
x
1.35 × 10−3 m
= 0.675 × 10−3
2.00 m
θ1 = 0.675 × 10−3 rad
Figure 36.1 λ = a sin θ1 = (0.750 × 10−3 m)sin(0.675 × 10−3 rad) = 506 nm
EVALUATE: θ1 is small so the approximation used to obtain Eq.(36.3) is valid and this equation could have been
used.
36.2. IDENTIFY: The angle is small, so ym = x
SET UP: y1 = 10.2 mm xλ
xλ (0.600 m)(5.46 × 10−7 m)
!a
=
= 3.21 × 10−5 m.
a
y1
10.2 × 10−3 m
EVALUATE: The diffraction pattern is observed at a distance of 60.0 cm from the slit.
mλ
IDENTIFY: The dark fringes are located at angles θ that satisfy sin θ =
, m = ±1, ± 2, ....
a
SET UP: The largest value of sin θ is 1.00.
EXECUTE: 36.3. mλ
.
a y1 = EXECUTE: (a) Solve for m that corresponds to sin θ = 1 : m = a λ = 0.0666 × 10−3 m
= 113.8. The largest value m
585 × 10−9 m can have is 113. m = ±1 , ±2 , …, ±113 gives 226 dark fringes.
" 585 × 10−9 m #
(b) For m = ±113, sin θ = ±113 $
% = ±0.9926 and θ = ±83.0° .
−3
& 0.0666 × 10 m ' 36.4. EVALUATE: When the slit width a is decreased, there are fewer dark fringes. When a < λ there are no dark
fringes and the central maximum completely fills the screen.
mλ
IDENTIFY and SET UP: λ / a is very small, so the approximate expression ym = R
is accurate. The distance
a
between the two dark fringes on either side of the central maximum is 2 y1 . 36.5. λR (633 × 10−9 m)(3.50 m)
= 2.95 × 10 −3 m = 2.95 mm . 2 y1 = 5.90 mm .
a
0.750 × 10−3 m
EVALUATE: When a is decreased, the width 2 y1 of the central maximum increases.
mλ
IDENTIFY: The minima are located by sin θ =
a
SET UP: a = 12.0 cm . x = 40.0 cm .
EXECUTE: y1 = = 361 362 36.6. Chapter 36 " 9.00 cm #
"λ#
EXECUTE: The angle to the first minimum is θ = arcsin $ % = arcsin $
% = 48.6°.
a'
&
& 12.00 cm '
So the distance from the central maximum to the first minimum is just y1 = x tan θ =
(40.0 cm) tan(48.6°) = ±45.4 cm.
EVALUATE: 2λ / a is greater than 1, so only the m = 1 minimum is seen.
IDENTIFY: The angle that locates the first diffraction minimum on one side of the central maximum is given by
λ
1
v
sin θ = . The time between crests is the period T. f = and λ = .
a
T
f
SET UP: The time between crests is the period, so T = 1.0 h .
1
1
v 800 km/h
EXECUTE: (a) f = =
= 1.0 h −1 . λ = =
= 800 km .
T 1.0 h
f
1.0 h −1
800 km
and θ = 10.2° .
4500 km
800 km
AustraliaAntarctica: sin θ =
and θ = 12.5° .
3700 km
EVALUATE: Diffraction effects are observed when the wavelength is about the same order of magnitude as the
dimensions of the opening through which the wave passes.
IDENTIFY: We can model the hole in the concrete barrier as a single slit that will produce a singleslit diffraction
pattern of the water waves on the shore.
SET UP: For singleslit diffraction, the angles at which destructive interference occurs are given by sinθm = mλ/a,
where m = 1, 2, 3, ….
EXECUTE: (a) The frequency of the water waves is f = 75.0 min −1 = 1.25 s −1 = 1.25 Hz, so their wavelength is λ = v/f =
(15.0 cm/s)/(1.25 Hz) = 12.0 cm.
At the first point for which destructive interference occurs, we have
tan θ = (0.613 m)/(3.20 m) ! θ = 10.84°. a sin θ = λ and
a = λ/sin θ = (12.0 cm)/(sin 10.84°) = 63.8 cm.
(b) First find the angles at which destructive interference occurs.
sin θ2 = 2λ/a = 2(12.0 cm)/(63.8 cm) → θ2 = ±22.1° (b) AfricaAntarctica: sin θ = 36.7. sin θ3 = 3λ/a = 3(12.0 cm)/(63.8 cm) → θ3 = ±34.3°
sin θ4 = 4λ/a = 4(12.0 cm)/(63.8 cm) → θ4 = ±48.8° 36.8. sin θ5 = 5λ/a = 5(12.0 cm)/(63.8 cm) → θ5 = ±70.1°
EVALUATE: These are large angles, so we cannot use the approximation that θm mλ/a.
mλ
IDENTIFY: The minima are located by sin θ =
. For part (b) apply Eq.(36.7).
a
SET UP: For the first minimum, m = 1 . The intensity at θ = 0 is I 0 .
mλ
mλ λ
= sin 90.0° = 1 =
= . Thus a = λ = 580 nm = 5.80 × 10−4 mm.
a
a
a
(b) According to Eq.(36.7), EXECUTE: (a) sinθ = (
I * sin [π a (sin θ ) λ ] ) ( sin [π (sin π / 4)] )
**
*
=+
, =+
, = 0.128.
I 0 * π a (sin θ ) λ * * π (sin π / 4) *
..
2 2 I ( sin [ (π / 2)(sin π / 4)] )
*
*
=+
, = 0.81. As a / λ decreases,
I 0 * (π / 2)(sin π / 4) *
.
2 EVALUATE: If a = λ / 2 , for example, then at θ = 45° ,
36.9. the screen becomes more uniformly illuminated.
IDENTIFY and SET UP: v = f λ gives λ . The person hears no sound at angles corresponding to diffraction
minima. The diffraction minima are located by sin θ = mλ / a, m = ±1, ± 2,… Solve for θ .
EXECUTE: λ = v / f = (344 m/s) /(1250 Hz) = 0.2752 m; a = 1.00 m. m = ±1, θ = ±16.0°; m = ±2,
θ = ±33.4°; m = ±3, θ = ±55.6°; no solution for larger m
EVALUATE: λ / a = 0.28 so for the large wavelength sound waves diffraction by the doorway is a large effect.
Diffraction would not be observable for visible light because its wavelength is much smaller and λ / a V 1. Diffraction 36.10. IDENTIFY: Compare E y to the expression E y = Emax sin( kx − ωt ) and determine k, and from that calculate λ .
mλ
.
a
SET UP: c = 3.00 × 108 m/s . The first dark band corresponds to m = 1.
2π
2π
2π
EXECUTE: (a) E = Emax sin( kx − ωt ) . k =
!λ =
=
= 5.24 × 10−7 m .
λ
k 1.20 × 107 m −1
c 3.0 × 108 m s
fλ =c! f = =
= 5.73 × 1014 Hz .
λ 5.24 × 10−7 m
λ
5.24 × 10−7 m
a=
=
= 1.09 × 10−6 m .
(b) a sin θ = λ .
sin θ
sin 28.6°
−7
(c) a sin θ = mλ (m = 1, 2, 3, . . .). sin θ 2 = ±2 λ = ±2 5.24 × 10−6 m and θ 2 = ±74! .
a
1.09 × 10 m
mλ
EVALUATE: For m = 3 ,
is greater than 1 so only the first and second dark bands appear.
a
IDENTIFY and SET UP: sinθ = λ / a locates the first minimum. y = x tan θ .
EXECUTE: tanθ = y x = (36.5 cm) (40.0 cm) and θ = 42.38° . f = c / λ . The dark bands are located by sin θ = 36.11. 36.12. a = λ sinθ = (620 × 10 −9 m) (sin 42.38°) = 0.920 µ m
EVALUATE: θ = 0.74 rad and sin θ = 0.67, so the approximation sin θ ≈ θ would not be accurate.
mλ
IDENTIFY: The angle is small, so ym = x
applies.
a
SET UP: The width of the central maximum is 2 y1 , so y1 = 3.00 mm . xλ
xλ (2.50 m)(5.00 × 10−7 m)
!a=
=
= 4.17 × 10−4 m.
a
y1
3.00 × 10−3m
xλ (2.50 m)(5.00 × 10−5 m)
=
= 4.17 × 10−2 m = 4.2 cm.
(b) a =
y1
3.00 × 10−3 m
xλ (2.50 m)(5.00 × 10−10 m)
=
= 4.17 × 10−7 m.
(c) a =
y1
3.00 × 10−3 m
EVALUATE: The ratio a / λ stays constant, so a is smaller when λ is smaller.
IDENTIFY: Calculate the angular positions of the minima and use y = x tanθ to calculate the distance on the
screen between them.
(a) SET UP: The central bright fringe is shown in Figure 36.13a.
EXECUTE: The first
minimum is located by
EXECUTE: (a) y1 = 36.13. sin θ1 = λ =
a
633 × 10−9 m
= 1.809 × 10−3
0.350 × 10−3 m
θ1 = 1.809 × 10−3 rad
Figure 36.13a
y1 = x tan θ1 = (3.00 m) tan(1.809 × 10−3 rad) = 5.427 × 10−3 m
w = 2 y1 = 2(5.427 × 10−3 m) = 1.09 × 10 −2 m = 10.9 mm
(b) SET UP: The first bright fringe on one side of the central maximum is shown in Figure 36.13b.
EXECUTE: w = y2 − y1
y1 = 5.427 × 10−3 m (part (a))
2λ
sin θ 2 =
= 3.618 × 10−3
a
θ 2 = 3.618 × 10−3 rad
y2 = x tan θ 2 = 1.085 × 10−2 m Figure 36.13b
w = y2 − y1 = 1.085 × 10 −2 m − 5.427 × 10 −3 m = 5.4 mm
EVALUATE: The central bright fringe is twice as wide as the other bright fringes. 363 364 Chapter 36 36.14. IDENTIFY: 2 SET UP:
EXECUTE: " sin( β / 2) #
2π
I = I0 $
a sin θ .
% . β=
λ
& β /2 '
The angle θ is small, so sin θ ≈ tan θ ≈ y / x . β= 2π a λ (a) y = 1.00 × 10−3 m : sin θ ≈ β
2 = 2π a y
2π (4.50 × 10−4 m)
y = (1520 m −1 ) y.
=
λ x (6.20 × 10−7 m)(3.00 m) (1520 m −1 )(1.00 × 10−3 m)
= 0.760.
2
2 2 " sin( β 2) #
" sin(0.760) #
! I = I0 $
% = I0 $
% = 0.822 I 0
& 0.760 '
& β2 ' (b) y = 3.00 × 10−3 m : β
2 = (1520 m −1 )(3.00 × 10−3 m)
= 2.28.
2
2 2 " sin( β 2) #
" sin(2.28) #
! I = I0 $
% = I0 $
% = 0.111I 0 .
β2 '
& 2.28 '
& (c) y = 5.00 × 10−3 m : β
2 = (1520 m −1 )(5.00 × 10−3 m)
= 3.80.
2
2 2 " sin( β 2) #
" sin(3.80) #
! I = I0 $
% = I0 $
% = 0.0259 I 0 .
β2 '
& 3.80 '
& 36.15. λx (6.20 × 10−7 m)(3.00 m)
= 4.1 mm . The distances in parts (a)
a
4.50 × 10−4 m
and (b) are within the central maximum. y = 5.00 mm is within the first secondary maximum.
(a) IDENTIFY: Use Eq.(36.2) with m = 1 to locate the angular position of the first minimum and then use
y = x tan θ to find its distance from the center of the screen.
SET UP: The diffraction pattern is sketched in Figure 36.15.
EVALUATE: The first minimum occurs at y1 = = sin θ1 = λ =
a
540 × 10−9 m
= 2.25 × 10−3
0.240 × 10−3 m
θ1 = 2.25 × 10−3 rad Figure 36.15
y1 = x tan θ1 = (3.00 m) tan(2.25 × 10−3 rad) = 6.75 × 10−3 m = 6.75 mm
(b) IDENTIFY and SET UP: Use Eqs.(36.5) and (36.6) to calculate the intensity at this point.
EXECUTE: Midway between the center of the central maximum and the first minimum implies
1
y = (6.75 mm) = 3.375 × 10−3 m.
2
y 3.375 × 10−3 m
tan θ = =
= 1.125 × 10−3 ; θ = 1.125 × 10−3 rad
x
3.00 m
The phase angle β at this point on the screen is
2π
" 2π #
a sin θ =
(0.240 × 10−3 m)sin(1.125 × 10−3 rad) = π .
540 × 10 −9 m
λ%
&
' β =$ 2 2 " sin β / 2 #
2 " sin π / 2 #
−6
Then I = I 0 $
% = (6.00 × 10 W/m ) $
%
& π /2 '
& β /2 '
"4#
I = $ 2 % (6.00 × 10−6 W/m 2 ) = 2.43 × 10 −6 W/m 2 .
&π '
EVALUATE: The intensity at this point midway between the center of the central maximum and the first
minimum is less than half the maximum intensity. Compare this result to the corresponding one for the twoslit
pattern, Exercise 35.23. Diffraction 36.16. 365 IDENTIFY: In the singleslit diffraction pattern, the intensity is a maximum at the center and zero at the dark
spots. At other points, it depends on the angle at which one is observing the light.
SET UP: Dark fringes occur when sin θm = mλ/a, where m = 1, 2, 3, …, and the intensity is given by
2 " sin β / 2 #
π a sin θ
I0 $
.
% , where β / 2 =
λ
& β /2 '
EXECUTE: (a) At the maximum possible angle, θ = 90°, so mmax = (asin90°)/λ = (0.0250 mm)/(632.8 nm) = 39.5
Since m must be an integer and sin θ must be 1, mmax = 39. The total number of dark fringes is 39 on each side of
the central maximum for a total of 78.
(b) The farthest dark fringe is for m = 39, giving
sinθ39 = (39)(632.8 nm)/(0.0250 mm) ! θ39 = ±80.8°
(c) The next closer dark fringe occurs at sinθ38 = (38)(632.8 nm)/(0.0250 mm) !θ38 = 74.1°.
The angle midway these two extreme fringes is (80.8° + 74.1°)/2 = 77.45°, and the intensity at this angle is I =
2 " sin β / 2 #
π a sin θ π (0.0250 mm)sin(77.45°)
=
= 121.15 rad, which
I0 $
% , where β / 2 =
λ
632.8 nm
β /2 '
&
2 36.17. / sin(121.15 rad) 0
4
2
gives I = ( 8.50 W/m 2 ) 1
2 = 5.55 × 10 W/m
3 121.15 rad 4
EVALUATE: At the angle in part (c), the intensity is so low that the light would be barely perceptible.
IDENTIFY and SET UP: Use Eq.(36.6) to calculate λ and use Eq.(36.5) to calculate I. θ = 3.25°,
β = 56.0 rad, a = 0.105 × 10−3 m.
" 2π #
% a sin θ so
&λ' β =$ (a) EXECUTE: λ= 2π a sin θ β = 2π (0.105 × 10−3 m)sin 3.25°
= 668 nm
56.0 rad
2 " sin β / 2 #
"4#
4
2
(b) I = I 0 $
[sin(28.0 rad)]2 = 9.36 × 10−5 I 0
% = I 0 $ 2 % (sin( β / 2)) = I 0
(56.0 rad)2
& β /2 '
&β '
EVALUATE: At the first minimum β = 2π rad and at the point considered in the problem β = 17.8π rad, so the
point is well outside the central maximum. Since β is close to mπ with m = 18, this point is near one of the
minima. The intensity here is much less than I 0 . 36.18. IDENTIFY: Use β = 2π a λ sin θ to calculate β . SET UP: The total intensity is given by drawing an arc of a circle that has length E0 and finding the length of the
chord which connects the starting and ending points of the curve.
E
2
2π a
2π a λ
EXECUTE: (a) β =
= π . From Figure 36.18a, π p = E0 ! E p = E0 .
sin θ =
λ
λ 2a
2
π
2 4I
"2#
The intensity is I = $ % I 0 = 20 = 0.405I 0 . This agrees with Eq.(36.5).
π
&π '
2π a
2π a λ
(b) β =
= 2π . From Figure 36.18b, it is clear that the total amplitude is zero, as is the intensity.
sin θ =
λ
λa
This also agrees with Eq.(36.5).
E
2
2π a
2π a 3λ
E0 . The intensity is
(c) β =
= 3π . From Figure 36.18c, 3π p = E0 ! E p =
sin θ =
λ
λ 2a
2
3π
2 4
"2#
I = $ % I 0 = 2 I 0 . This agrees with Eq.(36.5).
9π
& 3π ' 366 Chapter 36 EVALUATE: In part (a) the point is midway between the center of the central maximum and the first minimum.
In part (b) the point is at the first maximum and in (c) the point is approximately at the location of the first
secondary maximum. The phasor diagrams help illustrate the rapid decrease in intensity at successive maxima. Figure 36.18
36.19. IDENTIFY: The space between the skyscrapers behaves like a single slit and diffracts the radio waves.
SET UP: Cancellation of the waves occurs when a sin θ = mλ, m = 1, 2, 3, …, and the intensity of the waves is
2 " sin β / 2 #
π a sin θ
given by I 0 $
.
% , where β / 2 =
λ
β /2 '
&
EXECUTE: (a) First find the wavelength of the waves:
λ = c/f = (3.00 × 108 m/s)/(88.9 MHz) = 3.375 m
For no signal, a sin θ = mλ.
m = 1: sin θ1 = (1)(3.375 m)/(15.0 m) ! θ1 = ±13.0°
m = 2: sin θ2 = (2)(3.375 m)/(15.0 m) ! θ2 = ±26.7°
m = 3: sin θ3 = (3)(3.375 m)/(15.0 m) ! θ3 = ±42.4°
m = 4: sin θ4 = (4)(3.375 m)/(15.0 m) ! θ4 = ±64.1°
2 " sin β / 2 #
π a sin θ π (15.0 m)sin(5.00°)
=
= 1.217 rad
(b) I 0 $
% , where β / 2 =
λ
3.375 m
β /2 '
&
2 / sin(1.217 rad) 0
2
I = ( 3.50 W/m 2 ) 1
2 = 2.08 W/m
3 1.217 rad 4 36.20. EVALUATE: The wavelength of the radio waves is very long compared to that of visible light, but it is still
considerably shorter than the distance between the buildings.
IDENTIFY: The net intensity is the product of the factor due to singleslit diffraction and the factor due to double
slit interference.
2 φ#
" sin β / 2 #
"
SET UP: The doubleslit factor is I DS = I 0 $ cos 2 % and the singleslit factor is ISS = $
%.
2'
&
& β /2 '
EXECUTE: (a) d sinθ = mλ ! sinθ = mλ/d.
sinθ1 = λ/d, sinθ2 = 2λ/d, sinθ3 = 3λ/d, sinθ4 = 4λ/d
π a sin θ π (d / 3)sin θ
(b) At the interference bright fringes, cos2φ/2 = 1 and β / 2 =
=
. λ
λ
π ( d / 3)(λ / d )
= π / 3 . The intensity is therefore
At θ1, sin θ1 = λ/d, so β / 2 =
λ
2 2 φ # " sin β / 2 #
" sin π / 3 #
"
I1 = I 0 $ cos 2 % $
% = I 0 (1) $
% = 0.684 I0
2 '& β / 2 '
&
& π /3 '
At θ2, sin θ2 = 2λ/d, so β / 2 = π ( d / 3)(2λ / d )
= 2π / 3 . Using the same procedure as for θ1, we have I2 =
λ 2 " sin 2π / 3 #
I 0 (1) $
% = 0.171 I0
& 2π / 3 '
At θ3, we get β / 2 = π , which gives I3 = 0 since sin = 0.
2 " sin 4π / 3 #
At θ4, sin θ4 = 4λ/d, so β / 2 = 4π / 3 , which gives I 4 = I 0 $
% = 0.0427 I0
& 4π / 3 '
(c) Since d = 3a, every third interference maximum is missing.
(d) In Figure 36.12c in the textbook, every fourth interference maximum at the sides is missing because d = 4a. Diffraction 36.21. 367 EVALUATE: The result in this problem is different from that in Figure 36.12c because in this case d = 3a, so
every third interference maximum at the sides is missing. Also the “envelope” of the intensity function decreases
more rapidly here than in Figure 36.12c because the first diffraction minimum is reached sooner, and the decrease
in intensity from one interference maximum to the next is faster for a = d/3 than for a = d/4.
(a) IDENTIFY and SET UP: The interference fringes (maxima) are located by d sin θ = mλ , with
2 " sin β / 2 #
" 2π #
m = 0, ± 1, ± 2, …. The intensity I in the diffraction pattern is given by I = I 0 $
% , with β = $
% a sin θ .
β /2 '
&λ'
&
We want m = ±3 in the first equation to give θ that makes I = 0 in the second equation.
" 2π # " 3λ #
EXECUTE: d sin θ = mλ gives β = $
% a $ % = 2π (3a / d ).
&λ' &d'
sin β / 2
= 0 so β = 2π and then 2π = 2π (3a / d ) and ( d / a ) = 3.
β /2
(b) IDENTIFY and SET UP: Fringes m = 0, ± 1, ± 2 are within the central diffraction maximum and the m = ± 3
fringes coincide with the first diffraction minimum. Find the value of m for the fringes that coincide with the
second diffraction minimum.
EXECUTE: Second minimum implies β = 4π . I = 0 says " 2π #
" 2π # " mλ #
β = $ % a sin θ = $ % a $
% = 2π m(a / d ) = 2π (m / 3)
&λ'
&λ' &d '
Then β = 4π says 4π = 2π ( m / 3) and m = 6. Therefore the m = ±4 and m = +5 fringes are contained within the
36.22. 36.23. first diffraction maximum on one side of the central maximum; two fringes.
EVALUATE: The central maximum is twice as wide as the other maxima so it contains more fringes.
IDENTIFY and SET UP: Use Figure 36.14b in the textbook. There is totally destructive interference between slits
whose phasors are in opposite directions.
EXECUTE: By examining the diagram, we see that every fourth slit cancels each other.
EVALUATE: The total electric field is zero so the phasor diagram corresponds to a point of zero intensity. The
first two maxima are at φ = 0 and φ = π , so this point is not midway between two maxima.
(a) IDENTIFY and SET UP: If the slits are very narrow then the central maximum of the diffraction pattern for
each slit completely fills the screen and the intensity distribution is given solely by the twoslit interference. The
maxima are given by
d sin θ = mλ so sin θ = mλ / d . Solve for θ .
λ 580 × 10−9 m
EXECUTE: 1st order maximum: m = 1, so sin θ = =
= 1.094 × 10−3 ; θ = 0.0627°
d 0.530 × 10−3 m
2λ
2nd order maximum: m = 2, so sin θ =
= 2.188 × 10−3 ; θ = 0.125°
d
2 " sin β / 2 #
(b) IDENTIFY and SET UP: The intensity is given by Eq.(36.12): I = I 0 cos (φ / 2) $
% . Calculate φ and
& β /2 '
β at each θ from part (a).
2 " 2π d #
" 2π d #" mλ #
sin θ = $
= 2π m, so cos 2 (φ / 2) = cos 2 ( mπ ) = 1
λ%
λ %$ d %
&
'
&
'&
'
(Since the angular positions in part (a) correspond to interference maxima.)
" 2π a #
" 2π a #" mλ #
" 0.320 mm #
β =$
% sin θ = $
%$
% = 2π m(a / d ) = m 2π $
% = m(3.794 rad)
&λ'
& λ '& d '
& 0.530 mm '
EXECUTE: φ =$ 2 " sin(3.794 / 2) rad #
1st order maximum: m = 1, so I = I 0 (1) $
% = 0.249 I 0
& (3.794 / 2) rad '
2 36.24. " sin 3.794 rad #
2nd order maximum: m = 2, so I = I 0 (1) $
% = 0.0256I 0
& 3.794 rad '
EVALUATE: The first diffraction minimum is at an angle θ given by sin θ = λ / a so θ = 0.104°. The first order
fringe is within the central maximum and the second order fringe is inside the first diffraction maximum on one
side of the central maximum. The intensity here at this second fringe is much less than I 0 .
IDENTIFY: A doubleslit bright fringe is missing when it occurs at the same angle as a doubleslit dark fringe.
SET UP: Singleslit diffraction dark fringes occur when a sin θ = mλ, and doubleslit interference bright fringes
occur when d sin θ = m′ λ. 368 Chapter 36 EXECUTE: (a) The angles are the same for cancellation, so dividing the equations gives
d/a = m′ /m ! m′ /m = 7 ! m′ = 7m 36.25. When m = 1, m′ = 7; when m = 2, m′ = 14, and so forth, so every 7th bright fringe is missing from the doubleslit
interference pattern.
EVALUATE: (b) The result is independent of the wavelength, so every 7th fringe will be cancelled for all
wavelengths. But the bright interference fringes occur when d sin θ = mλ, so the location of the cancelled fringes
does depend on the wavelength.
IDENTIFY and SET UP: The phasor diagrams are similar to those in Fig.36.14. An interference minimum occurs
when the phasors add to zero.
EXECUTE: (a) The phasor diagram is given in Figure 36.25a Figure 36.25a There is destructive interference between the light through slits 1 and 3 and between 2 and 4.
(b) The phasor diagram is given in Figure 36.25b. Figure 36.25b There is destructive interference between the light through slits 1 and 2 and between 3 and 4.
(c) The phasor diagram is given in Figure 36.25c. Figure 36.25c 36.26. 36.27. There is destructive interference between light through slits 1 and 3 and between 2 and 4.
EVALUATE: Maxima occur when φ = 0, 2π , 4π , etc. Our diagrams show that there are three minima between
the maxima at φ = 0 and φ = 2π . This agrees with the general result that for N slits there are N − 1 minima
between each pair of principal maxima.
IDENTIFY: A doubleslit bright fringe is missing when it occurs at the same angle as a doubleslit dark fringe.
SET UP: Singleslit diffraction dark fringes occur when a sin θ = mλ, and doubleslit interference bright fringes
occur when d sin θ = m′ λ.
EXECUTE: (a) The angle at which the first bright fringe occurs is given by
tan θ1 = (1.53 mm)/(2500 mm) ! θ1 = 0.03507°. d sin θ1 = λ and
d = λ/(sinθ1) = (632.8 nm)/sin(0.03507°) = 0.00103 m = 1.03 mm
(b) The 7th doubleslit interference bright fringe is just cancelled by the 1st diffraction dark fringe, so sinθdiff = λ/a
and sinθinterf = 7λ/d
The angles are equal, so λ/a = 7λ/d → a = d/7 = (1.03 mm)/7 = 0.148 mm.
EVALUATE: We can generalize that if d = na, where n is a positive integer, then every nth doubleslit bright fringe
will be missing in the pattern.
mλ
IDENTIFY: The diffraction minima are located by sin θ = d and the twoslit interference maxima are located
a
miλ
. The third bright band is missing because the first order single slit minimum occurs at the same
by sinθ =
d
angle as the third order double slit maximum.
3 cm
SET UP: The pattern is sketched in Figure 36.27. tan θ =
, so θ = 1.91° .
90 cm λ
500 nm
=
= 1.50 × 10 4 nm = 15.0 µ m (width)
sinθ sin1.91°
3λ
3(500 nm)
Doubleslit bright fringe: d sin θ = 3λ and d =
=
= 4.50 × 104 nm = 45.0 µ m (separation) .
sinθ
sin1.91°
EXECUTE: Singleslit dark spot: a sin θ = λ and a = Diffraction 369 EVALUATE: Note that d / a = 3.0 . Figure 36.27
36.28. 36.29. IDENTIFY: The maxima are located by d sin θ = mλ .
SET UP: The order corresponds to the values of m.
EXECUTE: Firstorder: d sin θ1 = λ . Fourthorder: d sin θ 4 = 4λ .
d sin θ 4 4λ
, sinθ 4 = 4sin θ1 = 4sin8.94° and θ 4 = 38.4° .
=
d sin θ1
λ
EVALUATE: We did not have to solve for d.
IDENTIFY and SET UP: The bright bands are at angles θ given by d sin θ = mλ. Solve for d and then solve for θ
for the specified order.
EXECUTE: (a) θ = 78.4° for m = 3 and λ = 681 nm, so d = mλ / sin θ = 2.086 × 10−4 cm
The number of slits per cm is 1 / d = 4790 slits/cm
(b) 1st order: m = 1, so sin θ = λ / d = (681× 10−9 m) /(2.086 × 10−6 m) and θ = 19.1° 36.30. 2nd order: m = 2, so sin θ = 2λ / d and θ = 40.8°
(c) For m = 4, sinθ = 4λ / d is greater than 1.00, so there is no 4thorder bright band.
EVALUATE: The angular position of the bright bands for a particular wavelength increases as the order increases.
IDENTIFY: The bright spots are located by d sinθ = mλ .
SET UP: Thirdorder means m = 3 and secondorder means m = 2 .
mλ
mλ
mλ
= d = constant , so r r = v v .
EXECUTE:
sin θ
sin θ r sin θ v 36.31. " m #" λ #
" 2 #" 400 nm #
sin θ v = sin θ r $ v %$ v % = (sin 65.0°) $ %$
% = 0.345 and θ v = 20.2° .
mr '& λr '
& 3 '& 700 nm '
&
EVALUATE: The thirdorder line for a particular λ occurs at a larger angle than the secondorder line. In a given
order, the line for violet light (400 nm) occurs at a smaller angle than the line for red light (700 nm).
IDENTIFY and SET UP: Calculate d for the grating. Use Eq.(36.13) to calculate θ for the longest wavelength in
the visible spectrum and verify that θ is small. Then use Eq.(36.3) to relate the linear separation of lines on the
screen to the difference in wavelength.
"1#
−5
EXECUTE: (a) d = $
% cm = 1.111× 10 m
& 900 '
For λ = 700 nm, λ / d = 6.3 × 10−2. The firstorder lines are located at sin θ = λ / d ; sin θ is small enough for
sin θ ≈ θ to be an excellent approximation.
(b) y = xλ / d , where x = 2.50 m.
The distance on the screen between 1st order bright bands for two different wavelengths is ∆y = x( ∆λ ) / d , so 36.32. ∆λ = d ( ∆y ) / x = (1.111× 10−5 m)(3.00 × 10−3 m) /(2.50 m) = 13.3 nm
EVALUATE: The smaller d is (greater number of lines per cm) the smaller the ∆λ that can be measured.
IDENTIFY: The maxima are located by d sin θ = mλ .
1
SET UP: 350 slits mm ! d =
= 2.86 × 10−6 m
3.50 × 105 m −1 3610 Chapter 36 " 4.00 × 10−7 m #
"λ #
!
m = 1: θ 400 = arcsin $ % = arcsin $
% = 8.05 .
2.86 × 10−6 m '
&d'
&
" 7.00 × 10−7 m #
"λ #
!
= arcsin $ % = arcsin $
% = 14.18 . ∆θ1 = 14.18° − 8.05° = 6.13°.
d'
2.86 × 10−6 m '
&
& EXECUTE: θ 700 " 3(4.00 × 10−7 m) #
" 3λ #
m = 3 : θ 400 = arcsin $ % = arcsin $
% = 24.8° .
−6
&d'
& 2.86 × 10 m '
" 3(7.00 × 10−7 m) #
" 3λ #
% = 47.3° . ∆θ1 = 47.3° − 24.8° = 22.5°.
% = arcsin $
−6
&d'
& 2.86 × 10 m '
EVALUATE: ∆θ is larger in third order.
IDENTIFY: The maxima are located by d sin θ = mλ .
SET UP: d = 1.60 × 10 −6 m θ 700 = arcsin $ 36.33. EXECUTE: 36.34. " m[6.328 × 10−7 m] #
" mλ #
% = arcsin([0.396]m) . For m = 1, θ1 = 23.3° . For
% = arcsin $
−6
&d'
& 1.60 × 10 m ' θ = arcsin $ m = 2 , θ 2 = 52.3° . There are no other maxima.
EVALUATE: The reflective surface produces the same interference pattern as a grating with slit separation d.
IDENTIFY: The maxima are located by d sin θ = mλ .
1
SET UP: 5000 slits cm ! d =
= 2.00 × 10−6 m.
5.00 × 105 m −1
d sin θ (2.00 × 10−6 m)sin13.5!
=
= 4.67 × 10−7 m.
m
1
" 2(4.67 × 10−7 m) #
" mλ #
!
(b) m = 2 : θ = arcsin $
% = 27.8 .
% = arcsin $
d'
2.00 × 10−6 m '
&
&
EXECUTE: (a) λ = 36.35. EVALUATE: Since the angles are fairly small, the secondorder deviation is approximately twice the firstorder
deviation.
IDENTIFY: The maxima are located by d sin θ = mλ .
1
SET UP: 350 slits mm ! d =
= 2.86 × 10−6 m
3.50 × 105 m −1 " m(5.20 × 10−7 m) #
" mλ #
% = arcsin((0.182)m) .
% = arcsin $
−6
&d'
& 2.86 × 10 m '
m = 1 : θ = 10.5°; m = 2 : θ = 21.3°; m = 3 : θ = 33.1°.
EVALUATE: The angles are not precisely proportional to m, and deviate more from being proportional as the
angles increase.
EXECUTE: 36.36. θ = arcsin $ IDENTIFY: The resolution is described by R =
SET UP: λ
= Nm . Maxima are located by d sinθ = mλ .
∆λ For 500 slits/mm, d = (500 slits mm) −1 = (500,000 slits m) −1 . EXECUTE: (a) N = λ
m∆λ = 6.5645 × 10−7 m
= 1820 slits.
2(6.5645 × 10−7 m − 6.5627 × 10−7 m) " mλ #
−1
−7
−1
(b) θ = sin −1 $
% ! θ1 = sin ((2)(6.5645 × 10 m)(500,000 m )) = 41.0297° and
&d' θ 2 = sin −1 ((2)(6.5627 × 10−7 m)(500,000 m −1 )) = 41.0160° . ∆θ = 0.0137°
EVALUATE: d cosθ dθ = λ / N , so for 1820 slits the angular interval ∆θ between each of these maxima and the
λ
6.56 × 10−7 m
first adjacent minimum is ∆θ =
=
= 0.0137°. This is the same as the angular
Nd cosθ (1820)(2.0 × 10−6 m)cos 41°
36.37. separation of the maxima for the two wavelengths and 1820 slits is just sufficient to resolve these two wavelengths
in second order.
IDENTIFY: The resolving power depends on the line density and the width of the grating.
SET UP: The resolving power is given by R = Nm = = λ/∆λ.
EXECUTE: (a) R = Nm = (5000 lines/cm)(3.50 cm)(1) = 17,500
(b) The resolving power needed to resolve the sodium doublet is
R = λ/∆λ = (589 nm)/(589.59 nm – 589.00 nm) = 998 Diffraction 3611 so this grating can easily resolve the doublet.
(c) (i) R = λ/∆λ. Since R = 17,500 when m = 1, R = 2 × 17,500 = 35,000 for m = 2. Therefore
∆λ = λ/R = (587.8 nm)/35,000 = 0.0168 nm λmin = λ +∆λ = 587.8002 nm + 0.0168 nm = 587.8170 nm
(ii) λmax = λ − ∆λ = 587.8002 nm − 0.0168 nm = 587.7834 nm
EVALUATE: (iii) Therefore the range of resolvable wavelengths is 587.7834 nm < λ < 587.8170 nm.
36.38. IDENTIFY and SET UP: λ
587.8002 nm
587.8002
N
3302
slits
=
=
= 3302 slits .
=
= 2752
.
0.178
1.20 cm 1.20 cm
cm
m∆λ (587.9782 nm − 587.8002 nm)
EVALUATE: A smaller number of slits would be needed to resolve these two lines in higher order.
IDENTIFY and SET UP: The maxima occur at angles θ given by Eq.(36.16), 2d sin θ = mλ , where d is the
spacing between adjacent atomic planes. Solve for d.
EXECUTE: second order says m = 2.
mλ
2(0.0850 × 10−9 m)
d=
=
= 2.32 × 10−10 m = 0.232 nm
2sin θ
2sin 21.5°
EVALUATE: Our result is similar to d calculated in Example 36.5.
IDENTIFY: The maxima are given by 2d sinθ = mλ , m = 1 , 2, …
SET UP: d = 3.50 × 10−10 m .
2d sin θ
EXECUTE: (a) m = 1 and λ =
= 2(3.50 × 10−10 m)sin15.0° = 1.81 × 10−10 m = 0.181 nm . This is an x ray.
m
" 1.81 × 10−10 m #
"λ#
(b) sin θ = m $
% = m(0.2586) . m = 2 : θ = 31.1° . m = 3 : θ = 50.9° . The equation
% = m$
−10
& 2d '
& 2[3.50 × 10 m] '
EXECUTE: 36.39. 36.40. λ
= Nm
∆λ N= doesn't have any solutions for m > 3 .
EVALUATE: In this problem λ / d = 0.52.
36.41. IDENTIFY: Rayleigh's criterion says sin θ = 1.22 λ D
SET UP: The best resolution is 0.3 arcseconds, which is about (8.33 × 10−5 )° . 1.22λ 1.22(5.5 × 10−7 m)
=
= 0.46 m
sin θ
sin(8.33 × 10−5 °)
EVALUATE: (b) The Keck telescopes are able to gather more light than the Hale telescope, and
hence they can detect fainter objects. However, their larger size does not allow them to have greater
resolutionatmospheric conditions limit the resolution.
EXECUTE: (a) D = 36.42. IDENTIFY: Apply sin θ = 1.22
SET UP: 36.44. . θ = (1/ 60)° 1.22λ 1.22(5.5 × 10−7 m)
=
= 2.31 × 10−3 m = 2.3 mm
sin θ
sin(1/ 60)!
EVALUATE: The larger the diameter the smaller the angle that can be resolved.
λ
IDENTIFY: Apply sin θ = 1.22 .
D
W
SET UP: θ = , where W = 28 km and h = 1200 km . θ is small, so sin θ ≈ θ .
h
1.22λ
h
1.2 × 106 m
EXECUTE: D =
= 1.22λ = 1.22(0.036 m)
= 1.88 m
W
sin θ
2.8 × 104 m
EVALUATE: D must be significantly larger than the wavelength, so a much larger diameter is needed for
microwaves than for visible wavelengths.
λ
IDENTIFY: Apply sin θ = 1.22 .
D
SET UP: θ is small, so sin θ ≈ θ = 1.00 × 10 −8 rad .
D sin θ Dθ (8.00 × 106 m)(1.00 × 10−8 )
EXECUTE: λ =
≈
=
= 0.0656 m = 6.56 cm
1.22
1.22
1.22
EVALUATE: λ corresponds to microwaves.
EXECUTE: 36.43. λ D D= 3612 Chapter 36 36.45. IDENTIFY and SET UP: The angular size of the first dark ring is given by sin θ1 = 1.22λ / D (Eq.36.17). Calculate θ1 , and then the diameter of the ring on the screen is 2(4.5 m) tan θ1.
" 620 × 10−9 m #
sin θ1 = 1.22 $
% = 0.1022; θ1 = 0.1024 rad
−6
& 7.4 × 10 m '
The radius of the Airy disk (central bright spot) is r = (4.5 m) tan θ1 = 0.462 m. The diameter is
2r = 0.92 m = 92 cm.
EVALUATE: λ / D = 0.084. For this small D the central diffraction maximum is broad.
IDENTIFY: Rayleigh’s criterion limits the angular resolution.
SET UP: Rayleigh’s criterion is sin θ ! θ = 1.22 λ/D.
EXECUTE: (a) Using Rayleigh’s criterion
sinθ ! θ = 1.22 λ/D = (1.22)(550 nm)/(135/4 mm) = 1.99 × 10–5 rad
On the bear this angle subtends a distance x. θ = x/R and
x = Rθ = (11.5 m)(1.99 × 10–5 rad) = 2.29 × 10–4 m = 0.23 mm
(b) At f/22, D is 4/22 times as large as at f/4. Since θ is proportional to 1/D, and x is proportional to θ, x is
1/(4/22) = 22/4 times as large as it was at f/4. x = (0.229 mm)(22/4) = 1.3 mm
EVALUATE: A wideangle lens, such as one having a focal length of 28 mm, would have a much smaller opening
at f/22 and hence would have an even less resolving ability.
IDENTIFY and SET UP: Resolved by Rayleigh’s criterion means angular separation θ of the objects equals
1.22λ / D. The angular separation θ of the objects is their linear separation divided by their distance from the
telescope.
250 × 103 m
EXECUTE: θ =
, where 5.93 × 1011 m is the distance from earth to Jupiter. Thus θ = 4.216 × 10−7.
5.93 × 1011 m
λ
1.22λ 1.22(500 × 10−9 m)
Then θ = 1.22 and D =
=
= 1.45 m
θ
D
4.216 × 10−7
EVALUATE: This is a very large telescope mirror. The greater the angular resolution the greater the diameter the
lens or mirror must be.
EXECUTE: 36.46. 36.47. 36.48. IDENTIFY: Rayleigh’s criterion says θ res = 1.22 . y
, where s is the distance of the object from the lens and y = 4.00 mm .
s
y
λ
yD
(4.00 × 10 −3 m)(7.20 × 10−2 m)
EXECUTE:
= 1.22 . s =
=
= 429 m .
s
D
1.22λ
1.22(550 × 10−9 m)
EVALUATE: The focal length of the lens doesn’t enter into the calculation. In practice, it is difficult to achieve
resolution that is at the diffraction limit.
IDENTIFY and SET UP: Let y be the separation between the two points being resolved and let s be their distance
λy
from the telescope. Then the limit of resolution corresponds to 1.22 = .
Ds
EXECUTE: (a) Let the two points being resolved be the opposite edges of the crater, so y is the diameter of the
crater. For the moon, s = 3.8 × 108 m. y = 1.22λ s D .
SET UP: 36.49. λ
D D = 7.20 cm . θ res = Hubble: D = 2.4 m and λ = 400 nm gives the maximum resolution, so y = 77 m
Arecibo: D = 305 m and λ = 0.75 m; y = 1.1 × 106 m
yD
. Let y ≈ 0.30 (the size of a license plate). s = (0.30 m)(2.4 m) [(1.22)(400 × 10−9 m)] = 1500 km.
1.22λ
EVALUATE: D / λ is much larger for the optical telescope and it has a much larger resolution even though the
diameter of the radio telescope is much larger. (b) s = 36.50. IDENTIFY: Apply sin θ = 1.22 λ D . θ is small, so sin θ ≈ θ . Smallest resolving angle is for shortwavelength light (400 nm).
λ
400 × 10−9 m
10,000 mi
EXECUTE: θ ≈ 1.22 = (1.22)
, where R is the distance to the star.
= 9.61× 10−8 rad . θ =
SET UP: R= R
D
5.08 m
16,000 km
=
= 1.7 × 1011 km .
9.6 × 10−8 rad
This is less than a light year, so there are no stars this close. 10,000 mi θ
EVALUATE: 3613 Diffraction 36.51. IDENTIFY: Let y be the separation between the two points being resolved and let s be their distance from the
telescope. The limit of resolution corresponds to 1.22 λ D = y s .
s = 4.28 ly = 4.05 × 1016 m . Assume visible light, with λ = 400 m . SET UP: y = 1.22 λ s D = 1.22(400 × 10−9 m)(4.05 × 1016 m (10.0 m) = 2.0 × 109 m EXECUTE: 36.52. EVALUATE: The diameter of Jupiter is 1.38 × 108 m, so the resolution is insufficient, by about one order of
magnitude.
IDENTIFY: If the apparatus of Exercise 36.4 is placed in water, then all that changes is the wavelength λ → λ′ = λ .
n
SET UP: For y << x , the distance between the two dark fringes on either side of the central maximum is
D′ = 2 y′ . Let D = 2 y be the separation of 5.91 × 10 −3 m found in Exercise 36.4. 2 xλ ′ 2 xλ D 5.91 × 10−3 m
=
==
= 4.44 × 10−3 m = 4.44 mm.
a
an
n
1.33
EVALUATE: The water shortens the wavelength and this decreases the width of the central maximum.
EXECUTE: ′
2y1 = 2 36.53. " sin β / 2 #
(a) IDENTIFY and SET UP: The intensity in the diffraction pattern is given by Eq.(36.5): I = I 0 $
% , where
& β /2 ' 1
" 2π #
a sin θ . Solve for θ that gives I = I 0 . The angles θ + and θ − are shown in Figure 36.53.
λ%
2
&
'
1
sin β / 2
1
EXECUTE: I = I 0 so
=
2
β /2
2 β =$ sin x
1
=
= 0.7071.
x
2
Use trial and error to find the value of x that is a solution to this equation.
x
(sin x) / x
1.0 rad
0.841
Let x = β / 2; the equation for x is 1.5 rad
1.2 rad 0.665
0.777 1.4 rad
1.39 rad 0.7039
0.7077; thus x = 1.39 rad and β = 2 x = 2.78 rad
∆θ = θ + − θ − = 2θ + sin θ + = λβ
=
2π a λ " 2.78 rad #
"λ#
$
% = 0.4425 $ %
a & 2π rad '
&a'
Figure 36.53 "1#
= 2, sin θ + = 0.4425 $ % = 0.2212; θ + = 12.78°; ∆θ = 2θ + = 25.6°
& 2'
a
"1#
(ii) For = 5, sin θ + = 0.4425 $ % = 0.0885; θ + = 5.077°; ∆θ = 2θ + = 10.2°
λ
&5' (i) For a λ (iii) For "1#
= 10, sin θ + = 0.4425 $ % = 0.04425; θ + = 2.536°; ∆θ = 2θ + = 5.1°
λ
& 10 ' a (b) IDENTIFY and SET UP: sin θ 0 = λ
a locates the first minimum. Solve for θ 0 . 1
= 2, sin θ 0 = ;θ 0 = 30.0°; 2θ 0 = 60.0°
2
a
1
(ii) For = 5, sin θ 0 = ;θ 0 = 11.54°; 2θ 0 = 23.1°
5
λ EXECUTE: (i) For a λ 3614 Chapter 36 "1#
= 10, sin θ 0 = $ % ;θ 0 = 5.74°; 2θ 0 = 11.5°
& 10 '
EVALUATE: Either definition of the width shows that the central maximum gets narrower as the slit gets wider.
IDENTIFY: The two holes behave like double slits and cause the sound waves to interfere after they pass through
the holes. The motion of the speakers causes a Doppler shift in the wavelength of the sound.
SET UP: The wavelength of the sound that strikes the wall is λ = λ0 – vsTs, and destructive interference first
occurs where sin θ = λ/2.
EXECUTE: (a) First find the wavelength of the sound that strikes the openings in the wall.
λ = λ0 – vsTs = v/ fs – vs/ fs = (v – vs)/ fs = (344 m/s – 80.0 m/s)/(1250 Hz) = 0.211 m
Destructive interference first occurs where d sin θ = λ/2, which gives
d = λ/(2 sinθ) = (0.211 m)/(2 sin 12.7°) = 0.480 m
(b) λ = v/f = (344 m/s)/(1250 Hz) = 0.275 m
sinθ = λ/2d = (0.275 m)/[2(0.480 m)] → θ = ±16.7°
EVALUATE: The moving source produces sound of shorter wavelength than the stationary source, so the angles at
which destructive interference occurs are smaller for the moving source than for the stationary source.
IDENTIFY and SET UP: sin θ = λ / a locates the first dark band. In the liquid the wavelength changes and this
changes the angular position of the first diffraction minimum. (iii) For 36.54. 36.55. a λ EXECUTE: sin θ air = λair ; sin θ liquid = λliquid a
a
" sin θ liquid #
λliquid = λair $
% = 0.4836
& sin θ air '
λ = λair / n (Eq.33.5), so n = λair / λliquid = 1/ 0.4836 = 2.07 36.56. 36.57. EVALUATE: Light travels faster in air and n must be > 1.00. The smaller λ in the liquid reduces θ that located
the first dark band.
1
IDENTIFY: d = , so the bright fringes are located by 1 sin θ = λ
N
N
SET UP: Red: 1 sin λR = 700 nm . Violet: 1 sin λV = 400 nm .
N
N
sin θ R 7
sin(θ V + 15°) 7
EXECUTE:
= . θ R − θ V = 15° → θ R = θ V + 15° .
= . Using a trig identify from Appendix B,
sin θ V 4
sin θ V
4
sin θ V cos15° + cosθ V sin15°
= 7 4 . cos15° + cot θ V sin15° = 7 4 .
sin θ V
tan θ V = 0.330 ! θ V = 18.3° and θ R = θ V + 15° = 18.3° + 15° = 33.3°. Then 1 sin θ R = 700 nm gives
N
sin θ R
sin 33.3!
=
= 7.84 × 105 lines m = 7840 lines cm . The spectrum begins at 18 .3! and ends at 33.3! .
N=
700 nm 700 × 10−9 m
EVALUATE: As N is increased, the angular range of the visible spectrum increases.
(a) IDENTIFY and SET UP: The angular position of the first minimum is given by a sinθ = mλ (Eq.36.2), with
m = 1. The distance of the minimum from the center of the pattern is given by y = x tan θ . λ 540 × 10−9 m
= 1.50 × 10 −3 ; θ = 1.50 × 10−3 rad
a 0.360 × 10−3 m
y1 = x tan θ = (1.20 m) tan(1.50 × 10−3 rad) = 1.80 × 10−3 m = 1.80 mm.
(Note that θ is small enough for θ ≈ sin θ ≈ tan θ , and Eq.(36.3) applies.)
(b) IDENTIFY and SET UP: Find the phase angle β where I = I 0 / 2. Then use Eq.(36.6) to solve for θ and
y = x tan θ to find the distance.
sin θ = = 1
From part (a) of Problem 36.53, I = I 0 when β = 2.78 rad.
2
2π #
βλ
"
β = $ % a sin θ (Eq.(36.6)), so sin θ =
.
2π a
&λ'
βλ x (2.78 rad)(540 × 10−9 m)(1.20 m)
y = x tan θ ≈ x sin θ ≈
=
= 7.96 × 10−4 m = 0.796 mm
2π a
2π (0.360 × 10−3 m)
EVALUATE: The point where I = I 0 / 2 is not midway between the center of the central maximum and the first
minimum; see Exercise 36.15.
EXECUTE: Diffraction 3615 2 36.58. " sin γ #
I = I0 $
% . The maximum intensity occurs when the derivative of the intensity function with
&γ'
respect to γ is zero. IDENTIFY: SET UP: d "1#
1
d sin γ
= cos γ .
$ %=− 2 .
dγ & γ '
γ
dγ
2 " sin γ #" cos γ sin γ #
dI
d " sin γ #
cos γ sin γ
= I0
− 2 % =0.
− 2 ! γ cos γ = sin γ ! γ = tan γ .
$
% = 2$
%$
dγ
dγ & γ '
γ '& γ
γ'
γ
γ
&
(b) The graph in Figure 36.58 is a plot of f( γ ) = γ − tan γ . When f (γ ) equals zero, there is an intensity
maximum. Getting estimates from the graph, and then using trial and error to narrow in on the value, we find that
the three smallest γ values are γ = 4.49 rad 7.73 rad, and 10.9 rad.
EVALUATE: γ = 0 is the central maximum. The three values of γ we found are the locations of the first three
secondary maxima. The first four minima are at γ = 3.14 rad , 6.28 rad, 9.42 rad, and 12.6 rad. The maxima are
between adjacent minima, but not precisely midway between them. EXECUTE: Figure 36.58
36.59. IDENTIFY and SET UP: Relate the phase difference between adjacent slits to the sum of the phasors for all slits. The
2π d
2π dθ
λφ
when θ is small and sin θ ≈ θ . Thus θ =
.
sin θ ≈
phase difference between adjacent slits is φ =
2π d
λ
λ
EXECUTE: A principal maximum occurs when φ = φ max = m 2π , where m is an integer, since then all the phasors add. The first minima on either side of the mth principal maximum occur when φ = φ ± in = m 2π ± (2π / N ) and the
m
phasor diagram for N slits forms a closed loop and the resultant phasor is zero. The angular position of a principal
"λ#
"λ#±
±
maximum is θ = $
%φ max . The angular position of the adjacent minimum is θ min = $
%φ min.
& 2π d '
& 2π d '
2π #
λ
" λ #"
" λ #" 2π #
θ + in = $
m
%$ φ max +
% =θ +$
%$
% =θ +
2π d '&
N'
2π d '& N '
Nd
&
&
2π #
λ
" λ #"
%$ φ max −
% =θ −
N'
Nd
& 2π d '& θ − in = $
m 2λ
, as was to be shown.
Nd
EVALUATE: The angular width of the principal maximum decreases like 1/ N as N increases.
IDENTIFY: The change in wavelength of the Hα line is due to a Doppler shift in the wavelength due to the motion
of the galaxy.
c−v
SET UP: From Equation 16.30, the Doppler effect formula for light is f R =
fS .
c+v
EXECUTE: First find the wavelength of the light using the grating information.
λ = d sin θ1 = [1/(575,800 lines/m)] sin 23.41° = 6.900 × 10–7 m = 690.0 nm
The angular width of the principal maximum is θ 36.60. +
min −θ −
min = c−v
fS . In this case, fR is the frequency of the 690.0nm light that the
c+v
cosmologist measures, and fS is the frequency of the 656.3nm light of the Hα line obtained in the laboratory.
Using Equation 16.30, we have f R = 3616 Chapter 36 Solving for v gives v = 1 − ( f R / fS ) 1 + ( f R / fS ) 2
2 c. Since fλ = c, f = c/λ, which gives fR/fS = λS/λR. Substituting this into the equation for v, we get
"λ
1− $ S
λ
v= & R
"λ
1+ $ S
& λR 36.61. 2 2
#
" 656.3 nm #
1− $
%
%
& 690.0 nm ' 3.00 × 108 m/s = 1.501 × 107 m/s,
' c=
)
2(
2
#
" 656.3 nm #
1+ $
%
%
& 690.0 nm '
' which is 5.00% the speed of light.
EVALUATE: Since v is positive, the galaxy is moving away from us. We can also see this because the wavelength
has increased due to the motion.
IDENTIFY and SET UP: Draw the specified phasor diagrams. There is totally destructive interference between
two slits when their phasors are in opposite directions.
EXECUTE: (a) For eight slits, the phasor diagrams must have eight vectors. The diagrams for each specified
value of φ are sketched in Figure 36.61a. In each case the phasors all sum to zero.
(b) The additional phasor diagrams for φ = 3π / 2 and 3π / 4 are sketched in Figure 36.61b.
3π
5π
7π
3π
,φ =
, and φ =
, totally destructive interference occurs between slits four apart. For φ =
,
For φ =
4
4
4
2
totally destructive interference occurs with every second slit.
EVALUATE: At a minimum the phasors for all slits sum to zero. Figure 36.61
36.62. IDENTIFY: Maxima are given by 2d sinθ = mλ .
SET UP: d is the separation between crystal planes.
"
0.125 nm #
" mλ #
EXECUTE: (a) θ = arcsin $
% = arcsin(0.2216m) .
% = arcsin $ m
& 2d '
& 2(0.282 nm) '
For m = 1: θ = 12.8°, m = 2 : θ = 26.3°, m = 3: θ = 41.7°, and m = 4 : θ = 62.4°. No larger m values yield answers.
" 2mλ #
a
, then θ = arcsin $
$ 2a % = arcsin(0.3134m).
%
2
&
'
So for m = 1: θ = 18.3°, m = 2: θ = 38.8°, and m = 3: θ = 70.1°. No larger m values yield answers.
EVALUATE: In part (b), where d is smaller, the maxima for each m are at larger θ
IDENTIFY and SET UP: In each case consider the relevant phasor diagram.
EXECUTE: (a) For the maxima to occur for N slits, the sum of all the phase differences between the slits must
add to zero (the phasor diagram closes on itself). This requires that, adding up all the relative phase shifts,
2π m
N φ = 2π m, for some integer m . Therefore φ =
, for m not an integer multiple of N , which would give a
N
maximum.
2π m
(b) The sum of N phase shifts φ =
brings you full circle back to the maximum, so only the N − 1 previous
N
phases yield minima between each pair of principal maxima.
EVALUATE: The N − 1 minima between each pair of principal maxima cause the maxima to become sharper as N
increases.
IDENTIFY: Set d = a in the expressions for φ and β and use the results in Eq.(36.12).
SET UP: Figure 36.64 shows a pair of slits whose width and separation are equal (b) If the separation d = 36.63. 36.64. Diffraction 3617 EXECUTE: Figure 36.64 shows that the two slits are equivalent to a single slit of width 2a .
2π d
2π a
φ=
sin θ , so β =
sin θ = φ . So then the intensity is λ λ " sin 2 ( β /2) #
(2sin( β /2)cos(β /2))2
sin 2 β
sin 2 ( β ′/2)
2π (2a)
= I0
= I0
, where β ′ =
I = I 0 cos 2 ( β /2) $
sin θ ,
% = I0
2
2
2
2
λ
β
β
( β ′/2)
& ( β /2) '
which is Eq. (35.5) with double the slit width.
EVALUATE: In Chapter 35 we considered the limit where a << d . a > d is not possible. Figure 36.64
36.65. 36.66. IDENTIFY and SET UP: The condition for an intensity maximum is d sin θ = mλ , m = 0, ± 1, ± 2,… Third order
means m = 3. The longest observable wavelength is the one that gives θ = 90° and hence θ = 1.
1
EXECUTE: 6500 lines/cm so 6.50 × 105 lines/m and d =
m = 1.538 × 10−6 m
6.50 × 105
d sin θ (1.538 × 10−6 m)(1)
λ=
=
= 5.13 × 10−7 m = 513 nm
m
3
EVALUATE: The longest wavelength that can be obtained decreases as the order increases.
IDENTIFY and SET UP: As the rays first reach the slits there is already a phase difference between adjacent slits of
2π d sin θ ′
. This, added to the usual phase difference introduced after passing through the slits, yields the λ condition for an intensity maximum. For a maximum the total phase difference must equal 2π m .
2π d sin θ 2π d sin θ ′
EXECUTE:
+
= 2π m ! d (sin θ + sin θ ′) = mλ λ (b) 600 slits mm ! d = λ 1
= 1.67 × 10−6 m.
6.00 × 105 m −1 For θ ′ = 0! ,
m = 0 : θ = arcsin(0) = 0.
" 6.50 × 10−7 m #
"λ#
!
m = 1: θ = arcsin $ % = arcsin $
% = 22.9 .
1.67 × 10−6 m '
d'
&
&
" 6.50 × 10−7 m #
" λ#
!
m = −1: θ = arcsin $ − % = arcsin $ −
% = −22.9 .
1.67 × 10−6 m '
& d'
& For θ ′ = 20.0! ,
m = 0 : θ = arcsin(− sin 20.0! ) = −20.0!. 36.67. " 6.50 × 10−7 m
#
− sin 20.0! % = 2.71!.
m = 1: θ = arcsin $
1.67 × 10−6 m
&
'
" 6.50 × 10−7 m
#
− sin 20.0! % = −47.0!.
m = −1: θ = arcsin $ −
−6
& 1.67 × 10 m
'
EVALUATE: When θ ′ > 0 , the maxima are shifted downward on the screen, toward more negative angles.
mλ
IDENTIFY: The maxima are given by d sinθ = mλ . We need sin θ =
≤ 1in order for all the visible
d
wavelengths are to be seen.
1
SET UP: For 650 slits mm ! d =
= 1.53 × 10−6 m.
6.50 × 105 m −1 3618 Chapter 36 λ1
2λ
3λ
= 0.26; m = 2 : 1 = 0.52; m = 3 : 1 = 0.78.
d
d
d
λ2
2λ2
3λ2
= 0.92; m = 3 :
= 1.37. So, the third order does not contain the violet
λ2 = 7.00 × 10 −7 m : m = 1: = 0.46; m = 2 :
d
d
d
end of the spectrum, and therefore only the first and second order diffraction patterns contain all colors of the spectrum.
EVALUATE: θ for each maximum is larger for longer wavelengths.
λ
IDENTIFY: Apply sin θ = 1.22 .
D
∆x
SET UP: θ is small, so sin θ ≈ , where ∆x is the size of the detail and R = 7.2 ×108 ly. 1 ly = 9.41×1012 km . λ = c/f
R
λ ∆x
1.22λ R (1.22)cR (1.22)(3.00 × 105 km s)(7.2 × 108 ly)
EXECUTE: sin θ = 1.22 ≈
! ∆x =
=
=
= 2.06 ly .
DR
D
Df
(77.000 × 103 km)(1.665 × 109 Hz)
(9.41 × 1012 km ly)( 2.06 ly) = 1.94 × 1013 km.
∆x
EVALUATE: λ = 18 cm . λ / D is very small, so
is very small. Still, R is very large and ∆x is many orders of
R
magnitude larger than the diameter of the sun.
IDENTIFY and SET UP: Add the phases between adjacent sources.
EXECUTE: (a) d sin θ = mλ . Place 1st maximum at ∞ or θ = 90!. d = λ . If d < λ , this puts the first maximum
“beyond ∞. ” Thus, if d < λ there is only a single principal maximum.
(b) At a principal maximum when δ = 0 , the phase difference due to the path difference between adjacent slits
" d sin θ #
is Φ path = 2π $
% . This just scales 2π radians by the fraction the wavelength is of the path difference between
&λ'
adjacent sources. If we add a relative phase δ between sources, we still must maintain a total phase difference of
zero to keep our principal maximum.
2π d sin θ
" δλ #
Φ path ± δ = 0 !
= ±δ or θ = sin −1 $
%
λ
& 2π d '
0.280 m
(c) d =
= 0.0200 m (count the number of spaces between 15 points). Let θ = 45!. Also recall f λ = c , so
14
2π (0.0200 m)(8.800 × 109 Hz)sin 45!
δ max = ±
= ±2.61 radians.
(3.00 × 108 m s)
EXECUTE: 36.68. 36.69. 36.70. 36.71. λ1 = 4.00 × 10 −7 m : m = 1: EVALUATE: δ must vary over a wider range in order to sweep the beam through a greater angle.
IDENTIFY: The wavelength of the light is smaller under water than it is in air, which will affect the resolving
power of the lens, by Rayleigh’s criterion.
SET UP: The wavelength under water is λ = λ0/n, and for small angles Rayleigh’s criterion is θ = 1.22λ/D.
EXECUTE: (a) In air the wavelength is λ0 = c/f = (3.00 × 108 m/s)/(6.00 × 1014 Hz) = 5.00 × 10–7 m. In water the
wavelength is λ = λ0/n = (5.00 × 10–7 m)/1.33 = 3.76 × 10–7 m. With the lens open all the way, we have D = f/2.8 =
(35.0 mm)/2.80 = (0.0350 m)/2.80. In the water, we have
sin θ θ = 1.22 λ/D = (1.22)(3.76 × 10–7 m)/[(0.0350 m)/2.80] = 3.67 × 10–5 rad
Calling w the width of the resolvable detail, we have
θ = w/R → w = Rθ = (2750 mm)(3.67 × 10–5 rad) = 0.101 mm
(b) θ = 1.22 λ/D = (1.22)(5.00 × 10–7 m)/[(0.0350 m)/2.80] = 4.88 × 10–5 rad
w = Rθ = (2750 mm)(4.88 × 10–5 rad) = 0.134 mm
EVALUATE: Due to the reduced wavelength underwater, the resolution of the lens is better under water than in air.
IDENTIFY and SET UP: Resolved by Rayleigh’s criterion means the angular separation θ of the objects is given
by θ = 1.22λ / D. θ = y / s, where y = 75.0 m is the distance between the two objects and s is their distance from
the astronaut (her altitude).
y
λ
= 1.22
EXECUTE:
s
D
yD
(75.0 m)(4.00 × 10−3 m)
s=
=
= 4.92 × 105 m = 492 km
1.22λ
1.22(500 × 10−9 m)
EVALUATE: In practice, this diffraction limit of resolution is not achieved. Defects of vision and distortion by the
earth’s atmosphere limit the resolution more than diffraction does. Diffraction 36.72. IDENTIFY: Apply sin θ = 1.22 3619 λ .
D
∆x
, where ∆x is the size of the details and R is the distance to the earth.
SET UP: θ is small, so sin θ ≈
R
1 ly = 9.41× 1015 m .
EXECUTE: (a) R = D∆x (6.00 × 106 m)(2.50 × 105 m)
=
= 1.23 × 1017 m = 13.1 ly
1.22λ
(1.22)(1.0 × 10−5 m) 1.22λ R (1.22)(1.0 × 10−5 m)(4.22 ly)(9.41× 1015 m ly)
=
= 4.84 × 108 km . This is about 10,000 times the
1.0 m
D
diameter of the earth! Not enough resolution to see an earthlike planet! ∆x is about 3 times the distance from the
earth to the sun.
(1.22)(1.0 × 10−5 m)(59 ly)(9.41× 1015 m ly)
= 1.13 × 106 m = 1130 km.
(c) ∆x =
6.00 × 106 m
(b) ∆x = ∆x
1130 km
=
= 8.19 × 10−3 ; ∆x is small compared to the size of the planet.
Dplanet 1.38 × 105 km
36.73. EVALUATE: The very large diameter of Planet Imager allows it to resolve planetsized detail at great distances.
IDENTIFY and SET UP: Follow the steps specified in the problem.
EXECUTE: (a) From the segment dy′, the fraction of the amplitude of E0 that gets through is " dy′ #
" dy′ #
E0 $
% ! dE = E0 $
% sin( kx − ωt ).
a'
&
&a'
(b) The path difference between each little piece is
E dy′
y′ sin θ ! kx = k ( D − y′ sin θ ) ! dE = 0 sin(k ( D − y′ sin θ ) − ωt ). This can be rewritten as
a
E0 dy′
dE =
(sin( kD − ωt )cos(ky′ sin θ ) + sin( ky′ sin θ )cos( kD − ωt )).
a
(c) So the total amplitude is given by the integral over the slit of the above.
a2
E a2
! E = 5 dE = 0 5 dy′ (sin(kD − ωt ) cos(ky′ sin θ ) + sin(ky′ sin θ ) cos( kD − ωt )).
−a 2
a −a 2
But the second term integrates to zero, so we have:
a2 E= a2
/" sin(ky′ sin θ ) # 0
E0
sin( kD − ωt ) 5 dy′(cos( ky′ sin θ )) = E0 sin ( kD − ωt ) 1$
%2
−a 2
a
3& ka sin θ 2 ' 4 − a 2 " sin( ka (sin θ ) 2) #
" sin(π a (sin θ ) λ ) #
! E = E0 sin( kD − ωt ) $
% = E0 sin( kD − ωt ) $
%.
ka(sin θ ) 2 '
&
& π a (sin θ ) λ '
sin [. . .]
At θ = 0,
= 1 ! E = E0 sin(kD − ωt ).
[. . .]
2 36.74. 2 " sin(ka (sin θ )/2) #
" sin( β 2) #
2
2
(d) Since I ∝ E 2 ! I = I 0 $
% = I0 $
% , where we have used I 0 = E0 sin ( kx − ωt ).
ka(sin θ )/ 2 '
β2 '
&
&
EVALUATE: The same result for I (θ ) is obtained as was obtained using phasors.
IDENTIFY and SET UP: Follow the steps specified in the problem.
EXECUTE: (a) Each source can be thought of as a traveling wave evaluated at x = R with a maximum amplitude
of E0 . However, each successive source will pick up an extra phase from its respective pathlength to point
" d sin θ #
P . φ = 2π $
% which is just 2π , the maximum phase, scaled by whatever fraction the path difference,
&λ'
d sin θ , is of the wavelength, λ . By adding up the contributions from each source (including the accumulating
phase difference) this gives the expression provided.
(b) ei ( kR −ωt + nφ ) = cos( kR − ωt + nφ ) + i sin(kR − ωt + nφ ). The real part is just cos ( kR − ωt + nφ ). So,
/ N −1
0 N −1
Re 1 6 E0ei ( kR −ω t + nφ ) 2 = 6 E0 cos(kR − ωt + nφ ). (Note: Re means “the real part of . . . .”). But this is just
3 n=0
4 n =0
E0 cos( kR − ωt ) + E0 cos(kR − ωt + φ ) + E0 cos( kR − ωt + 2φ ) + " + E0 cos(kR − ωt + ( N − 1)φ ) 3620 Chapter 36 (c) N −1 6E
n=0 0 ei ( kR −ωt + nφ ) = E0 N −1 6e − iω t n=0 e + ikR einφ = E0ei ( kR −ω t ) N −1 ∞ N −1 N −1 6 e φ . 6 e φ = 6 (e φ ) . But recall 6 x
n=0 in n=0 in n =0 i n n=0 n = xN −1
.
x −1 eiN φ − 1
eiNφ − 1 eiNφ / 2 (eiNφ / 2 − e −iNφ / 2 )
(eiN φ / 2 − e −iN φ / 2 )
Let x = e so 6 (e ) = iφ
(nice trick!). But iφ
= iφ / 2 iφ / 2 − iφ / 2 = ei ( N −1)φ / 2 iφ / 2 −iφ / 2 .
e −1
e −1
e (e − e
)
(e − e
)
n=0
Putting everything together:
N −1
(eiNφ / 2 − e − iNφ / 2 )
6 E0ei ( kR −ωt + nφ ) = E0ei ( kR −ωt + ( N −1)φ / 2) (eiφ / 2 − e−iφ / 2 )
n=0
N −1 iφ iφ n / cos Nφ /2 + i sin Nφ /2 − cos Nφ /2 + i sin Nφ /2 0
= E0 [ cos( kR − ωt + ( N − 1)φ /2) + i sin( kR − ωt + ( N − 1)φ / 2)] 1
2
cosφ /2 + i sin φ /2 − cosφ /2 + i sin φ /2
3
4
sin( Nφ /2)
Taking only the real part gives ! E0 cos( kR − ωt + ( N − 1)φ /2)
= E.
sin φ /2
2 (d) I = E av = I 0 I0 ∝ sin 2 ( Nφ / 2)
. (The cos 2 term goes to
sin 2 (φ / 2) in the time average and is included in the definition of I 0 .) E02
.
2 EVALUATE: (e) N = 2. I = I 0 sin 2 (2φ / 2) I 0 (2sin φ / 2cos φ / 2)2
φ
=
= 4 I 0 cos 2 . Looking at Eq.(35.9),
sin 2 φ / 2
sin 2 φ / 2
2 ′
I 0 ∝ 2 E02 but for us I 0 ∝
36.75. 1
2 ′
E02 I 0
=.
2
4 IDENTIFY and SET UP: From Problem 36.74, I = I 0 sin 2 ( Nφ / 2)
. Use this result to obtain each result specified
sin 2 φ / 2 in the problem.
" N 2 # cos( Nφ / 2)
0
sin ( Nφ / 2)
ˆ
. Use l'Hopital's rule: lim
= lim $
= N . So lim I = N 2 I 0 .
%
φ →0
φ →0
φ →0
sin φ / 2
1 2 ' cos(φ / 2)
0
&
N
2π
. The
(b) The location of the first minimum is when the numerator first goes to zero at φmin = π or φmin =
2
N
1
width of the central maximum goes like 2φmin , so it is proportional to .
N
Nφ
(c) Whenever
= nπ where n is an integer, the numerator goes to zero, giving a minimum in intensity. That is,
2
2nπ
. This is true assuming that the denominator doesn’t go to zero as well, which
I is a minimum wherever φ =
N
EXECUTE: (a) lim I →
φ →0 occurs when φ
2 = mπ , where m is an integer. When both go to zero, using the result from part(a), there is a n
is an integer, there will be a maximum.
N
n
n
(d) From part (c), if
is an integer we get a maximum. Thus, there will be N − 1 minima. (Places where
is
N
N
not an integer for fixed N and integer n .) For example, n = 0 will be a maximum, but n = 1, 2. . ., N − 1 will be
minima with another maximum at n = N .
φ
" π 3π
#
, etc.) % and if N is odd then
(e) Between maxima
is a halfinteger multiple of π $ i.e. ,
22
2
&
'
maximum. That is, if sin 2 ( Nφ / 2)
→ 1, so I → I 0 .
sin 2 φ / 2
EVALUATE: These results show that the principal maxima become sharper as the number of slits is increased. 37 RELATIVITY 37.1. IDENTIFY and SET UP: Consider the distance A to O′ and B to O′ as observed by an observer on the ground
(Figure 37.1). Figure 37.1 37.2. EXECUTE: Simultaneous to observer on train means light pulses from A′ and B′ arrive at O′ at the same time.
To observer at O light from A′ has a longer distance to travel than light from B′ so O will conclude that the pulse
from A( A′) started before the pulse at B ( B′). To observer at O bolt A appeared to strike first.
EVALUATE: Section 37.2 shows that if they are simultaneous to the observer on the ground then an observer on
the train measures that the bolt at B′ struck first.
1
(a) =
= 2.29. t = τ = (2.29) (2.20 × 10−6 s) = 5.05 × 10−6 s.
1 − (0.9) 2
(b) d = vt = (0.900) (3.00 × 108 m s) (5.05 × 10−6 s) = 1.36 × 103 m = 1.36 km. 37.3. 1
IDENTIFY and SET UP: The problem asks for u such that ∆t0 / ∆t = .
2 37.4. 37.5. 2 ∆t0 u
2
"1#
gives u = c 1 − ( ∆t0 / ∆t ) = (3.00 × 108 m/s) 1 − $ % = 2.60 × 108 m/s ; = 0.867
c
& 2'
1 − u 2 / c2
Jet planes fly at less than ten times the speed of sound, less than about 3000 m/s. Jet planes fly at much lower
speeds than we calculated for u.
IDENTIFY: Time dilation occurs because the rocket is moving relative to Mars.
SET UP: The time dilation equation is ∆t = γ∆t0 , where t0 is the proper time.
EXECUTE: (a) The two time measurements are made at the same place on Mars by an observer at rest there, so
the observer on Mars measures the proper time.
1
(75.0 µ s) = 435 µ s
(b) ∆t = γ∆t0 =
1 − (0.985) 2
EXECUTE: ∆t = EVALUATE: The pulse lasts for a shorter time relative to the rocket than it does relative to the Mars observer.
(a) IDENTIFY and SET UP: ∆t0 = 2.60 × 10 −8 s; ∆t = 4.20 × 10−7 s. In the lab frame the pion is created and decays
at different points, so this time is not the proper time.
2
∆t0
u 2 " ∆t #
EXECUTE: ∆t =
says 1 − 2 = $ 0 %
c
& ∆t '
1 − u 2 / c2
2 2 " 2.60 × 10−8 s #
u
" ∆t #
= 1− $ 0 % = 1− $
% = 0.998; u = 0.998c
−7
c
& ∆t '
& 4.20 × 10 s '
EVALUATE: u < c, as it must be, but u/c is close to unity and the time dilation effects are large.
(b) IDENTIFY and SET UP: The speed in the laboratory frame is u = 0.998c; the time measured in this frame is
∆t , so the distance as measured in this frame is d = u ∆t
EXECUTE: d = (0.998)(2.998 × 108 m/s)(4.20 × 10−7 s) = 126 m
EVALUATE: The distance measured in the pion’s frame will be different because the time measured in the pion’s
frame is different (shorter).
371 372 37.6. Chapter 37 = 1.667
1.20 × 108 m
= 0.300 s.
(0.800c)
(b) (0.300 s) (0.800c ) = 7.20 × 107 m.
(c) ∆t0 = 0.300 s = 0.180 s. (This is what the racer measures your clock to read at that instant.) At your origin (a) ∆t0 = ∆t = you read the original
37.7. 37.8. 37.9. 37.10. order of events!
IDENTIFY and SET UP: A clock moving with respect to an observer appears to run more slowly than a clock at rest
in the observer’s frame. The clock in the spacecraft measurers the proper time ∆t0 . ∆t = 365 days = 8760 hours.
EXECUTE: The clock on the moving spacecraft runs slow and shows the smaller elapsed time.
∆t0 = ∆t 1 − u 2 / c 2 = (8760 h) 1 − (4.80 × 106 / 3.00 × 108 )2 = 8758.88 h . The difference in elapsed times is
8760 h − 8758.88 h = 1.12 h .
IDENTIFY and SET UP: The proper time is measured in the frame where the two events occur at the same point.
EXECUTE: (a) The time of 12.0 ms measured by the first officer on the craft is the proper time.
∆t0
gives u = c 1 − ( ∆t0 / ∆t ) 2 = c 1 − (12.0 × 10−3 / 0.190)2 = 0.998c .
(b) ∆t =
2
2
1− u / c
EVALUATE: The observer at rest with respect to the searchlight measures a much shorter duration for the event.
IDENTIFY and SET UP: l = l0 1 − u 2 / c 2 . The length measured when the spacecraft is moving is l = 74.0 m; l0 is
the length measured in a frame at rest relative to the spacecraft.
l
74.0 m
=
= 92.5 m.
EXECUTE: l0 =
2
2
1− u / c
1 − (0.600c / c ) 2
EVALUATE: l0 > l. The moving spacecraft appears to an observer on the planet to be shortened along the
direction of motion.
IDENTIFY and SET UP: When the meterstick is at rest with respect to you, you measure its length to be 1.000 m,
and that is its proper length, l0 . l = 0.3048 m .
EXECUTE: 37.11. l = l0 1 − u 2 / c 2 gives u = c 1 − (l / l0 ) 2 = c 1 − (0.3048 /1.00)2 = 0.9524c = 2.86 × 108 m/s . IDENTIFY and SET UP: The 2.2 µs lifetime is ∆t0 and the observer on earth measures ∆t. The atmosphere is
moving relative to the muon so in its frame the height of the atmosphere is l and l0 is 10 km.
EXECUTE: (a) The greatest speed the muon can have is c, so the greatest distance it can travel in 2.2 × 10 −6 s is
d = vt = (3.00 × 108 m/s)(2.2 × 10−6 s) = 660 m = 0.66 km .
(b) ∆t = 37.12. 1.20 × 108m
= 0.5 s. Clearly the observers (you and the racer) will not agree on the
(0.800) (3 × 108 m s) ∆t0
1 − u 2 / c2 = 2.2 × 10−6 s
1 − (0.999)2 = 4.9 × 10−5 s d = vt = (0.999)(3.00 × 108 m/s)(4.9 × 10−5 s) = 15 km
In the frame of the earth the muon can travel 15 km in the atmosphere during its lifetime.
(c) l = l0 1 − u 2 / c 2 = (10 km) 1 − (0.999)2 = 0.45 km
In the frame of the muon the height of the atmosphere is less than the distance it moves during its lifetime.
IDENTIFY and SET UP: The scientist at rest on the earth’s surface measures the proper length of the separation
between the point where the particle is created and the surface of the earth, so l0 = 45.0 km . The transit time measured in the particle’s frame is the proper time, ∆t0 .
EXECUTE: (a) t = l0
45.0 × 103 m
=
= 1.51 × 10−4 s
v (0.99540)(3.00 × 108 m/s) (b) l = l0 1 − u 2 / c 2 = (45.0 km) 1 − (0.99540)2 = 4.31 km
(c) time dilation formula: ∆t0 = ∆t 1 − u 2 / c 2 = (1.51 × 10−4 s) 1 − (0.99540)2 = 1.44 × 10−5 s l
4.31 × 103 m
=
= 1.44 × 10−5 s
v (0.99540)(3.00 × 108 m/s)
The two results agree.
(a) l0 = 3600 m .
from ∆l : t = 37.13. l = l0 1 − u2
(4.00 × 107 m s) 2
= l0 (3600 m) 1 −
= (3600 m)(0.991) = 3568 m.
c2
(3.00 × 108 m s)2 Relativity 373 l0
3600 m
=
= 9.00 × 10 −5 s.
u 4.00 × 107 m s
l
3568 m
(c) ∆t = =
= 8.92 × 10−5 s.
u 4.00 × 107 m s
Multiplying the last equation of (37.21) by u and adding to the first to eliminate t gives
(b) ∆t0 = 37.14. " u2 # 1
x′ + ut ′ = γ x $1 − 2 % = x,
& c' γ
and multiplying the first by u
and adding to the last to eliminate x gives
c2
t′ + 37.15. 37.16. " u2 # 1
u
x′ = t $1 − 2 % = t ,
2
c
& c' so x = ( x′ + ut ′) and t = (t ′ + ux′ c 2 ), which is indeed the same as Eq. (37.21) with the primed coordinates
replacing the unprimed, and a change of sign of u.
v′ + u
0.400c + 0.600c
(a) v =
=
= 0.806c
1 + uv′ c 2 1 + (0.400) (0.600)
v′ + u
0.900c + 0.600c
(b) v =
=
= 0.974c
1 + uv′ c 2 1 + (0.900)(0.600)
v′ + u
0.990c + 0.600c
(c) v =
=
= 0.997c.
′ c 2 1 + (0.990)(0.600)
1 + uv
= 1.667( = 5 3 if u = (4 5)c ).
(a) In Mavis’s frame the event “light on” has spacetime coordinates x′ = 0 and t ′ = 5.00 s, so from the result of
ux′ #
"
Exercise 37.14 or Example 37.7, x = ( x′ + ut′) and t = $ t ′ + 2 % ! x = ut ′ = 2.00 × 109 m, t = t ′ = 8.33 s .
c'
& (b) The 5.00s interval in Mavis’s frame is the proper time ∆t0 in Eq.(37.6), so ∆t = ∆t0 = 8.33 s, as in part (a).
37.17. (c) (8.33 s) (0.800c ) = 2.00 × 109 m, which is the distance x found in part (a).
IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light.
v −u
.
SET UP: The relativistic velocity addition formula is v′ = x
x
uv
1 − 2x
c
EXECUTE: (a) For the pursuit ship to catch the cruiser, the distance between them must be decreasing, so the
velocity of the cruiser relative to the pursuit ship must be directed toward the pursuit ship.
(b) Let the unprimed frame be Tatooine and let the primed frame be the pursuit ship. We want the velocity v′ of the
cruiser knowing the velocity of the primed frame u and the velocity of the cruiser v in the unprimed frame (Tatooine).
v −u
0.600c − 0.800c
=
= −0.385c
v′ = x
x
uvx 1 − (0.600) (0.800)
1− 2
c 37.18. The result implies that the cruiser is moving toward the pursuit ship at 0.385c .
EVALUATE: The nonrelativistic formula would have given –0.200c, which is considerably different from the
correct result.
Let u y be the ycomponent of the velocity of S ′ relative to S. Following the steps used in the derivation of
v′y + u y
.
Eq.(37.23) we get v y =
1 + u y v′y / c 2 37.19. IDENTIFY and SET UP: Reference frames S and S ′ are shown in Figure 37.19. Frame S is at rest in the
laboratory. Frame S ′ is
attached to particle 1.
Figure 37.19
u is the speed of S ′ relative to S; this is the speed of particle 1 as measured in the laboratory. Thus u = +0.650c.
The speed of particle 2 in S ′ is 0.950c. Also, since the two particles move in opposite directions, 2 moves in the
− x′ direction and v′ = −0.950c. We want to calculate vx , the speed of particle 2 in frame S; use Eq.(37.23).
x 374 Chapter 37 v′ + u
−0.950c + 0.650c
−0.300c
x
=
=
= −0.784c. The speed of the second particle,
′ / c 2 1 + (0.950c)(−0.650c) / c 2 1 − 0.6175
1 + uvx
as measured in the laboratory, is 0.784c.
EVALUATE: The incorrect Galilean expression for the relative velocity gives that the speed of the second particle
in the lab frame is 0.300c. The correct relativistic calculation gives a result more than twice this.
IDENTIFY and SET UP: Let S be the laboratory frame and let S ′ be the frame of one of the particles, as shown in
Figure 37.20. Let the positive x direction for both frames be from particle 1 to particle 2. In the lab frame particle 1
is moving in the +x direction and particle 2 is moving in the − x direction. Then u = 0.9520c and v = −0.9520c .
v′ is the velocity of particle 2 relative to particle 1.
v−u
−0.9520c − 0.9520c
EXECUTE: v′ =
=
= −0.9988c . The speed of particle 2 relative to particle 1
1 − uv / c 2 1 − (0.9520c )( −0.9520c) / c 2
is 0.9988c . v′ < 0 shows particle 2 is moving toward particle 1.
EXECUTE: 37.20. vx = Figure 37.20
37.21. 37.22. 37.23. IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light.
v −u
SET UP: The relativistic velocity addition formula is v′ = x
.
x
uv
1 − 2x
c
EXECUTE: In the relativistic velocity addition formula for this case, vx′ is the relative speed of particle 1 with
respect to particle 2, v is the speed of particle 2 measured in the laboratory, and u is the speed of particle 1
measured in the laboratory, u = – v.
v′
v − (−v)
2v
v′ =
=
. x v 2 − 2v + vx′ = 0 and (0.890c)v 2 − 2c 2v + (0.890c 3 ) = 0 .
x
1 − ( − v )v c 2 1 + v 2 c 2 c 2
This is a quadratic equation with solution v = 0.611c (v must be less than c).
EVALUATE: The nonrelativistic result would be 0.445c, which is considerably different from this result.
IDENTIFY and SET UP: Let the starfighter’s frame be S and let the enemy spaceship’s frame be S ′ . Let the
positive x direction for both frames be from the enemy spaceship toward the starfighter. Then u = + 0.400c .
v′ = +0.700c . v is the velocity of the missile relative to you.
v′ + u
0.700c + 0.400c
=
= 0.859c
EXECUTE: (a) v =
1 + uv′ / c 2 1 + (0.400)(0.700)
(b) Use the distance it moves as measured in your frame and the speed it has in your frame to calculate the time it
8.00 × 109 m
takes in your frame. t =
= 31.0 s .
(0.859)(3.00 × 108 m/s)
IDENTIFY and SET UP: The reference frames are shown in Figure 37.23. S = Arrakis frame
S ′ = spaceship frame
The object is the rocket.
Figure 37.23 u is the velocity of the spaceship relative to Arrakis.
vx = +0.360c; v′ = +0.920c
x
(In each frame the rocket is moving in the positive coordinate direction.) Relativity Use the Lorentz velocity transformation equation, Eq.(37.22): v′ =
x 375 vx − u
.
1 − uvx / c 2 vx − u
" v v′ #
" v v′ #
so v′ − u $ x 2 x % = vx − u and u $ 1 − x 2 x % = vx − v′
x
x
c'
1 − uvx / c 2
&c '
&
vx − v′
0.360c − 0.920c
0.560c
x
u=
=
=−
= −0.837c
1 − vx v′ / c 2 1 − (0.360c)(0.920c ) / c 2
0.6688
x EXECUTE: 37.24. v′ =
x The speed of the spacecraft relative to Arrakis is 0.837c = 2.51 × 108 m/s. The minus sign in our result for u means
that the spacecraft is moving in the − xdirection, so it is moving away from Arrakis.
EVALUATE: The incorrect Galilean expression also says that the spacecraft is moving away from Arrakis, but
with speed 0.920c – 0.360c = 0.560c.
IDENTIFY: We need to use the relativistic Doppler shift formula.
SET UP: The relativistic Doppler shift formula, Eq.(37.25), is f = c+u
f0 .
c−u c+u 2
f 0 . (c − u ) f 2 = (c + u ) f 02 . cf 2 − uf 2 = cf 02 + uf 02 . cf 2 − cf 02 = uf 2 + uf 02 and
c−u
c( f 2 − f 02 ) ( f / f 0 ) 2 − 1
=
u=
c.
f 2 + f 02
( f / f0 )2 + 1
(a) For f/f0 = 0.95, u = – 0.051c moving away from the source.
(b) For f/f0 = 5.0, u = 0.923c moving towards the source.
EVALUATE: Note that the speed required to achieve a 10 times greater Doppler shift is not 10 times the original
speed.
EXECUTE: 37.25. f2= IDENTIFY and SET UP: Source and observer are approaching, so use Eq.(37.25): f = c+u
f 0 . Solve for u, the
c−u speed of the light source relative to the observer.
"c+u# 2
(a) EXECUTE: f 2 = $
% f0
& c −u '
(c − u ) f 2 = (c + u ) f 02 and u = " ( f / f0 )2 − 1 #
c( f 2 − f 02 )
= c$
%
2
f 2 + f 02
& ( f / f0 ) + 1 ' λ0 = 675 nm, λ = 575 nm
" (675 nm/575 nm)2 − 1 #
8
7
u =$
% c = 0.159c = (0.159)(2.998 × 10 m/s) = 4.77 × 10 m/s; definitely speeding
(675 nm/575 nm)2 + 1 '
&
(b) 4.77 × 107 m/s = (4.77 × 107 m/s)(1 km/1000 m)(3600 s/1 h) = 1.72 × 108 km/h. Your fine would be $1.72 × 108
(172 million dollars).
EVALUATE: The source and observer are approaching, so f > f 0 and λ < λ0 . Our result gives u < c, as it must.
37.26. Using u = −0.600c = − ( 3 5 ) c in Eq.(37.25) gives 1 − ( 3 5)
25
f0 =
f 0 = f 0 2.
1 + ( 3 5)
85
#
#
#
#
IDENTIFY and SET UP: If F is parallel to v then F changes the magnitude of v and not its direction.
#
dp d "
mv
=$
F=
%
dt dt & 1 − v 2 / c 2 '
d
df dv
Use the chain rule to evaluate the derivative:
f (v (t )) =
.
dt
dv dt
m
mv
" dv #
" 1 #" 2v #" dv #
EXECUTE: (a) F =
$ %+
$ − %$ − %$ %
(1 − v 2 / c 2 )1/ 2 & dt ' (1 − v 2 / c 2 )3 / 2 & 2 '& c 2 '& dt '
f= 37.27. F= " v 2 v 2 # dv
dv
m
m
$1 − + % =
dt (1 − v 2 / c 2 )3 / 2 & c 2 c 2 ' dt (1 − v 2 / c 2 )3 / 2 dv
= a, so a = ( F / m)(1 − v 2 / c 2 )3 / 2 .
dt
EVALUATE: Our result agrees with Eq.(37.30). But 376 37.28. Chapter 37 #
#
#
#
(b) IDENTIFY and SET UP: If F is perpendicular to v then F changes the direction of v and not its magnitude.
#
# d"
#
mv
F= $
%.
2
2
dt & 1 − v / c '
#
#
a = dv / dt but the magnitude of v in the denominator of Eq.(37.29) is constant.
ma
EXECUTE: F =
and a = ( F / m)(1 − v 2 / c 2 )1/ 2 .
2
2
1− v / c
EVALUATE: This result agrees with Eq.(37.33).
1
IDENTIFY and SET UP: γ =
. If γ is 1.0% greater than 1 then γ = 1.010 , if γ is 10% greater than 1
1 − v2 / c2
then γ = 1.10 and if γ is 100% greater than 1 then γ = 2.00 . EXECUTE: v = c 1 − 1/ γ 2 (a) v = c 1 − 1/(1.010)2 = 0.140c
(b) v = c 1 − 1/(1.10) 2 = 0.417c
(c) v = c 1 − 1/(2.00) 2 = 0.866c
37.29. (a) p = mv = 2mv . 1 − v2 c2 1
v2
3
3
= 1 − 2 ! v2 = c2 ! v =
c = 0.866c.
4
c
4
2
1
v
(b) F = 3ma = 2ma ! 3 = 2 ! = (2)1/ 3 so
= 22 / 3 ! = 1 − 2−2 / 3 = 0.608
v2
c
1− 2
c
The force is found from Eq.(37.32) or Eq.(37.33).
(a) Indistinguishable from F = ma = 0.145 N.
! 1 = 2 1 − v2 c2 ! 37.30. (b) 3 ma = 1.75 N. (c) ma = 51.7 N.
(d) ma = 0.145 N, 0.333 N, 1.03 N.
3 37.31. (a) K = mc 2
1− v c
2 ! (b) K = 5mc 2 !
37.32.
37.33. 2 − mc 2 = mc 2 1
1− v c
2 2 1
1 − v2 c2 =2! v2
1
3
=1− 2 ! v =
c = 0.866c.
c
4
4 =6! v2
1
35
=1− 2 ! v =
c = 0.986c.
c
36
36 E = 2mc 2 = 2(1.67 × 10−27 kg)(3.00 × 108 m s) 2 = 3.01× 10−10 J = 1.88 × 109 eV.
IDENTIFY and SET UP: Use Eqs.(37.38) and (37.39).
EXECUTE: (a) E = mc 2 + K , so E = 4.00mc 2 means K = 3.00mc 2 = 4.50 × 10−10 J (b) E 2 = (mc 2 ) 2 + ( pc) 2 ; E = 4.00mc 2 , so 15.0(mc 2 ) 2 = ( pc )2 p = 15mc = 1.94 × 10−18 kg ⋅ m/s
(c) E = mc 2 / 1 − v 2 / c 2 37.34. 37.35. E = 4.00mc 2 gives 1 − v 2 / c 2 = 1/16 and v = 15/16c = 0.968c
EVALUATE: The speed is close to c since the kinetic energy is greater than the rest energy. Nonrelativistic
expressions relating E, K, p and v will be very inaccurate.
(a) W = ∆K = ( f − 1) mc 2 = (4.07 × 10−3 ) mc 2 .
(b) ( f − i ) mc 2 = 4.79mc 2 .
(c) The result of part (b) is far larger than that of part (a).
IDENTIFY: Use E = mc 2 to relate the mass increase to the energy increase.
(a) SET UP: Your total energy E increases because your gravitational potential energy mgy increases. Relativity 377 ∆E = mg ∆y EXECUTE: ∆E = ( ∆m)c so ∆m = ∆E / c 2 = mg (∆y ) / c 2
2 ∆m / m = ( g ∆y ) / c 2 = (9.80 m/s 2 )(30 m)/(2.998 × 108 m/s) 2 = 3.3 × 10−13%
This increase is much, much too small to be noticed.
(b) SET UP: The energy increases because potential energy is stored in the compressed spring.
EXECUTE: ∆E = ∆U = 1 kx 2 = 1 (2.00 × 104 N/m)(0.060 m)2 = 36.0 J
2
2 37.36. ∆m = ( ∆E ) / c 2 = 4.0 × 10−16 kg
Energy increases so mass increases. The mass increase is much, much too small to be noticed.
EVALUATE: In both cases the energy increase corresponds to a mass increase. But since c 2 is a very large
number the mass increase is very small.
m0
= 2m0 .
(a) E0 = m0c 2 . 2 E = mc 2 = 2m0c 2 . Therefore, m = 2m0 !
1 − v2 / c2 1
v2
v2 3
= 1 − 2 ! 2 = ! v = c 3 4 = 0.866c = 2.60 × 108 m s
4
c
c
4
m0
(b) 10 m0c 2 = mc 2 =
c2 .
1 − v2 c2
1−
37.37. v2
1
v 2 99
99
. v=c
=
! 2=
= 0.995c = 2.98 × 108 m s .
2
c 100
c 100
100 IDENTIFY and SET UP: The energy equivalent of mass is E = mc 2 . ρ = 7.86 g/cm3 = 7.86 × 103 kg/m3 . For a cube, V = L3 .
1.0 × 1020 J
E
=
= 1.11 × 103 kg
2
(3.00 × 108 m/s) 2
c
m
m
1.11 × 103 kg
(b) ρ = so V = =
= 0.141 m3 . L = V 1 / 3 = 0.521 m = 52.1 cm
V
ρ 7.86 × 103 kg/m3
EVALUATE: Particle/antiparticle annihilation has been observed in the laboratory, but only with small quantities
of antimatter.
(5.52 × 10−27 kg)(3.00 × 108 m s)2 = 4.97 × 10−10 J = 3105 MeV.
IDENTIFY and SET UP: The total energy is given in terms of the momentum by Eq.(37.39). In terms of the total
energy E, the kinetic energy K is K = E − mc 2 (from Eq.37.38). The rest energy is mc 2 .
EXECUTE: (a) m = 37.38.
37.39. EXECUTE: (a) E = (mc 2 ) 2 + ( pc) 2 = [(6.64 × 10−27 )(2.998 × 108 )2 ]2 + [(2.10 × 10−18 )(2.998 × 108 )]2 J
E = 8.67 × 10−10 J
(b) mc 2 = (6.64 × 10 −27 kg)(2.998 × 108 m/s)2 = 5.97 × 10−10 J 37.40. 37.41. K = E − mc 2 = 8.67 × 10−10 J − 5.97 × 10−10 J = 2.70 × 10−10 J
K
2.70 × 10−10 J
(c)
=
= 0.452
mc 2 5.97 × 10−10 J
EVALUATE: The incorrect nonrelativistic expressions for K and p give K = p 2 / 2m = 3.3 × 10 −10 J; the correct
relativistic value is less than this.
12
" " p #2 #
24
2 2 12
2
E = ( m c + p c ) = mc $1 + $
$ & mc % %
'%
&
'
2
2
" 1p#
p
1
2
E ≈ mc 2 $ 1 +
= mc 2 + mv 2 , the sum of the rest mass energy and the classical kinetic energy.
% = mc +
2 m 2c 2 '
2m
2
&
1
1
(a) v = 8 × 107 m s ! =
= 1.0376 . For m = mp , K nonrel = mv 2 = 5.34 × 10−12 J .
2
1 − v2 c2
K rel = ( − 1) mc 2 = 5.65 × 10−12 J. K rel
= 1.06.
K nonrel (b) v = 2.85 × 108 m s; = 3.203.
1
K rel = mv 2 = 6.78 × 10−11 J; K rel = ( − 1)mc 2 = 3.31 × 10−10 J; K rel K nonrel = 4.88.
2 378 Chapter 37 37.42. IDENTIFY: Since the speeds involved are close to that of light, we must use the relativistic formula for kinetic energy.
"
#
1
SET UP: The relativistic kinetic energy is K = (γ − 1) mc 2 = $
− 1% mc 2 .
2
2
& 1− v / c
'
"
#
"
#
1
1
− 1% mc 2 = (1.67 × 10−27 kg)(3.00 × 108 m s) 2 $
− 1%
(a) K = (γ − 1)mc 2 = $
2
2
$ 1 − 0.100c / c 2
& 1− v / c
'
(
)%
&
' 37.43. 37.44. 1
"
#
K = (1.50 × 10−10 J) $
− 1% = 7.56 × 10−13 J = 4.73 MeV
& 1 − 0.0100 '
"
#
1
(b) K = (1.50 × 10−10 J) $
− 1% = 2.32 × 10−11 J = 145 MeV
$ 1 − (0.500) 2
%
&
'
"
#
1
(c) K = (1.50 × 10−10 J) $
− 1% = 1.94 × 10−10 J = 1210 MeV
$ 1 − (0.900)2
%
&
'
(d) ∆E = 2.32 × 10−11 J − 7.56 × 10−13 J = 2.24 × 10−11 J = 140 MeV
(e) ∆E = 1.94 × 10−10 J − 2.32 × 10−11 J = 1.71× 10−10 J = 1070 MeV
1
(f) Without relativity, K = mv 2 . The work done in accelerating a proton from 0.100c to 0.500c in the
2
1
1
nonrelativistic limit is ∆E = m(0.500c) 2 − m(0.100c) 2 = 1.81× 10−11 J = 113 MeV .
2
2
The work done in accelerating a proton from 0.500c to 0.900c in the nonrelativistic limit is
1
1
∆E = m(0.900c) 2 − m(0.500c)2 = 4.21× 10−11 J = 263 MeV .
2
2
EVALUATE: We see in the first case the nonrelativistic result is within 20% of the relativistic result. In the second
case, the nonrelativistic result is very different from the relativistic result since the velocities are closer to c.
IDENTIFY and SET UP: Use Eq.(23.12) and conservation of energy to relate the potential difference to the kinetic
energy gained by the electron. Use Eq.(37.36) to calculate the kinetic energy from the speed.
EXECUTE: (a) K = q∆V = e∆V
"
#
1
K = mc 2 $
− 1% = 4.025mc 2 = 3.295 × 10−13 J = 2.06 MeV
2
2
& 1− v / c
'
6
∆V = K / e = 2.06 × 10 V
(b) From part (a), K = 3.30 × 10−13 J = 2.06 MeV
EVALUATE: The speed is close to c and the kinetic energy is four times the rest mass.
(a) According to Eq.(37.38) and conservation of massenergy
m
9.75
2 Mc 2 + mc 2 = γ 2 Mc 2 ! γ = 1 +
=1+
= 1.292.
2M
2(16.7)
Note that since γ = v
1
1
1
= 0.6331.
, we have that = 1 − 2 = 1 −
2
2
γ
c
(1.292) 2
1− v c (b) According to Eq.(37.36), the kinetic energy of each proton is
" 1.00 MeV #
K = (γ − 1) Mc 2 = (1.292 − 1)(1.67 × 10−27 kg)(3.00 × 108 m s )2 $
% = 274 MeV.
−13
& 1.60 × 10 J ' " 1.00 MeV #
(c) The rest energy of η 0 is mc 2 = (9.75 × 10−28 kg)(3.00 × 108 m s) 2 $
% = 548 MeV.
−13
& 1.60 × 10 J '
(d) The kinetic energy lost by the protons is the energy that produces the η 0 ,
548 MeV = 2(274 MeV).
37.45. IDENTIFY: The relativistic expression for the kinetic energy is K = (γ − 1) mc 2 , where γ = 1
The Newtonian expression for the kinetic energy is K N = mv 2 .
2
3
SET UP: Solve for v such that K = K N .
2 1
and x = v 2 / c 2 .
1− x Relativity 379 2 3
1
3
1
" 3#
(γ − 1) mc 2 = mv 2 .
−1 = x.
= $ 1 + x % . After a little algebra this becomes
4
4
1− x & 4 '
1− x
1
−15 ± (15)2 + 4(9)(8) . The positive root is x = 0.425 . x = v 2 / c 2 , so
9 x 2 + 15 x − 8 = 0. x =
18
v = x c = 0.652c .
EVALUATE: The fractional increase of the relativistic expression above the nonrelativistic one increases as v increases.
(4.0015 u)
The fraction of the initial mass (a) that becomes energy is 1 −
= 6.382 × 10−3 , and so the energy released
2(2.0136 u)
per kilogram is (6.382 × 10−3 )(1.00 kg)(3.00 × 108 m s)2 = 5.74 × 1014 J.
EXECUTE: ) ( 37.46. 1.0 × 1019 J
= 1.7 × 104 kg.
5.74 × 1014 J kg
(a) E = mc 2 , m = E c 2 = (3.8 × 1026 J) (2.998 × 108 m s )2 = 4.2 × 109 kg . (b)
37.47. 1 kg is equivalent to 2.2 lbs, so m = 4.6 × 106 tons
(b) The current mass of the sun is 1.99 × 1030 kg, so it would take it
37.48. (1.99 × 1030 kg) (4.2 × 109 kg s) = 4.7 × 1020s = 1.5 × 1013 years to use up all its mass.
IDENTIFY: Since the final speed is close to the speed of light, there will be a considerable difference between the
relativistic and nonrelativistic results.
1
12
SET UP: The nonrelativistic workenergy theorem is F ∆x = mv 2 − mv0 , and the relativistic formula for a
2
2
constant force is F ∆x = (γ − 1)mc 2 .
(a) Using the classical workenergy theorem and solving for ∆x , we obtain ∆x = 2
m(v 2 − v0 ) (0.100 × 10−9 kg)[(0.900)(3.00 × 108 m s)]2
=
= 3.65 m.
2F
2(1.00 × 106 N) (b) Using the relativistic workenergy theorem for a constant force, we obtain
(γ − 1) mc 2
∆x =
.
F
1
For the given speed, γ =
= 2.29, thus
1 − 0.9002 ∆x = 37.49. (2.29 − 1)(0.100 × 10−9 kg)(3.00 × 108 m s)2
= 11.6 m.
(1.00 × 106 N) EVALUATE: (c) The distance obtained from the relativistic treatment is greater. As we have seen, more energy is
required to accelerate an object to speeds close to c, so that force must act over a greater distance.
(a) IDENTIFY and SET UP: ∆t0 = 2.60 × 10−8 s is the proper time, measured in the pion’s frame. The time
measured in the lab must satisfy d = c∆t , where u ≈ c. Calculate t and then use Eq.(37.6) to calculate u.
d
1.20 × 103 m
= 4.003 × 10−6 s
EXECUTE: ∆t = =
c 2.998 × 108 m/s
2
∆t0
∆t
" ∆t #
∆t =
so (1 − u 2 / c 2 )1/ 2 = 0 and (1 − u 2 / c 2 ) = $ 0 %
∆t
& ∆t '
1 − u 2 / c2
Write u = (1 − ∆ )c so that (u / c) 2 = (1 − ∆ ) 2 = 1 − 2∆ + ∆ 2 ≈ 1 − 2∆ since ∆ is small. " ∆t #
Using this in the above gives 1 − (1 − 2∆ ) = $ 0 %
& ∆t '
2 2 2 1 " ∆t # 1 " 2.60 × 10−8 s #
−5
∆= $ 0% = $
% = 2.11 × 10
2 & ∆t ' 2 & 4.003 × 10−6 s '
EVALUATE: An alternative calculation is to say that the length of the tube must contract relative to the moving
pion so that the pion travels that length before decaying. The contracted length must be
l = c∆t0 = (2.998 × 108 m/s)(2.60 × 10−8 s) = 7.79 m.
"l#
l = l0 1 − u 2 / c 2 so 1 − u 2 / c 2 = $ %
& l0 ' 2 3710 Chapter 37
2 2 1 " l # 1 " 7.79 m #
−5
Then u = (1 − ∆)c gives ∆ = $ % = $
% = 2.11× 10 , which checks.
2 & l0 ' 2 & 1.20 × 103 m '
(b) IDENTIFY and SET UP: E = γ mc 2 (Eq.(37.38).
1
1
1
EXECUTE: γ =
=
=
= 154
2
2
2∆
1− u /c
2(2.11 × 10−5 ) 37.50. 37.51. E = 154(139.6 MeV) = 2.15 × 104 MeV = 21.5 GeV
EVALUATE: The total energy is 154 times the rest energy.
IDENTIFY and SET UP: The proper length of a side is l0 = a . The side along the direction of motion is shortened to l = l0 1 − v 2 / c 2 . The sides in the two directions perpendicular to the motion are unaffected by the motion and
still have a length a.
EXECUTE: V = a 2l = a 3 1 − v 2 / c 2
IDENTIFY and SET UP: There must be a length contraction such that the length a becomes the same as b;
l0 = a, l = b. l0 is the distance measured by an observer at rest relative to the spacecraft. Use Eq.(37.16) and solve
for u.
b
l
EXECUTE:
= 1 − u 2 / c 2 so = 1 − u 2 / c 2 ;
a
l0
a = 1.40b gives b /1.40b = 1 − u 2 / c 2 and thus 1 − u 2 / c 2 = 1/(1.40) 2 37.52. u = 1 − 1/(1.40) 2 c = 0.700c = 2.10 × 108 m/s
EVALUATE: A length on the spacecraft in the direction of the motion is shortened. A length perpendicular to the
motion is unchanged.
IDENTIFY and SET UP: The proper time ∆t0 is the time that elapses in the frame of the space probe. ∆t is the
time that elapses in the frame of the earth. The distance traveled is 42.2 light years, as measured in the earth frame.
c
"
#
EXECUTE: (a) Light travels 42.2 light years in 42.2 yr, so ∆t = $
% ( 42.2 yr) = 42.6 yr .
0.9910c '
& ∆t0 = ∆t 1 − u 2 / c 2 = (42.6 yr) 1 − (0.9910)2 = 5.7 yr . She measures her biological age to be
19 yr + 5.7 yr = 24.7 yr.
(b) Her age measured by someone on earth is 19 yr + 42.6 yr = 61.6 yr . 37.53. (a) E = γ mc 2 and γ = 10 = 1
1 − (v c ) 2 " " v #2
(b) ( pc) 2 = m 2v 2 2c 2 , E 2 = m 2c 4 $ $ %
$& c '
&
! 37.54. !
2 v
=
c 2 −1
2 ! 99
v
=
= 0.995.
100
c #
+ 1%
%
' E 2 − ( pc) 2
=
E2 1
"v#
1+ γ 2 $ %
&c' 2 = 1
= 0.01 = 1%.
1 + (10 (0.995)) 2 IDENTIFY and SET UP: The clock on the plane measures the proper time ∆t0 .
∆t = 4.00 h = 4.00 h (3600 s/1 h) = 1.44 × 104 s.
∆t0
∆t =
and ∆t0 = ∆t 1 − u 2 / c 2
2
2
1− u / c
" 1 u2 #
u
1 u2
; thus ∆t0 = ∆t $ 1 −
EXECUTE:
small so 1 − u 2 / c 2 = (1 − u 2 / c 2 )1/ 2 ≈ 1 −
2%
2 c2
c
& 2c '
2 The difference in the clock readings is ∆t − ∆t0 = 37.55. 1 u2
1"
250 m/s
#
−9
4
∆t = $
% (1.44 × 10 s) = 5.01 × 10 s. The
2 c2
2 & 2.998 × 108 m/s ' clock on the plane has the shorter elapsed time.
EVALUATE: ∆t0 is always less than ∆t ; our results agree with this. The speed of the plane is much less than the
speed of light, so the difference in the reading of the two clocks is very small.
IDENTIFY: Since the speed is very close to the speed of light, we must use the relativistic formula for kinetic
energy. Relativity 3711 "
#
1
SET UP: The relativistic formula for kinetic energy is K = mc 2 $
− 1% and the relativistic mass is
$ 1 − v2 c2
%
&
'
m
mrel =
.
1 − v2 c2
EXECUTE: (a) K = 7 × 1012 eV = 1.12 × 10−6 J . Using this value in the relativistic kinetic energy formula and
"
#
1
substituting the mass of the proton for m, we get K = mc 2 $
− 1%
$ 1 − v2 c2
%
&
'
2
v
1
1
which gives
= 7.45 × 103 and 1 − 2 =
. Solving for v gives
2
2
(7.45 × 103 ) 2
c
1− v c 1− v 2 (c + v )(c − v) 2(c − v)
=
=
, since c + v ! 2c. Substituting v = (1 − ∆ )c , we have.
c2
c2
c
1 3
1 − v 2 / c 2 ( 7.45 × 10 )
v 2 2(c − v) 2 [ c − (1 − ∆ )c ]
1− 2 =
=
= 2∆ . Solving for ∆ gives ∆ =
=
= 9 × 10−9 , to one
2
2
c
c
c
significant digit.
1
(b) Using the relativistic mass formula and the result that
= 7.45 × 103 , we have
2
2
1− v c
2 "
#
1
% = (7 × 103 )m , to one significant digit.
= m$
2
2%
$ 1− v c
1− v c
&
'
EVALUATE: At such high speeds, the proton’s mass is over 7000 times as great as its rest mass.
E
"1#
IDENTIFY and SET UP: The energy released is E = ( ∆m)c 2 . ∆m = $ 4 % (8.00 kg) . Pav = . The change in
t
& 10 '
gravitational potential energy is mg ∆y .
m mrel = 37.56. 2 2 "1#
EXECUTE: (a) E = ( ∆m)c 2 = $ 4 % (8.00 kg)(3.00 × 108 m/s)2 = 7.20 × 1013 J
& 10 '
E 7.20 × 1013 J
=
= 1.80 × 1019 W
t 4.00 × 10−6 s
E
7.20 × 1013 J
=
= 7.35 × 109 kg
(c) E = ∆U = mg ∆y . m =
g ∆y (9.80 m/s 2 )(1.00 × 103 m)
(b) Pav = 37.57. c
IDENTIFY and SET UP: In crown glass the speed of light is v = . Calculate the kinetic energy of an electron that
n
has this speed.
2.998 × 108 m/s
= 1.972 × 108 m/s.
EXECUTE: v =
1.52
K = mc 2 (γ − 1)
mc 2 = (9.109 × 10 −31 kg)(2.998 × 108 m/s)2 = 8.187 × 10−14 J(1 eV/1.602 × 10−19 J) = 0.5111 MeV γ= 37.58. 37.59. 1
1− v /c
2 2 = 1
1 − ((1.972 × 10 m/s)/(2.998 × 108 m/s))2
8 = 1.328 K = mc 2 (γ − 1) = (0.5111 MeV)(1.328 − 1) = 0.168 MeV
EVALUATE: No object can travel faster than the speed of light in vacuum but there is nothing that prohibits an
object from traveling faster than the speed of light in some material.
p ( E c) E
, where the atom and the photon have the same magnitude of momentum, E c .
(a) v = =
=
m
m
mc
E
(b) v =
$ c, so E $ mc 2 .
mc
IDENTIFY and SET UP: Let S be the lab frame and S ′ be the frame of the proton that is moving in the +x direction,
so u = + c / 2 . The reference frames and moving particles are shown in Figure 37.59. The other proton moves in 3712 Chapter 37 the − x direction in the lab frame, so v = −c / 2 . A proton has rest mass mp = 1.67 × 10−27 kg and rest energy mp c 2 = 938 MeV .
EXECUTE: (a) v′ = −c / 2 − c / 2
v −u
4c
=
=−
1 − uv / c 2 1 − (c / 2)( −c / 2) / c 2
5 4
c.
5
(b) In nonrelativistic mechanics the speeds just add and the speed of each relative to the other is c.
mc 2
(c) K =
− mc 2
1 − v2 / c2
(i) Relative to the lab frame each proton has speed v = c / 2 . The total kinetic energy of each proton is
938 MeV
K=
− (938 MeV) = 145 MeV .
2
"1#
1− $ %
& 2'
4
(ii) In its rest frame one proton has zero speed and zero kinetic energy and the other has speed c . In this frame
5
938 MeV
the kinetic energy of the moving proton is K =
− (938 MeV) = 625 MeV
2
" 4#
1− $ %
&5'
(d) (i) Each proton has speed v = c / 2 and kinetic energy
mc 2 938 MeV
1
2
"1 #
K = mv 2 = $ m % ( c / 2 ) =
=
= 117 MeV
2
8
8
&2 '
(ii) One proton has speed v = 0 and the other has speed c. The kinetic energy of the moving proton
1
938 MeV
is K = mc 2 =
= 469 MeV
2
2
EVALUATE: The relativistic expression for K gives a larger value than the nonrelativistic expression. The kinetic
energy of the system is different in different frames.
The speed of each proton relative to the other is Figure 37.59
37.60. IDENTIFY and SET UP: Let S be the lab frame and let S ′ the frame of the proton that is moving in the +x direction
in the lab frame, as shown in Figure 37.60. In S ′ the other proton moves in the − x′ direction with speed c / 2 , so
v′ = −c / 2 . In the lab frame each proton has speed α c , where α is a constant that we need to solve for.
v′ + u
−0.50c + α c
and
EXECUTE: (a) v =
with v = −α c , u = +α c and v′ = −0.50c gives −α c =
2
1 + uv′ / c
1 + (α c)(−0.50c) / c 2 −0.50 + α
. α 2 − 4α + 1 = 0 and α = 0.268 or α = 3.73 . Can’t have v > c , so only α = 0.268 is physically
1 − 0.50α
allowed. The speed measured by the observer in the lab is 0.268c.
(b) (i) v = 0.269c . γ = 1.0380 . K = (γ − 1)mc 2 = 35.6 MeV .
−α = Relativity 3713 (ii) v = 0.500c . γ = 1.1547 . K = (γ − 1)mc = 145 MeV .
2 37.61. 37.62. 37.63. x′ = c t ′ ! ( x − ut ) γ = c γ ( t − ux c
2 22 2 2 2 2 ) Figure 37.60 22 " u# 1
! x − ut = c(t − ux c 2 ) ! x $ 1 + % = x(u + c ) = t (u + c) ! x = ct ! x 2 = c 2t 2 .
& c' c
IDENTIFY and SET UP: Let S be the lab frame and let S ′ be the frame of the nucleus. Let the +x direction be the
direction the nucleus is moving. u = 0.7500c .
v′ + u
0.9995c + 0.7500c
EXECUTE: (a) v′ = +0.9995c . v =
=
= 0.999929c
1 + uv′ / c 2 1 + (0.7500)(0.9995)
−0.9995c + 0.7500c
(b) v′ = −0.9995c . v =
= −0.9965c
1 + (0.7500)( −0.9995)
(c) emitted in same direction:
"
#
"
#
1
1
− 1% mc 2 = (0.511 MeV) $
− 1% = 42.4 MeV
(i) K = $
2
2
2
$ 1 − (0.999929)
%
& 1− v / c
'
&
'
"
#
"
#
1
1
(ii) K ′ = $
− 1% mc 2 = (0.511 MeV) $
− 1% = 15.7 MeV
2
2
$ 1 − (0.9995) 2
%
& 1− v /c
'
&
'
(d) emitted in opposite direction:
"
#
"
#
1
1
(i) K = $
− 1% mc 2 = (0.511 MeV) $
− 1% = 5.60 MeV
2
2
$ 1 − (0.9965) 2
%
& 1− v / c
'
&
'
"
#
"
#2
1
1
(ii) K ′ = $
− 1% mc = (0.511 MeV) $
− 1% = 15.7 MeV
2
2
$ 1 − (0.9995) 2
%
& 1− v /c
'
&
'
IDENTIFY and SET UP: Use Eq.(37.30), with a = dv / dt , to obtain an expression for dv / dt. Separate the
variables v and t and integrate to obtain an expression for v (t ). In this expression, let t → ∞.
dv F
EXECUTE: a =
= (1 − v 2 / c 2 )3 / 2 . (Onedimensional motion is assumed, and all the F, v, and a refer to xdt m
components.)
dv
"F#
= $ % dt
(1 − v 2 / c 2 )3 / 2 & m '
Integrate from t = 0, when v = 0, to time t, when the velocity is v.
v
t" F #
dv
5 0 (1 − v 2 / c 2 )3 / 2 = 5 0 $ m % dt
&'
t" F #
Ft
Since F is constant, 5 $ % dt = . In the velocity integral make the change of variable y = v / c; then dy = dv / c.
0m
m
&'
v/c v/c
/
0
dv
dy
y
v
5 0 (1 − v 2 / c 2 )3 / 2 = c 5 0 (1 − y 2 )3 / 2 = c 1 (1 − y 2 )1/ 2 2 = 1 − v 2 / c 2
3
40
v
Ft
Thus
=.
1 − v2 / c2 m
v 3714 Chapter 37 37.64. Solve this equation for v:
2
2
v2
" Ft #
" Ft #
= $ % and v 2 = $ % (1 − v 2 / c 2 )
1 − v2 / c2 & m '
&m'
" " Ft #2 # " Ft # 2
( Ft / m)
Ft
so v =
=
v 2 $1 + $
=c
2
22
$ & mc % % $ m %
'% & '
1 + ( Ft / mc)
m c + F 2t 2
&
'
Ft
Ft
→
→ 1, so v → c.
As t → ∞,
22
22
m c +F t
F 2t 2
Ft
is always less than 1, so v < c always and v approaches c only when t → ∞.
EVALUATE: Note that
22
m c + F 2t 2
Setting x = 0 in Eq.(37.21), the first equation becomes x′ = −γ ut and the last, upon multiplication by c, becomes 37.65. ct ′ = γ ct.Squaring and subtracting gives c 2t′2 − x′2 = 2 (c 2t 2 − u 2t 2 ) = c 2t 2 , or x′ = c t ′2 − t 2 = 4.53 × 108 m.
(a) IDENTIFY and SET UP: Use the Lorentz coordinate transformation (Eq.37.21) for ( x1 , t1 ) and ( x2 , t2 ) :
x1 − ut1
x − ut2
′
′
x1 =
, x2 = 2
2
2
1− u /c
1 − u 2 / c2
2
t − ux1 / c
t − ux2 / c 2
′
′
t1 = 1
, t2 = 2
1 − u 2 / c2
1 − u 2 / c2
′2
′′
Same point in S ′ implies x1 = x′ . What then is ∆t ′ = t2 − t1 ?
′2
EXECUTE: x1 = x′ implies x1 − ut1 = x2 − ut2
x − x ∆x
u (t2 − t1 ) = x2 − x1 and u = 2 1 =
t2 − t1 ∆t
From the time transformation equations,
1
′′
( ∆t − u∆x / c 2 )
∆t ′ = t2 − t1 =
1 − u 2 / c2
∆x
gives
Using the result that u =
∆t
1
∆t ′ =
( ∆t − ( ∆x) 2 /(( ∆t )c 2 ))
2
1 − ( ∆x ) /(( ∆t ) 2 c 2 ) ∆t ′ = ∆t ′ = ∆t
( ∆t ) − ( ∆x ) 2 / c 2
2 ( ∆t ) 2 − ( ∆x ) 2 / c 2
( ∆t ) 2 − ( ∆x ) 2 / c 2 ( ∆t − ( ∆x) 2 /((∆t )c 2 )) = ( ∆t ) 2 − ( ∆x / c ) 2 , as was to be shown. This equation doesn’t have a physical solution (because of a negative square root) if ( ∆x / c ) 2 > (∆t ) 2 or ∆x ≥ c∆t.
′′
(b) IDENTIFY and SET UP: Now require that t2 = t1 (the two events are simultaneous in S ′ ) and use the Lorentz
coordinate transformation equations.
′′
EXECUTE: t2 = t1 implies t1 − ux1 / c 2 = t2 − ux2 / c 2
c 2 ∆t
"x −x #
" ∆x #
t2 − t1 = $ 2 2 1 % u so ∆t = $ 2 % u and u =
∆x
&c '
&c '
From the Lorentz transformation equations,
"
#
1
′
∆x′ = x′ − x1 = $
% (∆x − u ∆t ).
2
2
2
& 1− u / c '
Using the result that u = c 2 ∆t / ∆x gives
1
∆x′ =
(∆x − c 2 ( ∆t ) 2 / ∆x)
2
1 − c (∆t ) 2 /( ∆x) 2 ∆x′ =
∆x′ = ∆x
( ∆x) 2 − c 2 ( ∆t ) 2
(∆x) 2 − c 2 ( ∆t ) 2
( ∆x ) 2 − c 2 ( ∆t ) 2 (∆x − c 2 (∆t ) 2 / ∆x)
= (∆x) 2 − c 2 (∆t ) 2 (c) IDENTIFY and SET UP: The result from part (b) is ∆x′ = ( ∆x) 2 − c 2 (∆t ) 2 Relativity 3715 Solve for ∆t : ( ∆x′) 2 = ( ∆x) 2 − c 2 ( ∆t ) 2
( ∆x) 2 − ( ∆x′) 2
(5.00 m) 2 − (2.50 m)2
=
= 1.44 × 10−8 s
c
2.998 × 108 m/s
EVALUATE: This provides another illustration of the concept of simultaneity (Section 37.2): events observed to
be simultaneous in one frame are not simultaneous in another frame that is moving relative to the first.
1
(a) 80.0 m s is nonrelativistic, and K = mv 2 = 186 J.
2
(b) (γ − 1) mc 2 = 1.31 × 1015 J.
∆t = EXECUTE: 37.66. (c) In Eq. (37.23), c) v′ = 2.20 × 108 m s, u = −1.80 × 108 m s,and so v = 7.14 × 107 m s. 20.0 m (d)
(e) γ 20.0 m
= 9.09 × 10−8 s.
2.20 × 108 m s (f) t ′ =
37.67. = 13.6 m. t
13.6 m
= 6.18 × 10−8 s, or t ′ =
= 6.18 × 10−8 s.
γ
2.20 × 108 m s IDENTIFY and SET UP: An increase in wavelength corresponds to a decrease in frequency ( f = c / λ ), so the atoms are moving away from the earth. Receding, so use Eq.(37.26): f = c−u
f0
c+u " 1 − ( f / f0 )2 #
EXECUTE: Solve for u: ( f / f 0 ) 2 (c + u ) = c − u and u = c $
2%
& 1 + ( f / f0 ) '
f = c / λ , f 0 = c / λ0 so f / f 0 = λ0 / λ 37.68. " 1 − (λ0 / λ ) 2 #
" 1 − (656.3/ 953.4) 2 #
u = c$
= 0.357c = 1.07 × 108 m/s
% = c$
2
2%
& 1 + (656.3/ 953.4) '
& 1 + (λ0 / λ ) '
EVALUATE: The relative speed is large, 36% of c. The cosmological implication of such observations will be
discussed in Section 44.6.
The baseball had better be moving nonrelativistically, so the Doppler shift formula (Eq.(37.25)) becomes
f ≅ f 0 (1 − (u c)). In the baseball’s frame, this is the frequency with which the radar waves strike the baseball, and
the baseball reradiates at f. But in the coach’s frame, the reflected waves are Doppler shifted again, so the detected
frequency is f (1 − (u c)) = f 0 (1 − (u c)) 2 ≈ f 0 (1 − 2(u c)), so ∆f = 2 f 0 (u c ) and the fractional frequency shift is ∆f
= 2(u c ). In this case,
f0
u=
37.69. ∆f
(2.86 × 10 −7 )
c=
(3.00 × 108 m) = 42.9 m s = 154 km h = 92.5 mi h.
2 f0
2 IDENTIFY and SET UP: 500 light years = 4.73 × 1018 m . The proper distance l0 to the star is 500 light years. The
energy needed is the kinetic energy of the rocket at its final speed.
d
4.73 × 1018 m
EXECUTE: (a) u = 0.50c . ∆t = =
= 3.2 × 1010 s = 1000 yr
u (0.50)(3.00 × 108 m/s) The proper time is measured by the astronauts. ∆t0 = ∆t 1 − u 2 / c 2 = 866 yr "
#
1
− mc 2 = (1000 kg)(3.00 × 108 m/s)2 $
− 1% = 1.4 × 1019 J
$ 1 − (0.500) 2
%
1 − v2 / c2
&
'
This is 140% of the U.S. yearly use of energy.
d
4.73 × 1018 m
(b) u = 0.99c . ∆t = =
= 1.6 × 1010 s = 505 yr , ∆t0 = 71 yr
u (0.99)(3.00 × 108 m/s)
K= mc 2 "
#
1
K = (9.00 × 1019 J) $
− 1% = 5.5 × 1020 J
$ 1 − (0.99) 2
%
&
'
This is 55 times the U.S. yearly use. 3716 Chapter 37 (c) u = 0.9999c . ∆t = 37.70. d
4.73 × 1018 m
=
= 1.58 × 1010 s = 501 yr , ∆t0 = 7.1 yr
u (0.9999)(3.00 × 108 m/s) "
#
1
K = (9.00 × 1019 J) $
− 1% = 6.3 × 1021 J
2
$ 1 − (0.9999)
%
&
'
This is 630 times the U.S. yearly use.
The energy cost of accelerating a rocket to these speeds is immense.
(a) As in the hint, both the sender and the receiver measure the same distance. However, in our frame, the ship has
moved between emission of successive wavefronts, and we can use the time T = 1 f as the proper time, with the
result that f = γ f 0 > f 0 .
1/ 2 (b) Toward: f = f 0 c+u
" 1 + 0.758 #
= 345 MHz $
%
c−u
& 1 − 0.758 ' = 930 MHz f − f 0 = 930 MHz − 345 MHz = 585 MHz.
1/ 2 Away: f = f 0 37.71. c−u
" 1 − 0.758 #
= 345 MHz $
%
c+u
& 1 + 0.758 ' = 128 MHz and f − f 0 = −217 MHz. (c) f 0 = 1.53 f 0 = 528 MHz, f − f 0 = 183 MHz. The shift is still bigger than f 0 , but not as large as the approaching
frequency.
The crux of this problem is the question of simultaneity. To be “in the barn at one time” for the runner is different
than for a stationary observer in the barn. The diagram in Figure 37.71a shows the rod fitting into the barn at time
t = 0 , according to the stationary observer. The diagram in Figure 37.71b is in the runner’s frame of reference. The
front of the rod enters the barn at time t1 and leaves the back of the barn at time t2 . However, the back of the rod
does not enter the front of the barn until the later time t3 . Figure 37.71
37.72. (c/n) + V (c/n) + V
=
. For V nonrelativistic, this is
cV
1 + (V/nc)
1+ 2
nc
c"
1#
1#
"
v ≈ ((cn) + V )(1 − (V/nc )) = (nc/n) + V − (V/n 2 ) − (V 2 /nc) ≈ + $1 − 2 %V , so k = $1 − 2 % . For water, n = 1.333
n & n'
& n'
and k = 0.437.
In Eq.(37.23), u = V , v′ = (c n), and so v = 3717 Relativity 37.73. dv
dv
v−u
u
dv
. dt ′ = γ ( dt − udx c 2 ) . dv′ =
+
2
22
dt ′
(1 − uv c ) (1 − uv c ) c 2
1
dv′
v −u
"u#
=
+
$ %.
dv 1 − uv c 2 (1 − uv c 2 ) 2 & c 2 ' (a) a′ = "
" 1 − u 2 c2 #
1
(v − u ) u c 2 #
dv′ = dv $
+
= dv $
2
2 2%
2 2%
(1 − uv c ) '
& 1 − uv c
& (1 − uv c ) '
(1 − u 2 c 2 )
1
(1 − uv c 2 ) 2 dv (1 − u 2 c 2 )
a′ =
=
dt − uγ dx c 2 dt (1 − uv c 2 ) 2 (1 − uv c 2 ) dv = a (1 − u 2 c 2 )3 2 (1 − uv c 2 ) −3 .
−3 37.74. " uv′ #
(b) Changing frames from S ′ → S just involves changing a → a′, v → − v′ ! a = a′(1 − u 2 c 2 )3 2 $1 + 2 % .
c'
&
(a) The speed v′ is measured relative to the rocket, and so for the rocket and its occupant, v′ = 0. The acceleration
as seen in the rocket is given to be a′ = g , and so the acceleration as measured on the earth is
32 " u2 #
du
= g $1 − 2 % .
dt
& c'
(b) With v1 = 0 when t = 0 ,
t1
1
du
1 v1
du
v1
dt =
. 5 dt = 5
. t1 =
.
2
2 32
0
g (1 − u c )
g 0 (1 − u 2 c 2 )3 2
g 1 − v12 c 2 a= (c) dt ′ = γ dt = dt / 1 − u 2 c 2 , so the relation in part (b) between dt and du, expressed in terms of dt ′ and du, is
1
du
1
du
dt ′ = γ dt =
=
.
2
2 g (1 − u 2 c 2 ) 3 2
g (1 − u 2 c 2 ) 2
1− u c c
"v #
arctanh $ 1 % . For those who wish to
g
&c'
avoid inverse hyperbolic functions, the above integral may be done by the method of partial fractions;
c " c + v1 #
du
1 / du
du 0
′
gdt ′ =
1n $
=1
+
%.
2 , which integrates to t1 =
2 g & c − v1 '
(1 + u c)(1 − u c) 2 31 + u c 1 − uc 4
′
(d) Solving the expression from part (c) for v1 in terms of t1 , (v1 c ) = tanh( gt1 c), so that
′
Integrating as above (perhaps using the substitution z = u c ) gives t1 = ′
1 − (v1 c) 2 = 1 cosh ( gt1 c), using the appropriate indentities for hyperbolic functions. Using this in the expression ′
gt
c tanh( gt1 c)
c
" gt ′ #
′
= sinh( gt1 c), which may be rearranged slightly as 1 = sinh $ 1 % . If
′
c
g 1 cosh( gt1 c ) g
&c'
′
′
gt1 /c
− gt1 /c
v e −e
′
hyperbolic functions are not used, v1 in terms of t1 is found to be 1 = gt1′/c
which is the same as
c e + e− gt1′/c
c gt1′ c − gt1′ c
′
(e
−e
),
tanh( gt1 c ). Inserting this expression into the result of part (b) gives, after much algebra, t1 =
2g
which is equivalent to the expression found using hyperbolic functions.
(e) After the first acceleration period (of 5 years by Stella’s clock), the elapsed time on earth is
c
′
′
t1 = sinh( gt1 c) = 2.65 × 109 s = 84.0 yr.
g found in part (b), t1 = 37.75. The elapsed time will be the same for each of the four parts of the voyage, so when Stella has returned, Terra has
aged 336 yr and the year is 2436. (Keeping more precision than is given in the problem gives February 7 of that
year.)
(a) f 0 = 4.568110 × 1014 Hz; f + = 4.568910 × 1014 Hz; f − = 4.567710 × 1014 Hz
c + (u + v ) )
f0 *
c − (u + v) *
f +2 (c − (u + v)) = f 02 (c + (u + v))
,! 2
f − (c − (u − v)) = f 02 (c + (u − v))
c + (u − v) *
f− =
f0 *
c − (u − v ) .
f+ = 3718 Chapter 37 where u is the velocity of the center of mass and v is the orbital velocity. ! (u + v ) = ( f + f0 ) 2 − 1
( f 2 f 2) −1
c and (u − v) = −2 02
c
( f + f0 )2 + 1
( f − f0 ) + 1 ! u + v = 5.25 × 10 4 m s and u − v = −2.63 × 104 m s . This gives u = +1.31× 104 m s (moving toward at 13.1 km s) and v = 3.94 × 10 4 m/s .
(b) v = 3.94 × 104 m s; T = 11.0 days. 2π R = vt ! (3.94 × 104 m s)(11.0 days)(24 hrs day)(3600 sec hr)
= 5.96 × 109 m . This is about
2"
0.040 times the earthsun distance.
Also the gravitational force between them (a distance of 2R) must equal the centripetal force from the center of
mass:
(Gm 2 ) mv 2
4 Rv 2 4(5.96 × 109 m)(3.94 × 104 m s) 2
!m=
=
= 5.55 × 1029 kg = 0.279 msun .
=
2
(2 R)
R
G
6.672 × 10−11 N ⋅ m 2 kg 2
For any function f = f ( x, t ) and x = x( x′, t′), t = t ( x′, t′), let F ( x′, t ′) = f ( x( x′, t ′), t ( x′, t ′)) and use the standard
(but mathematically improper) notation F ( x′, t ′) = f ( x′, t ′). The chain rule is then
R= 37.76. ∂f ( x′, t ′) ∂f ( x, t ) ∂x′ ∂f ( x′, t ′) ∂t′
,
=
+
∂x
∂x′ ∂x
∂t ′
∂x
∂f ( x′, t ′) ∂f ( x, t ) ∂x′ ∂f ( x′, t ′) ∂t ′
.
=
+
∂t
∂x′ ∂t
∂t ′
∂t
In this solution, the explicit dependence of the functions on the sets of dependent variables is suppressed, and the
∂f ∂f ∂x′ ∂f ∂t′ ∂f ∂f ∂x′ ∂f ∂t ′
=
+
,
=
+
.
above relations are then
∂x ∂x′ ∂x ∂t′ ∂x ∂t ∂x′ ∂t ∂t ′ ∂t
∂x′
∂x′
∂t ′
∂t ′
∂E ∂E
∂2E ∂2E
(a)
= 1,
= −v, = 0 and
= 1. Then,
=
, and 2 = 2 . For the time derivative,
∂x
∂t
∂x
∂t
∂x ∂x′
∂x
∂x′
∂E
∂E ∂E
= −v
+
. To find the second time derivative, the chain rule must be applied to both terms; that is,
∂t
∂x′ ∂t′ ∂
∂t
∂
∂t ∂E
∂2E ∂2E
= −v 2 +
,
∂x′
∂x′ ∂t ′∂x′
∂E
∂2E ∂2E
= −v
+
.
∂t′
∂x′∂t ′ ∂t ′2 ∂2E
, collecting terms and equating the mixed partial derivatives gives
∂t 2
∂2E
∂2E
∂2E
∂2E ∂2E
= v 2 2 − 2v
+ 2 , and using this and the above expression for
gives the result.
∂t 2
∂x′
∂x′∂t ′ ∂t ′
∂x′2
∂x′
∂x′
∂t′
∂t ′
=,
= γ v,
= γ v / c 2 and
=.
(b) For the Lorentz transformation,
∂x
∂t
∂x
∂t
The first partials are then
Using these in ∂E
∂E
v ∂E ∂E
∂E
∂E
=
−2
=− v
+
,
∂x
∂x′
c ∂t′ ∂t
∂x′
∂t′
and the second partials are (again equating the mixed partials)
∂2E
=
∂x 2
∂2E
=
∂t 2 ∂2E
v2 ∂2E
v ∂2E
+24
−2 2 2
∂x′2
c ∂t ′2
c ∂x′∂t ′
2
2
∂2E
2 2∂ E
2∂ E
2
v
+
−2 v
.
2
2
∂x′
∂t ′
∂x′∂t′
2 Substituting into the wave equation and combining terms (note that the mixed partials cancel), ∂2E 1 ∂2E
−
=
∂x 2 c 2 ∂t 2 2 2
" v2 # ∂2E
1 # ∂2E ∂2E 1 ∂2E
2"v
= 0.
$1 − 2 % 2 + γ $ 4 − 2 % 2 = 2 − 2
∂x′ c ∂t ′2
& c ' ∂x′
& c c ' ∂t ′ Relativity 37.77. 3719 (a) In the center of momentum frame, the two protons approach each other with equal velocities (since the protons
have the same mass). After the collision, the two protons are at rest but now there are kaons as well. In this
situation the kinetic energy of the protons must equal the total rest energy of the two kaons ! 2( cm − 1) mp c 2 =
2mk c 2 !
vcm = c cm =1+ −1 2
cm
2
cm mk
= 1.526. The velocity of a proton in the center of momentum frame is then
mp = 0.7554c. To get the velocity of this proton in the lab frame, we must use the Lorentz velocity transformations. This is the
same as “hopping” into the proton that will be our target and asking what the velocity of the projectile proton is.
Taking the lab frame to be the unprimed frame moving to the left, u = vcm and v′ = vcm (the velocity of the projectile
proton in the center of momentum frame).
v′ + u
2vcm
1
vlab =
=
= 0.9619c ! lab =
= 3.658 ! K lab = ( lab − 1)mpc 2 = 2494 MeV.
2
2
uv′
vcm
vlab
1+ 2 1+ 2
1− 2
c
c
c K lab
2494 MeV
=
= 2.526.
2mk 2(493.7 MeV)
(c) The center of momentum case considered in part (a) is the same as this situation. Thus, the kinetic energy
required is just twice the rest mass energy of the kaons. K cm = 2(493.7 MeV) = 987.4 MeV. This offers a
substantial advantage over the fixed target experiment in part (b). It takes less energy to create two kaons in the
proton center of momentum frame.
(b) PHOTONS, ELECTRONS, AND ATOMS 38.1. 38 IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by V0 = h
φ
f − . The
e
e slope of V0 versus f is h/e. The value fth of f when V0 = 0 is related to φ by φ = hf th .
EXECUTE: (a) From the graph, f th = 1.25 × 1015 Hz . Therefore, with the value of h from part (b), φ = hf th = 4.8 eV .
(b) From the graph, the slope is 3.8 × 10−15 V ⋅ s . h = (e)(slope) = (1.60 × 10−16 C)(3.8 × 10−15 V ⋅ s) = 6.1 × 10−34 J ⋅ s
(c) No photoelectrons are produced for f < f th . 38.2. (d) For a different metal fth and φ are different. The slope is h/e so would be the same, but the graph would be
shifted right or left so it has a different intercept with the horizontal axis.
EVALUATE: As the frequency f of the light is increased above fth the energy of the photons in the light increases
and more energetic photons are produced. The work function we calculated is similar to that for gold or nickel.
IDENTIFY and SET UP: c = f λ relates frequency and wavelength and E = hf relates energy and frequency for a photon. c = 3.00 × 108 m/s . 1 eV = 1.60 × 10−16 J .
c 3.00 × 108 m/s
= 5.94 × 1014 Hz
EXECUTE: (a) f = =
λ 505 × 10−9 m
(b) E = hf = (6.626 × 10 −34 J ⋅ s)(5.94 × 1014 Hz) = 3.94 × 10−19 J = 2.46 eV
(c) K = 1 mv 2 so v =
2
38.3. 38.4. 3.00 × 108 m s
= 5.77 × 1014 Hz
5.20 × 10−7 m
h 6.63 × 10−34 J ⋅ s
= 1.28 × 10−27 kg ⋅ m s
p= =
5.20 × 10−7 m
f= c = E = pc = (1.28 × 10−27 kg ⋅ m s) (3.00 × 108 m s) = 3.84 × 10−19 J = 2.40 eV.
energy
hc
. 1 eV = 1.60 × 10−19 J . For a photon, E = hf = . h = 6.63 × 10−34 J ⋅ s.
IDENTIFY and SET UP: Pav =
t
λ
EXECUTE: (a) energy = Pavt = (0.600 W)(20.0 × 10−3 s) = 1.20 × 10 −2 J = 7.5 × 1016 eV (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
= 3.05 × 10−19 J = 1.91 eV
652 × 10−9 m
λ
(c) The number of photons is the total energy in a pulse divided by the energy of one photon:
1.20 × 10−2 J
= 3.93 × 1016 photons .
3.05 × 10−19 J/photon
EVALUATE: The number of photons in each pulse is very large.
IDENTIFY and SET UP: Eq.(38.2) relates the photon energy and wavelength. c = f λ relates speed, frequency and
wavelength for an electromagnetic wave.
E (2.45 × 106 eV)(1.602 × 10−19 J/1 eV)
= 5.92 × 1020 Hz
EXECUTE: (a) E = hf so f = =
6.626 × 10−34 J ⋅ s
h
c 2.998 × 108 m/s
= 5.06 × 10−13 m
(b) c = f λ so λ = =
f
5.92 × 1020 Hz
(c) EVALUATE: λ is comparable to a nuclear radius. Note that in doing the calculation the energy in MeV was
converted to the SI unit of Joules.
(b) E = 38.5. 2K
2(3.94 × 10−19 J)
=
= 9.1 mm/s
m
9.5 × 10−15 kg hc = 381 382 38.6. Chapter 38 λth = 272 nm . c = f λ . IDENTIFY and SET UP: h = 4.136 × 10−15 eV ⋅ s .
EXECUTE: (a) f th = 38.7. c λth 12
mvmax = hf − φ . At the threshold frequency, f th , vmax → 0.
2 3.00 × 108 m/s
= 1.10 × 1015 Hz .
272 × 10−9 m
eV ⋅ s)(1.10 × 1015 Hz) = 4.55 eV . = (b) φ = hf th = (4.136 × 10−15
12
(c) mvmax = hf − φ = (4.136 × 10−15 eV ⋅ s)(1.45 × 1015 Hz) − 4.55 eV = 6.00 eV − 4.55 eV = 1.45 eV
2
EVALUATE: The threshold wavelength depends on the work function for the surface.
hc
12
IDENTIFY and SET UP: Eq.(38.3): mvmax = hf − φ =
− φ . Take the work function φ from Table 38.1. Solve
2
λ
for vmax . Note that we wrote f as c / λ .
EXECUTE: 12
(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
− (5.1 eV)(1.602 × 10−19 J/1 eV)
mvmax =
2
235 × 10−9 m 12
mvmax = 8.453 × 10−19 J − 8.170 × 10−19 J = 2.83 × 10−20 J
2
2(2.83 × 10−20 J)
vmax =
= 2.49 × 105 m/s
9.109 × 10−31 kg 38.8. EVALUATE: The work function in eV was converted to joules for use in Eq.(38.3). A photon with λ = 235 nm
has energy greater then the work function for the surface.
hc
IDENTIFY and SET UP: φ = hf th =
. The minimum φ corresponds to the minimum λ . λth EXECUTE:
38.9. φ= hc λth = (4.136 × 10 −15 eV ⋅ s)(3.00 × 108 m/s)
= 1.77 eV
700 × 10−9 m
c = f λ . The source emits (0.05)(75 J) = 3.75 J of energy as visible light each second. IDENTIFY and SET UP: E = hf , with h = 6.63 × 10−34 J ⋅ s.
EXECUTE: (a) f = 38.10. c λ = 3.00 × 108 m/s
= 5.00 × 1014 Hz
600 × 10−9 m
J ⋅ s)(5.00 × 1014 Hz) = 3.32 × 10−19 J . The number of photons emitted per second is (b) E = hf = (6.63 × 10−34
3.75 J
= 1.13 × 1019 photons .
3.32 × 10−19 J/photon
(c) No. The frequency of the light depends on the energy of each photon. The number of photons emitted per
second is proportional to the power output of the source.
IDENTIFY: In the photoelectric effect, the energy of the photon is used to eject an electron from the surface, and
any excess energy goes into kinetic energy of the electron.
SET UP: The energy of a photon is E = hf, and the work function is given by φ = hf0, where f0 is the threshold frequency.
EXECUTE: (a) From the graph, we see that Kmax = 0 when λ = 250 nm, so the threshold wavelength is 250 nm.
Calling f0 the threshold frequency, we have
f0 = c/λ0 = (3.00 × 108 m/s)/(250 nm) = 1.2 × 1015 Hz.
(b) φ = hf0 = (4.136 × 10–15 eV ⋅ s )(1.2 × 1015 Hz) = 4.96 eV = 5.0 eV
(c) The graph (see Figure 38.10) is linear for λ < λ0 (1/λ > 1/λ0), and linear graphs are easier to interpret than curves.
EVALUATE: If the wavelength of the light is longer than the threshold wavelength (that is, if 1/λ < 1/λ0), the
kinetic energy of the electrons is really not defined since no photoelectrons are ejected from the metal. Figure 38.10 Photons, Electrons, and Atoms 38.11. 383 IDENTIFY: Protons have mass and photons are massless.
(a) SET UP: For a particle with mass, K = p 2 / 2m.
EXECUTE: p2 = 2 p1 means K 2 = 4 K1.
(b) SET UP: For a photon, E = pc. 38.12. EXECUTE: p2 = 2 p1 means E2 = 2 E1.
EVALUATE: The relation between E and p is different for particles with mass and particles without mass.
12
IDENTIFY and SET UP: eV0 = mvmax , where V0 is the stopping potential. The stopping potential in volts equals
2
12
eV0 in electron volts. mvmax = hf − φ .
2
12
EXECUTE: (a) eV0 = mvmax so
2
(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
− 2.3 eV = 4.96 eV − 2.3 eV = 2.7 eV . The stopping potential
eV0 = hf − φ =
250 × 10 −9 m
is 2.7 electron volts.
12
(b) mvmax = 2.7 eV
2
(c) vmax = 38.13. 2(2.7 eV)(1.60 × 10 −19 J/eV)
= 9.7 × 105 m/s
9.11 × 10−31 kg First use Eq.(38.4) to find the work function φ .
hc
eV0 = hf − φ so φ = hf − eV0 =
− eV0 (a) IDENTIFY:
SET UP: λ (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
EXECUTE: φ =
− (1.602 × 10−19 C)(0.181 V)
254 × 10−9 m
φ = 7.821 × 10−19 J − 2.900 × 10−20 J = 7.531× 10−19 J(1 eV/1.602 × 10−19 J) = 4.70 eV
IDENTIFY and SET UP: The threshold frequency f th is the smallest frequency that still produces photoelectrons. It corresponds to K max = 0 in Eq.(38.3), so hf th = φ .
f= EXECUTE: λ says hc λth =φ (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
= 2.64 × 10 −7 m = 264 nm
φ
7.531 × 10−19 J
(b) EVALUATE: As calculated in part (a), φ = 4.70 eV. This is the value given in Table 38.1 for copper.
IDENTIFY and SET UP: A photon has zero rest mass, so its energy and momentum are related by Eq.(37.40).
Eq.(38.5) then relates its momentum and wavelength.
EXECUTE: (a) E = pc = (8.24 × 10−28 kg ⋅ m/s)(2.998 × 108 m/s) = 2.47 × 10−19 J = λth = 38.14. c hc = (2.47 × 10 −19 J)(1 eV/1.602 × 10−19 J) = 1.54 eV h
6.626 × 10−34 J ⋅ s
=
= 8.04 × 10 −7 m = 804 nm
λ
p 8.24 × 10 −28 kg ⋅ m/s
EVALUATE: This wavelength is longer than visible wavelengths; it is in the infrared region of the
electromagnetic spectrum. To check our result we could verify that the same E is given by Eq.(38.2), using the λ
we have calculated.
1
"1 1#
IDENTIFY and SET UP: Balmer’s formula is = R $ 2 − 2 % . For the Hγ spectral line n = 5. Once we have λ ,
λ
&2 n '
calculate f from f = c / λ and E from Eq.(38.2).
(b) p = 38.15. h so λ = "1 1#
" 25 − 4 #
" 21 #
= R$ 2 − 2 % = R$
% = R$
%.
2 5'
100 '
&
&
& 100 '
100
100
m = 4.341 × 10−7 m = 434.1 nm.
=
Thus λ =
21R 21(1.097 × 107 ) EXECUTE: (a) (b) f = c λ = 1 λ 2.998 × 108 m/s
= 6.906 × 1014 Hz
4.341 × 10−7 m 384 Chapter 38 (c) E = hf = (6.626 × 10−34 J ⋅ s)(6.906 × 1014 Hz) = 4.576 × 10−19 J = 2.856 eV
EVALUATE: Section 38.3 shows that the longest wavelength in the Balmer series (Hα ) is 656 nm and the shortest is 365 nm. Our result for Hγ falls within this range. The photon energies for hydrogen atom transitions are
38.16. in the eV range, and our result is of this order.
IDENTIFY and SET UP: For the Lyman series the final state is n = 1 and the wavelengths are given by
1
"1 1 #
= R $ 2 − 2 % , n = 2,3,.... For the Paschen series the final state is n = 3 and the wavelengths are given by
λ
&1 n ' "1 1#
= R $ 2 − 2 % , n = 4,5,.... R = 1.097 × 107 m −1 . The longest wavelength is for the smallest n and the shortest
λ
&3 n '
wavelength is for n → ∞ .
1
4
" 1 1 # 3R
EXECUTE: Lyman Longest:
= R$ 2 − 2 % =
. λ=
= 121.5 nm .
λ
3(1.097 × 107 m −1 )
&1 2 ' 4
1 Shortest: 1 λ 1
"1 1 #
= R$ 2 − 2 % = R . λ =
= 91.16 nm
1 ∞'
1.097 × 107 m −1
& Paschen Longest: 1# R
"1
= R$ 2 − 2 % = .
λ
&3 ∞ ' 9
hc (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s)
=
= 2.31 × 10−19 J = 1.44 eV.
(a) Eγ =
8.60 × 10−7 m
So the internal energy of the atom increases by 1.44 eV to E = −6.52 eV + 1.44 eV = −5.08 eV.
Shortest: 38.17. 38.19. 1 (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s)
= 4.74 × 10 −19 J = 2.96 eV.
4.20 × 10−7 m
So the final internal energy of the atom decreases to E = −2.68 eV − 2.96 eV = −5.64 eV.
IDENTIFY and SET UP: The ionization threshold is at E = 0 . The energy of an absorbed photon equals the
energy gained by the atom and the energy of an emitted photon equals the energy lost by the atom.
EXECUTE: (a) ∆E = 0 − (−20 eV) = 20 eV
(b) When the atom in the n = 1 level absorbs a 18 eV photon, the final level of the atom is n = 4 . The possible
transitions from n = 4 and corresponding photon energies are n = 4 → n = 3, 3 eV ; n = 4 → n = 2, 8 eV ;
n = 4 → n = 1, 18 eV . Once the atom has gone to the n = 3 level, the following transitions can occur:
n = 3 → n = 2, 5 eV ; n = 3 → n = 1, 15 eV . Once the atom has gone to the n = 2 level, the following transition
can occur: n = 2 → n = 1, 10 eV . The possible energies of emitted photons are: 3 eV, 5 eV, 8 eV, 10 eV, 15 eV,
and 18 eV.
(c) There is no energy level 8 eV higher in energy than the ground state, so the photon cannot be absorbed.
(d) The photon energies for n = 3 → n = 2 and for n = 3 → n = 1 are 5 eV and 15 eV. The photon
energy for n = 4 → n = 3 is 3 eV. The work function must have a value between 3 eV and 5 eV.
IDENTIFY and SET UP: The wavelength of the photon is related to the transition energy Ei − Ef of the atom by
(b) Eγ = 38.18. 144
" 1 1 # 7R
= R$ 2 − 2 % =
. λ=
= 1875 nm .
λ
7(1.097 × 107 m −1 )
& 3 4 ' 144
1 Ei − Ef = hc hc λ = where hc = 1.240 × 10−6 eV ⋅ m . EXECUTE: (a) The minimum energy to ionize an atom is when the upper state in the transition has E = 0 , so
1.240 × 10−6 eV ⋅ m
= 16.79 eV .
E1 = −17.50 eV . For n = 5 → n = 1 , λ = 73.86 nm and E5 − E1 =
73.86 × 10−9 m
E5 = −17.50 eV + 16.79 eV = −0.71 eV . For n = 4 → n = 1 , λ = 75.63 nm and E4 = −1.10 eV . For
n = 3 → n = 1 , λ = 79.76 nm and E3 = −1.95 eV . For n = 2 → n = 1 , λ = 94.54 nm and E2 = −4.38 eV .
hc
1.240 × 10−6 eV ⋅ m
=
= 378 nm
Ei − Ef
3.28 eV
EVALUATE: The n = 4 → n = 2 transition energy is smaller than the n = 4 → n = 1 transition energy so the
wavelength is longer. In fact, this wavelength is longer than for any transition that ends in the n = 1 state.
(b) Ei − Ef = E4 − E2 = −1.10 eV − ( −4.38 eV) = 3.28 eV and λ = Photons, Electrons, and Atoms 38.20. 38.21. 385 (a) Equating initial kinetic energy and final potential energy and solving for the separation radius r,
1 (92e) (2e)
1 (184) (1.60 × 10−19 C)
=
= 5.54 × 10−14 m.
r=
4π P0
4π P0 (4.78 × 106 J C)
K
(b) The above result may be substituted into Coulomb’s law, or, the relation between the magnitude of the force
and the magnitude of the potential energy in a Coulombic field is
K (4.78 × 106 eV) (1.6 × 10−19 J ev)
= 13.8 N.
F= =
(5.54 × 10−14 m)
r
1 q1q2
.
(a) IDENTIFY: If the particles are treated as point charges, U =
4π P0 r
SET UP: q1 = 2e (alpha particle); q2 = 82e (gold nucleus); r is given so we can solve for U.
EXECUTE: (2)(82)(1.602 × 10−19 C)2
= 5.82 × 10−13 J
6.50 × 10 −14 m
J) = 3.63 × 106 eV = 3.63 MeV U = (8.987 × 109 N ⋅ m 2 /C2 ) U = 5.82 × 10 −13 J(1 eV/1.602 × 10−19 (b) IDENTIFY: Apply conservation of energy: K1 + U1 = K 2 + U 2 .
SET UP: Let point 1 be the initial position of the alpha particle and point 2 be where the alpha particle
momentarily comes to rest. Alpha particle is initially far from the lead nucleus implies r1 ≈ ∞ and U1 = 0. Alpha particle stops implies K 2 = 0.
EXECUTE: Conservation of energy thus says K1 = U 2 = 5.82 × 10−13 J = 3.63 MeV. 1
2K
2(5.82 × 10−13 J)
=
= 1.32 × 107 m/s
(c) K = mv 2 so v =
2
m
6.64 × 10−27 kg 38.22.
38.23. 38.24. EVALUATE: v / c = 0.044, so it is ok to use the nonrelativistic expression to relate K and v. When the alpha
particle stops, all its initial kinetic energy has been converted to electrostatic potential energy.
h
(a), (b) For either atom, the magnitude of the angular momentum is
= 1.05 × 10−34 kg ⋅ m 2 s.
2π
IDENTIFY and SET UP: Use the energy to calculate n for this state. Then use the Bohr equation, Eq.(38.10), to
calculate L.
EXECUTE: En = −(13.6 eV)/n 2 , so this state has n = 13.6 /1.51 = 3. In the Bohr model. L = nU so for this state
L = 3U = 3.16 × 10−34 kg ⋅ m 2 /s.
EVALUATE: We will find in Section 41.1 that the modern quantum mechanical description gives a different
result.
13.6 eV
hc
IDENTIFY and SET UP: For a hydrogen atom En = −
. ∆E =
, where ∆E is the magnitude of the
λ
n2
energy change for the atom and λ is the wavelength of the photon that is absorbed or emitted.
" 1 1#
EXECUTE: ∆E = E4 − E1 = −(13.6 eV) $ 2 − 2 % = +12.75 eV .
&4 1 ' hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
c
=
= 97.3 nm . f = = 3.08 × 1015 Hz .
∆E
λ
12.75 eV
1 Ze 2
, where Z = 4 is the nuclear
IDENTIFY: The force between the electron and the nucleus in Be3+ is F =
4π P0 r 2 λ= 38.25. charge. All the equations for the hydrogen atom apply to Be3+ if we replace e 2 by Ze 2 .
(a) SET UP: Modify Eq.(38.18).
1 me 4
EXECUTE: En = −
(hydrogen) becomes
P0 8n 2h 2
" 1 me 4 #
1 m( Ze2 ) 2
2 " 13.60 eV #
3+
= Z2$−
% = Z $−
% (for Be )
22
n2
P0 8n h
P0 8n 2h 2 '
&
'
&
" 13.60 eV #
The groundlevel energy of Be3+ is E1 = 16 $ −
% = −218 eV.
12
&
'
En = − EVALUATE: The groundlevel energy of Be3+ is Z 2 = 16 times the groundlevel energy of H.
(b) SET UP: The ionization energy is the energy difference between the n → ∞ level energy and the n = 1 level
energy. 386 Chapter 38 EXECUTE: The n → ∞ level energy is zero, so the ionization energy of Be3+ is 218 eV.
EVALUATE: This is 16 times the ionization energy of hydrogen.
"1 1#
1
(c) SET UP:
= R $ 2 − 2 % just as for hydrogen but now R has a different value.
λ
& n1 n2 '
EXECUTE: RH = me4
= 1.097 × 107 m −1 for hydrogen becomes
8P0 h3c me4
= 16(1.097 × 107 m −1 ) = 1.755 × 108 m −1 for Be3+ .
8P0 h3c
1
"1 1#
For n = 2 to n = 1, = RBe $ 2 − 2 % = 3R/4.
λ
&1 2 '
RBe = Z 2 λ = 4 /(3R ) = 4 /(3(1.755 × 108 m −1 )) = 7.60 × 10−9 m = 7.60 nm.
EVALUATE: This wavelength is smaller by a factor of 16 compared to the wavelength for the corresponding
transition in the hydrogen atom.
n 2h 2
(d) SET UP: Modify Eq.(38.12): rn = P0
(hydrogen).
π me2
22
nh
EXECUTE: rn = P0
(Be3+ ).
π m( Ze 2 ) 38.26. 38.27. EVALUATE: For a given n the orbit radius for Be3+ is smaller by a factor of Z = 4 compared to the
corresponding radius for hydrogen.
(a) We can find the photon’s energy from Eq. 38.8
"1 1#
"1 1#
E = hcR $ 2 − 2 % = (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s ) (1.097 × 107 m −1 ) $ 2 − 2 % = 4.58 × 10−19 J. The
&2 n '
&2 5 '
E
corresponding wavelength is =
= 434 nm.
hc
(b) In the Bohr model, the angular momentum of an electron with principal quantum number n is given by
h
Eq. 38.10: L = n . Thus, when an electron makes a transition from n = 5 to n = 2 orbital, there is the following
2π
loss in angular momentum (which we would assume is transferred to the photon):
h
3(6.63 × 10−34 J ⋅ s)
∆L = (2 − 5)
=−
= −3.17 × 10−34 J ⋅ s.
2π
2"
However, this prediction of the Bohr model is wrong (as shown in Chapter 41).
1 e2
(1.60 × 10−19 C) 2
: n = 1 ! v1 =
= 2.18 × 106 m/s
(a) vn =
P0 2nh
P0 2 (6.63 × 10−34 J ⋅ s) h = 2 ! v2 =
(b) Orbital period = v1
v
= 1.09 × 106 m s. h = 3 ! v3 = 1 = 7.27 × 105 m s.
2
3 2π rn 2P0 n 2h 2 me 2 4P02n3h3
=
=
vn
1 P0 ⋅ e2 2nh
me4
n = 1 ! T1 = 4P02 (6.63 × 10−34 J ⋅ s)3
= 1.53 × 10−16 s
(9.11 × 10−31 kg) (1.60 × 10−19 C)4 n = 2 : T2 = T1 (2)3 = 1.22 × 10−15 s. n = 3 : T3 = T1 (3)3 = 4.13 × 10−15 s.
(c) number of orbits =
38.28. 1.0 × 10−8 s
= 8.2 × 106.
1.22 × 10−15 s IDENTIFY and SET UP:
EXECUTE: (a) En = − En = − 13.6 eV
n2 13.6 eV
13.6 eV
and En +1 = −
n2
( n + 1) 2 /1
10
n 2 − (n + 1) 2
∆E = En +1 − En = (−13.6 eV) 1
− 2 2 = −(13.6 eV) 2
2
n4
(n )(n + 1) 2
3 (n + 1)
2n
2
2n + 1
∆E = (13.6 eV) 2
As n becomes large, ∆E → (13.6 eV) 4 = (13.6 eV) 3
n
n
(n )(n + 1) 2 Photons, Electrons, and Atoms 38.29. 387 Thus ∆E becomes small as n becomes large.
(b) rn = n 2r1 so the orbits get farther apart in space as n increases.
IDENTIFY and SET UP: The number of photons emitted each second is the total energy emitted divided by the
energy of one photon. The energy of one photon is given by Eq.(38.2). E = Pt gives the energy emitted by the
laser in time t.
EXECUTE: In 1.00 s the energy emitted by the laser is (7.50 × 10−3 W)(1.00 s) = 7.50 × 10−3 J.
The energy of each photon is E = hc λ = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
= 1.874 × 10−20 J.
10.6 × 10−6 m −3 7.50 × 10 J/s
= 4.00 × 1017 photons/s
1.874 × 10−20 J/photon
EVALUATE: The number of photons emitted per second is extremely large.
IDENTIFY and SET UP: Visible light has wavelengths from about 400 nm to about 700 nm. The energy of each
hc 1.99 × 10−25 J ⋅ m
photon is E = hf =
. The power is the total energy per second and the total energy Etot is the
= Therefore
38.30. λ λ number of photons N times the energy E of each photon.
EXECUTE: (a) 193 nm is shorter than visible light so is in the ultraviolet.
hc
(b) E =
= 1.03 × 10−18 J = 6.44 eV λ 38.31. E
NE
Pt (1.50 × 10−3 W)(12.0 × 10−9 s)
(c) P = tot =
so N =
=
= 1.75 × 107 photons
t
t
E
1.03 × 10−18 J
EVALUATE: A very small amount of energy is delivered to the lens in each pulse, but this still corresponds to a
large number of photons.
n
− ( E − E ) / kT
IDENTIFY: Apply Eq.(38.21): 5 s = e 5 s 3 p
n3 p
SET UP:
EXECUTE: From Fig.38.24a in the textbook, E5 s = 20.66 eV and E3 p = 18.70 eV E5 s − E3 p = 20.66 eV − 18.70 eV = 1.96 eV(1.602 × 10−19 J/1 eV) = 3.140 × 10−19 J (a)
(b) −19
−23
n5 s
= e− (3.140×10 J)/[(1.38×10 J/K)(600 K)] = e −37.90 = 3.5 × 10−17
n3 p (c) 38.32. −19
−23
n5 s
= e − (3.140×10 J)/[(1.38×10 J/K)(300 K)] = e −75.79 = 1.2 × 10−33
n3 p −19
−23
n5 s
= e− (3.140×10 J)/[(1.38×10 J/K)(1200 K)] = e−18.95 = 5.9 × 10−9
n3 p (d) EVALUATE: At each of these temperatures the number of atoms in the 5s excited state, the initial state for the
transition that emits 632.8 nm radiation, is quite small. The ratio increases as the temperature increases.
n2 P3 2
) KT
−( E
−E
= e 2 P3 2 2 P1 2 .
n2 P1/ 2 (6.626 × 10−34 J)(3.000 × 108 m s)
= 3.375 × 10−19 J.
5.890 × 10−7 m
1
hc (6.626 × 10−34 J)(3.000 × 108 m s)
=
=
= 3.371 × 10−19 J. so ∆E3 / 2 −1/ 2 = 3.375 × 10−19 J − 3.371 × 10−19 J =
5.896 × 10−7 m
2 From the diagram ∆E3 / 2 − g =
∆E1 2 − g 4.00 × 10−22 J. 38.33. eVAC = hf max = n2 P3 / 2
n2 P1/ 2 hc = e − (4.00 × 10 = −22 J) (1.38 × 10−23 J / K ⋅500 K). = 0.944. So more atoms are in the 2 p1 2 state. hc
min ! min = hc
(6.63 × 10−34 J ⋅ s)(3.00 × 108 m s)
=
= 3.11× 10−10 m
eVAC
(1.60 × 10−19 C)(4000 V) This is the same answer as would be obtained if electrons of this energy were used. Electron beams are much more
easily produced and accelerated than proton beams. 388 Chapter 38 38.34. IDENTIFY and SET UP:
EXECUTE: (a) V = hc λ = eV , where λ is the wavelength of the x ray and V is the accelerating voltage. hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
= 8.29 kV
eλ (1.60 × 10−19 C)(0.150 × 10−9 m) hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
= 4.14 × 10−11 m = 0.0414 nm
eV
(1.60 × 10−19 C)(30.0 × 103 V)
(c) No. A proton has the same magnitude of charge as an electron and therefore gains the same amount of kinetic
energy when accelerated by the same magnitude of potential difference.
IDENTIFY: The initial electrical potential energy of the accelerated electrons is converted to kinetic energy which
is then given to a photon.
SET UP: The electrical potential energy of an electron is eVAC, where VAC is the accelerating potential, and the
energy of a photon is hf. Since the energy of the electron is all given to a photon, we have eVAC = hf. For any wave,
fλ = v.
EXECUTE: (a) eVAC = hfmin gives
(b) λ = 38.35. fmin = eVAC/h = (1.60 × 10–19 C)(25,000 V)/(6.626 × 10–34 J ⋅ s ) = 6.037 × 1018 Hz 38.36. = 6.04 × 1018 Hz, rounded to three digits
(b) λmin = c/fmax = (3.00 × 108 m/s)/(6.037 × 1018 Hz) = 4.97 × 10–11 m = 0.0497 nm
(c) We assume that all the energy of the electron produces only one photon on impact with the screen.
EVALUATE: These photons are in the xray and γray part of the electromagnetic spectrum (see Figure 32.4 in the
textbook) and would be harmful to the eyes without protective glass on the screen to absorb them.
hc
IDENTIFY and SET UP: The wavelength of the x rays produced by the tube is give by
= eV . λ h
h
hc
λ′ = λ +
(1 − cosφ ) .
= 2.426 × 10−12 m . The energy of the scattered x ray is
.
λ′
mc
mc
hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
= 6.91 × 10−11 m = 0.0691 nm
EXECUTE: (a) λ =
(1.60 × 10−19 C)(18.0 × 103 V)
eV
h
(1 − cos φ ) = 6.91× 10−11 m + (2.426 × 10−12 m)(1 − cos 45.0°) .
mc
λ ′ = 6.98 × 10 −11 m = 0.0698 nm .
hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
(c) E =
=
= 17.8 keV
6.98 × 10−11 m
λ′
EVALUATE: The incident x ray has energy 18.0 keV. In the scattering event, the photon loses energy and its
wavelength increases.
h
IDENTIFY: Apply Eq.(38.23): λ ′ − λ =
(1 − cosφ ) = λC (1 − cos φ )
mc
SET UP: Solve for λ ′ : λ ′ = λ + λC (1 − cos φ )
The largest λ ′ corresponds to φ = 180°, so cosφ = −1.
(b) λ ′ = λ + 38.37. 38.38. EXECUTE: λ ′ = λ + 2λC = 0.0665 × 10−9 m + 2(2.426 × 10−12 m) = 7.135 × 10−11 m = 0.0714 nm. This wavelength
occurs at a scattering angle of φ = 180°.
EVALUATE: The incident photon transfers some of its energy and momentum to the electron from which it
scatters. Since the photon loses energy its wavelength increases, λ ′ > λ .
∆
(a) From Eq. (38.23), cosφ = 1 −
, and so ∆ = 0.0542 nm − 0.0500 nm,
( h mc) cos φ = 1 − 0.0042 nm
= −0.731, and φ = 137°.
0.002426 nm 0.0021 nm
= 0.134. φ = 82.3°.
0.002426 nm
(c) ∆ = 0, the photon is undeflected, cosφ = 1 and φ = 0. (b) ∆ = 0.0521 nm − 0.0500 nm. cosφ = 1 − 38.39. IDENTIFY and SET UP: The shift in wavelength of the photon is λ ′ − λ = wavelength after the scattering and h
(1 − cos φ ) where λ ′ is the
mc h
= λc = 2.426 × 10−12 m . The energy of a photon of wavelength λ is
mc Photons, Electrons, and Atoms E= 38.40. hc λ 38.42. 1.24 × 10 eV ⋅ m λ 389 . Conservation of energy applies to the collision, so the energy lost by the photon equals the energy gained by the electron.
EXECUTE: (a) λ ′ − λ = λc (1 − cos φ ) = (2.426 × 10−12 m)(1 − cos35.0°) = 4.39 × 10−13 m = 4.39 × 10−4 nm
(b) λ ′ = λ + 4.39 × 10−4 nm = 0.04250 nm + 4.39 × 10−4 nm = 0.04294 nm
hc
hc
(c) Eλ =
= 2.918 × 104 eV and Eλ ′ =
= 2.888 × 104 eV so the photon loses 300 eV of energy.
λ
λ′
(d) Energy conservation says the electron gains 300 eV of energy.
The change in wavelength of the scattered photon is given by Eq. 38.23
h
h
=
(1 − cos φ ) ! =
(1 − cosφ ).
mc
"
#
mc $
%
&
'
(6.63 × 10−34 J ⋅ s)
(1 + 1) = 2.65 × 10−14 m.
(1.67 × 10−27 kg)(3.00 × 108 m/s)(0.100)
The derivation of Eq.(38.23) is explicitly shown in Equations (38.24) through (38.27) with the final substitution of
h
p′ = h ′ and p = h yielding ′ − =
(1 − cosφ ).
mc
2.898 × 10−3 m ⋅ K
c
From Eq. (38.30), (a) m =
= 0.966 mm, and f =
= 3.10 × 1011 Hz. Note that a more precise
3.00 K
m
value of the Wien displacement law constant has been used.
(b) A factor of 100 increase in the temperature lowers #m by a factor of 100 to 9.66 µ m and raises the frequency Thus, 38.41. = −6 = by the same factor, to 3.10 × 1013 Hz.
(c) Similarly,
38.43. m
4 = 966 nm and f = 3.10 × 1014 Hz. (a) H = Ae$T ; A = π r 2l
14 14 "
#
100 W
"H#
T =$
% =$
2
4%
−3
−8
& Ae$ '
& 2π (0.20 × 10 m)(0.30 m)(0.26)(5.671 × 10 W m ⋅ K ) '
3
T = 2.06 × 10 K 38.44.
38.45. (b) mT = 2.90 × 10−3 m ⋅ K; m = 1410 nm
Much of the emitted radiation is in the infrared.
2.90 × 10−3 m ⋅ K 2.90 × 10−3 m ⋅ K
T=
=
= 7.25 × 103 K.
#m
400 × 10−9 m
IDENTIFY and SET UP: The wavelength λm where the Planck distribution peaks is given by Eq.(38.30).
2.90 × 10−3 m ⋅ K
= 1.06 × 10 −3 m = 1.06 mm.
2.728 K
EVALUATE: This wavelength is in the microwave portion of the electromagnetic spectrum. This radiation is often
referred to as the “microwave background” (Section 44.7). Note that in Eq.(38.30), T must be in kelvins.
IDENTIFY: Since the stars radiate as blackbodies, they obey the StefanBoltzmann law and Wien’s displacement
law.
SET UP: The StefanBoltzmann law says that the intensity of the radiation is I = σT 4, so the total radiated power
is P = σAT 4. Wien’s displacement law tells us that the peakintensity wavelength is λm = (constant)/T.
EXECUTE: (a) The hot and cool stars radiate the same total power, so the StefanBoltzmann law gives σAhTh4 = EXECUTE: 38.46. λm = σAcTc4 ! 4 Rh2Th4 = 4 Rc2Tc4 = 4 (3Rh)2Tc4 ! Th4 = 9T 4 ! Th = T 3 = 1.7T, rounded to two significant digits.
(b) Using Wien’s law, we take the ratio of the wavelengths, giving
λm (hot) Tc
T
1
==
=
= 0.58, rounded to two significant digits.
λm (cool) Th T 3
3
38.47. EVALUATE: Although the hot star has only 1/9 the surface area of the cool star, its absolute temperature has to be
only 1.7 times as great to radiate the same amount of energy.
(a) Let α = hc / kT . To find the maximum in the Planck distribution: dI
d"
=$
d
d& 5 ! − 5(eα − 1) = α ! − 5eα + 5 = α 2π hc 2 #
(2π hc 2 )
2π hc 2 ( −α # 2 )
− 5 α#
% = 0 = −5 5 α λ
αλ
(e − 1) '
(e − 1)
(e − 1) 2 ! Solve 5 − x = 5e x where x = α = hc
.
kT 3810 Chapter 38 Its root is 4.965, so = 4.965 ! = hc
.
(4.965) kT hc
(6.63 × 10−34 J ⋅ s)(3.00 × 108 m s)
=
= 2.90 × 10−3 m ⋅ K.
(4.965)k
(4.965)(1.38 × 10−23 J K)
IDENTIFY: Since the stars radiate as blackbodies, they obey the StefanBoltzmann law.
SET UP: The StefanBoltzmann law says that the intensity of the radiation is I = σT 4, so the total radiated power
is P = σAT 4.
EXECUTE: (a) I = σT 4 = (5.67 × 10–8 W/m 2 ⋅ K 4 )(24,000 K)4 = 1.9 × 1010 W/m2
(b) Wien’s law gives λm = (0.00290 m ⋅ K )/(24,000 K) = 1.2 × 10–7 m = 120 nm
This is not visible since the wavelength is less than 400 nm.
(c) P = AI ! 4 R2 = P/I = (1.00 × 1025 W)/(1.9 × 1010 W/m2)
which gives RSirius = 6.51 × 106 m = 6510 km.
RSirius/Rsun = (6.51 × 106 m)/(6.96 × 109 m) = 0.0093, which gives
RSirius = 0.0093 Rsun ! 1% Rsun
(d) Using the StefanBoltzmann law, we have
(b) 38.48. % T= m 2 38.49. x V1 ! I ( #) ≈
38.50. 4 2 4 4
2
4
"
# " 5800 K #
"R #"T #
Psun
σ AsunTsun
4π RsunTsun
P
Rsun
=
=
= $ sun % $ sun % ⋅ sun = $
%$
% = 39
4
2
4
PSirius σ ASiriusTSirius 4π RSiriusTSirius & RSirius ' & TSirius ' PSirius & 0.00935Rsun ' & 24,000 K '
EVALUATE: Even though the absolute surface temperature of Sirius B is about 4 times that of our sun, it radiates
only 1/39 times as much energy per second as our sun because it is so small.
2π hc 2
x2
but e x = 1 + x + + " ≈ 1 + x for
Eq. (38.32): I ( #) = 5 hc #kT
(e
− 1)
2 5 2π hc 2
2π ckT
=
= Eq. (38.31), which is Rayleigh’s distribution.
4
(hc kT ) 2.90 × 10−3 K ⋅ m
= 9.7 × 10−8 m = 97 nm
30,000 K
This peak is in the ultraviolet region, which is not visible. The star is blue because the largest part of the visible
light radiated is in the blue violet part of the visible spectrum
(a) Wien’s law: m =k.
T m = (b) P = $AT 4 (StefanBoltzmann law) W#
"
(100, 000)(3.86 × 1026 W) = $ 5.67 × 10−8 2 4 % (4π R 2 )(30,000 K)4
mK '
&
9
R = 8.2 × 10 m Rstar Rsun = 38.51. 8.2 × 109 m
= 12
6.96 × 108 m (c) The visual luminosity is proportional to the power radiated at visible wavelengths. Much of the power is
radiated nonvisible wavelengths, which does not contribute to the visible luminosity.
IDENTIFY and SET UP: Use c = f λ to relate frequency and wavelength and use E = hf to relate photon energy
and frequency.
EXECUTE: (a) One photon dissociates one AgBr molecule, so we need to find the energy required to dissociate a
single molecule. The problem states that it requires 1.00 × 105 J to dissociate one mole of AgBr, and one mole
contains Avogadro’s number (6.02 × 1023 ) of molecules, so the energy required to dissociate one AgBr is 1.00 × 105 J/mol
= 1.66 × 10−19 J/molecule.
6.02 × 1023 molecules/mol
The photon is to have this energy, so E = 1.66 × 10−19 J(1eV/1.602 × 10−19 J) = 1.04 eV.
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 1.20 × 10−6 m = 1200 nm
E
λ
1.66 × 10−19 J
c 2.998 × 108 m/s
= 2.50 × 1014 Hz
(c) c = f λ so f = =
λ 1.20 × 10−6 m
(d) E = hf = (6.626 × 10−34 J ⋅ s)(100 × 106 Hz) = 6.63 × 10−26 J
(b) E = hc so λ = E = 6.63 × 10−26 J(1 eV/1.602 × 10−19 J) = 4.14 × 10−7 eV Photons, Electrons, and Atoms 38.52. 3811 (e) EVALUATE: A photon with frequency f = 100 MHz has too little energy, by a large factor, to dissociate a
AgBr molecule. The photons in the visible light from a firefly do individually have enough energy to dissociate
AgBr. The huge number of 100 MHz photons can’t compensate for the fact that individually they have too little
energy.
h
h
(a) Assume a nonrelativistic velocity and conserve momentum ! mv = ! v =
.
m
2
1
1 "h#
h2
=
.
(b) K = mv 2 = m $
%
2
2 & m ' 2m 2
K
h2
h
=
⋅
=
. Recoil becomes an important concern for small m and small λ since this ratio
(c)
2
E 2m
hc 2mc
becomes large in those limits.
hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s )
(d) E = 10.2 eV ! =
=
= 1.22 × 10−7 m = 122 nm.
E
(10.2 eV)(1.60 × 10−19 J eV) K= (6.63 × 10−34 J ⋅ s)2
= 8.84 × 10−27 J = 5.53 × 10−8 eV.
2(1.67 × 10−27 kg)(1.22 × 10−7 m)2 K 5.53 × 10−8 eV
=
= 5.42 × 10−9. This is quite small so recoil can be neglected.
E
10.2 eV
38.53. IDENTIFY and SET UP: f= c λ . The ( f ,V0 ) values are: (8.20 × 1014 Hz, 1.48 V) , (7.41× 1014 Hz, 1.15 V) , (6.88 × 1014 Hz, 0.93 V) , (6.10 1014 Hz, 0.62 V) , (5.49 × 1014 Hz, 0.36 V) , (5.18 × 1014 Hz, 0.24 V) . The graph of V0 versus f is given in Figure 38.53.
EXECUTE: (a) The threshold frequency, f th , is f where V0 = 0 . From the graph this is f th = 4.56 × 1014 Hz .
c 3.00 × 108 m/s
=
= 658 nm
f th 4.56 × 1014 Hz
(c) φ = hf th = (4.136 × 10 −15 eV ⋅ s)(4.56 × 1014 Hz) = 1.89 eV
(b) λth = h
"h#
(d) eV0 = hf − φ so V0 = $ % f − φ . The slope of the graph is .
e
&e'
h"
1.48 V − 0.24 V
#
−15
=$
% = 4.11× 10 V/Hz and
e & 8.20 × 1014 Hz − 5.18 × 1014 Hz '
h = (4.11 × 10−15 V/Hz)(1.60 × 10−19 C) = 6.58 × 10−34 J ⋅ s . Figure 38.53
38.54. dN
( dE dt )
P
(200 W)(0.10)
=
=
=
= 6.03 × 1019 photons sec.
dt ( dE dN ) hf h(5.00 × 1014 Hz)
(dN dt )
(b) Demand
= 1.00 × 1011 photons sec ⋅ cm 2 .
4π r 2
(a) 1/ 2 38.55. "
#
6.03 × 1019 photons sec
Therefore, r = $
% = 6930 cm = 69.3 m.
11
2
& 4π (1.00 × 10 photons sec ⋅ cm ) '
(a) IDENTIFY: Apply the photoelectric effect equation, Eq.(38.4).
SET UP: eV0 = hf − φ = ( hc / λ ) − φ . Call the stopping potential V01 for λ1 and V02 for λ2 . Thus
eV01 = (hc / λ1 ) − φ and eV02 = (hc / λ2 ) − φ . Note that the work function φ is a property of the material and is
independent of the wavelength of the light.
"λ −λ #
EXECUTE: Subtracting one equation from the other gives e(V02 − V01 ) = hc $ 1 2 % .
& λ 1λ 2 ' 3812 Chapter 38 (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) " 295 × 10−9 m − 265 × 10−9 m #
$
% = 0.476 V.
−9
−9
1.602 × 10−19 C
& (295 × 10 m)(265 × 10 m) '
EVALUATE: e∆V0 , which is 0.476 eV, is the increase in photon energy from 295 nm to 265 nm. The stopping
potential increases when λ deceases because the photon energy increases when the wavelength decreases.
IDENTIFY: The photoelectric effect occurs, so the energy of the photon is used to eject an electron, with any
excess energy going into kinetic energy of the electron.
SET UP: Conservation of energy gives hf = hc/λ = Kmax + φ.
EXECUTE: (a) Using hc/λ = Kmax + φ, we solve for the work function:
(b) ∆V0 = 38.56. φ = hc/λ – Kmax = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(124 nm) – 4.16 eV = 5.85 eV 38.57. (b) The number N of photoelectrons per second is equal to the number of photons per second that strike the metal
per second. N × (energy of a photon) = 2.50 W. N(hc/λ) = 2.50 W.
N = (2.50 W)(124 nm)/[(6.626 × 10–34 J ⋅ s )(3.00 × 108 m/s)] = 1.56 × 1018 electrons/s
(c) N is proportional to the power, so if the power is cut in half, so is N, which gives
N = (1.56 × 1018 el/s)/2 = 7.80 × 1017 el/s
(d) If we cut the wavelength by half, the energy of each photon is doubled since E = hc/λ. To maintain the same
power, the number of photons must be half of what they were in part (b), so N is cut in half to 7.80 × 1017 el/s. We
could also see this from part (b), where N is proportional to λ. So if the wavelength is cut in half, so is N.
EVALUATE: In part (c), reducing the power does not reduce the maximum kinetic energy of the photons; it only
reduces the number of ejected electrons. In part (d), reducing the wavelength does change the maximum kinetic
energy of the photoelectrons because we have increased the energy of each photon.
IDENTIFY and SET UP: The energy added to mass m of the blood to heat it to Tf = 100 °C and to vaporize it is Q = mc (Tf − Ti ) + mLv , with c = 4190 J/kg ⋅ K and Lv = 2.256 × 106 J/kg . The energy of one photon is E= 38.58. 38.59. hc λ = 1.99 × 10−25 J ⋅ m λ . EXECUTE: (a) Q = (2.0 × 10−9 kg)(4190 J/kg ⋅ K)(100°C − 33°C) + (2.0 × 10−9 kg)(2.256 × 106 J/kg) = 5.07 × 10−3 J
The pulse must deliver 5.07 mJ of energy.
energy 5.07 × 10−3 J
(b) P =
=
= 11.3 W
t
450 × 10−6 s
hc 1.99 × 10−25 J ⋅ m
(c) One photon has energy E =
=
= 3.40 × 10−19 J . The number N of photons per pulse is the
585 × 10−9 m
λ
5.07 × 10−3 J
energy per pulse divided by the energy of one photon: N =
= 1.49 × 1016 photons
3.40 × 10−19 J/photon
hc
(a) 0 = , and the wavelengths are: cesium: 590 nm, copper: 264 nm, potassium: 539 nm, zinc: 288 nm.
E
b) The wavelengths of copper and zinc are in the ultraviolet, and visible light is not energetic enough to overcome
the threshold energy of these metals.
207 me mp
mm
(a) IDENTIFY and SET UP: Apply Eq.(38.20): mr = 1 2 =
m1 + m2 207 me + mp
EXECUTE: mr = 207(9.109 × 10−31 kg)(1.673 × 10−27 kg)
= 1.69 × 10−28 kg
207(9.109 × 10−31 kg) + 1.673 × 10−27 kg We have used me to denote the electron mass.
(b) IDENTIFY: In Eq.(38.18) replace m = me by mr : En = − 1 mr e 4
.
P02 8n 2 h 2 " m # " 1 m e4 #
1 m e4
SET UP: Write as En = $ r % $ − 2 H 2 % , since we know that 2 H 2 = 13.60 eV. Here mH denotes the
2
P0 8h
& mH '& P0 8n h '
−31
reduced mass for the hydrogen atom; mH = 0.99946(9.109 × 10 kg) = 9.104 × 10−31 kg.
" m # " 13.60 eV #
En = $ r % $ −
%
n2
'
& mH ' &
1.69 × 10−28 kg
E1 =
( −13.60 eV) = 186( −13.60 eV) = −2.53 keV
9.104 × 10−31 kg EXECUTE: Photons, Electrons, and Atoms 3813 " m # " R ch #
From part (b), En = $ r % $ − H 2 % , where RH = 1.097 × 107 m −1 is the Rydberg constant for the
& mH ' & n '
hc
hydrogen atom. Use this result in
= Ei − E f to find an expression for 1 / λ . The initial level for the transition is
(c) SET UP: λ the ni = 2 level and the final level is the n f = 1 level.
EXECUTE: hc λ = mr " RHch " RHch # #
− $ − 2 %%
$−
%
mH $ ni2
& nf ' '
& "1 1#
mr
RH $ 2 − 2 %
λ mH & nf ni '
1 1.69 × 10−28 kg
"1 1#
(1.097 × 107 m −1 ) $ 2 − 2 % = 1.527 × 109 m −1
=
λ 9.104 × 10−31 kg
&1 2 '
λ = 0.655 nm
EVALUATE: From Example 38.6 the wavelength of the radiation emitted in this transition in hydrogen is 122 nm.
m
The wavelength for muonium is H = 5.39 × 10 −3 times this. The reduced mass for hydrogen is very close to the
mr
1 = electron mass because the electron mass is much less then the proton mass: mp / me = 1836. The muon mass is 38.60. 207 me = 1.886 × 10−28 kg. The proton is only about 10 times more massive than the muon, so the reduced mass is
somewhat smaller than the muon mass. The muonproton atom has much more strongly bound energy levels and
much shorter wavelengths in its spectrum than for hydrogen.
(a) The change in wavelength of the scattered photon is given by Eq. 38.23 h
h
(1 − cos φ ) ! = ′ −
(1 − cosφ ) =
mc
mc
(6.63 × 10−34 J ⋅ s)
(0.0830 × 10−9 m) −
(1 + 1) = 0.0781 nm.
(9.11× 10−31 kg)(3.00 × 108 m s)
′− = (b) Since the collision is onedimensional, the magnitude of the electron’s momentum must be equal to the
magnitude of the change in the photon’s momentum. Thus, 1 # 9 −1
" 1 −1 #
"1
pe = h $ − % = (6.63 × 10−34 J ⋅ s) $
+
% (10 m )
′'
0.0781 0.0830 '
&
&
= 1.65 × 10−23 kg ⋅ m s ≈ 2 × 10−23 kg ⋅ m s.
(c) Since the electron is non relativistic ( β = 0.06), K e =
38.61. IDENTIFY and SET UP: λ′ = λ + pe2
= 1.49 × 10−16 J ≈ 10−16 J.
2m h
(1 − cosφ )
mc 2h
= 0.09485 m. Use Eq.(38.5) to calculate the momentum of the scattered photon. Apply
mc
conservation of energy to the collision to calculate the kinetic energy of the electron after the scattering. The
energy of the photon is given by Eq.(38.2),
EXECUTE: (a) p′ = h / λ ′ = 6.99 × 10−24 kg ⋅ m/s. φ = 180° so λ ′ = λ + (b) E = E′ + Ee ; hc / λ = hc / λ ′ + Ee 38.62. λ′ − λ
"1 1 #
Ee = hc $ − % = (hc)
= 1.129 × 10−16 J = 705 eV
λλ′
& λ λ′ '
EVALUATE: The energy of the incident photon is 13.8 keV, so only about 5% of its energy is transferred to the
electron. This corresponds to a fractional shift in the photon’s wavelength that is also 5%.
2h
(a) φ = 180° so (1 − cosφ ) = 2 ! ∆ =
= 0.0049 nm, so ′ = 0.1849 nm.
mc
"1 1 #
(b) ∆E = hc $ − % = 2.93 × 10−17 J = 183 eV. This will be the kinetic energy of the electron.
′'
&
(c) The kinetic energy is far less than the rest mass energy, so a nonrelativistic calculation is adequate;
v = 2 K m = 8.02 × 106 m s. 3814 Chapter 38 38.63. IDENTIFY and SET UP: The Hα line in the Balmer series corresponds to the n = 3 to n = 2 transition. En = − 13.6 eV hc
.
= ∆E .
λ
n2 "1 1#
EXECUTE: (a) The atom must be given an amount of energy E3 − E1 = −(13.6 eV) $ 2 − 2 % = 12.1 eV .
&3 1 '
hc
(b) There are three possible transitions. n = 3 → n = 1 : ∆E = 12.1 eV and λ =
= 103 nm ;
∆E
"1 1#
n = 3 → n = 2 : ∆E = −(13.6 eV) $ 2 − 2 % = 1.89 eV and λ = 657 nm ; n = 2 → n = 1 :
&3 2 ' 38.64. " 1 1#
∆E = −(13.6 eV) $ 2 − 2 % = 10.2 eV and λ = 122 nm .
&2 1 '
−( Eex − Eg )
n2
− ( E − E ) kT
!T =
.
= e ex g
n1
kln ( n2 / n1 ) Eex = E2 = −13.6 eV
= −3.4 eV. Eg = −13.6 eV. Eex − Eg = 10.2 eV = 1.63 × 10−18 J.
4 (a) −(1.63 × 10−18 J)
n2
= 10 −12. T =
= 4275 K.
(1.38 × 10−23 J K ) ln(10−12 )
n1 (b) −(1.63 × 10−18 J)
n2
= 10 −8. T =
= 6412 K.
(1.38 × 10−23 J K ) ln(10−8 )
n1 −(1.63 × 10−18 J)
n2
= 10 −4. T =
= 12824 K.
(1.38 × 10−23 J K ) ln(10−4 )
n1
(d) For absorption to take place in the Balmer series, hydrogen must start in the n = 2 state. From part (a), colder
stars have fewer atoms in this state leading to weaker absorption lines.
(a) IDENTIFY and SET UP: The photon energy is given to the electron in the atom. Some of this energy
overcomes the binding energy of the atom and what is left appears as kinetic energy of the free electron. Apply
hf = Ef − Ei , the energy given to the electron in the atom when a photon is absorbed.
(c) 38.65. EXECUTE: The energy of one photon is hc λ = 2.323 × 10 −18 J(1 eV/1.602 × 10 −19 hc λ = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
85.5 × 10−9 m J) = 14.50 eV. The final energy of the electron is Ef = Ei + hf . In the ground state of the hydrogen atom the energy of the electron 38.66. is Ei = −13.60 eV. Thus Ef = −13.60 eV + 14.50 eV = 0.90 eV.
(b) EVALUATE: At thermal equilibrium a few atoms will be in the n = 2 excited levels, which have an energy of
−13.6 eV/4 = −3.40 eV, 10.2 eV greater than the energy of the ground state. If an electron with E = −3.40 eV
gains 14.5 eV from the absorbed photon, it will end up with 14.5 eV − 3.4 eV = 11.1 eV of kinetic energy.
IDENTIFY: The diffraction grating allows us to determine the peakintensity wavelength of the light. Then
Wien’s displacement law allows us to calculate the temperature of the blackbody, and the StefanBoltzmann law
allows us to calculate the rate at which it radiates energy.
SET UP: The bright spots for a diffraction grating occur when d sin θ = mλ. Wien’s displacement law is
2.90 × 10 −3 m ⋅ K
λpeak =
, and the StefanBoltzmann law says that the intensity of the radiation is I = σT 4, so the
T
total radiated power is P = σAT 4.
EXECUTE: (a) First find the wavelength of the light:
λ = d sin θ = [1/(385,000 lines/m)] sin(11.6°) = 5.22 × 10–7 m
Now use Wien’s law to find the temperature: T = (2.90 × 10–3 m ⋅ K )/(5.22 × 10–7 m) = 5550 K.
(b) The energy radiated by the blackbody is equal to the power times the time, giving
U = Pt = IAt = σAT 4t, which gives
t = U/(σAT 4) = (12.0 × 106 J)/[(5.67 × 10–8 W/m 2 ⋅ K 4 )(4 )(0.0750 m)2(5550 K)4] = 3.16 s.
EVALUATE: By ordinary standards, this blackbody is very hot, so it does not take long to radiate 12.0 MJ of
energy. Photons, Electrons, and Atoms 38.67. 3815 IDENTIFY: Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the StefanBoltzmann law apply to its radiation.
2.90 × 10−3 m ⋅ K
, and the StefanBoltzmann law says that the
SET UP: Wien’s displacement law is λpeak =
T
4
intensity of the radiation is I = σT , so the total radiated power is P = σAT 4.
EXECUTE: (a) First use Wien’s law to find the peak wavelength:
λm = (2.90 × 10–3 m ⋅ K )/(3000 K) = 9.667 × 10–7 m
Call N the number osf photons/second radiated. N × (energy per photon) = IA = σAT 4.
λ σ AT 4
N (hc/λm) = σAT 4. N = m
.
hc
(9.667 × 10−7 m)(5.67 × 10−8 W/m2 ⋅ K 4 )(4π )(600 × 6.96 × 108 m)2 (3000 K)4
.
N=
(6.626 × 10−34 J ⋅ s)(3.00 × 108 m/s)
N = 5 × 1049 photons/s.
2 4 2
I B AB σ ABTB4 4π RBTB4 " 600 RS # " 3000 K #
4
=
=
=$
%$
% = 3 × 10
2
I S AS σ ASTS4 4π RS TS4 & RS ' & 5800 K '
EVALUATE: Betelgeuse radiates 30,000 times as much energy per second as does our sun!
IDENTIFY: The blackbody radiates heat into the water, but the water also radiates heat back into the blackbody.
The net heat entering the water causes evaporation. Wien’s law tells us the peak wavelength radiated, but a
thermophile in the water measures the wavelength and frequency of the light in the water.
dQ
4
4
SET UP: By the StefanBoltzman law, the net power radiated by the blackbody is
= σ A (Tsphere − Twater ) . Since
dt
dQ
dm
= Lv
. Wien’s displacement law is
this heat evaporates water, the rate at which water evaporates is
dt
dt
2.90 × 10−3 m ⋅ K
λm =
, and the wavelength in the water is λw = λ0/n.
T
dQ
dQ
dm
4
4
EXECUTE: (a) The net radiated heat is
= σ A (Tsphere − Twater ) and the evaporation rate is
= Lv
, where
dt
dt
dt
dm
4
4
= σ A (Tsphere − Twater ) .
dm is the mass of water that evaporates in time dt. Equating these two rates gives Lv
dt (b) 38.68. dm σ ( 4π R
=
dt 2 )(T 4
sphere Lv 4
− Twater ) . −8
2
4
2
4
4
dm ( 5.67 × 10 W/m ⋅ K ) ( 4π ) (0.120 m) / (498 K) − (373 K) 0
3
4 = 1.92 × 10−4 kg/s = 0.193 g/s
=
dt
2256 × 103 J/Kg
(b) (i) Wien’s law gives λm = (0.00290 m ⋅ K )/(498 K) = 5.82 × 10–6 m
But this would be the wavelength in vacuum. In the water the thermophile organism would measure λw = λ0/n =
(5.82 × 10–6 m)/1.333 = 4.37 × 10–6 m = 4.37 µm
(ii) The frequency is the same as if the wave were in air, so
f = c/λ0 = (3.00 ×108 m/s)/(5.82 × 10–6 m) = 5.15 × 1013 Hz EVALUATE: An alternative way is to use the quantities in the water: f = 38.69. c/n λ0 / n = c/λ0, which gives the same answer for the frequency. An organism in the water would measure the light coming to it through the water, so the
wavelength it would measure would be reduced by a factor of 1/n.
IDENTIFY: The energy of the peakintensity photons must be equal to the energy difference between the n = 1
and the n = 4 states. Wien’s law allows us to calculate what the temperature of the blackbody must be for it to
radiate with its peak intensity at this wavelength.
13.6 eV
SET UP: In the Bohr model, the energy of an electron in shell n is En = −
, and Wien’s displacement law
n2
2.90 × 10−3 m ⋅ K
is λm =
. The energy of a photon is E = hf = hc/λ.
T
EXECUTE: First find the energy (∆E) that a photon would need to excite the atom. The ground state of the atom
is n = 1 and the third excited state is n = 4. This energy is the difference between the two energy levels. Therefore 3816 Chapter 38 " 1 1#
∆E = ( −13.6 eV ) $ 2 − 2 % = 12.8 eV. Now find the wavelength of the photon having this amount of energy.
&4 1 '
hc/λ = 12.8 eV and λ = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(12.8 eV) = 9.73 ×10–8 m 38.70. 38.71. Now use Wien’s law to find the temperature. T = (0.00290 m ⋅ K )/(9.73 × 10–8 m) = 2.98 × 104 K.
EVALUATE: This temperature is well above ordinary room temperatures, which is why hydrogen atoms are not in
excited states during everyday conditions.
IDENTIFY and SET UP: Electrical power is VI. Q = mc∆T .
EXECUTE: (a) (0.010)VI = (0.010)(18.0 × 103 V)(60.0 × 10−3 A) = 10.8 W = 10.8 J/s
(b) The energy in the electron beam that isn’t converted to x rays stays in the target and appears as thermal energy.
Q
1.07 × 103 J
For t = 1.00 s , Q = (0.990)VI (1.00 s) = 1.07 × 103 J and ∆T =
=
= 32.9 K . The
mc (0.250 kg)(130 J/kg ⋅ K)
temperature rises at a rate of 32.9 K/s.
EVALUATE: The target must be made of a material that has a high melting point.
IDENTIFY: Apply conservation of energy and conservation of linear momentum to the system of atom plus
photon.
(a) SET UP: Let Etr be the transition energy, Eph be the energy of the photon with wavelength λ ′, and Er be the kinetic energy of the recoiling atom. Conservation of energy gives Eph + Er = Etr .
Eph = hc
hc
hc
so
= Etr − Er and λ =
.
λ′
λ′
Etr − Er EXECUTE: If the recoil energy is neglected then the photon wavelength is λ = hc / Etr . "1
#
1 # " hc #"
1
∆λ = λ ′ − λ = hc $
−
− 1%
%=$
%$
Etr − Er Etr ' & Etr '& 1 − Er / Etr
&
'
−1 "
1
E#
E
E
= $ 1 − r % ≈ 1 + r since r V 1
1 − Er / Etr & E tr '
Etr
Etr
(We have used the binomial theorem, Appendix B.)
hc " Er #
" Er # 2
Thus ∆λ =
$
% , or since Etr = hc / λ , ∆λ = $ % λ .
Etr & Etr '
& hc '
SET UP: Use conservation of linear momentum to find Er : Assuming that the atom is initially at rest, the momentum pr of the recoiling atom must be equal in magnitude and opposite in direction to the momentum
pph = h / λ of the emitted photon: h / λ = pr . pr2
h2
, where m is the mass of the atom, so Er =
.
2m
2mλ 2
" h 2 #" λ 2 #
h
"E #
Use this result in the above equation: ∆λ = $ r % λ 2 = $
;
%$ % =
2mλ 2 '& hc ' 2mc
& hc '
&
EXECUTE: Er = note that this result for ∆λ is independent of the atomic transition energy.
6.626 × 10−34 J ⋅ s
h
(b) For a hydrogen atom m = mp and ∆λ =
=
= 6.61 × 10−16 m
2mpc 2(1.673 × 10−27 kg)(2.998 × 108 m/s) 38.72. EVALUATE: The correction is independent of n. The wavelengths of photons emitted in hydrogen atom
transitions are on the order of 100 nm = 10−7 m, so the recoil correction is exceedingly small.
(a) ∆ 1 = (h mc )(1 − cos &1 ), ∆ 2 = (h mc)(1 − cos &2 ), and so the overall wavelength shift is ∆ = (h mc )(2 − cos &1 − cos &2 ).
(b) For a single scattering through angle & , ∆ = ( h mc)(1 − cos & ). For two successive scatterings through an s angle of θ 2 for each scattering, ∆ t = 2(h mc )(1 − cosθ 2). 1 − cos & = 2(1 − cos 2 (& 2)) and ∆ s = ( h mc)2(1 − cos 2 (& 2)) cos(θ 2) ≤ 1 so 1 − cos 2 (θ 2) ≥ (1 − cos(θ 2)) and ∆ s ≥∆ t Photons, Electrons, and Atoms 3817 Equality holds only when θ = 180°.
(c) ( h mc)2(1 − cos30.0°) = 0.268(h mc).
38.73. (d) ( h mc)(1 − cos 60°) = 0.500(h mc), which is indeed greater than the shift found in part (c).
IDENTIFY and SET UP: Find the average change in wavelength for one scattering and use that in ∆λ in
Eq.(38.23) to calculate the average scattering angle φ .
EXECUTE: (a) The wavelength of a 1 MeV photon is
hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s)
= 1 × 10−12 m
λ= =
E
1 × 106 eV
The total change in wavelength therefore is 500 × 10 −9 m − 1 × 10−12 m = 500 × 10−9 m.
If this shift is produced in 10 26 Compton scattering events, the wavelength shift in each scattering event is
500 × 10−9 m
∆λ =
= 5 × 10−33 m.
1 × 1026
h
(b) Use this ∆λ in ∆λ =
(1 − cos φ ) and solve for φ . We anticipate that φ will be very small, since ∆λ is
mc
much less than h / mc, so we can use cosφ ≈ 1 − φ 2 / 2.
h
h2
∆λ =
(1 − (1 − φ 2 / 2)) =
φ
mc
2mc φ= 2∆λ
2(5 × 10−33 m)
=
= 6.4 × 10−11 rad = (4 × 10−9 )°
( h / mc )
2.426 × 10−12 m φ in radians is much less than 1 so the approximation we used is valid.
(c) IDENTIFY and SET UP: We know the total transit time and the total number of scatterings, so we can calculate
the average time between scatterings.
EXECUTE: The total time to travel from the core to the surface is (106 y)(3.156 × 107 s/y) = 3.2 × 1013 s. There are
3.2 × 1013 s
= 3.2 × 10−13 s.
1026
The distance light travels in this time is d = ct = (3.0 × 108 m/s)(3.2 × 10−13 s) = 0.1 mm
EVALUATE: The photons are on the average scattered through a very small angle in each scattering event. The
average distance a photon travels between scatterings is very small.
hc
(a) The final energy of the photon is E ′ = , and E = E ′ + K , where K is the kinetic energy of the electron after
′
the collision. Then,
′
hc
hc
hc
=
=
=
=
.
′mc /
0
E ′ + K ( hc ′) + K ( hc ′) + (γ − 1) mc 2
1
1+
−1
h 1 (1 − v 2 c 2 )1 2 2
3
4
10 26 scatterings during this time, so the average time between scatterings is t = 38.74. ( K = mc 2 (γ − 1) since the relativistic expression must be used for threefigure accuracy). (b) φ = arccos(1 − ∆ ( h mc)).
(c) γ − 1 = (1 − ( 1 ) 12
1.80 2
3.00 ) − 1 = 1.25 − 1 = 0.250, !=
1+ (5.10 × 10 −12 5.10 × 10−3 mm
= 3.34 × 10−3 nm .
m)(9.11 × 10−31 kg)(3.00 × 108 m s)(0.250)
(6.63 × 10−34 J ⋅ s) " φ = arccos $1 −
& 38.75. h
= 2.43 × 10−12 m
mc (5.10 × 10−12 m − 3.34 × 10−12 m) #
% = 74.0°.
2.43 × 10−12 m
' (a) IDENTIFY and SET UP: Conservation of energy applied to the collision gives Eλ = Eλ ′ + Ee , where Ee is the kinetic energy of the electron after the collision and Eλ and Eλ ′ are the energies of the photon before and after the
collision. The energy of a photon is related to its wavelength according to Eq.(38.2). 3818 Chapter 38 "1 1 #
" λ′ − λ #
Ee = hc $ − % = hc $
%
& λ λ′ '
& λλ ′ ' EXECUTE: "
#
0.0032 × 10−9 m
Ee = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s) $
%
−9
−9
& (0.1100 × 10 m)(0.1132 × 10 m) '
Ee = 5.105 × 10−17 J = 319 eV 1
2 Ee
2(5.105 × 10−17 J)
Ee = mv 2 so v =
=
= 1.06 × 107 m/s
2
m
9.109 × 10−31 kg
(b) The wavelength λ of a photon with energy Ee is given by Ee = hc / λ so hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 3.89 nm
Ee
5.105 × 10−17 J
EVALUATE: Only a small portion of the incident photon’s energy is transferred to the struck electron; this is why
the wavelength calculated in part (b) is much larger than the wavelength of the incident photon in the Compton
scattering.
h
IDENTIFY: Apply the Compton scattering formula λ ′ − λ = ∆λ =
(1 − cos φ ) = λc (1 − cos φ )
mc
(a) SET UP: Largest ∆λ is for φ = 180°.
EXECUTE: For φ = 180°, ∆λ = 2λc = 2(2.426 pm) = 4.85 pm. λ= 38.76. λ ′ − λ = λc (1 − cos φ )
Wavelength doubles implies λ ′ = 2λ so λ ′ − λ = λ . Thus λ = λC (1 − cosφ ). λ is related to E by Eq.(38.2).
EXECUTE: E = hc / λ , so smallest energy photon means largest wavelength photon, so φ = 180° and
(b) SET UP: λ = 2λc = 4.85 pm. Then E = 38.77. hc λ = (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
= 4.096 × 10−14 J(1 eV/1.602 × 10−19 J) =
4.85 × 10−12 m 0.256 MeV.
EVALUATE: Any photon Compton scattered at φ = 180° has a wavelength increase of 2λc = 4.85 pm. 4.85 pm is
near the shortwavelength end of the range of xray wavelengths.
2π hc 2
c
(a) I ( ) = 5 hc kT
but =
f
− 1)
(e
! I( f ) =
(b) =5 5
∞
0 ∞
0 2π hc 2
2π hf 5
= 3 hf kT
5
hf kT
− 1) c (e
− 1)
(c f ) (e 0
" −c #
I ( ) d λ = 5 I ( f ) df $ 2 %
∞
&f ' 2π hf 3 df
2π (kT ) 4 ∞ x3
2π ( kT ) 4 1
(2π )5 (kT ) 4 2π 5k 4T 4
4
=
23
hf kT
5 0 e x − 1 dx = c 2h3 240 (2π ) = 240h3c 2 = 15c 2h3
c (e
ch
− 1)
2 2π 5k 4T 4
= σ as shown in Eq. (38.36). Plugging in the values for the constants we get
15h3c 2
$ = 5.67 × 10−8 W m 2 ⋅ K 4 . (c) The expression 38.78. I = $T 4 , P = IA, and ∆E = Pt ; combining, t=
38.79. ∆E
(100 J)
=
= 8.81 × 103 s = 2.45 hrs.
Aσ T 4 (4.00 × 10−6 m 2 )(5.67 × 10−8 W m 2 ⋅ K 4 )(473 K)4 (a) The period was found in Exercise 38.27b: T = 4P02n 3h3
1
me 4
and frequency is just f = = 2 3 3 .
4
me
T 4P0 n h 1
me4 " 1 1 #
(b) Eq. (38.6) tells us that f = ( E2 − E1 ). So f = 2 3 $ 2 − 2 % (from Eq. (38.18)).
h
8P0 h & n2 n1 '
If n2 = n and n1 = n + 1, then
= 11
1
1
−=−
2
n2 n12 n 2 ( n + 1) 2 # 1" " 2
me 4
1"
1
## 2
1−
≈ 2 $ 1 − $1 − + " % % = 3 for large n ! f ≈ 2 3 3 .
2$
2%
n & (1 + 1 n) ' n & & n
4P0 n h
'' n Photons, Electrons, and Atoms 38.80. 3819 h
Each photon has momentum p = , and if the rate at which the photons strike the surface is ( dN dt ) , the force
on the surface is ( h )( dN dt ), and the pressure is ( h )( dN dt ) A. The intensity is 38.81. I = (dN dt )( E ) A = ( dN dt )(hc ) A , and comparison of the two expressions gives the pressure as ( I c).
####
Momentum: p + P = p′ + P ′ ! p − P = − p′ − P′ ! p′ = P − ( p + P′)
energy: pc + E = p′c + E′ = p′c + ( P′c ) 2 + (mc 2 ) 2
! ( pc − p′c + E ) 2 = ( P′c) 2 + ( mc 2 ) 2 = (Pc) 2 + ((p + p′)c) 2 − 2 P (p + p′)c 2 + (mc 2 ) 2
( pc − p′c) 2 + E 2 = E 2 + ( pc + p′c) 2 − 2( Pc 2 )( p + p′) + 2 Ec( p − p′) − 4 pp′c 2 + 2 Ec( p − p′)
+2( Pc 2 )( p + p′) = 0 ! p′( Pc 2 − 2 pc 2 − Ec) = p ( − Ec − Pc 2 )
! p′ = p Ec + Pc 2
E + Pc
=p
2 pc 2 + Ec − Pc 2
2 pc + ( E − Pc ) 2hc
" 2 hc + ( E − Pc) #
" E − Pc #
! ′= $
%= $
%+
E + Pc
&
'
& E + Pc ' E + Pc
( ( E − Pc) + 2hc)
! ′=
E + Pc
" mc 2 #
If E W mc 2 , Pc = E 2 − ( mc 2 ) 2 = E 1 − $
%
&E' " 1 " mc 2 # 2
#
%
≈ E $1 − $
% + "%
$ 2& E '
&
' #( mc 2 ) 2 hc hc " m 2c 4 #
+
= $1 +
%
2 E (2 E ) E
E&
4hcE '
= 10.6 × 10−6 m, E = 1.00 × 1010 eV = 1.60 × 10−9 J ! E − Pc ≈
(b) If 2 1 ( mc 2 ) 2
!
2E 1 ≈ ! ′≈ hc
1.60 × 10−9 " (9.11 × 10−31 kg) 2c 4 (10.6 × 10−6 m) #
$1 +
%
J&
4hc (1.6 × 10−9 J)
' = (1.24 × 10−16 m)(1 + 56.0) = 7.08 × 10−15 m. (c) These photons are gamma rays. We have taken infrared radiation and converted it into gamma rays! Perhaps
useful in nuclear medicine, nuclear spectroscopy, or high energy physics: wherever controlled gamma ray sources
might be useful. 39 THE WAVE NATURE OF PARTICLES 39.1. IDENTIFY and SET UP:
EXECUTE: (a) λ = h
h
=
. For an electron, m = 9.11 × 10 −31 kg . For a proton, m = 1.67 × 10 −27 kg .
p mv 6.63 × 10−34 J ⋅ s
= 1.55 × 10−10 m = 0.155 nm
(9.11 × 10−31 kg)(4.70 × 106 m/s) (b) λ is proportional to
39.2. λ= IDENTIFY and SET UP: "m #
" 9.11 × 10−31 kg #
1
−14
, so λp = λe $ e % = (1.55 × 10−10 m) $
% = 8.46 × 10 m .
−27
$ mp %
m
& 1.67 × 10 kg '
&'
For a photon, E = hc λ . For an electron or proton, p = h λ and E = p2
h2
, so E =
.
2m
2mλ 2 (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
EXECUTE: (a) E =
=
= 6.2 keV
0.20 × 10−9 m
λ
hc 2 (b) E = 39.3. "m #
" 9.11× 10−31 kg #
(c) Ep = Ee $ e % = (38 eV) $
% = 0.021 eV
−27
$ mp %
& 1.67 × 10 kg '
&'
EVALUATE: For a given wavelength a photon has much more energy than an electron, which in turn has more
energy than a proton.
h
h (6.63 × 10−34 J ⋅ s)
= 2.37 × 10−24 kg ⋅ m s.
(a) = ! p = =
p
(2.80 × 10−10 m)
(b) K = = 39.4.
= 39.5. " 6.63 × 10−34 J ⋅ s #
h2
1
=$
= 6.03 × 10−18 J = 38 eV
%
2mλ 2 & 0.20 × 10−9 m ' 2(9.11 × 10−31 kg) p 2 (2.37 × 10−24 kg ⋅ m s) 2
=
= 3.08 × 10−18 J = 19.3 eV.
2m
2(9.11 × 10−31 kg) h
h
=
p
2mE
(6.63 × 10−34 J ⋅ s)
2(6.64 × 10−27 kg) (4.20 × 106 eV) (1.60 × 10−19 J e V) = 7.02 × 10−15 m. h
h
=
. In the Bohr model, mvrn = n( h / 2π ),
p mv
so mv = nh /(2π rn ). Combine these two expressions and obtain an equation for λ in terms of n. Then IDENTIFY and SET UP: The de Broglie wavelength is λ = " 2π rn # 2π rn
.
%=
n
& nh '
EXECUTE: (a) For n = 1, λ = 2π r1 with r1 = a0 = 0.529 × 10−10 m, so λ = 2π (0.529 × 10−10 m) = 3.32 × 10−10 m λ = h$ λ = 2π r1; the de Broglie wavelength equals the circumference of the orbit.
(b) For n = 4, λ = 2π r4 / 4.
rn = n 2a0 so r4 = 16a0 . λ = 2π (16a0 ) / 4 = 4(2π a0 ) = 4(3.32 × 10−10 m) = 1.33 × 10−9 m
11
= times the circumference of the orbit.
n4
EVALUATE: As n increases the momentum of the electron increases and its de Broglie wavelength decreases. For
any n, the circumference of the orbits equals an integer number of de Broglie wavelengths. λ = 2π r4 / 4; the de Broglie wavelength is 392 39.6. 39.7. 39.8. Chapter 39 (a) For a nonrelativistic particle, K = (b) (6.63 × 10−34 J ⋅ s) 2(800 eV)(1.60 × 1019 J/eV)(9.11 × 1031 kg) = 4.34 × 10−11 m.
IDENTIFY: A person walking through a door is like a particle going through a slit and hence should exhibit wave
properties.
SET UP: The de Broglie wavelength of the person is λ = h/mv.
EXECUTE: (a) Assume m = 75 kg and v = 1.0 m/s.
λ = h/mv = (6.626 × 10–34 J ⋅ s)/[(75 kg)(1.0 m/s)] = 8.8 × 10–36 m
EVALUATE: (b) A typical doorway is about 1 m wide, so the person’s de Broglie wavelength is much too small
to show wave behavior through a “slit” that is about 1035 times as wide as the wavelength. Hence ordinary objects
do not show wave behavior in everyday life. Combining Equations 37.38 and 37.39 gives p = mc ' 2 − 1.
(a) λ = 39.9. p2
h
h
, so λ = =
.
2m
p
2 Km h
= (h mc)
p γ 2 − 1 = 4.43 × 10−12 m. (The incorrect nonrelativistic calculation gives 5.05 × 10−12 m.) (b) ( h mc) γ 2 − 1 = 7.07 × 10−13 m.
IDENTIFY and SET UP: A photon has zero mass and its energy and wavelength are related by Eq.(38.2). An
electron has mass. Its energy is related to its momentum by E = p 2 / 2m and its wavelength is related to its
momentum by Eq.(39.1).
EXECUTE: (a) photon
hc
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
E=
so λ =
=
= 62.0 nm
E
(20.0 eV)(1.602 × 10−19 J/eV)
λ
electron E = p 2 /(2m) so p = 2mE =
λ = h / p = 0.274 nm 2(9.109 × 10−31 kg)(20.0 eV)(1.602 × 10−19 J/eV) = 2.416 × 10−24 kg ⋅ m/s (b) photon E = hc / λ = 7.946 × 10−19 J = 4.96 eV
electron λ = h / p so p = h / λ = 2.650 × 10−27 kg ⋅ m/s
E = p 2 /(2m) = 3.856 × 10 −24 J = 2.41 × 10 −5 eV
(c) EVALUATE: You should use a probe of wavelength approximately 250 nm. An electron with λ = 250 nm has
much less energy than a photon with λ = 250 nm, so is less likely to damage the molecule. Note that λ = h / p
applies to all particles, those with mass and those with zero mass. E = hf = hc / λ applies only to photons and 39.10. E = p 2 / 2m applies only to particles with mass.
IDENTIFY: Any moving particle has a de Broglie wavelength. The speed of a molecule, and hence its de Broglie
wavelength, depends on the temperature of the gas.
SET UP: The average kinetic energy of the molecule is Kav = 3/2 kT, and the de Broglie wavelength is λ =
h/mv = h/p. EXECUTE: (a) Combining Kav = 3/2 kT and K = p2/2m gives 3/2 kT = pav2/2m and pav = 3mkT . The de Broglie
6.626 × 10−34 J ⋅ s
h
h
wavelength is λ = =
=
= 1.08 × 10−10 m .
p
3mkT
3( 2 × 1.67 × 10−27 kg )(1.38 × 10−23 J/K ) ( 273 K )
(b) For an electron, λ = h/p = h/mv gives
v= h
6.626 × 10−34 J ⋅ s
=
= 6.75 × 106 m/s
mλ ( 9.11 × 10 −31 kg )(1.08 × 10−10 m ) This is about 2% the speed of light, so we do not need to use relativity.
(c) For photon:
E = hc/λ = (6.626 × 10–34 J ⋅ s)(3.00 × 108 m/s)/(1.08 × 10–10 m) = 1.84 × 10–15 J
For the H2 molecule: Kav = (3/2)kT = 3/2 (1.38 × 10–23 J/K)(273 K) = 5.65 × 10–21 J
For the electron: K = ½ mv2 = ½ (9.11 × 10–31 kg)(6.73 × 106 m/s)2 = 2.06 × 10–17 J
EVALUATE: The photon has about 100 times more energy than the electron and 300,000 times more energy than
the H2 molecule. This shows that photons of a given wavelength will have much more energy than particles of the
same wavelength. The Wave Nature of Particles 39.11. 39.12. 393 IDENTIFY and SET UP: Use Eq.(39.1).
h
h
6.626 × 10−34 J ⋅ s
EXECUTE: λ = =
=
= 3.90 × 10−34 m
p mv (5.00 × 10−3 kg)(340 m/s)
EVALUATE: This wavelength is extremely short; the bullet will not exhibit wavelike properties.
(a) λ = h mv → v = h m λ 1
Energy conservation: e∆V = mv 2
2
2 "h#
m$
%
mv
h2
(6.626 × 10−34 J ⋅ s) 2
mλ '
∆V =
=&
=
=
= 66.9 V
2
−19
2e
2e
2emλ
2(1.60 × 10 C) (9.11× 10−31 kg) (0.15 × 10−9 m)2
2 (b) Ephoton = hf = hc λ = (6.626 × 10−34 J ⋅ s) (3.0 × 108 m s)
= 1.33 × 10−15 J
0.15 × 10−9 m e∆V = K = Ephoton and ∆V = 39.13. 39.14. Ephoton
e = 1.33 × 10−15 J
= 8310 V
1.6 × 10−19 C (a) λ = 0.10 nm . p = mv = h λ so v = h ( mλ ) = 7.3 × 106 m s . 1
(b) E = mv 2 = 150 eV
2
(c) E = hc / λ = 12 KeV
(d) The electron is a better probe because for the same λ it has less energy and is less damaging to the structure
being probed.
IDENTIFY: The electrons behave like waves and are diffracted by the slit.
SET UP: We use conservation of energy to find the speed of the electrons, and then use this speed to find their de
Broglie wavelength, which is λ = h/mv. Finally we know that the first dark fringe for singleslit diffraction occurs
when a sin θ = λ.
EXECUTE: (a) Use energy conservation to find the speed of the electron: ½ mv2 = eV.
v= 2 (1.60 10−19 C ) (100 V)
2eV
=
= 5.93 × 106 m/s
9.11 × 10−31 kg
m which is about 2% the speed of light, so we can ignore relativity.
(b) First find the de Broglie wavelength: λ= h
6.626 × 10−34 J ⋅ s
=
= 1.23 × 10–10 m = 0.123 nm
mv ( 9.11 × 10−31 kg )( 5.93 × 106 m/s ) For the first single slit dark fringe, we have a sin θ = λ, which gives a= 39.15. 39.16. λ
sin θ = 1.23 × 10−10 m
= 6.16 × 10–10 m = 0.616 nm
sin(11.5°) EVALUATE: The slit width is around 5 times the de Broglie wavelength of the electron, and both are much
smaller than the wavelength of visible light.
h
For m =1, λ = d sin & =
.
2mE
h2
(6.63 × 10−34 J ⋅ s) 2
=
= 6.91 × 10−20 J = 0.432 eV.
E=
2
2
−27
2md sin & 2(1.675 × 10 kg) (9.10 × 10−11 m)2 sin 2 (28.6°)
h
h
mh
Intensity maxima occur when d sin & = m . = =
so d sin & =
. (Careful! Here, m is the order
p
2ME
2ME
of the maxima, whereas M is the mass of the incoming particle.)
mh
(2)(6.63 × 10−34 J ⋅ s)
(a) d =
=
2 ME sin θ
2(9.11 × 10−31 kg)(188 eV)(1.60 × 10−19 J/eV) sin(60.6°)
= 2.06 × 10−10 m = 0.206 nm. (b) m = 1 also gives a maximum. "
#
(1) (6.63 × 10−34 J ⋅ s)
% = 25.8° .
& = arcsin $
$ 2(9.11× 10−31 kg) (188 eV) (1.60 × 10−19 J e V) (2.06 × 10−10 m) %
&
' 394 Chapter 39 This is the only other one. If we let m ≥ 3, then there are no more maxima.
(c) E = m2h2
(1) 2 (6.63 × 10−34 J ⋅ s) 2
=
2 Md 2 sin 2 & 2(9.11 × 10−31 kg) (2.60 × 10−10 m) 2 sin 2 (60.6°) = 7.49 × 10−18 J = 46.8 eV.
Using this energy, if we let m = 2, then sin & > 1. Thus, there is no m = 2 maximum in this case. 39.17. h
h
" mh #
=
, so θ = arcsin $
% . (Careful! Here, m is the order of
p Mv
& dMv '
the maximum, whereas M is the incoming particle mass.)
"h#
(a) m = 1 ! &1 = arcsin $
%
& dMv '
The condition for a maximum is d sinθ = mλ . λ = "
#
6.63 × 10−34 J ⋅ s
= arcsin $
% = 2.07°.
−6
(1.60 × 10 m) (9.11× 10−31 kg) (1.26 × 104 m s) '
&
"
#
(2) (6.63 × 10−34 J ⋅ s)
m = 2 ! &2 = arcsin $
% = 4.14°.
−6
(1.60 × 10 m) (9.11× 10−31 kg) (1.26 × 104 m s) '
&
" " radians #
(b) For small angles (in radians!) y ≅ D& , so y1 ≈ (50.0 cm) (2.07°) $
% = 1.81 cm ,
& 180° ' 39.18. " " radians #
y2 ≈ (50.0 cm) (4.14°) $
% = 3.61 cm and y2 − y1 = 3.61 cm − 1.81 cm = 1.81 cm.
& 180° '
IDENTIFY: Since we know only that the mosquito is somewhere in the room, there is an uncertainty in its
position. The Heisenberg uncertainty principle tells us that there is an uncertainty in its momentum.
SET UP: The uncertainty principle is ∆ x∆px ≥ % .
EXECUTE: (a) You know the mosquito is somewhere in the room, so the maximum uncertainty in its horizontal
position is ∆ x = 5.0 m.
(b) The uncertainty principle gives ∆ x∆px ≥ % , and ∆px = m∆vx since we know the mosquito’s mass. This gives ∆ x m∆vx ≥ % , which we can solve for ∆vx to get the minimum uncertainty in vx.
∆vx = 39.19. which is hardly a serious impediment!
EVALUATE: For something as “large” as a mosquito, the uncertainty principle places a negligible limitation on
our ability to measure its speed.
(a) IDENTIFY and SET UP: Use ∆ x∆px ≥ h / 2π to calculate ∆ x and obtain ∆vx from this.
EXECUTE:
∆vx = 39.20. %
1.055 × 10−34 J ⋅ s
= 1.4 × 10–29 m/s
=
m∆x (1.5 × 106 kg)(5.0 m) ∆px ≥ h
6.626 × 10−34 J ⋅ s
=
= 1.055 × 10−28 kg ⋅ m/s
2π∆ x 2π (1.00 × 10−6 m) ∆p x 1.055 × 10−28 kg ⋅ m/s
=
= 8.79 × 10−32 m/s
m
1200 kg (b) EVALUATE: Even for this very small ∆ x the minimum ∆vx required by the Heisenberg uncertainty principle
is very small. The uncertainty principle does not impose any practical limit on the simultaneous measurements of
the positions and velocities of ordinary objects.
IDENTIFY: Since we know that the marble is somewhere on the table, there is an uncertainty in its position. The
Heisenberg uncertainty principle tells us that there is therefore an uncertainty in its momentum.
SET UP: The uncertainty principle is ∆ x∆px ≥ % .
EXECUTE: (a) Since the marble is somewhere on the table, the maximum uncertainty in its horizontal position is
∆ x = 1.75 m.
(b) Following the same procedure as in part (b) of problem 39.18, the minimum uncertainty in the horizontal
velocity of the marble is
∆vx = %
1.055 × 10−34 J ⋅ s
= 6.03 × 10–33 m/s
=
m∆x ( 0.0100 kg ) (1.75 m) (c) The uncertainty principle tells us that we cannot know that the marble’s horizontal velocity is exactly zero, so
the smallest we could measure it to be is 6.03 × 10–33 m/s, from part (b). The longest time it could remain on the The Wave Nature of Particles table is the time to travel the full width of the table (1.75 m), so t = x/vx = (1.75 m)/(6.03 × 10
s = 9.20 × 1024 years
Since the universe is about 14 × 109 years old, this time is about –33 395 m/s) = 2.90 × 1032 9.0 × 1024 yr
≈ 6 × 1014 times the age of the universe! Don’t hold your breath!
14 × 109 yr 39.21. 39.22. EVALUATE: For household objects, the uncertainty principle places a negligible limitation on our ability to
measure their speed.
h
Heisenberg’s Uncertainty Principles tells us that ∆ x∆px ≥ . We can treat the standard deviation as a direct
2"
h
−10
= 1.05 × 10−34 J ⋅ s
measure of uncertainty. Here ∆ x∆px = (1.2 × 10 m) (3.0 × 10−25 kg ⋅ m s) = 3.6 × 10−35 J ⋅ s but
2"
h
Therefore ∆ x∆px <
so the claim is not valid .
2"
(a) ( ∆ x) (m∆vx ) ≥ h 2" , and setting ∆vx = (0.010)vx and the product of the uncertainties equal to h / 2" (for the
minimum uncertainty) gives vx = h ( 2"m(0.010)∆ x ) = 57.9 m s.
(b) Repeating with the proton mass gives 31.6 mm s. h
(6.63 × 10−34 J ⋅ s)
=
= 2.03 × 10−32 J = 1.27 × 10−13 eV.
2" ∆t
2" (5.2 × 10−3 s) 39.23. ∆E > 39.24. IDENTIFY and SET UP: The Heisenberg Uncertainty Principle says ∆ x∆px ≥ ∆ x∆px is h / 2π . ∆px = m∆vx . h
. The minimum allowed
2π h
h
6.63 × 10−34 J ⋅ s
. ∆vx =
=
= 3.2 × 104 m/s
2π m∆ x 2π (1.67 × 10−27 kg)(2.0 × 10−12 m)
2π
h
6.63 × 10−34 J ⋅ s
(b) ∆ x =
=
= 4.6 × 10−4 m
2π m∆vx 2π (9.11× 10−31 kg)(0.250 m/s)
EXECUTE: (a) m∆ x∆vx = 39.25. ∆E ∆t = h
(6.63 × 10−34 J ⋅ s)
h
=
= 1.39 × 10−14 J = 8.69 × 104 eV = 0.0869 MeV.
. ∆E =
2" ∆t 2" (7.6 × 10−21 s)
2"
∆E 0.0869 MeV c 2
=
= 2.81× 10−5.
E
3097 MeV c 2 h
. ∆E = ∆mc 2 . ∆m = 2.06 × 109 eV c 2 = 3.30 × 10−10 J c 2 .
2"
h
6.63 × 10−34 J ⋅ s
∆t =
=
= 3.20 × 10−25 s.
2" ∆mc 2 2" (3.30 × 10−10 J) 39.26. ∆E ∆t = 39.27. IDENTIFY and SET UP: For a photon Eph = 39.29. λ = 1.99 × 10−25 J ⋅ m λ 2 . For an electron Ee = p2
1 "h#
h2
=
.
$ %=
2m 2m & λ ' 2mλ 2 −25 1.99 × 10 J ⋅ m
= 1.99 × 10−17 J
10.0 × 10−9 m
(6.63 × 10−34 J ⋅ s) 2
electron Ee =
= 2.41× 10−21 J
2(9.11 × 10−31 kg)(10.0 × 10−9 m)2
Eph 1.99 × 10−17 J
=
= 8.26 × 103
Ee 2.41 × 10−21 J
(b) The electron has much less energy so would be less damaging.
EVALUATE: For a particle with mass, such as an electron, E ~ λ −2 . For a massless photon E ~ λ −1 .
p 2 (h λ )2
(h λ )2
, so V =
=
= 419 V.
(a) eV = K =
2m
2m
2me
9.11× 10−31 kg
(b) The voltage is reduced by the ratio of the particle masses, (419 V)
= 0.229 V.
1.67 × 10−27 kg
EXECUTE: (a) photon Eph = 39.28. hc 2 ψ ( x ) = A sin kx. The position probability density is given by ψ ( x ) = A2 sin 2 kx.
EXECUTE: (a) The probability is highest where sin kx = 1 so kx = 2π x / λ = nπ / 2, n = 1, 3, 5,…
x = nλ / 4, n = 1, 3, 5,… so x = λ / 4, 3λ / 4, 5λ /4,…
IDENTIFY and SET UP: 396 Chapter 39 39.30. (b) The probability of finding the particle is zero where ψ = 0, which occurs where sin kx = 0 and
kx = 2π x / λ = nπ , n = 0, 1, 2,…
x = nλ / 2, n = 0, 1, 2,… so x = 0, λ / 2, λ , 3λ / 2,…
EVALUATE: The situation is analogous to a standing wave, with the probability analogous to the square of the
amplitude of the standing wave.
2
2
2
Ψ ∗ = ( ∗ sin )t , so Ψ = Ψ *Ψ = ( *( sin 2 )t = ( sin 2 )t . Ψ is not timeindependent, so Ψ is not the 2 39.31. wavefunction for a stationary state.
IDENTIFY: To describe a real situation, a wave function must be normalizable.
SET UP: ψ 2 dV is the probability that the particle is found in volume dV. Since the particle must be somewhere,
ψ must have the property that ψ 2 dV = 1 when the integral is taken over all space.
EXECUTE: (a) In one dimension, as we have here, the integral discussed above is of the form
(b) Using the result from part (a), we have 5 ( e ) dx = 5
∞ ax 2 −∞ ∞
−∞ e2 ax dx = e 2 ax
2a 5 ∞ −∞  ψ ( x) 2 dx = 1. ∞ = ∞ . Hence this wave function cannot
−∞ be normalized and therefore cannot be a valid wave function.
(c) We only need to integrate this wave function of 0 to because it is zero for x < 0. For normalization we have
∞ 1 = 5 ψ 2 dx =
−∞ 39.32. 5 ( Ae ) dx = 5
∞ ∞ − bx 2 0 0 A2e−2bx dx = A2e −2bx
−2b ∞ =
0 A2
A2
, which gives
= 1, so A = 2b .
2b
2b EVALUATE: If b were positive, the given wave function could not be normalized, so it would not be allowable.
(a) The uncertainty in the particle position is proportional to the width of ( ( x ) , and is inversely proportional to α . This can be seen by either plotting the function for different values of α , finding the expectation
value x 2 = 5 ( 2 x 2 dx for the normalized wave function or by finding the full width at halfmaximum. The particle’s uncertainty in position decreases with increasing α . The dependence of the expectation value 7 x 2 8 on α
may be found by considering
∞ 7 x2 8 = 5xe 2 −2α x 2 −∞
∞ 5e dx =− −2α x 2 dx ∞
1 ∂ / −2α x2 0
1∂ / 1
dx 2 = −
ln 1 5 e
ln 1
2 ∂α 3 −∞
2 ∂α 3 2α
4 ∞ 5e −∞ −u 2 01
du 2 =
,
4 4α −∞ 39.33.
39.34. where the substitution u = α x has been made.
(b) Since the uncertainty in position decreases, the uncertainty in momentum must increase.
" x − iy #
" x + iy #
" x − iy # " x + iy #
2
*
*
f ( x, y ) = $
% ! f = f f =$
% and f ( x, y ) = $
%⋅$
% = 1.
x + iy '
x − iy '
&
&
& x + iy ' & x − iy '
2 The same. ( ( x, y , z ) = ( * ( x, y , z )( ( x, y, z )
2 ( ( x, y, z )eiφ = (( * ( x, y, z )e− iφ )(( ( x, y, z )e + iφ ) = ( * ( x, y, z )( ( x, y, z ).
39.35. The complex conjugate means convert all i’s to–i’s and viceversa. eiφ ⋅ e −iφ = 1.
IDENTIFY: To describe a real situation, a wave function must be normalizable.
SET UP: ψ 2 dV is the probability that the particle is found in volume dV. Since the particle must be somewhere,
ψ must have the property that ψ 2 dV = 1 when the integral is taken over all space.
EXECUTE: (a) For normalization of the onedimensional wave function, we have
∞ 1 = 5 ψ 2 dx =
−∞ 5 ( Ae ) dx + 5 ( Ae ) dx = 5
0 ∞ bx 2 −∞ − bx 2 0 0
−∞ ∞ A2e 2bx dx + 5 A2e −2bx dx .
0 ∞
( e2bx 0
e −2bx ) A2
*
*
–1/2
−1
1 = A2 +
+
, = , which gives A = b = 2.00 m = 1.41 m
2b −∞ −2b 0 * b
*
.
(b) The graph of the wavefunction versus x is given in Figure 39.35. (c) (i) P = 5 + 5.00 m
− 0.500 m ψ 2 dx = 2 5 + 5.00 m
0 A2e −2bx dx , where we have used the fact that the wave function is an even function of x. Evaluating the integral gives
− A2 −2b (0.500 m)
−(2.00 m −1 ) −2.00
− 1) =
(e
( e − 1) = 0.865
b
2.00 m −1
There is a little more than an 86% probability that the particle will be found within 50 cm of the origin.
P= The Wave Nature of Particles
2 397 −1 A
2.00 m
1
=
= = 0.500
−1
2b 2(2.00 m ) 2
There is a 5050 chance that the particle will be found to the left of the origin, which agrees with the fact that the
wave function is symmetric about the yaxis.
(ii) P = 5 ( Ae ) dx = 5 (iii) P = 5 0 bx 2 −∞ 1.00 m
0.500 m 0 −∞ A2e 2bx dx = A2e −2bx dx ( ) A2 −2(2.00 m1 )(1.00 m) −2(2.00 m1 )(0.500 m)
1
e
−e
= − ( e −4 − e−2 ) = 0.0585
2
−2b
EVALUATE: There is little chance of finding the particle in regions where the wave function is small.
= 39.36. Figure 39.35
−% 2 d 2(
+ U( = E( . Let ψ = A(1 + B( 2
Eq. (39.18):
2m dx 2
−% 2 d 2
!
( A(1 + B( 2 ) + U ( A(1 + B( 2 ) = E ( A(1 + B( 2 )
2m dx 2
" % 2 d 2(1
#
" % 2 d 2( 2
#
! A$ −
+ U(1 − E(1 % + B $ −
+ U( 2 − E( 2 % = 0.
2
2
& 2m dx
'
& 2m dx
' But each of (1 and (2 satisfy Schrödinger’s equation separately so the equation still holds true, for any A or B.
39.37. 39.38. % 2 d 2(
+ U( = BE1(1 + CE2( 2 . If ( were a solution with energy E, then BE1(1 + CE2( 2 = BE(1 + CE(2 or
2m dx 2
B( E1 − E )(1 = C ( E − E2 )( 2 . This would mean that (1 is a constant multiple of ( 2 , and (1 and ( 2 would be wave
functions with the same energy. However, E1 ≠ E2 , so this is not possible, and ( cannot be a solution to Eq. (39.18).
− (a) λ =
(b) h
(6.63 × 10 −34 J ⋅ s)
= 1.94 × 10−10 m.
=
−31
−19
2mK
2(9.11 × 10 kg)(40 eV)(1.60 × 10 J eV) (2.5 m)(9.11 × 10−31 kg)1 2
R
R
=
=
= 6.67 × 10−7 s.
v
2E m
2(40 eV)(1.6 × 10−19 J eV) (c) The width w is w = 2R ' and w = ∆v yt = ∆p yt m, where t is the time found in part (b) and a is the slit width.
a
2mλ R
Combining the expressions for w, ∆p y =
= 2.65 × 10 −28 kg ⋅ m s.
at
h
= 0.40 *m, which is the same order of magnitude.
(d) ∆y =
2"∆p y
39.39. (a) E = hc λ = 12 eV
(b) Find E for an electron with λ = 0.10 × 10−6 m. λ = h p so p = h λ = 6.626 × 10 −27 kg ⋅ m s .
E = p 2 (2m) = 1.5 × 10−4 eV . E = q∆V so ∆V = 1.5 × 10−4 V
v = p m = (6.626 × 10 −27 kg ⋅ m s) (9.109 × 10 −31 kg) = 7.3 × 103 m s (c) Same λ so same p. E = p 2 /(2m) but now m = 1.673 × 10−27 kg so E = 8.2 × 10−8 eV and ∆V = 8.2 × 10−8 V.
v = p m = (6.626 × 10 −27 kg ⋅ m s) (1.673 × 10 −27 kg) = 4.0 m s 39.40. (a) Single slit diffraction: a sin & = mλ . λ = a sin & = (150 × 10−9 m)sin20° = 5.13 × 10−8 m λ = h mv → v = h mλ . v =
(b) a sin &2 = 2λ . sin &2 = ±2 λ 6.626 × 10−34 J ⋅ s
= 1.42 × 104 m s
(9.11× 10−31 kg)(5.13 × 10−8 m) " 5.13 × 10−8 m #
= ±2 $
% = ±0.684 . &2 = ±43.2°
−9
a
& 150 × 10 m ' 398 Chapter 39 39.41. IDENTIFY: The electrons behave like waves and produce a doubleslit interference pattern after passing through
the slits.
SET UP: The first angle at which destructive interference occurs is given by d sin θ = λ/2. The de Broglie
wavelength of each of the electrons is λ = h/mv.
EXECUTE: (a) First find the wavelength of the electrons. For the first dark fringe, we have d sin θ = λ/2, which gives
(1.25 nm)(sin 18.0°) = λ/2 , and λ = 0.7725 nm. Now solve the de Broglie wavelength equation for the speed of the
electron: v= 39.42. h
6.626 × 10−34 J ⋅ s
=
= 9.42 × 105 m/s
mλ (9.11× 10−31 kg)(0.7725 × 10−9 m) which is about 0.3% the speed of light, so they are nonrelativistic.
(b) Energy conservation gives eV = ½ mv2 and
V = mv2/2e = (9.11 × 10–31 kg)(9.42 × 105 m)2/[2(1.60 × 10–19 C)] = 2.52 V
EVALUATE: The hole must be much smaller than the wavelength of visible light for the electrons to show diffraction.
IDENTIFY: The alpha particles and protons behave as waves and exhibit circularaperture diffraction after passing
through the hole.
SET UP: For a round hole, the first dark ring occurs at the angle θ for which sinθ = 1.22λ /D, where D is the
diameter of the hole. The de Broglie wavelength for a particle is λ = h/p = h/mv.
EXECUTE: Taking the ratio of the sines for the alpha particle and proton gives
sin θα 1.22λα λα
=
=
sin θ p 1.22λp λp
The de Broglie wavelength gives λp = h/pp and λα = h/pα, so sin θα h / pα pp
. Using K = p2/2m, we have
=
=
sin θ p h / pp pα p = 2mK . Since the alpha particle has twice the charge of the proton and both are accelerated through the same
potential difference, Kα = 2Kp. Therefore pp = 2mp K p and pα = 2mα Kα = 2mα (2 K p ) = 4mα K p .
Substituting these quantities into the ratio of the sines gives
2mp K p
mp
sin θα pp
=
=
=
sin θ p pα
2mα
4mα K p
Solving for sin θα gives sin θα = 39.43. 1.67 × 10−27 kg
sin15.0° and θα = 5.3°.
2(6.64 × 10 −27 kg)
EVALUATE: Since sin θ is inversely proportional to the mass of the particle, the largermass alpha particles form
their first dark ring at a smaller angle than the ring for the lighter protons.
IDENTIFY: Both the electrons and photons behave like waves and exhibit singleslit diffraction after passing
through their respective slits.
SET UP: The energy of the photon is E = hc/λ and the de Broglie wavelength of the electron is λ = h/mv = h/p.
Destructive interference for a single slit first occurs when a sin θ = λ.
EXECUTE: (a) For the photon: λ = hc/E and a sinθ = λ. Since the a and θ are the same for the photons and
electrons, they must both have the same wavelength. Equating these two expressions for λ gives a sin θ = hc/E.
h
h
and a sin θ = λ. Equating these two expressions for λ gives a sin θ =
.
For the electron, λ = h/p =
2mK
2mK
h
, which gives E = c 2mK = (4.05 × 10−7 J1/2 ) K
Equating the two expressions for asinθ gives hc/E =
2mK E c 2mK
2mc 2
=
=
. Since v << c, mc2 > K, so the square root is >1. Therefore E/K > 1, meaning that the
K
K
K
photon has more energy than the electron.
EVALUATE: As we have seen in Problem 39.10, when a photon and a particle have the same wavelength, the
photon has more energy than the particle.
d sin & (40.0 × 10−6 m)sin(0.0300 rad)
According to Eq.(35.4) λ =
=
= 600 nm. The velocity of an electron with
m
2
this wavelength is given by Eq.(39.1)
(b) 39.44. v= p
h
(6.63 × 10−34 J ⋅ s)
=
=
= 1.21× 103 m s.
mm
(9.11 × 10−31 kg)(600 × 10−9 m) The Wave Nature of Particles 399 Since this velocity is much smaller than c we can calculate the energy of the electron classically 39.45. 1
1
K = mv 2 = (9.11 × 10−31 kg)(1.21 × 103 m s) 2 = 6.70 × 10−25 J = 4.19 *eV.
2
2
The de Broglie wavelength of the blood cell is = h
(6.63 × 10 −34 J ⋅ s)
=
= 1.66 × 10 −17 m.
mv (1.00 × 10−14 kg)(4.00 × 10−3 m s) We need not be concerned about wave behavior.
12 39.46. " v2 #
h $1 − 2 %
c'
h
(a) λ = = &
p
mv " v2 #
h 2v 2
v2
! λ 2 m 2 v 2 = h 2 $ 1 − 2 % = h 2 − 2 ! λ 2 m 2v 2 + h 2 2 = h 2
c
c
& c'
! v2 = (b) v = c
12 " " λ #2 #
$1 + $
%
$ & ( h mc) % %
''
& h2
" 2 2 h2 #
$λ m + 2 %
c'
& = c
c2
!v=
.
12
2 22
"λ m c
#
" " mcλ # 2 #
+ 1%
$
2
$1 + $
$ & h %%
&h
'
'%
&
' " 1 " mcλ #2 #
m 2 c 2λ 2
≈ c $1 − $
.
% = (1 − ∆ )c. ∆ =
$ 2& h % %
2h 2
''
& h
.
mc
(9.11× 10−31 kg)2 (3.00 × 108 m s)2 (1.00 × 10−15 m)2
∆=
= 8.50 × 10−8
2(6.63 × 10−34 J ⋅ s) 2
! v = (1 − )c = (1 − 8.50 × 10−8 )c. (c) λ = 1.00 × 10 −15 m << 39.47. (a) Recall λ = λ= h
h
h
=
=
. So for an electron:
p
2mE
2mq∆V
6.63 × 10−34 J ⋅ s 2(9.11 × 10 −31 kg)(1.60 × 10 (b) For an alpha particle: λ =
39.48. −19 C)(125 V) ! = 1.10 × 10−10 m. 6.63 × 10 −34 J ⋅ s
2(6.64 × 10 −27 kg)2(1.60 × 10−19 C)(125 V) IDENTIFY and SET UP: The minimum uncertainty product is ∆ x∆px = = 9.10 × 10−13 m. h
. ∆ x = r1 , where r1 is the radius of the
2π h
h
.
and p1 = mv1 =
2π
2π r1
h
h
6.63 × 10−34 J ⋅ s
=
=
= 2.0 × 10−24 kg ⋅ m/s . This is the same as the magnitude of
EXECUTE: ∆px =
2π∆ x 2π r1 2π (0.529 × 10−10 m)
the momentum of the electron in the n = 1 Bohr orbit.
EVALUATE: Since the momentum is the same order of magnitude as the uncertainty in the momentum, the
uncertainty principle plays a large role in the structure of atoms.
hc
IDENTIFY and SET UP: Combining the two equations in the hint gives PC = K ( K + 2mc 2 ) and λ =
.
K ( K + 2mc 2 )
hc
h
EXECUTE: (a) With K = 3mc 2 this becomes λ =
=
.
2
2
2
15mc
3mc (3mc + 2mc )
n = 1 Bohr orbit. In the n = 1 Bohr orbit, mv1r1 = 39.49. (b) (i) K = 3mc 2 = 3(9.109 × 10 −31 kg)(2.998 × 108 m/s) 2 = 2.456 × 10 −13 J = 1.53 MeV
h
6.626 × 10 −34 J ⋅ s
=
= 6.26 × 10−13 m
15mc
15(9.109 × 10−31 kg)(2.998 × 108 m/s)
(ii) K is proportional to m, so for a proton K = (mp / me )(1.53 MeV) = 1836(1.53 MeV) = 2810 MeV λ= λ is proportional to 1/m, so for a proton λ = (me / mp )(6.26 × 10 −13 m) = (1/1836)(6.626 × 10−13 m) = 3.41 × 10 −16 m
EVALUATE: The proton has a larger rest mass energy so its kinetic energy is larger when K = 3mc 2 . The proton
also has larger momentum so has a smaller λ . 3910 Chapter 39 39.50. (a) (6.626 × 10−34 J ⋅ s)
= 2.1 × 10−20 kg ⋅ m s.
2" (5.0 × 10−15 m) (b) K = ( pc ) 2 + (mc 2 ) 2 − mc 2 = 1.3 × 10−13 J = 0.82 MeV. 39.51. (c) The result of part (b), about 1 MeV = 1 × 106 eV , is many orders of magnitude larger than the potential energy
of an electron in a hydrogen atom.
(a) IDENTIFY and SET UP: ∆ x∆px ≥ h / 2π Estimate ∆ x as ∆ x ≈ 5.0 × 10 −15 m.
EXECUTE: Then the minimum allowed ∆px is ∆px ≈ h
6.626 × 10−34 J ⋅ s
=
= 2.1× 10−20 kg ⋅ m/s
2π∆x 2π (5.0 × 10−15 m) (b) IDENTIFY and SET UP: Assume p ≈ 2.1 × 10−20 kg ⋅ m/s. Use Eq.(37.39) to calculate E, and then K = E − mc 2 .
EXECUTE: E = (mc 2 ) 2 + ( pc) 2 mc 2 = (9.109 × 10 −31 kg)(2.998 × 108 m/s)2 = 8.187 × 10−14 J
pc = (2.1 × 10−20 kg ⋅ m/s)(2.998 × 108 m/s) = 6.296 × 10−12 J E = (8.187 × 10−14 J)2 + (6.296 × 10−12 J) 2 = 6.297 × 10−12 J 39.52. 39.53. 39.54. K = E − mc 2 = 6.297 × 10−12 J − 8.187 × 10−14 J = 6.215 × 10−12 J(1 eV/1.602 × 10−19 J) = 39 MeV
(c) IDENTIFY and SET UP: The Coulomb potential energy for a pair of point charges is given by Eq.(23.9). The
proton has charge +e and the electron has charge –e.
ke 2
(8.988 × 109 N ⋅ m 2 / C2 )(1.602 × 10−19 C)2
=−
= −4.6 × 10−14 J = −0.29 MeV
EXECUTE: U = −
5.0 × 10−15 m
r
EVALUATE: The kinetic energy of the electron required by the uncertainty principle would be much larger than
the magnitude of the negative Coulomb potential energy. The total energy of the electron would be large and
positive and the electron could not be bound within the nucleus.
(a) Take the direction of the electron beam to be the xdirection and the direction of motion perpendicular to the
∆p
h
6.626 × 10−34 J ⋅ s
=
= 0.23 m/s
beam to be the ydirection. ∆v y = y =
2π m∆y 2π (9.11 × 10−31 kg)(0.50 × 10−3 m)
m
(b) The uncertainty ∆r in the position of the point where the electrons strike the screen is
∆p x
h
x
∆r = ∆v yt = y =
= 9.56 × 10−10 m,
m vx 2"m∆y 2K m (c) This is far too small to affect the clarity of the picture.
h
h
IDENTIFY and SET UP: ∆E ∆t ≥
. Take the minimum uncertainty product, so ∆E =
, with
2π
2π∆t
∆E
∆t = 8.4 × 10−17 s . m = 264me . ∆m = 2 .
c
6.63 × 10−34 J ⋅ s
1.26 × 10−18 J
EXECUTE: ∆E =
= 1.26 × 10−18 J . ∆m =
= 1.4 × 10−35 kg .
−17
2π (8.4 × 10 s)
(3.00 × 108 m/s) 2
∆m
1.4 × 10−35 kg
=
= 5.8 × 10−8
m (264)(9.11× 10−31 kg)
IDENTIFY: The insect behaves like a wave as it passes through the hole in the screen.
SET UP: (a) For wave behavior to show up, the wavelength of the insect must be of the order of the diameter of
the hole. The de Broglie wavelength is λ = h/mv.
EXECUTE: The de Broglie wavelength of the insect must be of the order of the diameter of the hole in the screen,
so λ 5.00 mm. The de Broglie wavelength gives v= 39.55. h
6.626 × 10−34 J ⋅ s
=
= 1.33 × 10–25 m/s
mλ (1.25 × 10−6 kg ) ( 0.00400 m ) (b) t = x/v = (0.000500 m)/(1.33 × 10–25 m/s) = 3.77 × 1021 s = 1.4 × 1010 yr
The universe is about 14 billion years old (1.4 × 1010 yr), so this time would be about 85,000 times the age of the
universe.
EVALUATE: Don’t expect to see a diffracting insect! Wave behavior of particles occurs only at the very small scale.
IDENTIFY and SET UP: Use Eq.(39.1) to relate your wavelength and speed.
h
h
6.626 × 10−34 J ⋅ s
, so v =
=
= 1.1× 10−35 m/s
EXECUTE: (a) λ =
mv
mλ (60.0 kg)(1.0 m) The Wave Nature of Particles 3911 distance
0.80 m
=
= 7.3 × 1034 s(1 y/3.156 × 107 s) = 2.3 × 1027 y
velocity 1.1× 10−35 m/s
Since you walk through doorways much more quickly than this, you will not experience diffraction effects.
EVALUATE: A 1 kg object moving at 1 m/s has a de Broglie wavelength λ = 6.6 × 10−34 m, which is exceedingly
small. An object like you has a very, very small λ at ordinary speeds and does not exhibit wavelike properties.
hc
(a) E = 2.58 eV = 4.13 × 10−19 J, with a wavelength of λ =
= 4.82 × 10−7 m = 482 nm
E
h
(6.63 × 10−34 J ⋅ s)
=
= 6.43 × 10−28 J = 4.02 × 10−9 eV.
(b) ∆E =
2" ∆t 2" (1.64 × 10−7 s)
(c) λ E = hc, so ( ∆λ ) E + λ∆E = 0, and ∆E E = ∆λ λ , so
(b) t = 39.56. 39.57. 39.58. 39.59. " 6.43 × 10−28 J #
−16
−7
∆λ = λ ∆E E = (4.82 × 10−7 m) $
% = 7.50 × 10 m = 7.50 × 10 nm.
−19
& 4.13 × 10 J '
IDENTIFY: The electrons behave as waves whose wavelength is equal to the de Broglie wavelength.
SET UP: The de Broglie wavelength is λ = h/mv, and the energy of a photon is E = hf = hc/λ.
EXECUTE: (a) Use the de Broglie wavelength to find the speed of the electron.
h
6.626 × 10−34 J ⋅ s
v=
=
= 7.27 × 105 m/s
mλ ( 9.11 × 10−31 kg )(1.00 × 10−9 m )
which is much less than the speed of light, so it is nonrelativistic.
(b) Energy conservation gives eV = ½ mv2.
V = mv2/2e = (9.11 × 10–31 kg)(7.27 × 105 m/s)2/[2(1.60 × 10–19 C)] = 1.51 V
(c) K = eV = e(1.51 V) = 1.51 eV, which is about ¼ the potential energy of the NaCl crystal, so the electron would
not be too damaging.
(d) E = hc/λ = (4.136 × 10–15 eV s)(3.00 × 108 m/s)/(1.00 × 10–9 m) = 1240 eV
which would certainly destroy the molecules under study.
EVALUATE: As we have seen in Problems 39.10 and 39.43, when a particle and a photon have the same
wavelength, the photon has much more energy.
h
λ′
"
#
sin & ′ = sin & , and λ ′ = ( h p′) = (h 2mE ′ ), and so & ′ = arcsin $
sin & % .
λ
& λ 2mE ′
'
−34
"
#
(6.63 × 10 J ⋅ s)sin 35.8°
% = 20.9°
θ ′ = arcsin $
$ (3.00 × 10−11 m) 2(9.11× 10−31 kg)(4.50 × 10+3 )(1.60 × 10−19 J eV) %
&
'
(a) The maxima occur when 2d sin & = mλ as described in Section 38.7.
(b) λ = (6.63 × 10−34 J ⋅ s)
h
h
=
. λ=
= 1.46 × 10−10 m = 0.146 nm .
−37
−19
p
2mE
2(9.11× 10 kg)(71.0 eV) (1.60 × 10 J/eV ) " mλ #
& = sin −1 $
% (Note: This m is the order of the maximum, not the mass.)
& 2d '
" (1)(1.46 × 10−10 m) #
! sin −1 $
% = 53.3°.
−11
& 2(9.10 × 10 m) ' 39.60. 39.61. (c) The work function of the metal acts like an attractive potential increasing the kinetic energy of incoming
electrons by eφ . An increase in kinetic energy is an increase in momentum that leads to a smaller wavelength. A
smaller wavelength gives a smaller angle θ (see part (b)).
1
(a) Using the given approximation, E = ( (h x) 2 m + kx 2 ) , ( dE dx) = kx − ( h 2 mx 3 ), and the minimum energy
2
h
occurs when kx = ( h 2 mx3 ), or x 2 =
. The minimum energy is then h k m.
mk
(b) They are the same.
(a) IDENTIFY and SET UP: U = A x . Eq.(7.17) relates force and potential. The slope of the function A x is not continuous at x = 0 so we must consider the regions x > 0 and x < 0 separately.
d ( Ax)
EXECUTE: For x > 0, x = x so U = Ax and F = −
= − A. For x < 0, x = − x so U = − Ax and
dx
d (− Ax)
F =−
= + A. We can write this result as F = − A x / x, valid for all x except for x = 0.
dx 3912 Chapter 39 (b) IDENTIFY and SET UP: Use the uncertainty principle, expressed as ∆p∆ x ≈ h, and as in Problem 39.50
estimate ∆p by p and ∆ x by x. Use this to write the energy E of the particle as a function of x. Find the value of x
that gives the minimum E and then find the minimum E.
p2
EXECUTE: E = K + U =
+Ax
2m
px ≈ h, so p ≈ h / x h2
+Ax.
2mx 2
h2
+ Ax.
For x > 0, E =
2mx 2
Then E ≈ To find the value of x that gives minimum E set
0= dE
= 0.
dx −2h 2
+A
2mx 3
1/ 3 " h2 #
h2
x=
and x = $
%
mA
& mA '
With this x the minimum E is
3 h2 " mA #
E=
$
%
2m & h 2 ' 2/3 1/ 3 " h2 #
+ A$
%
& mA ' 1
= h 2 / 3m −1/ 3 A2 / 3 + h 2 / 3m −1/ 3 A2 / 3
2 1/ 3 39.62. " h 2 A2 #
E = 3$
%
2
&m'
EVALUATE: The potential well is shaped like a V. The larger A is the steeper the slope of U and the smaller the
region to which the particle is confined and the greater is its energy. Note that for the x that minimizes E, 2K = U.
∗
For this wave function, Ψ ∗ = (1∗ei)1t + ( 2 ei)2 t , so
∗
∗
∗
∗
*
∗
Ψ 2 = Ψ ∗Ψ = ((1 ei)1t + (2 ei)2 t )((1e −i)1t + ( 2e −i)2 t ) = (1 (1 + ( 2(2 + (1 ( 2ei ()1 − )2 )t + ( 2(1ei ()2 − )1 )t . The frequencies )1 and )2 are given as not being the same, so Ψ
39.63. 2 is not timeindependent, and Ψ is not the wave function for a stationary state.
The timedependent equation, with the separated form for Ψ ( x, t ) as given becomes
" % 2 d 2(
#
i%( (−i)) = $ −
+ U ( x)( % .
2m dx 2
&
' Since ( is a solution of the timeindependent solution with energy E , the term in parenthesis is E( , and so
)% = E , and ) = ( E %).
39.64. p 2 (%k ) 2
%k 2
=
!)=
.
2m
2m
2m
% 2 ∂ 2( ( x, t )
(b) From Problem 39.63 the timedependent Schrödinger’s equation is −
+
2m ∂x 2
∂( ( x, t )
∂ 2( ( x, t )
2mi ∂( ( x, t )
U ( x )( ( x, t ) = i%
. U ( x) = 0 for a free particle, so
.
=−
∂t
∂x 2
∂t
%
Try ( ( x, t ) = cos(kx − )t ) :
(a) ) = 2" f = 2" E E
2 " 2"
p
p= .
= . k=
=
h
h
%
λ
% %) = E = K = ∂(
( x, t ) = A) sin(kx − )t )
∂t
∂( ( x, t )
∂ 2(
= − Ak sin(kx − )t ) and 2 = Ak 2 cos( kx − )t ).
∂x
∂x
" 2mi #
Putting this into the Schrödinger’s equation, Ak 2 cos(kx − )t ) = − $
% A) sin(kx − )t ).
&%'
This is not generally true for all x and t so is not a solution. The Wave Nature of Particles 3913 (c) Try ( ( x, t ) = A sin( kx − )t ) : ∂( ( x, t )
= − A) cos( kx − )t )
∂t
∂( ( x, t )
∂ 2( ( x, t )
= Ak cos(kx − )t ) and
= − Ak 2 sin(kx − )t ).
∂x
∂x 2
" 2mi #
Again, − Ak 2 sin( kx − )t ) = − $
% A) cos(kx − )t ) is not generally true for all x and t so is not a solution.
&%'
(d) Try ( ( x, t ) = A cos( kx − )t ) + B sin(kx − )t ) : 39.65. ∂( ( x, t )
= + A) sin( kx − )t ) − B) cos(kx − )t )
∂t
∂( ( x, t )
∂ 2( ( x, t )
= − Ak sin(kx − )t ) + Bk cos(kx − )t ) and
= − Ak 2 cos((kx − )t ) − Bk 2 sin(kx − )t ).
∂x
∂x 2
Putting this into the Schrödinger’s equation,
2mi
− Ak 2 cos(kx − )t ) − Bk 2 sin( kx − )t ) = −
( + A) sin(kx − )t ) − B) cos(kx − )t )).
%
%k 2
. Collect sin and cos terms.
Recall that ) =
2m
( A + iB ) k 2 cos( kx − )t ) + (iA − B) k 2 sin (kx − )t ) = 0. This is only true if B = iA.
(a) IDENTIFY and SET UP: Let the ydirection be from the thrower to the catcher, and let the xdirection be
horizontal and perpendicular to the ydirection. A cube with volume V = 125 cm3 = 0.125 × 10−3 m3 has side length
l = V 1 / 3 = (0.125 × 10−3 m3 )1/ 3 = 0.050 m. Thus estimate ∆ x as ∆ x ≈ 0.050 m. Use the uncertainty principle to
estimate ∆px . h
0.0663 J ⋅ s
=
= 0.21 kg ⋅ m/s
2π∆ x 2π (0.050 m)
(The value of h in this other universe has been used.)
(b) IDENTIFY and SET UP: ∆ x = (∆vx )t is the uncertainty in the xcoordinate of the ball when it reaches the
EXECUTE: ∆ x∆px ≥ h / 2π then gives ∆px ≈ catcher, where t is the time it takes the ball to reach the second student. Obtain ∆vx from ∆ px .
EXECUTE: The uncertainty in the ball’s horizontal velocity is ∆vx = ∆px 0.21 kg ⋅ m/s
=
= 0.84 m/s
0.25 kg
m 12 m
= 2.0 s. The uncertainty in the xcoordinate
6.0 m/s
of the ball when it reaches the second student that is introduced by ∆vx is ∆ x = (∆vx )t = (0.84 m/s)(2.0 s) = 1.7 m.
The ball could miss the second student by about 1.7 m.
EVALUATE: A game of catch would be very different in this universe. We don’t notice the effects of the
uncertainty principle in everyday life because h is so small.
The time it takes the ball to travel to the second student is t = 39.66. (a) ( 2 = A2 x 2e−2( %x 2 + +y 2 + 'z 2 ) 2 . To save some algebra, let u = x 2 , so that ( = ue−2α u f ( y, z ) . ∂2
1
1
2
, x0 = ±
.
( = (1 − 2α u ) ( ; the maximum occurs at u0 =
∂u
2α
2α
(b) ( vanishes at x = 0, so the probability of finding the particle in the x = 0 plane is zero. The wave function
vanishes for x = ±∞.
39.67. 2 (a) IDENTIFY and SET UP: The probability is P = ψ dV with dV = 4π r 2 dr
EXECUTE: 2 2 2 ψ = A2e −2α r so P = 4π A2 r 2e −2α r dr (b) IDENTIFY and SET UP: P is maximum where
EXECUTE:
2 dP
=0
dr d 2 −2α r 2
(r e
)=0
dr
2 2re −2α r − 4α r 3e −2α r = 0 and this reduces to 2r − 4α r 3 = 0
r = 0 is a solution of the equation but corresponds to a minimum not a maximum. Seek r not equal to 0 so divide by
r and get 2 − 4α r 2 = 0 3914 Chapter 39 1
(We took the positive square root since r must be positive.)
2α This gives r = EVALUATE: This is different from the value of r, r = 0, where ψ 2 is a maximum. At r = 0, ψ 2 has a maximum but the volume element dV = 4π r dr is zero here so P does not have a maximum at r = 0.
2 39.68. (a) B (k ) = e −α 22 k B (0) = Bmax = 1
B (kh ) = 22
1
1
2
= e −α kh ! ln(1 2) = −α 2 kh ! kh =
ln(2) = )k .
2
α ∞ (b) Using integral tables: ( ( x) = 5 e −α
0 22 k cos kxdk = " − x2 / 4α 2
(e
). ( ( x) is a maximum when x = 0.
2α 1
− x2
! h2 = ln(1/2) ! xh = 2α ln2 = )x
4α
2
4α
h)k #
h "1
h
h ln 2
"
#
(d) ) p )x = $
ln2 % 2α ln2 =
(2ln2) =
.
% )x =
$
"
2" & α
2"
& 2" '
'
(c) ( ( xh ) = " 2 2 when e − xh / 4α = ( 39.69. ) k0 " 1 #
∞
sin kx
(a) ( ( x) = 5 B ( k )cos kxdk = 5 $ % cos kxdk =
0
0
k0 x
& k0 ' k0 =
0 sin k0 x
k0 x (b) ( ( x) has a maximum value at the origin x = 0. ( ( x0 ) = 0 when k0 x0 = " so x0 = "
. Thus the width of this
k0 2"
2"
. If k0 =
, wx = L. B( k ) versus k is graphed in Figure 39.69a. The graph of ( ( x) versus
L
k0
x is in Figure 39.69b. function wx = 2 x0 = (c) If k0 = π L wx = 2 L. hk
h
" hw # " 2" # hw
(d) wp wx = $ k % $ % = k = 0 = h. The uncertainty principle states that wp wx ≥ . For us, no matter what
2"
2" ' & k0 ' k0
k0
&
h
k0 is, w p wx = h, which is greater than
.
2" Figure 39.69
39.70. p 2 (h λ )2 n 2h 2
=
=
.
2m
2m
8mL2
m, E1 = 2.15 × 10−17 J = 134 eV. (a) For a standing wave, nλ = 2 L, and En =
(b) With L = a0 = 0.5292 × 10−10 The Wave Nature of Particles 39.71. Time of flight of the marble, from a freefall kinematic equation is just t = 2y
2(25.0 m)
=
= 2.26 s .
9.81 m s 2
g ht
" ∆p #
∆ x f = ∆ xi + ( ∆vx )t = ∆ xi + $ x % t =
+ ∆ xi
2π∆ xi m
&m'
d (∆ x f )
− ht
To minimize ∆ x f with respect to ∆ xi ,
=0=
+1
2"m( ∆ xi ) 2
d ( ∆ xi )
" ht #
! ∆xi (min) = $
%
& 2"m '
! ∆x f (min) = ht
ht
2ht
2(6.63 × 10−34 J ⋅ s)(2.26 s)
+
=
=
= 2.18 × 10−16 m = 2.18 × 10−7 nm.
2"m
2"m
"m
" (0.0200 kg) 3915 40 QUANTUM MECHANICS 40.1. IDENTIFY and SET UP: The energy levels for a particle in a box are given by En =
EXECUTE: (a) The lowest level is for n = 1, and E1 = n2h 2
.
8mL2 (1)(6.626 × 10−34 J ⋅ s) 2
= 1.2 × 10−67 J.
8(0.20 kg)(1.5 m)2 1
2E
2(1.2 × 10−67 J)
(b) E = mv 2 so v =
=
= 1.1 × 10−33 m/s. If the ball has this speed the time it would take it
2
m
0.20 kg
to travel from one side of the table to the other is t = 1.5 m
= 1.4 × 1033 s.
1.1 × 10−33 m/s h2
, E2 = 4 E1 , so ∆E = E2 − E1 = 3E1 = 3(1.2 × 10−67 J) = 3.6 × 10−67 J
8mL2
(d) EVALUATE: No, quantum mechanical effects are not important for the game of billiards. The discrete,
quantized nature of the energy levels is completely unobservable.
h
L=
8mE1
(c) E1 = 40.2. L= 40.3. 8(1.673 × 10 −27 kg)(5.0 × 10 eV)(1.602 × 10
6 −19 J eV ) = 6.4 × 10−15 m. IDENTIFY: An electron in the lowest energy state in this box must have the same energy as it would in the ground
state of hydrogen.
nh 2
SET UP: The energy of the nth level of an electron in a box is En =
.
8mL2
EXECUTE: An electron in the ground state of hydrogen has an energy of −13.6 eV, so find the width
corresponding to an energy of E1 = 13.6 eV. Solving for L gives
L= 40.4. (6.626 × 10−34 J ⋅ s) h
(6.626 × 10−34 J ⋅ s)
=
= 1.66 × 10−10 m.
8mE1
8(9.11 × 10−31 kg)(13.6 eV)(1.602 × 10−19 J eV ) EVALUATE: This width is of the same order of magnitude as the diameter of a Bohr atom with the electron in the
K shell.
(a) The energy of the given photon is
E = hf = h c λ = (6.63 × 10−34 J ⋅ s) (3.00 × 103 m/s)
= 1.63 × 10−18 J.
(122 × 10−9 m) The energy levels of a particle in a box are given by Eq.40.9
h2
h 2 ( n12 − n2 )
(6.63 × 10−34 J ⋅ s) 2 (22 − 12 )
( n 2 − n2 ). L =
=
= 3.33 × 10−10 m.
2
8mL
8m∆ E
8(9.11 × 10−31 kg)(1.63 × 10−20 J)
(b) The ground state energy for an electron in a box of the calculated dimensions is
h2
(6.63 × 10−34 J ⋅ s) 2
E=
=
= 5.43 × 10−19 J = 3.40 eV (onethird of the original photon energy),
2
8mL 8(9.11 × 10−31 kg)(3.33 × 10−10 m)2
which does not correspond to the −13.6 eV ground state energy of the hydrogen atom. Note that the energy levels for
a particle in a box are proportional to n 2 , whereas the energy levels for the hydrogen atom are proportional to − 12 .
n ∆E = 401 402 40.5. Chapter 40 IDENTIFY and SET UP: Eq.(40.9) gives the energy levels. Use this to obtain an expression for E2 − E1 and use the
value given for this energy difference to solve for L.
4h 2
h2
. The energy separation
EXECUTE: Ground state energy is E1 =
; first excited state energy is E2 =
2
8mL
8mL2
3h2
3
=
. This gives L = h
between these two levels is ∆ E = E2 − E1 =
8m∆ E
8mL2
3
= 6.1 × 10−10 m = 0.61 nm.
8(9.109 × 10 −31 kg)(3.0 eV)(1.602 × 10−19 J/1 eV)
EVALUATE: This energy difference is typical for an atom and L is comparable to the size of an atom.
(a) The wave function for n = 1 vanishes only at x = 0 and x = L in the range 0 ≤ x ≤ L.
(b) In the range for x, the sine term is a maximum only at the middle of the box, x = L / 2.
(c) The answers to parts (a) and (b) are consistent with the figure.
IDENTIFY and SET UP: For the n = 2 first excited state the normalized wave function is given by Eq.(40.13).
2 2 " 2π x #
2
2
2
" 2π x #
ψ 2 ( x) =
sin $
% . ψ 2 ( x) dx = sin $
% dx. Examine ψ 2 ( x) dx and find where it is zero and where it is
L
L'
L
L'
&
&
maximum.
2
" 2π x #
EXECUTE: (a) ψ 2 dx = 0 implies sin $
%=0
&L'
2π x
= mπ , m = 0, 1, 2, . . . ; x = m( L/2)
L
For m = 0, x = 0; for m = 1, x = L/2; for m = 2, x = L
The probability of finding the particle is zero at x = 0, L/2, and L.
2
" 2π x #
(b) ψ 2 dx is maximum when sin $
% = ±1
&L'
2π x
= m(π /2), m = 1, 3, 5, . . . ; x = m( L/4)
L
For m = 1, x = L/4; for m = 3, x = 3L/4
The probability of finding the particle is largest at x = L/ 4 and 3L/4.
L = 6.626 × 10 −34 J ⋅ s 40.6. 40.7. (c) EVALUATE: The answers to part (a) correspond to the zeros of ψ answers to part (b) correspond to the two values of x where ψ 2 2 shown in Fig.40.5 in the textbook and the in the figure is maximum. dψ
8" 2 m
2m
= − k 2ψ , and for ψ to be a solution of Eq.(40.3), k 2 = E 2 = E 2 .
2
dx
h
%
(b) The wave function must vanish at the rigid walls; the given function will vanish at x = 0 for any k , but to
vanish at x = L, kL = n" for integer n.
2 40.8. 40.9. (a) IDENTIFY and SET UP:
satisfied.
EXECUTE: Eq.(40.3): − ψ = A cos kx. Calculate dψ 2 /dx 2 and substitute into Eq.(40.3) to see if this equation is h2 d 2ψ
= Eψ
8π 2 m dx 2 dψ
= A( −k sin kx) = − Ak sin kx
dx
2
dψ
= − Ak ( k cos kx) = − Ak 2 cos kx
dx 2
h2
Thus Eq.(40.3) requires − 2 (− Ak 2 cos kx ) = E ( A cos kx).
8π m
22
hk
2mE
2mE
=
This says − 2 = E ; k =
8π m
( h/2π )
%
2mE
.
%
(b) EVALUATE: The wave function for a particle in a box with rigid walls at x = 0 and x = L must satisfy the
boundary conditions ψ = 0 at x = 0 and ψ = 0 at x = L. ψ (0) = A cos0 = A, since cos 0 = 1. Thus ψ is not 0 at
x = 0 and this wave function isn't acceptable because it doesn't satisfy the required boundary condition, even
though it is a solution to the Schrödinger equation. ψ = A cos kx is a solution to Eq.(40.3) if k = Quantum Mechanics 40.10. (a) The third excited state is n = 4, so ∆ E = (42 − 1)
(b) λ =
40.11. 403 hc (6.63 × 10−34 J ⋅ s)(3.0 × 108 m/s)
=
= 3.44 nm
∆E
5.78 × 10−17 J Recall λ =
(a) E1 = h
h
=
.
p
2mE h2
h
! λ1 =
= 2 L = 2(3.0 × 10−10 m) = 6.0 × 10−10 m. The wavelength is twice the width of
2
8mL
2mh 2 /8mL2 the box. p1 =
(b) E2 = p2 = h λ2 h2
15(6.626 × 10−34 J ⋅ s) 2
=
= 5.78 × 10−17 J = 361 eV.
2
8mL 8(9.11× 10−31 kg)(0.125 × 10−9 m)2 h λ1 = (6.63 × 10−34 J ⋅ s)
= 1.1 × 10−24 kg ⋅ m/s
6.0 × 10−10 m 2 4h
! λ2 = L = 3.0 × 10 −10 m. The wavelength is the same as the width of the box.
8mL2
= 2 p1 = 2.2 × 10−24 kg ⋅ m/s. 9h 2
2
! λ3 = L = 2.0 × 10−10 m. The wavelength is twothirds the width of the box.
8mL2
3
−24
p3 = 3 p1 = 3.3 × 10 kg ⋅ m/s.
IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we
substitute that wave function into the Schrödinger equation. (c) E3 =
40.12. SET UP: We must substitute the equation Ψ ( x, t ) = equation − 2
" nπ x # −iEn t/%
sin $
into the onedimensional Schrödinger
%e
L
&L' % 2 d 2ψ ( x)
+ U ( x)ψ ( x) = Eψ ( x).
2m dx 2
2 EXECUTE: Taking the second derivative of Ψ ( x, t ) with respect to x gives d 2 Ψ (x, t )
" nπ #
= −$
% Ψ (x, t )
dx 2
&L'
2 Substituting this result into − % 2 d 2ψ (x)
% 2 " nπ #
+ U (x)ψ (x) = Eψ (x), we get
$
% Ψ (x, t ) = E Ψ (x, t ) which
2m dx 2
2m & L ' 2 % 2 " nπ #
$
% , the energies of a particle in a box.
2m & L '
EVALUATE: Since this process gives us the energies of a particle in a box, the given wave function is a solution
to the Schrödinger equation.
−% 2 d 2(
(a) Eq.(40.1):
+ U( = E(.
2m dx 2
" % 2k 2
#
−% 2 d 2
% 2k 2
( A sin kx ) + U 0 A sin kx =
Lefthand side:
A sin kx + U 0 A sin kx = $
+ U 0 % (.
2
2m dx
2m
& 2m
'
gives En = 40.13. % 2k 2
% 2k 2
+ U 0 > U 0 > E for constant k . But
+ U 0 should equal E ! no solution.
2m
2m
% 2k 2
(b) If E > U 0 , then
+ U 0 = E is consistent and so ( = A sin kx is a solution of Eq.(40.1) for this case.
2m
According to Eq.(40.17), the wavelength of the electron inside of the square well is given by
h
2mE
k=
. By an analysis similar to that used to derive Eq.40.17, we can show that outside
! λin =
%
2m(3U 0 ) But 40.14. the box λout =
Thus, the ratio of the wavelengths is h
h
=
.
2 m( E − U 0 )
2m(2U 0 ) 2m(3U 0 )
λout
3
=
=
.
2
λin
2m(2U 0 ) 404 Chapter 40 40.15. E1 = 0.625E∞ = 0.625 " 2% 2
; E1 = 2.00 eV = 3.20 × 10−19 J
2mL2
1/ 2 40.16. "
#
0.625
−10
L = "% $
% = 3.43 × 10 m
2(9.109 × 10−31 kg)(3.20 × 10−19 J) '
&
Since U 0 = 6 E∞ we can use the result E1 = 0.625 E∞ from Section 40.3, so U 0 − E1 = 5.375E∞ and the maximum
wavelength of the photon would be λ=
λ=
40.17. Eq.(40.16): ( = Asin hc
hc
8mL2c
=
=
2
2
U 0 − E1 (5.375)(h /8mL ) (5.375)h
8(9.11 × 10−31 kg)(1.50 × 10−9 m)2 (3.00 × 108 m/s)
= 1.38 × 10−6 m.
(5.375)(6.63 × 10−34 J ⋅ s) 2mE
2mE
x + B cos
x
%
% d 2(
2mE
2mE
−2mE
" 2mE #
" 2mE #
= − A $ 2 % sin
x − B $ 2 % cos
x=
(( ) = Eq.(40.15).
2
dx
%
%
%2
&% '
&% '
40.18. 40.19. d(
d 2(
= κ (Ceκ x − De −κ x ),
= κ 2 (Ceκ x + De −κ x ) = κ 2( for all constants C and D. Hence ( is a solution to
dx
dx 2
%2 2
Eq.(40.1) for −
κ + U 0 = E , or κ = [2m(U 0 − E )]1/ 2 %, and κ is real for E < U 0 .
2m
IDENTIFY: Find the transition energy ∆ E and set it equal to the energy of the absorbed photon. Use E = hc/λ to
find the wavelength of the photon.
π 2% 2
SET UP: U 0 = 6 E∞ , as in Fig.40.8 in the textbook, so E1 = 0.625E∞ and E3 = 5.09E∞ with E∞ =
. In this
2mL2
problem the particle bound in the well is a proton, so m = 1.673 × 10 −27 kg.
EXECUTE: E∞ = π 2% 2
2 2mL = π 2 (1.055 × 10−34 J ⋅ s)2
2(1.673 × 10−27 kg)(4.0 × 10−15 m)2 = 2.052 × 10−12 J. The transition energy is ∆ E = E3 − E1 = (5.09 − 0.625) E∞ = 4.465E∞ . ∆ E = 4.465(2.052 × 10−12 J) = 9.162 × 10−12 J
The wavelength of the photon that is absorbed is related to the transition energy by ∆ E = hc/λ , so
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 2.2 × 10−14 m = 22 fm.
∆E
9.162 × 10−12 J
EVALUATE: The wavelength of the photon is comparable to the size of the box.
IDENTIFY: The longest wavelength corresponds to the smallest energy change.
h2
SET UP: The ground level energy level of the infinite well is E∞ =
, and the energy of the photon must be
8mL2
equal to the energy difference between the two shells.
EXECUTE: The 400.0 nm photon must correspond to the n = 1 to n = 2 transition. Since U 0 = 6 E∞ , we have λ= 40.20. E2 = 2.43E∞ and E1 = 0.625E∞ . The energy of the photon is equal to the energy difference between the two levels,
and E∞ = h2
hc
1.805 h 2
, which gives Eγ = E2 − E1 !
= ( 2.43 − 0.625) E∞ =
8mL2
8mL2
λ (1.805) hλ
(1.805)(6.626 × 10−34 J ⋅ s)(4.00 × 10−7 m)
=
= 4.68 × 10 −10 m = 0.468 nm.
8mc
8(9.11 × 10 −31 kg)(3.00 × 108 m s)
EVALUATE: This width is approximately half that of a Bohr hydrogen atom.
E"
E # −2 L 2 m (U 0 − E )/% E
6.0 eV
=
and E − U 0 = 5 eV = 8.0 × 10−19 J.
T = 16 $1 −
.
%e
U0 & U0 '
U 0 11.0 eV Solving for L gives L = 40.21. " 6.0 eV # "
6.0 ev # −2(0.80×10−9 m)
(a) L = 0.80 × 10−9 m: T = 16 $
% $1 −
%e
11.0 eV ' & 11.0 eV '
&
(b) L = 0.40 × 10−9 m: T = 4.2 × 10−4. 2(9.11×10−31 kg)(8.0×10−19 J) /1.055×10−34 J ⋅s = 4.4 × 10−8 Quantum Mechanics 40.22. The transmission coefficient is T = 16 E"
E # −2
$1 −
%e
U0 & U0 ' 2 m (U 0 − E ) L/% 405 , with E = 5.0 eV, L = 0.60 × 10−9 m, and m = 9.11 × 10 −31 kg (a) U 0 = 7.0 eV ! T = 5.5 × 10 −4 .
(b) U 0 = 9.0 eV ! T = 1.8 × 10−5
(c) U 0 = 13.0 eV ! T = 1.1 × 10−7.
40.23. IDENTIFY and SET UP: Use Eq.(39.1), where K = p 2 / 2m and E = K + U . λ = h/p = h/ 2mK , so λ K is constant EXECUTE: λ1 K1 = λ2 K 2 ; λ1 and K1 are for x > L where K1 = 2U 0 and λ2 and K 2 are for 0 < x < L where
K2 = E − U0 = U 0 λ1
1
K2
U0
=
=
=
λ2
K1
2U 0
2
40.24. EVALUATE: When the particle is passing over the barrier its kinetic energy is less and its wavelength is larger.
IDENTIFY: The probability of tunneling depends on the energy of the particle and the width of the barrier.
E"
E#
SET UP: The probability of tunneling is approximately T = Ge−2κ L , where G = 16 $ 1 −
% and
U0 & U0 ' κ= 2 m (U 0 − E )
% EXECUTE: κ= . G = 16 E"
E#
50.0 eV " 50.0 eV #
$1 −
% = 16
$1 −
% = 3.27.
U0 & U0 '
70.0 eV & 70.0 eV ' 2m(U 0 − E )
2(1.67 × 10−27 kg)(70.0 eV − 50.0 eV)(1.60 × 10−19 J/eV)
=
= 9.8 × 1011 m −1
(6.63 × 10−34 J ⋅ s) 2"
% 1
1
" 3.27 #
−12
ln(G / T ) =
ln $
% = 3.6 × 10 m = 3.6 pm
2κ
2(9.8 × 1011 m −1 ) & 0.0030 '
If the proton were replaced with an electron, the electron’s mass is much smaller so L would be larger.
EVALUATE: An electron can tunnel through a much wider barrier than a proton of the same energy.
Solving T = Ge−2κ L for L gives L = 40.25. IDENTIFY and SET UP: The probability is T = Ae−2κ L , with A = 16 2m(U 0 − E )
E"
E#
.
$1 −
% and κ =
U0 & U0 '
% E = 32 eV, U 0 = 41 eV, L = 0.25 × 10 −9 m. Calculate T. EXECUTE: (a) A = 16 E"
E#
32 " 32 #
$1 −
% = 16 $ 1 − % = 2.741.
U0 & U0 '
41 & 41 ' 2m(U 0 − E )
%
2(9.109 × 10−31 kg)(41 eV − 32 eV)(1.602 × 10−19 J/eV)
= 1.536 × 1010 m −1
κ=
1.055 × 10−34 J ⋅ s κ= 10 −1 −9 T = Ae−2κ L = (2.741)e −2(1.536×10 m )(0.25×10 m) = 2.741e −7.68 = 0.0013
(b) The only change in the mass m, which appears in κ .
2m(U 0 − E )
κ=
%
2(1.673 × 10−27 kg)(41 eV − 32 eV)(1.602 × 10−19 J/eV)
= 6.584 × 1011 m −1
κ=
1.055 × 10−34 J ⋅ s
11
1
−9
Then T = Ae−2κ L = (2.741)e −2(6.584×10 m )(0.25×10 m) = 2.741e −392.2 = 10−143
EVALUATE: The more massive proton has a much smaller probability of tunneling than the electron does. 406 Chapter 40 40.26. T = Ge −2κ L with G = 16 −2
2m(U 0 − E )
E"
E#
E"
E#
, so T = 16 $ 1 −
$1 −
% and κ =
%e
U0 & U0 '
%
U0 & U0 ' 2 m (U 0 − E )
% L . (a) If U 0 = 30.0 × 106 eV, L = 2.0 × 10−15 m, m = 6.64 × 10−27 kg and
U 0 − E = 1.0 × 106 eV (E = 29.0 × 106 eV), T = 0.090. (b) If U 0 − E = 10.0 × 106 eV (E = 20.0 × 106 eV), T = 0.014.
40.27. IDENTIFY and SET UP: The energy levels are given by Eq.(40.26), where ω =
k′
110 N/m
=
= 21.0 rad/s
m
0.250 kg
The ground state energy is given by Eq.(40.26):
1
1
E0 = %ω = (1.055 × 10−34 J ⋅ s)(21.0 rad/s) = 1.11× 10−33 J(1 eV/1.602 × 10−19 J) = 6.93 × 10−15 eV
2
2
1#
1#
"
"
En = $ n + % %ω ; E( n +1) = $ n + 1 + % %ω
2'
2'
&
&
The energy separation between these adjacent levels is
∆E = En +1 − En = %ω = 2 E0 = 2(1.11× 10−33 J) = 2.22 × 10−33 J = 1.39 × 10−14 eV
EVALUATE: These energies are extremely small; quantum effects are not important for this oscillator.
d(
d 2(
= (4 x 2, 2 − 2,)( , and ( is a solution of Eq.(40.21) if
Let mk ′ 2% = , , and so
= −2 xδ ( and
dx
dx 2
%2
1
1
E = , = % k ′/m = %).
2
2
m
IDENTIFY: We can model the molecule as a harmonic oscillator. The energy of the photon is equal to the energy
difference between the two levels of the oscillator.
SET UP: The energy of a photon is Eγ = hf = hc/λ , and the energy levels of a harmonic oscillator are given by
EXECUTE: 40.28. 40.29. k′
.
m ω= 1 # k′ "
1#
"
En = $ n + % %
= $ n + % %ω .
2' m &
2'
&
(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
= 0.21 eV
λ
5.8 × 10−6 m
k′
2" %c
k′
(b) The transition energy is ∆ E = En +1 − En = %ω = %
, which gives
. Solving for k ′, we get
=%
λ
m
m
4" 2c 2 m 4" 2 (3.00 × 108 m s )2 (5.6 × 10−26 kg)
=
= 5,900 N/m.
k′ =
λ2
(5.8 × 10−6 m) 2
EVALUATE: This would be a rather strong spring in the physics lab.
According to Eq.(40.26), the energy released during the transition between two adjacent levels is twice the ground
state energy E3 − E2 = %) = 2 E0 = 11.2 eV.
For a photon of energy E
c hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m s )
E = hf ! λ = =
=
= 111 nm.
f
E
(11.2 eV)(1.60 × 10−19 J/eV)
EXECUTE: (a) The photon’s energy is Eγ = 40.30. 40.31. hc = IDENTIFY and SET UP: Use the energies given in Eq.(40.26) to solve for the amplitude A and maximum speed
vmax of the oscillator. Use these to estimate ∆ x and ∆px and compute the uncertainty product ∆ x∆px .
EXECUTE: The total energy of a Newtonian oscillator is given by E = 1 k ′A2 where k ′ is the force constant and A
2 is the amplitude of the oscillator. Set this equal to the energy E = (n + 1 ) %ω of an excited level that has quantum
2 number n, where ω = k′
, and solve for A:
m 1
2 k ′A2 = ( n + 1 ) %ω
2 (2n + 1)%ω
k′
2
The total energy of the Newtonian oscillator can also be written as E = 1 mvmax . Set this equal to E = (n + 1 ) %ω and
2
2
A= solve for vmax :
vmax = 1
2 2
mvmax = ( n + 1 ) %ω
2 (2n + 1)%ω
m Quantum Mechanics 407 Thus the maximum linear momentum of the oscillator is pmax = mvmax = (2n + 1)%mω . Assume that A represents
the uncertainty ∆ x in position and that pmax is the corresponding uncertainty ∆ px in momentum. Then the
( 2n + 1)%ω
m
"1#
= (2n + 1)%ω $ % = (2n + 1)%.
(2n + 1)%mω = (2n + 1)%ω
k′
k′
&ω '
EVALUATE: For n = 1 this gives ∆ x∆ px = 3%, in agreement with the result derived in Section 40.4. The
uncertainty product ∆ x∆ px increases with n.
uncertainty product is ∆ x∆ px = 40.32. ( ( A) 2 ( (2 A) 2 "
#
mk ′
)#
"
= exp $ −
(2 A) 2 % = exp $ − mk ′ 4 % = e −4 = 1.83 × 10−2.
$
%
%
k′ '
&
( (0)
&
'
This figure cannot be read this precisely, but the qualitative decrease in amplitude with distance is clear.
IDENTIFY: We model the atomic vibration in the crystal as a harmonic oscillator.
1 # k′ "
1#
"
SET UP: The energy levels of a harmonic oscillator are given by En = $ n + % %
= $ n + % %ω .
2' m &
2'
&
EXECUTE: (a) The ground state energy of a simple harmonic oscillator is
(b) 40.33. 2 "
mk ′ 2 #
)#
"
A % = exp $ − mk ′ % = e −1 = 0.368.
= exp $ −
$
%
%
k′ '
( (0)
&
&
'
This is consistent with what is shown in Figure 40.20 in the textbook.
(a) 2 1
1 k ′ (1.055 × 10−34 J ⋅ s)
12.2 N/m
E0 = %ω = %
=
= 9.43 × 10−22 J = 5.89 × 10−3 eV
2
2m
2
3.82 × 10−26 kg
(b) E4 − E3 = %ω = 2 E0 = 0.0118 eV, so λ = 40.34. hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
= 106 µ m
E
1.88 × 10−21 J (c) En +1 − En = % = 2 E0 = 0.0118 eV
EVALUATE: These energy differences are much smaller than those due to electron transitions in the hydrogen
atom.
IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we
substitute that wave function into the Schrödinger equation.
SET UP: The given function is ψ ( x) = Aeikx , and the onedimensional Schrödinger equation is − % d 2ψ ( x)
+ U ( x)ψ ( x) = Eψ ( x).
2m dx 2 EXECUTE: Start with the given function and take the indicated derivatives: ψ ( x) = Aeikx . dψ ( x )
= Aikeikx .
dx d 2ψ ( x)
d 2ψ ( x)
% d 2ψ ( x) % 2 2
= Ai 2k 2eikx = − Ak 2eikx .
= − k 2ψ ( x). −
=
k ψ ( x). Substituting these results into the
2
2
dx
dx
2m dx 2
2m
% 2k 2
ψ ( x) + U 0ψ ( x) = E ψ ( x).
onedimensional Schrödinger equation gives
2m
% 2k 2
EVALUATE: ψ ( x ) = A eikx is a solution to the onedimensional Schrödinger equation if E − U 0 =
or
2m
2 m( E − U 0 )
. (Since U 0 < E was given, k is the square root of a positive quantity.) In terms of the particle’s
%2
momentum p: k = p/%, and in terms of the particle’s de Broglie wavelength λ : k = 2π /λ .
k= 40.35. IDENTIFY: Let I refer to the region x < 0 and let II refer to the region x > 0, so ψ I ( x) = Aeik1 x + Be− ik1 x and ψ II ( x) = Ceik x . Set ψ I (0) = ψ II (0) and
2 SET UP: dψ I dψ II
=
at x = 0.
dx
dx d ikx
(e ) = ikeikx .
dx dψ I dψ II
=
at x = 0 gives ik1 A − ik1B = ik2C. Solving this pair of
dx
dx
"k −k #
" 2k 2 #
equations for B and C gives B = $ 1 2 % A and C = $
% A.
& k1 + k2 '
& k1 + k2 '
EXECUTE: ψ I (0) = ψ II (0) gives A + B = C. 408 Chapter 40 EVALUATE: The probability of reflection is R = B 2 (k1 − k2 ) 2
=
. The probability of transmission is
A2 ( k1 + k2 ) 2 C2
4k12
=
. Note that R + T = 1.
A2 ( k1 + k2 ) 2
(n + 1) 2 − n 2 2n + 1 2 1
(a) Rn =
=
= + 2 . This is never larger than it is for n = 1, and R1 = 3.
n2
n2
nn
(b) R approaches zero; in the classical limit, there is no quantization, and the spacing of successive levels is
vanishingly small compared to the energy levels.
n2h 2
IDENTIFY and SET UP: The energy levels are given by Eq.(40.9): En =
. Calculate ∆ E for the transition
8mL2
and set ∆ E = hc/λ , the energy of the photon.
T= 40.36. 40.37. EXECUTE: (a) Ground level, n = 1, E1 = First excited level, n = 2, E2 = 4h 2
8mL2 The transition energy is ∆E = E2 − E1 = 3h 2
. Set the transition energy equal to the energy hc/λ of the emitted
8mL2 3h 2
.
λ 8mL2
2
−31
8mcL 8(9.109 × 10 kg)(2.998 × 108 m/s)(4.18 × 10−9 m)2
λ=
=
3h
3(6.626 × 10−34 J ⋅ s)
−5
λ = 1.92 × 10 m = 19.2 µ m.
9h 2
9h 2
4h 2
5h 2
(b) Second excited level has n = 3 and E3 =
. The transition energy is ∆ E = E3 − E2 =
−
=
.
2
2
2
8mL
8mL 8mL 8mL2
hc 5h 2
8mcL2 3
=
so λ =
= (19.2 µ m) = 11.5 µ m.
2
5h
5
λ 8mL
EVALUATE: The energy spacing between adjacent levels increases with n, and this corresponds to a shorter
wavelength and more energetic photon in part (b) than in part (a).
L4
2 L/4
"x
2 L/ 4 1 "
2"x #
1"
L
2 "x #
11
(a) 5
sin 2 dx = 5
sin
$1 − cos
% dx = $ x −
% = − , which is about 0.0908.
0
0
L
L
L
2&
L'
L&
2π
L '0
4 2"
photon. This gives 40.38. h2
8mL2 hc = L2 (b) Repeating with limits of L 4 and L 2 gives 1"
L
2"x #
11
,
$ x − sin
%=+
L ' L 4 4 2π
L&
2" about 0.409.
(c) The particle is much likely to be nearer the middle of the box than the edge.
(d) The results sum to exactly 1/2, which means that the particle is as likely to be between x = 0 and L 2 as it is to
be between x = L 2 and x = L.
(e) These results are represented in Figure 40.5b in the textbook.
40.39. IDENTIFY: The probability of the particle being between x1 and x2 is 5 x2
x1 2 ψ dx, where ψ is the normalized wave function for the particle.
2
"πx#
sin $
%.
L
&L'
EXECUTE: The probability P of the particle being between x = L / 4 and x = 3L / 4 is
3L / 4
2 3L / 4 2 " π x #
2
P=5
ψ 1 dx = 5
sin $
% dx. Let y = π x / L; dx = ( L / π ) dy and the integration limits become π / 4 and
L/4
L L/4
&L'
3π / 4.
3π / 4
2 " L # 3π / 4 2
2 /1
1
0
P = $ %5
sin y dy = 1 y − sin 2 y 2
L & π ' π /4
4
π 32
4 π /4
(a) SET UP: The normalized wave function for the ground state is ψ 1 = 2 / 3π π 1 " 3π # 1 " π # 0
− − sin
+ sin
π 1 8 8 4 $ 2 % 4 $ 2 %2
&'
& '4
3
2 "π 1
1#11
1
1
P = $ − ( −1) + (1) % = + = 0.818. (Note: The integral formula 5 sin 2 y dy = y − sin 2 y was used.)
4'2π
2
4
π&4 4
P= Quantum Mechanics (b) SET UP: The normalized wave function for the first excited state is ψ 2 =
EXECUTE: P=5 3L / 4
L/4 2 ψ 2 dx = 409 2
" 2π x #
sin $
%
L
&L' 2 3 L / 4 2 " 2π x #
sin $
% dx. Let y = 2π x / L; dx = ( L / 2π ) dy and the integration limits
L 5L/ 4
&L' become π / 2 and 3π / 2.
3π / 2
2 " L # 3π / 2 2
1 /1
1
1 " 3π π #
0
P= $
sin y dy = 1 y − sin 2 y 2
=$
− % = 0.500
%
L & 2π ' 5 π / 2
4
π 32
4 π /2 π & 4 4 '
(c) EVALUATE: These results are consistent with Fig.40.4b in the textbook. That figure shows that ψ 2 is more concentrated near the center of the box for the ground state than for the first excited state; this is consistent with the
answer to part (a) being larger than the answer to part (b). Also, this figure shows that for the first excited state half
the area under ψ
40.40. 2 curve lies between L/4 and 3L/4, consistent with our answer to part (b). Using the normalized wave function (1 = 2 L sin("x L) , the probabilities  ( 2 dx are
(a) (2 L) sin 2 ( " 4)dx = dx / L
(b) (2 L) sin 2 (π / 2) dx = 2dx / L
(c) (2 L )sin 2 (3" 4) = dx L . 40.41. IDENTIFY and SET UP: The normalized wave function for the n = 2 first excited level is ψ 2 = 2
" 2π x #
sin $
%.
L
&L' 2 P = ψ ( x) dx is the probability that the particle will be found in the interval x to x + dx.
EXECUTE: (a) x = L/4 2
" " 2π #" L # #
sin $ $
%$ % % =
L
& & L '& 4 ' '
P = ( 2/L) dx
(b) x = L/2 2
"π #
sin $ % =
L
&2' 2
" " 2π #" L # #
sin $ $
%$ % % =
L
& & L '& 2 ' ' 2
sin(π ) = 0
L ψ ( x) = ψ ( x) = 2
.
L P=0
(c) x = 3L/4
2
" " 2π #" 3L # #
sin $ $
%$ % % =
L
& & L '& 4 ' '
P = ( 2/L) dx ψ ( x) = 2
2
" 3π #
sin $ % = −
.
L
L
&2' EVALUATE: Our results are consistent with the n = 2 part of Fig.40.5 in the textbook. ψ
40.42. 40.43. 2 is zero at the center of the box and is symmetric about this point.
#
#
hn ˆ
##
#
%n" hn
∆ p = pfinal − pinitial . p = %k =
=
. At x = 0 the initial momentum at the wall is pinitial = − i and the final
L
2L
2L
#
#
hn ˆ
hn ˆ " hn ˆ #
hn ˆ
momentum, after turning around, is pfinal = + i . So, ∆p = + i − $ − i % = + i . At x = L the initial
2L
2L & 2L '
L
#
#
hn ˆ
hn ˆ
momentum is pinitial = + i and the final momentum, after turning around, is pfinal = − i . So,
2L
2L
#
hn ˆ hn ˆ
hn ˆ
∆p=− i −
i =− i
2L
2L
L
d 2( ( x )
2m
&&
(a) For a free particle, U ( x) = 0 so Schrodinger's equation becomes
= − 2 E( ( x ). The graph is given in
2
dx
h
Figure 40.43.
d( ( x)
d 2( ( x )
2m
% 2κ 2
.
(b) For x < 0: ( ( x) = e +κ x .
= κ e +κ x .
= κ 2e +κ x . So κ 2 = − 2 E ! E = −
dx
dx
2m
%
d( ( x)
d 2( ( x)
(c) For x > 0: ( ( x ) = e −κ x .
= − ke −κ x .
= κ 2e−κ x
dx
dx 4010 Chapter 40 So again κ 2 = − 2m
−% 2κ 2
&&
E!E=
. Parts (c) and (d) show ( ( x) satisfies the Schrodinger's equation, provided
2m
%2 −% 2κ 2
.
2m
d( ( x)
(d) Note
is discontinuous at x = 0. (That is, negative for x > 0 and positive for x < 0.)
dx E= 40.44. Figure 40.43
IDENTIFY: We start with the penetration distance formula given in the problem.
%
SET UP: The given formula is η =
.
2m(U 0 − E )
EXECUTE: (a) Substitute the given numbers into the formula:
%
1.055 × 10−34 J ⋅ s
η=
=
= 7.4 × 10 −11 m
2m (U 0 − E )
2(9.11 × 10 −31 kg)(20 eV − 13 eV)(1.602 × 10−19 J/eV) (b) η =
40.45. 1.055 × 10−34 J ⋅ s
2(1.67 × 10 −27 kg)(30 MeV − 20 MeV)(1.602 × 10 −13 J/MeV) = 1.44 × 10−15 m EVALUATE: The penetration depth varies widely depending on the mass and energy of the particle.
(a) We set the solutions for inside and outside the well equal to each other at the well boundaries, x = 0 and L.
x = 0 : A sin(0) + B = C ! B = C , since we must have D = 0 for x < 0. 2mEL
2mEL
+ B cos
= + De−κ L since C = 0 for x > L.
%
%
2mE
.
This gives A sin kL + B cos kL = De−κ L , where k =
%
(b) Requiring continuous derivatives at the boundaries yields
d(
x = 0:
= kA cos(k ⋅ 0) − kB sin(k ⋅ 0) = kA = κ Cek ⋅0 ! kA = κ C
dx
x = L: kA cos kL − kB sin kL = −κ De−κ L .
2m(U 0 − E )
1 "T #
E"
E#
T = Ge−2κ L with G = 16 $1 −
! L = − ln $ % .
% and κ =
%
2κ & G '
U0 & U0 '
x = L: A sin 40.46. If E = 5.5 eV, U 0 = 10.0 eV, m = 9.11 × 10−31 kg, and T = 0.0010. Then = 2(9.11× 10−31 kg)(4.5 eV)(1.60 × 10−19 J eV)
5.5 eV "
5.5 eV #
= 1.09 × 1010 m −1 and G = 16
$1 −
% = 3.96
−34
10.0 eV & 10.0 eV '
(1.054 × 10 J ⋅ s) so L = −
40.47. 1
" 0.0010 #
−10
ln $
% = 3.8 × 10 m = 0.38 nm.
2(1.09 × 1010 m −1 ) & 3.96 ' IDENTIFY and SET UP: When κ L is large, then eκ L is large and e −κ L is small. When κ L is small,
sinh κ L → κ L. Consider both κ L large and κ L small limits. / (U sinh κ L)2 0
EXECUTE: (a) T = 11 + 0
2
4E (U 0 − E ) 4
3 −1 Quantum Mechanics sinh κ L = e κL −e
2 4011 −κ L −1 2
/
eκ L
U 0 e 2κ L 0
16 E (U 0 − E )
and T → 11 +
2=
2
2
16 E (U 0 − E ) 4
16 E (U 0 − E ) + U 0 e 2κ L
3
2
2
For κ L W1, 16 E (U 0 − E ) + U 0 e2κ L → U 0 e 2κ L For κ L W1, sinhκ L → T→ " E #"
16 E (U 0 − E )
E # − 2κ L
= 16 $ %$ 1 −
% e , which is Eq.(40.21).
U 02e 2κ L
U 0 '& U 0 '
& L 2m(U 0 − E )
. So κ L W1 when L is large (barrier is wide) or U 0 − E is large. (E is small compared to U 0 . )
%
2m(U 0 − E )
; κ becomes small as E approaches U 0 . For κ small, sinh κ L → κ L and
(c) κ =
%
−1
−1
2
/
/ U 2 2m(U 0 − E ) L2 0
U 0 κ 2 L2 0
= 11 + 0 2
T → 11 +
2
2 (using the definition of κ )
% 4 E (U 0 − E ) 4
3 4 E (U 0 − E ) 4
3
(b) κ L = / 2U 2 L2 m 0
Thus T → 11 + 0 2 2
4 E% 4
3
U 0 → E so −1 2
/ 2 EL2 m 0
U0
→ E and T → 11 +
2
E
4% 2 4
3 −1 −1 / " kL # 2 0
2mE
But k = 2 , so T → 11 + $ % 2 , as was to be shown.
%
1 &2'2
3
4
EVALUATE: When κ L is large Eq.(40.20) applies and T is small. When E → U 0 , T does not approach unity.
1
(a) E = mv 2 = (n + (1 2))%) = ( n + (1 2))hf , and solving for n,
2
12
mv 1 (1/2)(0.020 kg)(0.360 m/s)2 1
n= 2
−=
− = 1.3 × 1030.
hf
2 (6.63 × 10−34 J ⋅ s)(1.50 Hz) 2
2 40.48. 40.49. 40.50. (b) The difference between energies is %) = hf = (6.63 × 10−34 J ⋅ s)(1.50 Hz) = 9.95 × 10−34 J. This energy is too
small to be detected with current technology
IDENTIFY and SET UP: Calculate the angular frequency ω of the pendulum and apply Eq.(40.26) for the energy levels.
2π
2π
EXECUTE: ω =
=
= 4π s −1
T
0.500 s
1
1
The groundstate energy is E0 = %ω = (1.055 × 10−34 J ⋅ s)(4π s −1 ) = 6.63 × 10−34 J.
2
2
E0 = 6.63 × 10−34 J(1 eV/1.602 × 10−19 J) = 4.14 × 10 −15 eV 1#
"
En = $ n + % %ω
2'
&
1#
"
En +1 = $ n + 1 + % %ω
2'
&
The energy difference between the adjacent energy levels is
∆ E = En +1 − En = %ω = 2 E0 = 1.33 × 10−33 J = 8.30 × 10−15 eV
EVALUATE: These energies are much too small to detect. Quantum effects are not important for ordinary size objects.
IDENTIFY: We model the electrons in the lattice as a particle in a box. The energy of the photon is equal to the
energy difference between the two energy states in the box.
n2h 2
SET UP: The energy of an electron in the nth level is En =
. We do not know the initial or final levels, but
8mL2
we do know they differ by 1. The energy of the photon, hc/λ , is equal to the energy difference between the two states.
hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
=
EXECUTE: The energy difference between the levels is ∆ E =
λ
1.649 × 10−7 m
−1 8
1.206 × 10 J. Using the formula for the energy levels in a box, this energy difference is equal to
h2
h2
∆ E = / n 2 − (n − 1) 2 0
3
4 8mL2 = (2n − 1) 8mL2 . 4012 40.51. Chapter 40 " ∆ E 8mL2 # 1 " (1.206 × 10−18 J)8(9.11× 10−31 kg)(0.500 × 10−9 m) 2 #
+ 1% = $
+ 1% = 3.
Solving for n gives n = $
2
(6.626 × 10−34 J ⋅ s) 2
&h
' 2&
'
The transition is from n = 3 to n = 2.
EVALUATE: We know the transition is not from the n = 4 to the n = 3 state because we let n be the higher state
and n − 1 the lower state.
IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we
substitute that wave function into the Schrödinger equation.
22
SET UP: The given wave function is ψ 0 ( x ) = A0 e −α x / 2 and the Schrödinger equation is
% d 2ψ ( x) k ′x 2
−
+
ψ ( x) = E ψ ( x).
2m dx 2
2
22
22
dψ 0 ( x )
EXECUTE: (a) Start by taking the derivatives: ψ 0 ( x ) = A0e −α x / 2 .
= −α 2 xA0 e −α x / 2 .
dx
2
d 2ψ 0 ( x)
−α 2 x 2 / 2 d ψ 0 ( x )
2 −α 2 x 2 / 2
22 2
2
22 2
.
= − A0α e
+ (α ) x A0 e
= [ −α + (α ) x ] ψ 0 ( x).
dx 2
dx 2
% d 2ψ 0 ( x)
%2
% d 2ψ ( x) k ′x 2
[ −α 2 + (α 2 ) 2 x 2 ] ψ 0 ( x). Equation (40.22) is −
+
−
=−
ψ ( x) = E ψ ( x). Substituting
2
2m dx
2m
2m dx 2
2
2
2
%
k ′x
mω
[ −α 2 + (α 2 ) 2 x 2 ] ψ 0 ( x) +
ψ 0 ( x ) = E ψ 0 ( x ). Since α 2 =
the above result into that equation gives −
and
%
2m
2
2
%2 2 2 k′
% 2 " mω # mω 2
k′
= 0.
(α ) + = −
ω=
, the coefficient of x2 is −
$
%+
2m
2
2m & % '
2
m
1/ 4 " mω #
(b) A0 = $
%
& %π ' %
2
. The probability density function ψ is
ωm
1/ 2
1/ 2
2
2
" mω #
" mω #
=$
e − mω x /% . At x = 0, ψ 0 = $
%
%.
& %π '
& %π ' (c) The classical turning points are at A = ±
2 ψ 0 ( x ) = A02e −α
2 40.52. 22 x 2 d ψ 0 ( x)
d ψ 0 ( x)
mω " mω #
" mω #
−α 2 x 2
−α 2 x 2
2
=$
= −2
= 0.
. At x = 0,
% (−α 2 x)e
$
% xe
dx
% & %π '
dx
& %π '
2
2
1/ 2
d 2 ψ 0 ( x)
d 2 ψ 0 ( x)
mω " mω #
−α 2 x 2
22
< 0. Therefore, at x = 0, the first derivative is
= −2
. At x = 0,
$
% [1 − 2α x ]e
2
2
dx
dx
% & %π '
zero and the second derivative is negative. Therefore, the probability density function has a maximum at x = 0.
22
%2
EVALUATE: ψ 0 ( x) = A0e−α x / 2 is a solution to equation (40.22) if −
(−α 2 )ψ 0 ( x ) = E ψ 0 ( x) or
2m
% 2α 2 %ω
%ω
E=
=
. E0 =
corresponds to n = 0 in Equation (40.26).
2
2m
2
IDENTIFY: If the given wave function is a solution to the Schrödinger equation, we will get an identity when we
substitute that wave function into the Schrödinger equation.
22
SET UP: The given wave function is ψ 1 ( x ) = A1 2 xe −α x / 2 and the Schrödinger equation is
% d 2ψ ( x ) k ′x 2
−
+
ψ ( x) = E ψ ( x).
2m dx 2
2
22
EXECUTE: (a) Start by taking the indicated derivatives: ψ 1 ( x ) = A1 2 xe−α x / 2 .
1/ 2 1/ 2 22
22
22
22
22
dψ 1 ( x )
d 2ψ 1 ( x)
= −2α 2 x 2 A1e −α x / 2 + 2 A1e −α x / 2 .
= −2 A1α 2 2 xe−α x / 2 − 2 A1α 2 x 2 (−α 2 x )e−α x / 2 + 2 A1 (−α 2 x)e−α x / 2 .
2
dx
dx
d 2ψ 1 ( x)
2
22 2
2
= [−2α + (α ) x − α ] ψ 1 ( x ) = [−3α 2 + (α 2 )2 x 2 ] ψ 1 ( x)
dx 2
% d 2ψ 1 ( x)
%2
[ −3α 2 + (α 2 ) 2 x 2 ] ψ 1 ( x )
−
=−
2m dx 2
2m Quantum Mechanics 4013 % d ψ ( x) k ′x
ψ ( x) = E ψ ( x). Substituting the above result into that equation gives
+
2m dx 2
2
mω
k ′x 2
%2
k′
[−3α 2 + (α 2 ) 2 x 2 ] ψ 1 ( x) +
ψ 1 ( x) = E ψ 1 ( x). Since α 2 =
and ω =
, the coefficient of x2 is
−
2m
2
%
m
2
%2 2 2 k′
% 2 " mω # mω 2
−
=0
(α ) + = −
$
%+
2m
2
2m & % '
2
2 Equation (40.22) is − 2 1/ 4 (b) A1 = 1 " mω #
$
%
2 & %π ' 1/ 2 2 2 (c) The probability density function ψ is ψ 1 ( x) = A12 4 x 2 e−α
d ψ 1 ( x) 2 At x = 0, ψ 1 = 0.
d ψ 1 ( x) At x = 0,
d 2 ψ 1 ( x)
2 = A12 8 xe−α 2 = 0. At x = ± 2 dx
2
d 2 ψ 1 ( x) 1 α , 22 x + A12 4 x 2 (−α 2 2 x )e −α d ψ 1 ( x) 1 " mω #
=$
%
2 & %π ' 4 x 2e − mω x 2
% 22 + A12 8 x( −α 2 2 x)e−α = A12 8e −α 22 − A1216 x 2α 2e −α 22 22 x 22 x = A12 8 xe −α 22 x − A12 8 x3α 2e −α 22 x 2 dx = A12 8e −α x x 2 dx dx 22 = 0. − A12 8(3 x 2 )α 2e −α − A12 24 x 2α 2e−α 22 22 x A12 8 x3α 2 ( −α 2 2 x)e−α 22 x 22 . + A1216 x 4 (α 2 ) 2 e −α x . At x = 0, d 2 ψ 1 ( x) 2 > 0. So at
dx 2
dx 2
x = 0, the first derivative is zero and the second derivative is positive. Therefore, the probability density function
x has a minimum at x = 0. At x = ± 1 α , x d 2 ψ 1 ( x)
dx 2 x 2 < 0. So at x = ± 1 α , the first derivative is zero and the second derivative is negative. Therefore, the probability density function has maxima at x = ±
classical turning points for n = 0 as found in the previous question.
EVALUATE: ψ 1 ( x ) = A1 2 xe −α 22 x /2 is a solution to equation (40.22) if − , corresponding to the %2
(−3α 2 )ψ 1 ( x) = E ψ 1 ( x) or
2m 3% 2α 2 3%ω
3%ω
. E1 =
=
corresponds to n = 1 in Equation (40.26).
2m
2
2
IDENTIFY and SET UP: Evaluate ∂ 2ψ / ∂x 2 , ∂ 2ψ / ∂y 2 , and ∂ 2ψ / ∂z 2 for the proposed ψ and put Eq.(40.29). Use
E= 40.53. 1 α that ψ nx , ψ ny , and ψ nz are each solutions to Eq.(40.22).
% 2 " ∂ 2ψ ∂ 2ψ ∂ 2ψ
+
+
$
2m & ∂x 2 ∂y 2 ∂z 2 #
% + Uψ = Eψ
'
2
% 2 d ψ nx 1 2
ψ nx , ψ ny , ψ nz are each solutions of Eq.(40.22), so −
+ k ′x ψ nx = Enxψ nx
2m dx 2
2
2
% 2 d ψ ny 1 2
−
+ k ′y ψ ny = Enyψ ny
2m dy 2
2
2
% 2 d ψ nz 1 2
−
+ k ′z ψ nz = Enzψ nz
2m dz 2
2
1
1
1
ψ = ψ nx ( x)ψ ny ( y )ψ nz ( z ), U = k ′x 2 + k ′y 2 + k ′z 2
2
2
2
2
2
2
" d 2ψ ny #
" d 2ψ nx #
∂ψ
∂ψ
∂ 2ψ " d ψ nz #
%ψ nxψ nz , 2 = $
ψψ .
=$
%ψ nyψ nz , 2 = $
$ dz 2 % nx ny
%
$ dy 2 %
∂x 2 $ dx 2 %
∂y
∂z
&
'
&
'
&
'
" % 2 d 2ψ nx 1 2 #
% 2 " ∂ 2ψ ∂ 2ψ ∂ 2ψ #
+ 2 + 2 % + Uψ = $ −
+ k ′x ψ nx %ψ nyψ nz
So −
$
$ 2m dx 2
%
∂z '
2m & ∂x 2 ∂y
2
&
' EXECUTE: (a) − " % 2 d 2ψ ny 1 2
+$ −
+ k ′y ψ ny
$ 2m dy 2
2
&
− % 2 " ∂ 2ψ ∂ 2ψ ∂ 2ψ
+
+
$
2m & ∂x 2 ∂y 2 ∂z 2 #
" % 2 d 2ψ nz 1 2 #
%ψ nxψ nz + $ −
+ k ′z ψ nz %ψ nxψ ny
$ 2m dz 2
%
%
2
&
'
' #
% + Uψ = ( Enx + Eny + Enz )ψ
' 4014 Chapter 40 Therefore, we have shown that this ψ is a solution to Eq.(40.29), with energy
3#
"
Enx ny nz = Enx + Eny + Enz = $ nx + n y + nz + % %ω
2'
&
3
(b) and (c) The ground state has nx = n y = nz = 0, so the energy is E000 = %ω. There is only one set of nx , n y and
2
nz that give this energy.
5
Firstexcited state: nx = 1, ny = nz = 0 or ny = 1, nx = nz = 0 or nz = 1, nx = ny = 0 and E100 = E010 = E001 = %ω
2
There are three different sets of nx , n y , nz quantum numbers that give this energy, so there are three different 40.54. quantum states that have this same energy.
EVALUATE: For the threedimensional isotropic harmonic oscillator, the wave function is a product of onedimensional harmonic oscillator wavefunctions for each dimension. The energy is a sum of energies for three onedimensional oscillators. All the excited states are degenerate, with more than one state having the same energy.
1#
"
′
ω1 = k1′ m , )2 = k2 m . Let (nx ( x) be a solution of Eq.(40.22) with Enx = $ nx + % %)1 , ( nx ( y ) be a similar
2'
&
solution, ( nz ( z ) be a solution of Eq.(40.22) but with z as the independent variable instead of x, and
1#
"
energy Enz = $ nz + % )2.
2'
&
(a) As in Problem 40.53, look for a solution of the form ( ( x, y , z ) = ( nx ( x )( ny ( y )( nz ( z ). Then,
% 2 ∂ 2( "
1
∂ 2(
∂ 2(
′#
= $ Enx − k1x 2 % ( with similar relations for
and 2 . Adding,
2
2
2m ∂x
2
∂y
∂z
&
'
2
" ∂ 2( ∂ 2( ∂ 2( # "
%
1 2 1 2 1 2#
′
−
+
+
$
% = $ En + Eny + Enz − k1′x − k1′ y − k2 z % (
2m & ∂x 2 ∂y 2 ∂z 2 ' & x
2
2
2
' − = ( Enx + Eny + Enz − U )( = ( E − U )( 40.55. /
1 # 20
"
where the energy E is E = Enx + Eny + Enz = % 1( nx + n y + 1))12 + $ nz + % )2 2 , with nx , n y and nz all nonnegative
2' 4
&
3
integers.
1#
"
(b) The ground level corresponds to nx = n y = nz = 0, and E = % $ )2 + ) 2 % . The first excited level corresponds to
1
2
2'
&
3#
"
nx = ny = 0and nz = 1, since )12 > ) 2 , and E = %) $ ω 2 + ) 2 % . There is only one set of quantum numbers for both
2
1
2
2'
&
the ground state and the first excited state.
(a) ( ( x ) = A sin kx and ( (− L 2) = 0 = ( (+ L 2)
2n" 2"
" + kL # + kL
! 0 = A sin $
= n" ! k =
=
%!
2
L
λ
&2' L
h nh
p 2 n 2 h 2 (2n) 2 h 2
! pn =
=
! En = n =
=
, where n = 1, 2...
λn L
n
2m 2mL2
8mL2
(b) ( ( x ) = A cos kx and ( ( − L / 2) = 0 = ψ ( + L / 2)
!λ = "
(2n + 1)" 2"
" kL # kL
! 0 = A cos $ % !
= (2n + 1) ! k =
=
2'
2
2
L
λ
&
2L
(2n + 1) h
!λ =
! pn =
(2n + 1)
2L
(2n + 1) 2 h 2
n = 0, 1, 2...
8mL2
(c) The combination of all the energies in parts (a) and (b) is the same energy levels as given in Eq.(40.9), where
n2 h 2
En =
.
8mL2
(d) Part (a)’s wave functions are odd, and part (b)’s are even.
(a) As with the particle in a box, ( ( x) = A sin kx, where A is a constant and k 2 = 2mE % 2 . Unlike the particle in a
box, however, k and hence E do not have simple forms.
! En = 40.56. Quantum Mechanics 4015 (b) For x > L, the wave function must have the form of Eq.(40.18). For the wave function to remain finite as
x → ∞, C = 0. The constant  2 = 2m (U 0 − E ) %, as in Eq.(14.17) and Eq.(40.18). (c) At x = L, A sin kL = De − L and kA cos kL = − De − L . Dividing the second of these by the first gives
k cot kL = −  , a transcendental equation that must be solved numerically for different values of the length L and the ratio E U 0 .
40.57. p2
h
h
+ U ( x) ! p = 2m( E − U ( x)). λ = ! λ ( x) =
.
2m
p
2m( E − U ( x))
(b) As U ( x ) gets larger (i.e., U ( x ) approaches E from below—recall k ≥ 0), E − U ( x)
gets smaller, so λ ( x) gets larger.
(c) When E = U ( x), E − U ( x) = 0, so λ ( x ) → ∞.
b dx
b
b
dx
1b
n
hn
2m( E − U ( x )) dx = ! 5 2m( E − U ( x)) dx = .
(d) 5
=5
=
a λ ( x)
a
a
2
2
h 2m( E − U ( x)) h 5 a
(e) U ( x ) = 0 for 0 < x < L with classical turning points at x = 0 and x = L. So,
(a) E = K + U ( x) = 5 b 2m( E − U ( x )) dx = 5 a L
0 L 2mEdx = 2mE 5 dx = 2mEL. So, from part (d),
0 2 22 hn
1 " hn #
hn
!E=
$
%=
2
2m & 2 L ' 8mL2 .
(f ) Since U ( x) = 0 in the region between the turning points at x = 0 and x = L, the results is the same as part (e).
2mE L = The height U 0 never enters the calculation. WKB is best used with smoothly varying potentials U ( x).
40.58. 12
2E
.
(a) At the turning points E = k ′xTP ! xTP = ±
2
k′
(b) + 2 E/k ′ 5 − 2 E/k ′ 1
nh
"
#
2m $ E − k ′x 2 % dx = . To evaluate the integral, we want to get it into a form that matches the
2
2
&
' standard integral given.
Letting A2 =
! mk ′ 5 40.59. b
a 1
2mE
2E
"
#
2m $ E − k ′x 2 % = 2mE − mk ′x 2 = mk ′
− x 2 = mk ′
− x2.
2
mk ′
k′
&
' 2E
2E
2E
,a=−
,and b = +
k′
k′
k′
" x #0
mk ′ /
2
2
2
A − x dx = 2
1 x A − x + A arcsin $ % 2
$ A%
21
& '2
3
4
2 b 2 0 / 2E 2E 2E 2E
" 2E k ′ #0
2E
m""#
arcsin $
arcsin (1) = 2 E
= mk ′ 1
−
+
= mk ′
$ %.
$ 2E k ′ %2
%2
k′ k′
k′
k′
k′
k′ & 2 '
1
&
'4
3
hn
m
hn
k′
h
" = . Recall ) =
, so E =
)n = %)n.
Using WKB, this is equal to , so E
2
k′
2
m
2"
%) "
1 ##
"
(c) We are missing the zeropointenergy offset of
$ recall E = %) $ n + % % . However, our approximation isn’t
2&
2 ''
&
bad at all!
E
(a) At the turning points E = A xTP ! xTP = ± .
A
(b) 5 +E / A
−E / A 2m( E − A x ) dx = 25 dy = −2mA dx when x =
25 E
A
0 2m( E − Ax)dx = − E/A
0 2m( E − Ax) dx. Let y = 2m( E − Ax) ! E
, y = 0, and when x = 0, y = 2mE. So
A
1 0 12
2 32
5 2mE y dy = − 3mA y
mA
23 2
hn
1 " 3mAh #
23
!E=
(2mE )3 2 =
$
%n.
3mA
2
2m & 4 ' 0 =
2 mE hn
2
. So,
(2mE )3 2 . Using WKB, this is equal to
2
3mA 4016 Chapter 40 (c) The difference in energy decreases between successive levels. For example:
12 3 − 02 3 = 1, 22 3 − 12 3 = 0.59, 33 2 − 23 2 = 0.49,... • A sharp ∞ step gave everincreasing level differences (~ n 2 ). • A parabola (~ x 2 ) gave evenly spaced levels (~n ). • Now, a linear potential (~ x ) gives everdecreasing level differences (~ n 2 3 ).
Roughly speaking, if the curvature of the potential (~ second derivative) is bigger than that of a parabola, then the
level differences will increase. If the curvature is less than a parabola, the differences will decrease. 41 ATOMIC STRUCTURE 41.1. IDENTIFY and SET UP: L = l (l + 1)% . Lz = ml % . l = 0, 1, 2,..., n − 1. ml = 0, ± 1, ± 2,..., ± l . cosθ = Lz / L . EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2% , Lz = %,0, −% . l = 2 : L = 6% , Lz = 2%, %,0, −%, −2% . 41.2. (b) In each case cosθ = Lz / L . L = 0 : θ not defined. L = 2% : 45.0°, 90.0°, 135.0° . L = 6% :
35.3°, 65.9°, 90.0°, 114.1°, 144.7° .
#
EVALUATE: There is no state where L is totally aligned along the z axis.
IDENTIFY and SET UP: L = l (l + 1)% . Lz = ml % . l = 0,1, 2,..., n − 1. ml = 0, ±1, ±2,..., ±l . cosθ = Lz / L .
EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2% , Lz = %,0, −% . l = 2 : L = 6% , Lz = 2%, %,0, −%, −2% . l = 3 :
L = 2 3% , Lz = 3%,2%, %,0, −%, −2%, −3% . l = 4 : L = 2 5% , Lz = 4%,3%,2%, %,0, −%, −2%, −3%, −4% .
(b) L = 0 : θ not defined. L = 2% : 45.0°,90.0°,135.0° . L = 6% : 35.3°,65.9°,90.0°,114.1°,144.7° . L = 2 3% : 41.3. 54.7°,73.2°,90.0°,106.8°,125.3°,150.0° . L = 2 5% : 26.6°,47.9°,63.4°,77.1°,90.0°,102.9°,116.6°,132.1°,153.4° .
(c) The minimum angle is 26.6° and occurs for l = 4 , ml = +4 . The maximum angle is 153.4° and occurs for
l = 4 , ml = −4 .
IDENTIFY and SET UP: The magnitude of the orbital angular momentum L is related to the quantum number l by
Eq.(41.4): L = l (l + 1)%, 1 = 0, 1, 2,…
2 −34
2
" L # " 4.716 × 10 kg ⋅ m /s #
l (l + 1) = $ % = $
% = 20
−34
& % ' & 1.055 × 10 J ⋅ s '
And then l (l + 1) = 20 gives that l = 4.
EVALUATE: l must be integer.
(a) ( ml ) max = 2, so (Lz ) max = 2%. EXECUTE: 41.4. (b) l (l + 1)% = 6% = 2.45%. "m #
"L #
(c) The angle is arccos $ z % = arccos $ l % , and the angles are, for ml = −2 to ml = 2, 144.7°,
L'
&
& 6'
114.1°, 90.0°, 65.9°, 35.3°. The angle corresponding to ml = l will always be larger for larger l .
41.5. IDENTIFY and SET UP: The angular momentum L is related to the quantum number l by Eq.(41.4), L = l (l + 1)%. The maximum l, lmax , for a given n is lmax = n − 1.
EXECUTE: For n = 2, lmax = 1 and L = 2% = 1.414%. For n = 20, lmax = 19 and L = (19)(20)% = 19.49%. 41.6. 41.7. For n = 200, lmax = 199 and L = (199)(200)% = 199.5%.
EVALUATE: As n increases, the maximum L gets closer to the value n% postulated in the Bohr model.
The (l , ml ) combinations are (0, 0), (1, 0), (1, ± 1) , (2, 0), ( 2, ± 1), ( 2, ± 2), (3, 0),
(3, ± 1), (3, ± 2), (3, ± 3), (4, 0), (4, ± 1), (4, ± 2), (4, ± 3), and (4, ± 4), a total of 25.
13.60 eV
= −0.544 eV.
(b) Each state has the same energy (n is the same), −
25
−19
2
1 q1q2
−1 (1.60 × 10 C)
=
= −2.3 × 10−18 J
U=
4"P0 r
4"P0 1.0 × 10−10 m
U= −2.3 × 10−18 J
= −14.4 eV.
1.60 × 10−19 J eV
411 412 41.8. Chapter 41 (a) As in Example 41.3, the probability is
a2 P=5 a/2
0  (1s 2 4"r 2 dr = 4 /" ar 2 a 2 r a 3 # −2 r a 0
5e −1
−
−
− %e
= 0.0803 .
2 =1−
3 1$
a 3& 2
2
4'
2
40 (b) The difference in the probabilities is (1 − 5e −2 ) − (1 − (5 2)e −1 ) = (5 2)(e−1 − 2e −2 ) = 0.243.
41.9. (a)  ( 2 = (*( = R (r ) 2  Θ(& ) 2 ( Ae− imlφ )( Ae + imlφ ) = A2  R( r ) 2  Θ(& ) 2 , which is independent of φ .
(b) 41.10. 5 2"
0 En = −  Φ (φ ) 2 dφ = A2 5 2"
0 dφ = 2"A2 = 1 ! A = 1
.
2" 1
mr e4
E1
∆ E12 = E2 − E1 = 2 − E1 = −(0.75) E1.
2
22
(4"P0 ) 2n %
2 (a) If mr = m = 9.11 × 10−31 kg
2
mr e 4
(9.109 × 10−31 kg)(1.602 × 10−19 C) 4
=
(8.988 ×109 N ⋅ m 2 C ) = 2.177 × 10−18 J = 13.59 eV
22
2
−34
(4"P0 ) %
2(1.055 × 10 J ⋅ s)
For 2 → 1 transition, the coefficient is (0.75)(13.59 eV) = 10.19 eV.
m
(b) If mr = , using the result from part (a),
2 mr e 4
" m 2 # " 13.59 eV #
= (13.59 eV) $
%=$
% = 6.795 eV.
(4"P0 ) 2 % 2
2
&m' &
'
" 10.19 eV #
Similarly, the 2 → 1 transition, ! $
% = 5.095 eV.
2
&
'
(c) If mr = 185.8m, using the result from part (a), mr e4
" 185.8m #
= (13.59 eV) $
% = 2525 eV,
22
(4"P0 ) %
&m'
41.11. and the 2 → 1 transition gives ! (10.19 eV)(185.8) = 1893 eV.
4π P0% 2
P h2
IDENTIFY and SET UP: Eq.(41.8) gives a =
= 0 2.
mr e 2
π mr e
EXECUTE: (a) mr = m a= P0 h 2
(8.854 × 10−12 C2 /N ⋅ m 2 )(6.626 × 10−34 J ⋅ s)2
=
= 0.5293 × 10−10 m
π mr e 2
π (9.109 × 10−31 kg)(1.602 × 10−19 C)2 (b) mr = m / 2 " P h2 #
a = 2 $ 0 2 % = 1.059 × 10−10 m
& π mr e '
(c) mr = 185.8m
a= 41.12.
41.13. 1 " P0 h 2 #
−13
$
% = 2.849 × 10 m
185.8 & π mr e 2 ' EVALUATE: a is the radius for the n = 1 level in the Bohr model. When the reduced mass mr increases, a
decreases. For positronium and muonium the reduced mass effect is large.
eiml φ = cos( mlφ ) + i sin( mlφ ), and to be periodic with period 2" , ml 2" must be an integer multiple of 2" , so ml
must be an integer.
a
a1
P (a ) = 5 (1s 2V = 5
e −2 r a (4"r 2dr ) .
0 "a 3
0
a 4a
4 /" − ar 2 a 2 r a 2 # −2 r a 0
4 /" − a 3 a 3 a 3 # −2 a 3 0 0
P (a ) = 3 5 r 2e −2 r a dr = 3 1$
−
− %e
− − %e + e 2
2 = 3 1$
2
4'
2 4'
44
ao
a 3& 2
4 0 a 3& 2
! P ( a ) = 1 − 5e −2 .
41.14. (a) ∆E = *B B = (5.79 × 10−5 e V T)(0.400 T) = 2.32 × 10−5 eV
(b) ml = −2 the lowest possible value of ml . Atomic Structure 413 (c) The energy level diagram is sketched in Figure 41.14. 41.15. Figure 41.14
IDENTIFY and SET UP: The interaction energy between an external magnetic field and the orbital angular
momentum of the atom is given by Eq.(41.18). The energy depends on ml with the most negative ml value having
the lowest energy.
EXECUTE: (a) For the 5g level, l = 4 and there are 2l + 1 = 9 different ml states. The 5g level is split into 9 levels
by the magnetic field.
(b) Each ml level is shifted in energy an amount given by U = ml µ B B. Adjacent levels differ in ml by one, so
∆U = µ B B.
e% (1.602 × 10−19 C)(1.055 × 10−34 J ⋅ s)
=
= 9.277 × 10−24 A ⋅ m 2
2m
2(9.109 × 10−31 kg)
∆U = µ B B = (9.277 × 10−24 A/m 2 )(0.600 T) = 5.566 × 10−24 J(1 eV/1.602 × 10−19 J) = 3.47 × 10−5 eV µB = (c) The level of highest energy is for the largest ml , which is ml = l = 4; U 4 = 4 µ B B. The level of lowest energy is for the smallest ml , which is ml = −l = −4; U −4 = −4 µ B B. The separation between these two levels is 41.16. U 4 − U −4 = 8µ B B = 8(3.47 × 10−5 eV) = 2.78 × 10−4 eV.
EVALUATE: The energy separations are proportional to the magnetic field. The energy of the n = 5 level in the
absence of the external magnetic field is ( −13.6 eV)/52 = −0.544 eV, so the interaction energy with the magnetic
field is much less than the binding energy of the state.
(a) According to Figure 41.11 in the textbook there are three different transitions that are consistent with the
selection rules. The initial ml values are 0, ±1; and the final ml value is 0.
(b) The transition from ml = 0 to ml = 0 produces the same wavelength (122 nm) that was seen without the magnetic field.
(c) The larger wavelength (smaller energy) is produced from the ml = − 1 to ml = 0 transition.
(d) The shorter wavelength (greater energy) is produced from the ml = + 1 to ml = 0 transition. 41.17. 3 p ! n = 3, l = 1, ∆U = *B B ! B = U
(2.71 × 10−5 eV)
=
= 0.468 T
*B (5.79 × 10−5 e V T) (b) Three: ml = 0, ± 1.
41.18. 41.19. (2.00232)
" e # " −% #
(a) U = + (2.00232) $
*B B
%$
%B = −
2
& 2m ' & 2 '
(2.00232)
U =−
(5.788 × 10−5 e V T)(0.480 T) = −2.78 × 10−5 eV.
2
(b) Since n = 1, l = 0 so there is no orbital magnetic dipole interaction. But if n ≠ 0 there could be since l < n
allows for l ≠ 0.
##
IDENTIFY and SET UP: The interaction energy is U = − µ ⋅ B , with µ z given by Eq.(41.22).
#
#
EXECUTE: U = − µ ⋅ B = + µ z B, since the magnetic field is in the negative zdirection.
"e#
"e#
% S z , so U = −(2.00232) $
% Sz B
2m '
&
& 2m '
" e% #
S z = ms %, so U = −2.00232 $
% ms B
& 2m '
e%
= µB = 5.788 × 10−5 eV/T
2m
U = −2.00232µ B ms B
1
The ms = + level has lower energy.
2
1#
1#
" 1 " 1 ##
"
"
∆U = U $ ms = − % − U $ ms = + % = −2.00232 µB B $ − − $ + % % = +2.00232µB B
2'
2'
&
&
& 2 & 2 '' µ z = −( 2.00232) $ ∆U = +2.00232(5.788 × 10−5 eV/T)(1.45 T) = 1.68 × 10−4 eV 414 41.20.
41.21. 41.22. 41.23. Chapter 41 EVALUATE: The interaction energy with the electron spin is the same order of magnitude as the interaction
energy with the orbital angular momentum for states with ml ≠ 0. But a 1s state has l = 0 and ml = 0, so there is no
orbital magnetic interaction.
" 1# " 1# " 3# " 3#
" 5#
The allowed (l , j ) combinations are $ 0, % , $1, % , $1, % , $ 2, % and $ 2, % .
2' & 2' & 2' & 2'
&
& 2'
IDENTIFY and SET UP: j can have the values l + 1/ 2 and l − 1/ 2.
EXECUTE: If j takes the values 7/2 and 9/2 it must be that l − 1/ 2 = 7 / 2 and l = 8/ 2 = 4. The letter that labels
this l is g.
EVALUATE: l must be an integer.
hc (4.136 × 10−15 eV ⋅ s)(300 × 108 m s )
c (3.00 × 108 m s )
(a) λ =
=
= 21 cm, f = =
= 1.4 × 109 Hz, a short radio
λ
∆E
(5.9 × 10−6 eV)
0.21 m
wave.
(b) As in Example 41.6, the effective field is B ≅ ∆ E 2 *B = 5.1 × 10−2 T, for smaller than that found in the example.
IDENTIFY and SET UP: For a classical particle L = I ω . For a uniform sphere with mass m and radius R,
2
"2
#
I = mR 2 , so L = $ mR 2 % ω. Solve for ω and then use v = rω to solve for v.
5
5
&
'
EXECUTE: (a) L = ω= 41.24.
41.25. 41.26.
41.27. 3
2
3
% so mR 2ω =
%
4
5
4 5 3/ 4%
5 3/ 4(1.055 × 10−34 J ⋅ s)
=
= 2.5 × 1030 rad/s
2
2mR
2(9.109 × 10 −31 kg)(1.0 × 10−17 m)2 (b) v = rω = (1.0 × 10−17 m)(2.5 × 1030 rad/s) = 2.5 × 1013 m/s.
EVALUATE: This is much greater than the speed of light c, so the model cannot be valid.
However the number of electrons is obtained, the results must be consistent with Table (41.3); adding two more
electrons to the zinc configuration gives 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 2 .
The ten lowest energy levels for electrons are in the n = 1 and n = 2 shells.
1
n = 1, l = 0, ml = 0,
ms = ± : 2 states.
2
1
n = 2, l = 0, ml = 0,
ms = ± : 2 states.
2
1
n = 2, l = 1, ml = 0, ± 1, ms = ± : 6 states.
2
For the outer electrons, there are more inner electrons to screen the nucleus.
IDENTIFY and SET UP: The energy of an atomic level is given in terms of n and Z eff by Eq.(41.27), " Z2 #
En = − $ eff % (13.6 eV). The ionization energy for a level with energy − En is + En .
2
&n '
(2.771) 2
EXECUTE: n = 5 and Zeff = 2.771 gives E5 = −
(13.6 eV) = −4.18 eV
52
The ionization energy is 4.18 eV.
2
EVALUATE: The energy of an atomic state is proportional to Z eff .
41.28. 41.29. For the 4s state, E = −4.339 eV and Z eff = 4 (−4.339) (−13.6) = 2.26. Similarly, Z eff = 1.79 for the 4p state and
1.05 for the 4d state. The electrons in the states with higher l tend to be further away from the filled subshells and
the screening is more complete.
IDENTIFY and SET UP: Use the exclusion principle to determine the groundstate electron configuration, as in
Table 41.3. Estimate the energy by estimating Z eff , taking into account the electron screening of the nucleus.
EXECUTE: (a) Z = 7 for nitrogen so a nitrogen atom has 7 electrons. N 2+ has 5 electrons: 1s 2 2 s 2 2 p.
(b) Z eff = 7 −= 3 for the 2p level.
4 " Z2 #
32
En = − $ eff % (13.6 eV) = − 2 (13.6 eV) = −30.6 eV
2
2
&n '
(c) Z = 15 for phosphorus so a phosphorus atom has 15 electrons.
P 2+ has 13 electrons: 1s 2 2 s 2 2 p 6 3s 2 3 p Atomic Structure 415 (d) Z eff = 15 − 12 = 3 for the 3p level. 41.30. 41.31. " Z2 #
32
En = − $ eff % (13.6 eV) = − 2 (13.6 eV) = −13.6 eV
2
3
&n '
EVALUATE: In these ions there is one electron outside filled subshells, so it is a reasonable approximation to
assume full screening by these innersubshell electrons.
13.6 eV 2
(a) E2 = −
Z eff , so Z eff = 1.26.
4
(b) Similarly, Z eff = 2.26.
(c) Z eff becomes larger going down the columns in the periodic table.
IDENTIFY and SET UP: Estimate Z eff by considering electron screening and use Eq.(41.27) to calculate the
energy. Z eff is calculated as in Example 41.8.
EXECUTE: (a) The element Be has nuclear charge Z = 4. The ion Be + has 3 electrons. The outermost electron
sees the nuclear charge screened by the other two electrons so Z eff = 4 −= 2.
2 " Z2 #
22
En = − $ eff % (13.6 eV) so E2 = − 2 (13.6 eV) = −13.6 eV
2
2
&n '
22
(13.6 eV) = −3.4 eV
42
EVALUATE: For the electron in the highest lstate it is reasonable to assume full screening by the other electrons,
as in Example 41.8. The highest lstates of Be + , Mg + , Ca + , etc. all have a Z eff = 2. But the energies are different
because for each ion the outermost sublevel has a different n quantum number.
7.46 × 103 eV
= 28.0, which corresponds to the element Nickel (Ni).
Ekx ≅ ( Z − 1) 2 (10.2 eV) . Z ≈ 1 +
10.2 eV (b) The outermost electron in Ca + sees a Z eff = 2. E4 = − 41.32.
41.33. (a) Z = 20 : f = (2.48 × 1015 Hz)(20 − 1) 2 = 8.95 × 1017 Hz .
c 3.00 × 108 m s
=
= 3.35 × 10−10 m.
f 8.95 × 1017 Hz
(b) Z = 27: f = 1.68 × 1018 Hz. E = 6.96 keV. λ = 1.79 × 10 −10 m.
E = hf = (4.14 × 10−15 eV ⋅ s) (8.95 × 1017 Hz) = 3.71 keV. λ = 41.34. (c) Z = 48 : f = 5.48 × 1018 Hz, E = 22.7 keV, λ = 5.47 × 10 −11 m.
IDENTIFY: The orbital angular momentum is limited by the shell the electron is in.
SET UP: For an electron in the n shell, its orbital angular momentum quantum number l is limited by 0 l < n,
and its orbital angular momentum is given by L = l (l + 1) % . The zcomponent of its angular momentum is Lz = ml %, where ml = 0, ±1, … , ±l, and its spin angular momentum is S = 3 / 4 % for all electrons. Its energy in
the nth shell is En = −(13.6 eV)/n 2 .
EXECUTE: (a) L = l (l + 1) % = 12% ! l = 3. Therefore the smallest that n can be is 4, so En = – (13.6 eV)/n2 =
– (13.6 eV)/42 = –0.8500 eV.
(b) For l = 3, ml = ±3, ±2, ±1, 0. Since Lz = ml %, the largest Lz can be is 3 % and the smallest it can be is –3 % . 41.35. (c) S = 3 / 4 % for all electrons.
(d) In this case, n = 3, so l = 2, 1, 0. Therefore the maximum that L can be is Lmax = 2(2 + 1) % = 6 % . The
minimum L can be is zero when l = 0.
EVALUATE: At the quantum level, electrons in atoms can have only certain allowed values of their angular momentum.
IDENTIFY: The total energy determines what shell the electron is in, which limits its angular momentum.
SET UP: The electron’s orbital angular momentum is given by L = l (l + 1) % , and its total energy in the nth shell is En = −(13.6 eV)/n 2 .
EXECUTE: (a) First find n: En = −(13.6 eV)/n 2 = − 0.5440 eV which gives n = 5, so l = 4, 3, 2, 1, 0. Therefore the possible values of L are given by L = l (l + 1) % , giving L = 0,
2 2 %, 6%, 12%, 20%. (b) E6 = – (13.6 eV)/6 = –0.3778 eV. ∆E = E6 – E5 = –0.3778 eV – (–0.5440 eV) = +0.1662 eV
This must be the energy of the photon, so ∆E = hc/λ, which gives
λ = hc/∆E = (4.136 × 10–15 eV ⋅ s )(3.00 ×108 m/s)/(0.1662 eV) = 7.47 × 10–6 m = 7470 nm, which is in the infrared
and hence not visible.
EVALUATE: The electron can have any of the five possible values for its angular momentum, but it cannot have
any others. 416 Chapter 41 41.36. IDENTIFY: For the N shell, n = 4, which limits the values of the other quantum numbers.
SET UP: In the nth shell, 0 l < n, ml = 0, ±1, … , ±l, and ms = ±1/2. The orbital angular momentum of the
electron is L = l (l + 1) % and its spin angular momentum is S = 3 / 4 % .
EXECUTE: (a) For l = 3 we can have ml = ±3, ±2±, ±1, 0 and ms = ±½; for l = 2 we can have ml = ±2, ±1, 0 and
ms = ±½; for l = 1, we can have ml = ±1, 0 and ms = ±1/2 ; for l = 0, we can have ml = 0 and ms = ±1/2.
(b) For the N shell, n = 4, and for an felectron, l = 3, giving L = l (l + 1) % = 3(3 + 1) % = 12% . Lz = ml % = ±3%, ±2%, ± %, 0, so the maximum value is 3% . S = 3 / 4 % for all electrons. 41.37. (c) For a dstate electron, l = 2, giving L = 2(2 + 1) % = 6% . Lz = ml %, and the maximum value of ml is 2, so the
maximum value of Lz is 2 % . The smallest angle occurs when Lz is most closely aligned along the angular
L
2%
2
=
and θmin = 35.3°. The largest
momentum vector, which is when Lz is greatest. Therefore cosθ min = z =
L
6%
6
angle occurs when Lz is as far as possible from the Lvector, which is when Lz is most negative. Therefore
−2%
2
cosθ max =
=−
and θ max = 144.7° .
6%
6
(d) This is not possible since l = 3 for an felectron, but in the M shell the maximum value of l is 2.
EVALUATE: The fact that the angle in part (c) cannot be zero tells us that the orbital angular momentum of the
electron cannot be totally aligned along any specified direction.
IDENTIFY: The inner electrons shield part of the nuclear charge from the outer electron.
Z2
SET UP: The electron’s energy in the nth shell, due to shielding, is En = − eff (13.6 eV) , where Zeff e is the
n2
effective charge that the electron “sees” for the nucleus.
2
Z eff
(4 2 )( −1.947 eV)
(13.6 eV) and n = 4 for the 4s state. Solving for Zeff gives Z eff = −
n2
13.6 eV
= 1.51. The nucleus contains a charge of +11e, so the average number of electrons that screen this nucleus must
be 11 – 1.51 = 9.49 electrons
(b) (i) The charge of the nucleus is +19e, but 17.2e is screened by the electrons, so the outer electron “sees” 19e –
17.2e = 1.8e and Zeff = 1.8.
Z2
(1.8)2
(ii) En = − eff (13.6 eV) = − 2 (13.6 eV) = −2.75 eV
n2
4
EVALUATE: Sodium has 11 protons, so the inner 10 electrons shield a large portion of this charge from the outer
electron. But they don’t shield 10 of the protons, since the inner electrons are not totally equivalent to a uniform
spherical shell. (They are lumpy.) EXECUTE: (a) En = − 2 41.38.
41.39. 2 See Example 41.3; r 2 ( = Cr 2e −2 r/a , d (r 2 ( ) = Ce −2 r/a (2r − (2r 2 /a)), and for a maximum, r = a, the distance of
dr
the electron from the nucleus in the Bohr model.
(a) IDENTIFY and SET UP: The energy is given by Eq.(38.18), which is identical to Eq.(41.3). The potential
energy is given by Eq.(23.9), with q = + Ze and q0 = −e.
EXECUTE: E1s = − E1s = U ( r ) gives − 1 me 4
1 e2
; U (r ) = −
2
2
(4π P0 ) 2%
4π P0 r 1 me 4
1 e2
=−
2
2
(4π P0 ) 2%
4π P0 r (4π P0 )2% 2
= 2a
me 2
EVALUATE: The turning point is twice the Bohr radius.
(b) IDENTIFY and SET UP: For the 1s state the probability that the electron is in the classically forbidden region
r= ∞ ∞ is P (r > 2a ) = 5 ψ 1s dV = 4π 5 ψ 1s r 2 dr. The normalized wave function of the 1s state of hydrogen is given in
2 2a Example 41.3: ψ 1s (r ) =
EXECUTE: 2 2a 1 π a3 e − r / a . Evaluate the integral; the integrand is the same as in Example 41.3. "1#∞
P (r > 2a ) = 4π $ 3 % 5 r 2e−2 r / a dr
& π a ' 2a Atomic Structure Use the integral formula 5r e 2 −α r 417 " r 2 2r 2 #
dr = −e −α r $ + 2 + 3 % , with α = 2 / a.
α'
&α α
∞ " ar 2 a 2 r a 3 # 0
4/
4
P (r > 2a ) = − 3 1 e −2 r / a $
+
+ % 2 = + 3 e −4 (2a 3 + a 3 + a3 / 4)
a3
2
4 '4 2a
a
&2 41.40. P (r > 2a ) = 4e −4 (13/ 4) = 13e−4 = 0.238.
EVALUATE: These is a 23.8% probability of the electron being found in the classically forbidden region, where
classically its kinetic energy would be negative.
(a) For large values of n, the inner electrons will completely shield the nucleus, so Z eff = 1 and the ionization energy would be 13.60 eV
.
n2 13.60 eV
= 1.11× 10−4 eV, r350 = (350) 2 a0 = (350)2 (0.529 × 10−10 m) = 6.48 × 10−6 m .
3502
13.60 eV
(c) Similarly for n = 650,
= 3.22 × 10−5 eV, r650 = (650) 2 (0.529 × 10−10 m) = 2.24 × 10−5 m.
(650) 2
(b) 41.41. ψ 2 s (r ) = r # − r/ 2 a
"
$ 2 − %e
a'
32π a &
1 3 ∞ ∞ (a) IDENTIFY and SET UP: Let I = 5 ψ 2 s dV = 4π 5 ψ 2 s r 2 dr. If ψ 2 s is normalized then we will find that
2 0 2 0 I = 1.
2 r#
1 ∞"
4r 3 r 4 # − r / a
" 1 # ∞"
I = 4π $
+ 2 % e dr
2 − % e − r/a r 2 dr = 3 5 $ 4r 2 −
3 % 50 $
a'
a
a'
8a 0 &
& 32π a ' &
∞
n!
Use the integral formula 5 x ne −α x dx = n +1 , with α = 1/ a EXECUTE: α 0 1"
4
1
#1
I = 3 $ 4(2!)(a 3 ) − (3!)(a )4 + 2 (4!)(a )5 % = (8 − 24 + 24) = 1; this ψ 2 s is normalized.
a
a
8a &
'8
(b) SET UP: For a spherically symmetric state such as the 2s, the probability that the electron will be found at
4a 4a r < 4a is P (r < 4a ) = 5 ψ 2 s dV = 4π 5 ψ 2 s r 2 dr.
2 0 EXECUTE: P (r < 4a ) = Let P (r < 4a ) = 2 0 1 4 a " 2 4r 3 r 4 # − r / a
+ 2 % e dr
$ 4r −
8a 3 5 0 &
a
a' 1
( I1 + I 2 + I 3 ).
8a 3 4a I1 = 4 5 r 2e − r / a dr
0 " r 2 2r 2 #
Use the integral formula 5 r 2e −α r dr = −e −α r $ + 2 + 3 % with α = 1/ a.
α'
&α α
−r / a
2
2
3 4a
−4
3
I1 = −4[e ( r a + 2ra + 2a )]0 = ( −104e + 8)a . I2 = − 4 4a 3 −r / a
r e dr
a 50 " r 3 3r 2 6r 6 #
Use the integral formula 5 r 3e −α r dr = −e −α r $ + 2 + 3 + 4 % with α = 1/ a.
a α'
&α α
4 −r / a 3
4
I 2 = [e (r a + 3r 2 a 2 + 6ra 3 + 6a 4 )] 0 a = (568e −4 − 24) a3 .
a
1 4a
I 3 = 2 5 r 4e − r / a dr
a0
" r 4 4r 3 12r 2 24r 24 #
Use the integral formula 5 r 4e −α r dr = −e −α r $ + 2 + 3 + 4 + 5 % with α = 1/ a.
α
a
a'
&α α I3 = − 1 −r / a 4
4
[e (r a + 4r 3a 2 + 12r 2 a 3 + 24ra 4 + 24a 5 )] 0 a = ( −824e −4 + 24)a3 .
a2 418 Chapter 41 Thus P (r < 4a ) = 1
1
( I1 + I 2 + I 3 ) = 3 a3 ([8 − 24 + 24] + e −4 [−104 + 568 − 824])
3
8a
8a 1
P (r < 4a ) = (8 − 360e −4 ) = 1 − 45e −4 = 0.176.
8
EVALUATE: There is an 82.4% probability that the electron will be found at r > 4a. In the Bohr model the
electron is for certain at r = 4a; this is a poor description of the radial probability distribution for this state.
41.42. (a) Since the given ( ( r ) is real, r 2  ( 2 = r 2( 2 . The probability density will be an extreme when d 22
d( #
d( #
"
"
( r ( ) = 2 $ r( 2 + r 2(
% = 2r( $ ( + r
% = 0. This occurs at r = 0, a minimum, and when ( = 0, also a
dr
dr '
dr '
&
&
dψ
= 0. Within a multiplicative constant, ( ( r ) = (2 − r a )e− r 2 a ,
minimum. A maximum must correspond to ( + r
dr
d(
1
= − (2 − r 2a)e − r 2 a , and the condition for a maximum is (2 − r a ) = (r a ) (2 − r 2a ), or r 2 − 6ra + 4a 2 = 0.
dr
a
The solutions to the quadratic are r = a(3 ± 5). The ratio of the probability densities at these radii is 3.68, with 41.43. the larger density at r = a(3 + 5) .
(b) ( = 0 at r = 2a
Parts (a) and (b) are consistent with Figure 41.5 in the textbook; note the two relative maxima, one on each side of
the minimum of zero at r = 2a.
L
"L #
IDENTIFY: Use Figure 41.2 in the textbook to relate θ L to Lz and L: cosθ L = z so θ L = arccos $ z %
L
&L'
(a) SET UP: The smallest angle (θ L ) min is for the state with the largest L and the largest Lz . This is the state with l = n − 1 and ml = l = n − 1.
EXECUTE: Lz = ml % = ( n − 1)% L = l (l + 1)% = (n − 1) n %
" (n − 1)% #
" ( n − 1) #
" n −1 #
(θ L ) min = arccos $
= arccos $
% = arccos $
% = arccos( 1 − 1/ n ).
$
%
$ ( n − 1)n% %
$ (n − 1) n %
%
n'
&
&
'
&
'
EVALUATE: Note that (θ L ) min approaches 0° as n → ∞.
(b) SET UP: The largest angle (θ L ) max is for l = n − 1 and ml = −l = −(n − 1).
EXECUTE: A similar calculation to part (a) yields (θ L ) max = arccos(− 1 − 1/ n )
EVALUATE: Note that (θ L ) max approaches 180° as n → ∞.
41.44. (a) L2 + L2 = L2 − L2 = l (l + 1)% 2 − ml2% 2 so L2 + L2 = l (l + 1) − ml2 %.
x
y
z
x
y
(b) This is the magnitude of the component of angular momentum perpendicular to the zaxis.
(c) The maximum value is l (l + 1)% = L, when ml = 0. That is, if the electron is known to have no zcomponent of angular momentum, the angular momentum must be perpendicular to the zaxis. The minimum is
ml = ±l.
41.45. l % when 4
r4
" 1 # 4 − r 2 a dP " 1 # " 3 r # − r 2 a dP
4r − % e
P( r ) = $
re
.
=$
.
= 0 when 4r 3 − = 0; r = 4a. In the Bohr
5 %$
5%
dr & 24a ' &
a'
dr
a
& 24a ' model, rn = n 2 a so r2 = 4a, which agrees.
41.46. The time required to transit the horizontal 50 cm region is t = ∆ x 0.500 m
=
= 0.952 ms. The force required to
vx 525 m s "
# 2(0.50 × 10−3 m)
2∆z
0.1079 kg mol
=±$
=
%
−3
2
23
2
t
& 6.022 × 10 atoms mol ' (0.952 × 10 s)
N. According to Eq.(41.22), the value of *z is  * z  = 9.28 × 10−24 A ⋅ m 2 . Thus, the required deflect each spin component by 0.50 mm is Fz = maz = ± m
±1.98 × 10−22 magneticfield gradient is dBz
F
1.98 × 10−22 N
= z=
= 21.3 T m.
*z 9.28 × 10−24 J T
dz Atomic Structure 41.47. 419 Decay from a 3d to 2 p state in hydrogen means that n = 3 → n = 2 and ml = ±2, ± 1, 0 → ml = ±1, 0. However
selection rules limit the possibilities for decay. The emitted photon carries off one unit of angular momentum so
l must change by 1 and hence ml must change by 0 or ±1. The shift in the transition energy from the zero field
e%B
( ml3 − ml2 ), where ml3 is the 3d ml value and ml2 is the 2 p ml value. Thus
2m
there are only three different energy shifts. They and the transitions that have them, labeled by the ml names, are: value is just U = ( ml3 − ml2 ) *B B = e%B
: 2 → 1,
2m
0 :1 → 1, 41.48. 1 → 0, 0 → −1 0 → 0, − 1 → −1 e%B
−
: 0 → 1, − 1 → 0, − 2 → −1
2m
IDENTIFY: The presence of an external magnetic field shifts the energy levels up or down, depending upon the
value of ml.
SET UP: The selection rules tell us that for allowed transitions, ∆l = 1 and ∆ml = 0 or ±1.
EXECUTE: (a) E = hc/λ = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(475.082 nm) = 2.612 eV.
(b) For allowed transitions, ∆l = 1 and ∆ml = 0 or ±1. For the 3d state, n = 3, l = 2, and ml can have the values 2, 1,
0, –1, –2. In the 2p state, n = 2, l = 1, and ml can be 1, 0, –1. Therefore the 9 allowed transitions from the 3d state
in the presence of a magnetic field are:
l = 1, ml = 1;
l = 2 , ml = 2
l = 2 , ml = 1
l = 1, ml = 0
l = 2 , ml = 1
l = 1, ml = 1
l = 2 , ml = 0
l = 1, ml = 0
l = 2 , ml = 0
l = 1, ml = 1
l = 2 , ml = 0
l = 1, ml = –1
l = 2 , ml = –1
l = 1, ml = 0
l = 2 , ml = –1
l = 1, ml = –1
l = 2 , ml = –2
l = 1, ml = –1
(c) ∆E = µ BB = (5.788 × 10–5 eV/T)(3.500 T) = 0.000203 eV
So the energies of the new states are –8.50000 eV + 0 and –8.50000 eV ± 0.000203 eV, giving energies of:
–8.50020 eV, –8.50000 eV, and –8.49980 eV
(d) The energy differences of the allowed transitions are equal to the energy differences if no magnetic field were
present (2.61176 eV, from part (a)), and that value ±∆E (0.000203 eV, from part (c)). Therefore we get the
following.
For E = 2.61176 eV: λ = 475.082 nm (which was given)
For E = 2.61176 eV + 0.000203 eV = 2.611963 eV: λ = hc/E = (4.136 ×10–15 eV ⋅ s )(3.00 × 108 m/s)/(2.611963 eV) = 475.045 nm
For E = 2.61176 eV – 0.000203 eV = 2.61156 eV: λ = hc/E = (4.136 ×10–15 eV ⋅ s )(3.00 × 108 m/s)/(2.61156 eV) = 475.119 nm
41.49. 41.50. EVALUATE: Even a strong magnetic field produces small changes in the energy levels, and hence in the
wavelengths of the emitted light.
IDENTIFY: The presence of an external magnetic field shifts the energy levels up or down, depending upon the
value of ml.
SET UP: The energy difference due to the magnetic field is ∆E = µ BB and the energy of a photon is E = hc/λ.
EXECUTE: For the p state, ml = 0 or ±1, and for the s state ml = 0. Between any two adjacent lines, ∆E = µ BB.
Since the change in the wavelength (∆λ) is very small, the energy change (∆E ) is also very small, so we can use
hc∆λ
hc∆λ
hc
hc∆λ
. Since ∆E = µ BB, we get µ B B =
and B =
.
differentials. E = hc/λ .  dE  = 2 d λ and ∆E =
2
2
λ
λ
λ
µ Bλ 2
B = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)(0.0462 nm)/(5.788 × 10–5 eV/T)(575.050 nm)2 = 3.00 T
EVALUATE: Even a strong magnetic field produces small changes in the energy levels, and hence in the
wavelengths of the emitted light.
(a) The energy shift from zero field is ∆U 0 = ml µ B B.
For ml = 2, ∆U 0 = (2) (5.79 × 10−5 e V T) (1.40 T) = 1.62 × 10−4 eV.
For ml = 1, ∆U 0 = (1)(5.79 × 10−5 e V T) (1.40 T) = 8.11× 10−5 eV. (b)  ∆λ = λ0 ∆E 
" 36 # 1
, where E0 = (13.6 eV)((1/ 4) − (1/ 9)), λ0 = $ % = 6.563 × 10−7 m
E0
& 5 'R 4110 41.51. Chapter 41 and ∆ E = 1.62 × 10−4 eV − 8.11 × 10−5 eV = 8.09 × 10−5 eV from part (a). Then,  ∆λ  = 2.81 × 10−11 m = 0.0281 nm .
The wavelength corresponds to a larger energy change, and so the wavelength is smaller.
n
IDENTIFY: The ratio according to the Boltzmann distribution is given by Eq.(38.21): 1 = e − ( E1 − E0 ) / kT , where 1 is
n0
the higher energy state and 0 is the lower energy state.
" e% #
SET UP: The interaction energy with the magnetic field is U = − µ z B = 2.00232 $
% ms B (Example 41.5.). The
& 2m '
1
1
energy of the ms = + level is increased and the energy of the ms = − level is decreased.
2
2
n1/ 2
− (U1 / 2 −U −1/ 2 ) / kT
=e
n−1/ 2 EXECUTE: " e% # " 1 " 1 # #
" e% #
U1 / 2 − U −1/ 2 = 2.00232 $
% B $ − $ − % % = 2.00232 $
% B = 2.00232µB B
2m ' & 2 & 2 ' '
&
& 2m ' n1/ 2
= e − ( 2.00232) µB B / kT
n−1/ 2
(a) B = 5.00 × 10−5 T
−24
2
−5
−23
n1/ 2
= e −2.00232(9.274×10 A/m )(5.00×10 T)/([1.381×10 J/K][300 K ])
n−1/ 2
−7
n1/ 2
= e −2.24×10 = 0.99999978 = 1 − 2.2 × 10−7
n−1/ 2 (b) B = 5.00 × 10−5 T, −3
n1/ 2
= e −2.24×10 = 0.9978
n−1/ 2 −2
n1/ 2
= e −2.24×10 = 0.978
n−1/ 2
EVALUATE: For small fields the energy separation between the two spin states is much less than kT for
T = 300 K and the states are equally populated. For B = 5.00 T the energy spacing is large enough for there to be
a small excess of atoms in the lower state. (c) B = 5.00 × 10 −5 T, 41.52. Using Eq.(41.4), L = mvr = l (l + 1)%, and the Bohr radius from Eq.(38.15), we obtain the following value for v : l (l + 1)%
2(6.63 × 10−34 J ⋅ s)
=
= 7.74 × 105 m s. The magnetic field generated by the
2
m( n a0 ) 2" (9.11 × 10−31 kg) (4) (5.29 × 10−11 m)
“moving” proton at the electrons position can be calculated from Eq.(28.1):
*  q  v sin φ
(1.60 × 10−19 C) (7.74 × 105 m s) sin(90°)
B= 0
= (10−7 T ⋅ m A)
= 0.277 T.
2
4"
(4) 2 (5.29 × 10−11 m)2
r
v= 41.53. 3
1
1
3
ms can take on 4 different values: ms = − , − , + , + . Each nlml state can have 4
2
2
2
2
electrons, each with one of the four different ms values. Apply the exclusion principle to determine the electron
configurations.
EXECUTE: (a) For a filled n = 1 shell, the electron configuration would be 1s 4 ; four electrons and Z = 4. For a
IDENTIFY and SET UP: filled n = 2 shell, the electron configuration would be 1s 4 2 s 4 2 p12 ; twenty electrons and Z = 20. 41.54. (b) Sodium has Z = 11; 11 electrons. The groundstate electron configuration would be 1s 4 2 s 4 2 p 3.
EVALUATE: The chemical properties of each element would be very different.
(a) Z 2 ( −13.6 eV) = (7) 2 (−13.6 eV) = −666 eV.
(b) The negative of the result of part (a), 666 eV.
(c) The radius of the ground state orbit is inversely proportional to the nuclear charge, and
a
= (0.529 × 10−10 m) 7 = 7.56 × 10−12 m.
Z
hc
hc
(d) λ =
=
, where E0 is the energy found in part (b), and λ = 2.49 nm.
∆E E 1 − 1
0
2
2
12 ( ) Atomic Structure 41.55. 4111 (a) IDENTIFY and SET UP: The energy of the photon equals the transition energy of the atom: ∆ E = hc / λ . The
energies of the states are given by Eq.(41.3).
13.60 eV
13.60 eV
13.60 eV
EXECUTE: En = −
so E2 = −
and E1 = −
n2
4
1
1 #3
"
∆ E = E2 − E1 = 13.60 eV $ − + 1% = (13.60 eV) = 10.20 eV = (10.20 eV)(1.602 × 10−19 J/eV) = 1.634 × 10−18 J
&4 '4 hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 1.22 × 10−7 m = 122 nm
∆E
1.634 × 10−18 J
(b) IDENTIFY and SET UP: Calculate the change in ∆E due to the orbital magnetic interaction energy,
Eq.(41.17), and relate this to the shift ∆λ in the photon wavelength.
EXECUTE: The shift of a level due to the energy of interaction with the magnetic field in the zdirection is
U = ml µ B B. The ground state has ml = 0 so is unaffected by the magnetic field. The n = 2 initial state has λ= ml = −1 so its energy is shifted downward an amount U = ml µ B B = (−1)(9.274 × 10−24 A/m 2 )(2.20 T) =
( −2.040 × 10−23 J)(1 eV /1.602 × 10−19 J) = 1.273 × 10−4 eV
Note that the shift in energy due to the magnetic field is a very small fraction of the 10.2 eV transition energy.
Problem 39.56c shows that in this situation ∆λ / λ = ∆ E / E . This gives " 1.273 × 10−4 eV #
−3
∆λ = λ ∆ E / E = 122 nm $
% = 1.52 × 10 nm = 1.52 pm.
10.2 eV
&
'
EVALUATE: The upper level in the transition is lowered in energy so the transition energy is decreased. A smaller
∆E means a larger λ ; the magnetic field increases the wavelength. The fractional shift in wavelength, ∆λ / λ is
41.56. 41.57. small, only 1.2 × 10 −5.
The effective field is that which gives rise to the observed difference in the energy level transition,
∆ E hc " λ1 − λ2 # 2"mc " λ1 − λ2 #
−3
=
B=
$
%=
$
% . Substitution of numerical values gives B = 3.64 × 10 T, much smaller
*B *B & λ1λ2 '
e & λ1λ2 '
than that for sodium.
IDENTIFY: Estimate the atomic transition energy and use Eq.(38.6) to relate this to the photon wavelength.
(a) SET UP: vanadium, Z = 23
minimum wavelength; corresponds to largest transition energy
EXECUTE: The highest occupied shell is the N shell ( n = 4). The highest energy transition is N → K , with
transition energy ∆E = EN − EK . Since the shell energies scale like 1/ n 2 neglect EN relative to EK , so
∆ E = EK = ( Z − 1) 2 (13.6 eV) = (23 − 1)2 (13.6 eV) = 6.582 × 103 eV = 1.055 × 10−15 J. The energy of the emitted
photon equals this transition energy, so the photon’s wavelength is given by ∆ E = hc / λ so λ = hc / ∆ E. (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
= 1.88 × 10−10 m = 0.188 nm.
1.055 × 10−15 J
SET UP: maximum wavelength; corresponds to smallest transition energy, so for the Kα transition
EXECUTE: The frequency of the photon emitted in this transition is given by Moseley’s law (Eq.41.29):
f = (2.48 × 1015 Hz)(Z − 1) 2 = (2.48 × 1015 Hz)(23 − 1)2 = 1.200 × 1018 Hz λ= c 2.998 × 108 m/s
=
= 2.50 × 10−10 m = 0.250 nm
f 1.200 × 1018 Hz
(b) rhenium, Z = 45
Apply the analysis of part (a), just with this different value of Z.
minimum wavelength
∆ E = EK = ( Z − 1) 2 (13.6 eV) = (45 − 1)2 (13.6 eV) = 2.633 × 104 eV = 4.218 × 10−15 J. λ= (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
= 4.71 × 10−11 m = 0.0471 nm.
4.218 × 10−15 J
maximum wavelength
f = (2.48 × 1015 Hz)(Z − 1) 2 = (2.48 × 1015 Hz)(45 − 1)2 = 4.801× 1018 Hz λ = hc / ∆ E = c 2.998 × 108 m/s
=
= 6.24 × 10−11 m = 0.0624 nm
f 4.801 × 1018 Hz
EVALUATE: Our calculated wavelengths have values corresponding to x rays. The transition energies increase
when Z increases and the photon wavelengths decrease. λ= 4112 Chapter 41 41.58. (a) ∆ E = (2.00232) 41.59. (a) To calculate the total number of states for the n th principal quantum number shell we must multiply all the
possibilities. The spin states multiply everything by 2. The maximum l value is (n –1), and each l value has
( 2l + 1)ml values. So the total number of states is e
e%
hc
2"mc
!B=
B∆S z ≈ B =
.
λ
λe
2m
m
2" (9.11 × 10−31 kg) (3.00 × 108 m s )
(b) B =
= 0.307 T.
(0.0350 m)(1.60 × 10−19 C) n −1 n −1 n −1 l =0 l =0 l =0 N = 26 (2l + 1) = 261 + 46 l = 2n +
41.60. 4( n − 1)(n)
= 2n + 2n 2 − 2n = 2n 2 .
2 (b) The n = 5 shell (Oshell) has 50 states.
IDENTIFY: We treat the Earth as an electron.
SET UP: The intrinsic spin angular momentum of an electron is S = 3
% , and the angular momentum of the
4 spinning Earth is S = I ω , where I = 2/5 mR2.
EXECUTE: (a) Using S = I ω = 41.61. 3
% and solving for ω gives
4 3
3
%
(1.055 ×10−34 J ⋅ s )
4=
4
ω=
= 9.40 × 10−73 rad/s
2
2
2
2
24
6
mR
( 5.97 ×10 kg )( 6.38 × 10 m )
5
5
(b) We could not use this approach on the electron because in quantum physics we do not view it in the classical
sense as a spinning ball.
EVALUATE: The angular velocity we have just calculated for the Earth would certainly be masked by its present
angular spin of one revolution per day.
1
The potential U ( x ) = k ′x 2 is that of a simple harmonic oscillator. Treated quantum mechanically (see Section 40.4)
2
each energy state has energy En = %) (n + 1 ). Since electrons obey the exclusion principle, this allows us to put two
2
electrons (one for each ms = ± 1 ) for every value of neach quantum state is then defined by the ordered pair of
2
quantum numbers (n, ms ). By placing two electrons in each energy level the lowest energy is then
N −1
" N −1 "
1 ##
10
" N −1 #
/ N −1
/ ( N − 1)( N ) N 0
+ 2=
2 $ 6 En % = 2 $ 6 %ω $ n + % % = 2%ω 1 6 n + 6 2 = 2%ω 1
2 ''
2
24
&
3
n =0 2 4
& n =0 '
3 n=0
& n=0 %ω [ N 2 − N + N ] = %ω N 2 = %N 2 41.62. k′
.
m Here we used the hint from Problem 41.59 to do the first sum, realizing that the first value of n is zero and the last
value of n is N – 1, giving us a total of N energy levels filled.
(a) Apply Coulomb’s law to the orbiting electron and set it equal to the centripetal force. There is an attractive
(+2e)( −e) ( −e)(−e)
force with charge +2e a distance r away and a repulsive force a distance 2r away. So,
+
=
4π P0 r 2
4π P0 (2r ) 2
− mv 2
%
. But, from the quantization of angular momentum in the first Bohr orbit, L = mvr = % ! v =
.
r
mr
2 "%#
−m $
2
2
2
%
e
−2e
−mv
& mr ' = − % ! −7 e = − 4"P0% .
So
+
=
=
4 r2
mr 3
4"P0 r 2 4"P0 (4r ) 2
r
r
mr 3
2 2 2 4 " 4"P0% 2 # 4
4
−10
−11
r= $
% = a0 = (0.529 × 10 m) = 3.02 × 10 m.
7 & me 2 ' 7
7
−% 7 %
7
(1.054 × 10−34 J ⋅ s)
=
=
= 3.83 × 106 m s.
And v =
mr 4 ma0 4 (9.11× 10−31 kg)(0.529 × 10−10 m)
"1
#
(b) K = 2 $ mv 2 % = 9.11 × 10−31 kg (3.83 × 106 m s) 2 = 1.34 × 10 −17 J = 83.5 eV.
&2
' Atomic Structure 4113 " −2e2 #
−4e2
−7 " e 2 #
e2
e2
−17
=
+
=
(c) U = 2 $
%+
$
% = −2.67 × 10 J = −166.9 eV
4"P0 r ' 4"P0 (2r ) 4"P0 r 4"E0 (2r ) 2 & 4"P0 r '
&
(d) E∞ = −[ −166.9 eV + 83.5 eV] = 83.4 eV, which is only off by about 5% from the real value of 79.0 eV.
41.63. (a) The radius is inversely proportional to Z, so the classical turning radius is 2a Z .
(b) The normalized wave function is (1s ( r ) = outside the classical turning point is P = 5 ∞ 2a Z 1
3 "a Z 3 e − Z r a and the probability of the electron being found 2 (1s 4"r 2 dr =
∞ 4
a Z3
3 5 ∞ 2a Z e −2 Zr a r 2dr. Making the change of variable u = Zr a , dr = (a Z ) du changes the integral to P = 45 e −2uu 2du , which is independent of Z. The probability is
2 that found in Problem 41.39, 0.238, independent of Z. 42 MOLECULES AND CONDENSED MATTER 42.1. 3
2 K 2(7.9 × 10−4 eV)(1.60 × 10−19 J eV)
(a) K = kT ! T =
=
= 6.1 K
2
3k
3(1.38 × 10−23 J K)
2(4.48 eV) (1.60 × 10 −19 J eV)
(b) T =
= 34,600 K.
3(1.38 × 10−23 J K) (c) The thermal energy associated with room temperature (300 K) is much greater than the bond energy of He 2
(calculated in part (a)), so the typical collision at room temperature will be more than enough to break up He2 . 42.2.
42.3. However, the thermal energy at 300 K is much less than the bond energy of H 2 , so we would expect it to remain
intact at room temperature.
1 e2
(a) U = −
= −5.0 eV.
4".0 r
(b) −5.0 eV + (4.3 eV − 3.5 eV) = −4.2 eV.
IDENTIFY: The energy given to the photon comes from a transition between rotational states.
%2
SET UP: The rotational energy of a molecule is E = l (l + 1)
and the energy of the photon is E = hc/λ.
2I
EXECUTE: Use the energy formula, the energy difference between the l = 3 and l = 1 rotational levels of the
%2
5% 2
molecule is ∆E = [3(3 + 1) − 1(1 + 1) ] =
. Since ∆E = hc/λ, we get hc/λ = 5 % 2 /I. Solving for I gives
2I
I I= −34
5%λ 5 (1.055 × 10 J ⋅ s ) (1.780 nm)
=
= 4.981× 10−52 kg ⋅ m 2 .
2π c
2π ( 3.00 × 108 m/s ) Using I = mr r02, we can solve for r0: r0 = I ( mN + mH )
=
mN mH ( 4.981×10 −52 kg ⋅ m 2 )( 2.33 × 10−26 kg + 1.67 × 10−27 kg ) ( 2.33 ×10 −26 kg )(1.67 × 10−27 kg ) 42.4. r0 = 5.65 × 10–13 m
EVALUATE: This separation is much smaller than the diameter of a typical atom and is not very realistic. But we
are treating a hypothetical NH molecule.
The energy of the emitted photon is 1.01 × 10−5 eV, and so its frequency and wavelength are 42.5. E (1.01 × 10−5 eV)(1.60 × 10−19 J eV)
c (3.00 × 108 m s)
=
= 2.44 GHz and λ = =
= 0.123 m. This frequency
h
(6.63 × 10−34 J ⋅ s)
f
(2.44 × 109 Hz)
corresponds to that given for a microwave oven.
Let 1 refer to C and 2 to O. m1 = 1.993 × 10−26 kg, m2 = 2.656 × 10−26 kg, r0 = 0.1128 nm .
f= " m2 #
" m1 #
r1 = $
% r0 = 0.0644 nm (carbon) ; r2 = $
% r0 = 0.0484 nm (oxygen)
& m1 + m2 '
& m1 + m2 '
(b) I = m1r12 + m2r22 = 1.45 × 10−46 kg ⋅ m 2 ; yes, this agrees with Example 42.2.
42.6. Each atom has a mass m and is at a distance L 2 from the center, so the moment of inertia is
2( m)( L 2)2 = mL2 2 = 2.21 × 10−44 kg ⋅ m 2 . 42.7. IDENTIFY and SET UP: Set K = E1 from Example 42.2. Use K = 1 I ω 2 to solve for ω and v = rω to solve for v.
2
EXECUTE: (a) From Example 42.2, E1 = 0.479 meV = 7.674 × 10−23 J and I = 1.449 × 10−46 kg ⋅ m 2
K = 1 I ω 2 and K = E gives ω = 2 E1 / I = 1.03 × 1012 rad/s
2
421 422 Chapter 42 (b) v1 = r1ω1 = (0.0644 × 10−9 m)(1.03 × 1012 rad/s) = 66.3 m/s (carbon)
v2 = r2ω2 = (0.0484 × 10−9 m)(1.03 × 1012 rad/s) = 49.8 m/s (oxygen) (c) T = 2π / ω = 6.10 × 10−12 s
EVALUATE: From the information in Example 42.3 we can calculate the vibrational period to be
T = 2π / ω = 2π mr / k ′ = 1.5 × 10−14 s. The rotational motion is over an order of magnitude slower than the
42.8.
42.9. vibrational motion.
2
hc
" 2 "c #
∆E =
= % k ′ mr , and solving for k ′, k ′ = $
% mr = 205 N m.
λ
&λ'
IDENTIFY and SET UP: The energy of a rotational level with quantum number l is El = l (l + 1)% 2 / 2 I (Eq.(42.3)).
I = mr r 2 , with the reduced mass mr given by Eq.(42.4). Calculate I and ∆ E and then use ∆ E = hc / λ to find λ . EXECUTE: (a) mr = m1m2
mLi mH
(1.17 × 10−26 kg)(1.67 × 10−27 kg)
=
=
= 1.461 × 10−27 kg
m1 + m2 mLi + mH 1.17 × 10−26 kg + 1.67 × 10−27 kg I = mr r 2 = (1.461 × 10−27 kg)(0.159 × 10−9 m)2 = 3.694 × 10−47 kg ⋅ m 2 " %2 #
" %2 #
l = 3 : E = 3(4) $ % = 6 $ %
& 2I '
&I'
" %2 #
" %2 #
l = 4 : E = 4(5) $ % = 10 $ %
& 2I '
&I' 42.10. " %2 #
" (1.055 × 10−34 J ⋅ s) 2 #
−21
−3
∆ E = E4 − E3 = 4 $ % = 4 $
% = 1.20 × 10 J = 7.49 × 10 eV
I'
3.694 × 10−47 kg ⋅ m 2 '
&
&
hc (4.136 × 10−15 eV)(2.998 × 108 m/s)
=
= 166 µ m
(b) ∆ E = hc / λ so λ =
∆E
7.49 × 10−3 eV
EVALUATE: LiH has a smaller reduced mass than CO and λ is somewhat smaller here than the λ calculated for
CO in Example 42.2
IDENTIFY: The vibrational energy of the molecule is related to its force constant and reduced mass, while the
rotational energy depends on its moment of inertia, which in turn depends on the reduced mass.
1#
1#
k′
%2
"
"
SET UP: The vibrational energy is En = $ n + % %ω = $ n + % %
and the rotational energy is El = l (l + 1) .
2I
2'
2 ' mr
&
&
EXECUTE: For a vibrational transition, we have ∆Ev = % rotational transition is ∆ER =
mr r02 = 2% 2
, which gives
∆ER mr = k′
, so we first need to find mr. The energy for a
mr %2
2% 2
[ 2(2 + 1) − 1(1 + 1)] = . Solving for I and using the fact that I = mrr02, we have
2I
I 2 (1.055 × 10−34 J ⋅ s )( 6.583 × 10−16 eV ⋅ s )
2% 2
= 2.0014 × 10–28 kg
=
2
r02 ∆ER
0.8860 × 10−9 m ) (8.841× 10−4 eV )
( Now look at the vibrational transition to find the force constant.
∆Ev = % 42.11. k′
mr ! k′ = 2
( 2.0014 ×10−28 kg ) (0.2560 eV)2 = 30.27 N/m
mr ( ∆Ev )
=
2
%2
( 6.583 ×10−16 eV ⋅ s ) EVALUATE: This would be a rather weak spring in the laboratory.
l (l + 1)% 2
l (l − 1)% 2
%2
l% 2
, El −1 =
(a) El =
! ∆E = (l 2 + l − l 2 + l ) =
2I
2I
2I
I
E
E
l%
=
=
.
(b) f =
h
2"% 2"I Molecules and Condensed Matter 42.12. IDENTIFY: 423 Find ∆ E for the transition and compute λ from ∆ E = hc / λ . !2
!2
, with
= 0.2395 × 10−3 eV. From Example 42.3, ∆ E = 0.2690 eV
2I
2I
is the spacing between vibrational levels. Thus En = ( n + 1 )%ω , with %ω = 0.2690 eV. By Eq.(42.9),
2 SET UP: From Example 42.2, El = l (l + 1) %2
.
2I
(a) n = 0 → n = 1 and l = 1 → l = 2 E = En + El = ( n + 1 )%ω + l (l + 1)
2 EXECUTE: " %2 #
For n = 0, l = 1, Ei = 1 %ω + 2 $ % .
2
& 2I '
" %2 #
For n = 1, l = 2, E f = 3 %ω + 6 $ % .
2
& 2I '
" %2 #
∆ E = E f − Ei = %ω + 4 $ % = 0.2690 eV + 4(0.2395 × 10−3 eV) = 0.2700 eV
& 2I '
hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s)
hc
=
= 4.592 × 10−6 m = 4.592 µ m
= ∆ E so λ =
∆E
0.2700 eV
λ
(b) n = 0 → n = 1 and l = 2 → l = 1
" %2 #
For n = 0, l = 2, Ei = 1 %ω + 6 $ % .
2
& 2I '
" %2 #
For n = 1, l = 1, E f = 3 %ω + 2 $ % .
2
& 2I '
" %2 #
∆ E = E f − Ei = %ω − 4 $ % = 0.2690 eV − 4(0.2395 × 10−3 eV) = 0.2680 eV
& 2I '
−15
hc (4.136 × 10 eV ⋅ s)(2.998 × 108 m/s)
=
= 4.627 × 10−6 m = 4.627 µ m
λ=
∆E
0.2680 eV
(c) n = 0 → n = 1 and l = 3 → l = 2
" %2 #
For n = 0, l = 3, Ei = 1 %ω + 12 $ % .
2
& 2I '
" %2 #
For n = 1, l = 2, E f = 3 %ω + 6 $ % .
2
& 2I '
" %2 #
∆ E = E f − Ei = %ω − 6 $ % = 0.2690 eV − 6(0.2395 × 10−3 eV) = 0.2676 eV
& 2I '
−15
hc (4.136 × 10 eV ⋅ s)(2.998 × 108 m/s)
=
= 4.634 × 10−6 m = 4.634 µ m
λ=
∆E
0.2676 eV
EVALUATE: All three transitions are for n = 0 → n = 1. The spacing between vibrational levels is larger than the
spacing between rotational levels, so the difference in λ for the various rotational transitions is small. When the
transition is to a larger l, ∆ E > %ω and when the transition is to a smaller l, ∆ E < %ω.
42.13. (a) IDENTIFY and SET UP: Use ω = k ′ / mr and ω = 2π f to calculate k ′. The atomic masses are used in Eq.(42.4) to calculate mr .
EXECUTE:
mr = f= ω
2π = 1
2π k′
, so k ′ = mr (2π f ) 2
mr m1m2
mH mF
(1.67 × 10−27 kg)(3.15 × 10−26 kg)
=
=
= 1.586 × 10−27 kg
m1 + m2 mH + mF 1.67 × 10−27 kg + 3.15 × 10−26 kg k ′ = mr (2π f ) 2 = (1.586 × 10−27 kg)(2π [1.24 × 1014 Hz])2 = 963 N/m (b) IDENTIFY and SET UP: The energy levels are given by Eq.(42.7). En = (n + 1 )%ω = ( n + 1 )hf , since
2
2
%ω = (h / 2π )ω and (ω / 2π ) = f . The energy spacing between adjacent levels is ∆ E = En +1 − En = (n + 1 + 1 − n − 1 ) hf = hf , independent of n.
2
2 424 Chapter 42 EXECUTE: ∆ E = hf = (6.626 × 10−34 J ⋅ s)(1.24 × 1014 Hz) = 8.22 × 10−20 J = 0.513 eV
(c) IDENTIFY and SET UP: The photon energy equals the transition energy so ∆ E = hc / λ .
c 2.998 × 108 m/s
=
= 2.42 × 10−6 m = 2.42 µ m
1.24 × 1014 Hz
f
EVALUATE: This photon is infrared, which is typical for vibrational transitions.
For an average spacing a, the density is / = m a 3 , where m is the average of the ionic masses, and so
EXECUTE:
42.14. hf = hc / λ so λ = a3 = 42.15. −26
−25
m ( 6.49 × 10 kg + 1.33 × 10 kg ) 2
=
= 3.60 × 10−29 m3 ,
(2.75 × 103 kg m3 )
/ and a = 3.30 × 10−10 m = 0.330 nm .
(b) The larger (higher atomic number) atoms have the larger spacing.
IDENTIFY and SET UP: Find the volume occupied by each atom. The density is the average mass of Na and Cl
divided by this volume.
EXECUTE: Each atom occupies a cube with side length 0.282 nm. Therefore, the volume occupied by each atom
is V = (0.282 × 10−9 m)3 = 2.24 × 10−29 m 3. In NaCl there are equal numbers of Na and Cl atoms, so the average
mass of the atoms in the crystal is m = 1 ( mNa + mCl ) = 1 (3.82 × 10−26 kg + 5.89 × 10−26 kg) = 4.855 × 10−26 kg
2
2
m 4.855 × 10−26 kg
=
= 2.17 × 103 kg/m3 .
2.24 × 10−29 m3
V
EVALUATE: The density of water is 1.00 × 103 kg/m 3 , so our result is reasonable. The density then is ρ = 42.16. −34
8
hc ( 6.63 × 10 J ⋅ s ) ( 3.00 × 10 m s )
=
= 0.200 nm.
E ( 6.20 × 103 eV ) (1.60 × 10−19 J eV ) (a) As a photon, λ = (b) As a matter wave, λ= ( 6.63 ×10−34 J ⋅ s )
h
h
=
=
= 0.200 nm
p
2mE
2 ( 9.11× 10−31 kg ) ( 37.6 eV ) (1.60 × 10−19 J eV ) (c) As a matter wave, λ=
42.17. ( 6.63 ×10−34 J ⋅ s )
h
=
= 0.200 nm .
2mE
2 (1.67 × 10−27 kg ) ( 0.0205 eV ) (1.60 × 10−19 J eV ) IDENTIFY: The energy gap is the energy of the maximumwavelength photon.
SET UP: The energy difference is equal to the energy of the photon, so ∆E = hc/λ.
EXECUTE: (a) Using the photon wavelength to find the energy difference gives
∆E = hc/λ = (4.136 × 10–15 eV ⋅ s )(3.00 × 108 m/s)/(1.11 × 10–6 m) = 1.12 eV 42.18. 42.19. (b) A wavelength of 1.11 µ m = 1110 nm is in the infrared, shorter than that of visible light.
EVALUATE: Since visible photons have more than enough energy to excite electrons from the valence to the
conduction band, visible light will be absorbed, which makes silicon opaque.
hc
(a)
= 2.27 × 10−7 m = 227 nm , in the ultraviolet.
∆E
(b) Visible light lacks enough energy to excite the electrons into the conduction band, so visible light passes
through the diamond unabsorbed.
(c) Impurities can lower the gap energy making it easier for the material to absorb shorter wavelength visible light.
This allows longer wavelength visible light to pass through, giving the diamond color.
hc (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s)
∆E =
=
= 2.14 × 10−13 J = 1.34 × 106 eV . So the number of electrons that can be
9.31 × 10−13 m
λ excited to the conduction band is n =
42.20. 1.34 × 106 eV
= 1.20 × 106 electrons
1.12 eV 1 = 5 ( dV
2 3
L
"L
" n "y # # " L
" n "x # # "
" n "z # #
"L#
= A2 $ 5 sin 2 $ x % dx % $ 5 sin 2 $ y % dy % $ 5 sin 2 $ z % dz % = A2 $ %
$
%
& L ' '&0
&L' '
&2'
& L ' '&0
&0 so A = ( 2 L ) 32 (assuming A to be real positive). Molecules and Condensed Matter 42.21. 425 Density of states:
g(E) = ( 2m ) 32 V 23 2" % E1 2 = (2(9.11 × 10−31 kg))3 2 (1.0 × 10−6 m3 )(5.0 eV)1 2 (1.60 × 10−19 J eV)1 2
2" 2 (1.054 × 10−34 J ⋅ s)3 g ( E ) = ( 9.5 × 1040 states J ) (1.60 × 10−19 J eV ) = 1.5 × 1022 states eV.
42.22. 42.23. vrms = 3kT m = 1.17 × 105 m s , as found in Example 42.9. The equipartition theorem does not hold for the
electrons at the Fermi energy. Although these electrons are very energetic, they cannot lose energy, unlike
electrons in a free electron gas.
% 2 " ∂ 2ψ ∂ 2ψ ∂ 2ψ #
+
+
(a) IDENTIFY and SET UP: The threedimensional Schrödinger equation is −
$
% + Uψ = Eψ
2m & ∂x 2 ∂y 2 ∂z 2 '
(Eq.40.29). For free electrons, U = 0. Evaluate ∂ 2ψ /∂x 2 , ∂ 2ψ /∂y 2 , and ∂ 2ψ / ∂z 2 for ψ as given by Eq.(42.10).
Put the results into Eq.(40.20) and see if the equation is satisfied.
∂ψ nxπ
" n πx# " n π y # " n πz#
A cos $ x % sin $ y % sin $ z %
EXECUTE:
=
L
∂x
&L' &L' &L'
∂ 2ψ
"nπ #
"n πx# "n π y# " n πz #
"nπ #
= − $ x % A sin $ x % sin $ y % sin $ z % = − $ x % ψ
L'
L' &L' &L'
∂x 2
&
&
&L'
2 2 2 Similarly "n π #
∂ 2ψ
∂ 2ψ
"nπ #
= − $ y % ψ and
= −$ z % ψ .
2
∂y
L'
∂z 2
&L'
&
2 2
(n 2 + ny + nz2 )π 2% 2
# %2 " π 2 # 2
2
(nx + ny + nz2 )ψ = x
=
ψ
%
$ 2%
2mL2
' 2m & L '
2
( n 2 + n y + nz2 )π 2% 2
This equals Eψ , with E = x
, which is Eq.(42.11).
2mL2
EVALUATE: ψ given by Eq.(42.10) is a solution to Eq.(40.29), with E as given by Eq.(42.11).
(b) IDENTIFY and SET UP: Find the set of quantum numbers nx , n y , and nz that give the lowest three values of Therefore, − % 2 " ∂ 2ψ ∂ 2ψ ∂ 2ψ
+
+
$
2m & ∂x 2 ∂y 2 ∂z 2 E. The degeneracy is the number of sets nx , n y , nz and ms that give the same E.
3π 2% 2
. No other combination of nx , n y , and nz
2mL2
gives this same E, so the only degeneracy is the degeneracy of two due to spin.
π 2% 2 6π 2% 2
=
First excited level: next lower E so one n equals 2 and the others equal 1. E = (22 + 12 + 12 )
2mL2 2mL2
There are three different sets of nx , n y , nz values that give this E: EXECUTE: Ground level: lowest E so nx = n y = nz = 1 and E = nx = 2, ny = 1, nz = 1; nx = 1, ny = 2, nz = 1; nx = 1, n y = 1, nz = 2
This gives a degeneracy of 3 so the total degeneracy, with the factor of 2 from spin, is 6.
Second excited level: next lower E so two of nx , n y , nz equal 2 and the other equals 1.
9π 2% 2
2mL2 2mL2
There are different sets of nx , n y , nz values that give this E:
E = (22 + 2 2 + 12 ) π 2% 2 = nx = 2, ny = 2, nz = 1; nx = 2, ny = 1, nz = 2; nx = 1, ny = 2, nz = 2. 42.24.
42.25. Thus, as for the first excited level, the total degeneracy, including spin, is 6.
EVALUATE: The wavefunction for the 3dimensional box is a product of the wavefunctions for a 1dimensional
box in the x, y, and z coordinates and the energy is the sum of energies for three 1dimensional boxes. All levels
except for the ground level have a degeneracy greater than two. Compare to the 3dimensional isotropic harmonic
oscillator treated in Problem 40.53.
12
Eq.(42.13) may be solved for nrs = ( 2mE ) ( L %" ), and substituting this into Eq. (42.12), using L3 = V , gives Eq.(42.14).
(a) IDENTIFY and SET UP: The electron contribution to the molar heat capacity at constant volume of a metal is
" π 2 KT #
CV = $
% R.
& 2 EF '
π 2 (1.381 × 10 −23 J/K)(300 K)
EXECUTE: CV =
R = 0.0233R.
2(5.48 eV)(1.602 × 10−19 J/eV) 426 Chapter 42 (b) EVALUATE: The electron contribution found in part (a) is 0.0233R = 0.194 J/mol ⋅ K. This is
0.194 / 25.3 = 7.67 × 10−3 = 0.767% of the total CV .
(c) Only a small fraction of CV is due to the electrons. Most of CV is due to the vibrational motion of the ions.
42.26. 3
(a) From Eq. (42.22), Eav = EF = 1.94 eV.
5
(b)
(c) 42.27. 42.28. 2E m = 2 (1.94 eV ) (1.60 × 10−19 J eV )
9.11× 10−31 kg = 8.25 × 105 m s. −19
EF ( 3.23 eV ) (1.60 × 10 J eV )
=
= 3.74 × 104 K.
k
(1.38 ×10−23 J K ) IDENTIFY: The probability is given by the FermiDirac distribution.
1
SET UP: The FermiDirac distribution is f ( E ) = ( E − EF ) / kT
.
e
+1
EXECUTE: We calculate the value of f (E), where E = 8.520 eV, EF = 8.500 eV,
k = 1.38 × 10–23 J/K = 8.625 × 10–5 eV/K, and T = 20°C = 293 K. The result is f (E) = 0.312 = 31.2%.
EVALUATE: Since the energy is close to the Fermi energy, the probability is quite high that the state is occupied
by an electron.
(a) See Example 42.10: The probabilities are 1.78 × 10−7 , 2.37 × 10−6 , and 1.51 × 10 −5 .
(b) The Fermi distribution, Eq.(42.17), has the property that f ( EF − E ) = 1 − f ( E ) (see Problem (42.48)), and so 42.29. 42.30. the probability that a state at the top of the valence band is occupied is the same as the probability that a state of the
bottom of the conduction band is filled (this result depends on having the Fermi energy in the middle of the gap).
1
IDENTIFY: Use Eq.(42.17), f ( E ) = ( E − EF ) / kT
. Solve for E − EF .
e
+1
1
SET UP: e( E − EF ) / kT =
−1
f (E)
The problem states that f ( E ) = 4.4 × 10−4 for E at the bottom of the conduction band.
1
− 1 = 2.272 × 103.
EXECUTE: e( E − EF ) / kT =
4.4 × 10−4
E − EF = kT ln(2.272 × 103 ) = (1.3807 × 10−23 J/T)(300 K)ln(2.272 × 103 ) = 3.201× 10−20 J = 0.20 eV
EF = E − 0.20 eV; the Fermi level is 0.20 eV below the bottom of the conduction band.
EVALUATE: The energy gap between the Fermi level and bottom of the conduction band is large compared to
kT at T = 300 K and as a result f ( E ) is small.
IDENTIFY: The current depends on the voltage across the diode and its temperature, so the resistance also
depends on these quantities.
SET UP: The current is I = IS (eeV/kT – 1) and the resistance is R = V/I.
V
V
eV
e(0.0850 V)
=
EXECUTE: (a) The resistance is R = =
=
. The exponent is
I IS ( eeV / kT − 1)
kT (8.625 × 10−5 eV/K ) (293 K)
3.3635, giving R = 85.0 mV
= 4.06 .
(0.750 mA) ( e3.3635 − 1) (b) In this case, the exponent is which gives R = 42.31. eV
e(−0.050 V)
=
= −1.979
kT ( 8.625 × 10−5 eV/K ) (293 K) −50.0 mV
= 77.4
(0.750 mA) ( e−1.979 − 1) EVALUATE: Reversing the voltage can make a considerable change in the resistance of a diode.
IDENTIFY and SET UP: The voltagecurrent relation is given by Eq.(42.23): I = Is (eeV / kT − 1). Use the current for V = +15.0 mV to solve for the constant I s .
EXECUTE: (a) Find Is : V = +15.0 × 10−3 V gives I = 9.25 × 10 −3 A
eV (1.602 × 10−19 C)(15.0 × 10−3 V)
=
= 0.5800
kT
(1.381 × 10 −23 J/K)(300 K)
I
9.25 × 10 −3 A
I s = eV / kT
=
= 1.177 × 10−2 = 11.77 mA
e
−1
e0.5800 − 1 427 Molecules and Condensed Matter Then can calculate I for V = 10.0 mV: −19 −3 eV (1.602 × 10 C)(10.0 × 10 V)
=
= 0.3867
kT
(1.381 × 10 −23 J/K)(300 K) I = Is (eeV / kT − 1) = (11.77 mA)(e0.3867 − 1) = 5.56 mA
eV
eV
has the same magnitude as in part (a) but not V is negative so
is negative.
kT
kT
eV
V = −15.0 mV :
= −0.5800 and I = I s (eeV / kT − 1) = (11.77 mA)(e −0.5800 − 1) = −5.18 mA
kT
eV
V = −10.0 mV :
= −0.3867 and I = I s (eeV / kT − 1) = (11.77 mA)(e−0.3867 − 1) = −3.77 mA
kT
EVALUATE: There is a directional asymmetry in the current, with a forwardbias voltage producing more current
than a reversebias voltage of the same magnitude, but the voltage is small enough for the asymmetry not be
pronounced. Compare to Example 42.11, where more extreme voltages are considered.
(a) Solving Eq.(42.23) for the voltage as a function of current,
(b) 42.32. V= 42.33. # kT " 40.0 mA #
kT " I
ln $ + 1% =
ln $
+ 1% = 0.0645 V.
e
& 3.60 mA '
& IS ' e (b) From part (a), the quantity eeV kT = 12.11 , so far a reversebias voltage of the same magnitude,
"1
#
I = I S ( e − eV kT − 1) = I S $
− 1 % = −3.30 mA .
& 12.11 '
IDENTIFY: During the transition, the molecule emits a photon of light having energy equal to the energy
difference between the two vibrational states of the molecule.
1#
1#
k′
"
"
SET UP: The vibrational energy is En = $ n + % %ω = $ n + % %
.
2'
2 ' mr
&
&
EXECUTE: (a) The energy difference between two adjacent energy states is ∆E = % k′
, and this is the energy of
mr
2 " ∆E #
the photon, so ∆E = hc/λ. Equating these two expressions for ∆E and solving for k ′ , we have k ′ = mr $
%=
&%'
2
∆E hc / λ 2π c
mH mO " ∆E #
=
=
with the appropriate numbers gives us
$
% , and using
%
%
λ
mH + mO & % ' k′ = ω
1
=
(b) f =
2π 2π (1.67 ×10 −27 1.67 × 10−27 k′
1
=
mr 2π kg )( 2.656 × 10−26 kg ) / 2π ( 3.00 × 108 m/s ) 0
1
2 = 977 N/m
kg + 2.656 × 10−26 kg 1 2.39 × 10−6 m 2
3
4
2 mH mO
mH + mO
. Substituting the appropriate numbers gives us
k′ (1.67 ×10
f= 42.34.
42.35. 1
2π −27 kg )( 2.656 × 10−26 kg ) 1.67 × 10−27 kg + 2.656 × 10−26 kg
= 1.25 × 1014 Hz
977 N/m EVALUATE: The frequency is close to, but not quite in, the visible range.
2% 2
hλ
I=
=
= 7.14 × 10−48 kg ⋅ m 2 .
∆E 2" 2c
IDENTIFY and SET UP: Eq.(21.14) gives the electric dipole moment as p = qd , where the dipole consists of
charges ± q separated by distance d.
EXECUTE: (a) Point charges +e and −e separated by distance d, so
p = ed = (1.602 × 10−19 C)(0.24 × 10−9 m) = 3.8 × 10−29 C ⋅ m
p 3.0 × 10−29 C ⋅ m
=
= 1.3 × 10−19 C
d
0.24 × 10−9 m
q
1.3 × 10−19 C
= 0.81
(c) =
e 1.602 × 10−19 C (b) p = qd so q = 428 Chapter 42 p 1.5 × 10 −30 C ⋅ m
=
= 9.37 × 10 −21 C
d
0.16 × 10 −9 m
q 9.37 × 10−21 C
=
= 0.058
e 1.602 × 10−19 C
EVALUATE: The fractional ionic character for the bond in HI is much less than the fractional ionic character for
the bond in NaCl. The bond in HI is mostly covalent and not very ionic.
1 e2
The electrical potential energy is U = −5.13 eV, and r = −
= 2.8 × 10−10 m.
4"P0 U (d) q = 42.36.
42.37. (a) IDENTIFY: E (Na) + E (Cl) = E (Na + ) + E (Cl− ) + U ( r ). Solving for U ( r ) gives U ( r ) = −[ E (Na + ) − E (Na)] + [ E (Cl) − E (Cl− )]. SET UP: [ E (Na + ) − E (Na)] is the ionization energy of Na, the energy required to remove one electron, and is equal to 5.1 eV. [ E (Cl) − E (Cl− )] is the electron affinity of Cl, the magnitude of the decrease in energy when an
electron is attached to a neutral Cl atom, and is equal to 3.6 eV.
1 e2
EXECUTE: U = −5.1 eV + 3.6 eV = −1.5 eV = −2.4 × 10−19 J, and −
= −2.4 × 10−19 J
4π P0 r
"1#
(1.602 × 10−19 C)2
e2
= (8.988 × 109 N ⋅ m 2 /C 2 )
r =$
%
−19
2.4 × 10 −19 J
& 4π P0 ' 2.4 × 10 J
r = 9.6 × 10−10 m = 0.96 nm
(b) ionization energy of K = 4.3 eV; electron affinity of Br = 3.5 eV
Thus U = −4.3 eV + 3.5 eV = −0.8 eV = −1.28 × 10−19 J, and − 42.38. 1 e2
= −1.28 × 10−19 J
4π P0 r "1#
(1.602 × 10−19 C) 2
e2
= (8.988 × 109 N ⋅ m 2 / C2 )
r =$
%
−19
1.28 × 10−19 J
& 4π P0 ' 1.28 × 10 J
r = 1.8 × 10−9 m = 1.8 nm
EVALUATE: K has a smaller ionization energy than Na and the electron affinities of Cl and Br are very similar,
so it takes less energy to make K + + Br − from K + Br than to make Na + + Cl− from Na + Cl. Thus, the
stabilization distance is larger for KBr than for NaCl.
The energies corresponding to the observed wavelengths are 3.29 × 10−21 J, 2.87 × 10−21 J, 2.47 × 10 −21 J,
2.06 × 10−21 J and 1.65 × 10−21 J. The average spacing of these energies is 0.410 × 10−21 J and these are seen to correspond to transition from levels 8, 7, 6, 5 and 4 to the respective next lower levels. Then,
42.39. %2
= 0.410 × 10−21 J ,
I from which I = 2.71 × 10−47 kg ⋅ m 2 .
(a) IDENTIFY: The rotational energies of a molecule depend on its moment of inertia, which in turn depends on
the separation between the atoms in the molecule.
SET UP: Problem 42.38 gives I = 2.71 × 10 −47 kg ⋅ m 2 . I = mr r 2 . Calculate mr and solve for r.
EXECUTE: mr = mH mCl
(1.67 × 10−27 kg)(5.81 × 10−26 kg)
=
= 1.623 × 10−27 kg
mH + mCl 1.67 × 10−27 kg + 5.81 × 10−26 kg I
2.71 × 10 −47 kg ⋅ m 2
=
= 1.29 × 10 −10 m = 0.129 nm
mr
1.623 × 10−27 kg
EVALUATE: This is a typical atomic separation for a diatomic molecule; see Example 42.2 for the corresponding
distance for CO.
(b) IDENTIFY: Each transition is from the level l to the level l − 1. The rotational energies are given by Eq.(42.3).
The transition energy is related to the photon wavelength by ∆ E = hc / λ .
r= " %2 # " %2 #
El = l (l + 1)% 2 / 2 I , so ∆ E = El − El −1 = [l (l + 1) − l (l − 1)]$ % = l $ % .
& 2I ' & I '
2
" % # hc
EXECUTE: l $ % =
&I' λ
SET UP: l= 2π cI 2π (2.998 × 108 m/s)(2.71× 10−47 kg ⋅ m 2 ) 4.843 × 10−4 m
=
=
(1.055 × 10−34 J ⋅ s)λ
λ
%λ Molecules and Condensed Matter 429 4.843 × 10−4 m
= 8.
60.4 × 10−6 m
4.843 × 10−4 m
= 7.
For λ = 69.0 µ m, l =
69.0 × 10−6 m
4.843 × 10−4 m
= 6.
For λ = 80.4 µ m, l =
80.4 × 10−6 m
4.843 × 10−4 m
= 5.
For λ = 96.4 µ m, l =
96.4 × 10−6 m
4.843 × 10−4 m
= 4.
For λ = 120.4 µ m, l =
120.4 × 10−6 m
EVALUATE: In each case l is an integer, as it must be.
(c) IDENTIFY and SET UP: Longest λ implies smallest ∆ E , and this is for the transition from l = 1 to l = 0.
For λ = 60.4 µ m, l = " %2 #
(1.055 × 10−34 J ⋅ s)2
∆ E = l $ % = (1)
= 4.099 × 10−22 J
2.71 × 10−47 kg ⋅ m 2
I'
&
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
λ=
=
= 4.85 × 10−4 m = 485 µ m.
∆E
4.099 × 10−22 J
EVALUATE: This is longer than any wavelengths in part (b).
(d) IDENTIFY: What changes is mr , the reduced mass of the molecule.
EXECUTE: " %2 #
hc
2π cI
SET UP: The transition energy is ∆ E = l $ % and ∆ E = , so λ =
(part (b)). I = mr r 2 , so λ is directly
l%
λ
&I'
λ (HCl) λ (DCl)
m (DCl)
proportional to mr .
=
so λ (DCl) = λ (HCl) r
mr (HCl) mr (DCl)
mr (HCl)
EXECUTE: The mass of a deuterium atom is approximately twice the mass of a hydrogen atom, so
mD = 3.34 × 10−27 kg.
mr (DCl) = m D m Cl
(3.34 × 10−27 kg)(5.81× 10−27 kg)
=
= 3.158 × 10−27 kg
m D + m Cl 3.34 × 10−27 kg + 5.81 × 10−26 kg " 3.158 × 10−27 kg #
% = (1.946)λ (HCl)
−27
& 1.623 × 10 kg '
l = 8 → l = 7; λ = (60.4 µ m)(1.946) = 118 µ m
l = 7 → l = 6; λ = (69.0 µ m)(1.946) = 134 µ m
l = 6 → l = 5; λ = (80.4 µ m)(1.946) = 156 µ m
l = 5 → l = 4; λ = (96.4 µ m)(1.946) = 188 µ m
l = 4 → l = 3; λ = (120.4 µ m)(1.946) = 234 µ m
EVALUATE: The moment of inertia increases when H is replaced by D, so the transition energies decrease and
the wavelengths increase. The larger the rotational inertia the smaller the rotational energy for a given l (Eq.42.3).
% 2l hl λ
From the result of Problem 42.11, the moment inertia of the molecule is I =
=
= 6.43 × 10−46 kg ⋅ m 2 and
∆E 4" 2c λ (DCl) = λ (HCl) $ 42.40. from Eq.(42.6) the separation is r0 =
42.41. I
= 0.193 nm.
mr L2 % 2l (l + 1)
=
. Eg = 0 (l = 0), and there is an additional multiplicative factor of 2l + 1 because for each l
2I
2I
2
n
state there are really (2l + 1) ml states with the same energy. So l = (2l + 1)e − % l ( l +1) /(2 IkT ) .
n0
(a) Eex = (b) T = 300 K, I = 1.449 × 10−46 kg ⋅ m 2 . % 2 (1) (1 + 1)
E
7.67 × 10−23 J
= 7.67 × 10−23 J. l =1 =
= 0.0185.
2
−46
2(1.449 × 10 kg ⋅ m )
kT (1.38 × 10−23 J K) (300 K)
n
( 2l + 1) = 3 , so l =1 = (3)e −0.0185 = 2.95.
n0 (i) El =1 = 4210 Chapter 42 El = 2
% 2 (2) (2 + 1)
=
= 0.0556.
2(1.449 × 10−46 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K)
kT
n
( 2l + 1) = 5 , so l =1 = (5)(e −0.0556 ) = 4.73.
n0 (ii) (iii) El =10
% 2 (10) (10 + 1)
=
= 1.02.
kT
2(1.449 × 10−46 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K) (2l + 1) = 21, so
(iv) nl =10
= (21) (e −1.02 ) = 7.57.
n0 El = 20
% 2 (20) (20 + 1)
=
= 3.89.
kT
2(1.449 × 10−46 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K) ( 2l + 1) = 41 , so (v) nl = 20
= (41)e −3.89 = 0.838.
n0 El = 50
% 2 (50) (50 + 1)
=
= 23.6.
−46
kT
2(1.449 × 10 kg ⋅ m 2 ) (1.38 × 10−23 J K) (300 K) nl =50
= (101)e −23.6 = 5.69 × 10−9.
n0
(c) There is a competing effect between the (2l + 1) term and the decaying exponential. The 2l + 1 term dominates
for small l, while the exponential term dominates for large l.
(a) I CO = 1.449 × 10−46 kg ⋅ m 2 .
( 2l + 1) = 101 , so 42.42. El =1 = % 2l (l + 1) (1.054 × 10−34 J ⋅ s)2 (1) (1 + 1)
=
= 7.67 × 10−23 J . El = 0 = 0.
2I
2(1.449 × 10−46 kg ⋅ m 2 ) ∆E = 7.67 × 10−23 J = 4.79 × 10 −4 eV. hc (6.63 × 10−34 J ⋅ s) (3.00 × 108 m s)
=
= 2.59 × 10−3 m = 2.59 mm.
∆E
(7.67 × 10−23 J)
(b) Let’s compare the value of kT when T = 20 K to that of ∆E for the l = 1 →= 0 rotational transition:
l λ= kT = (1.38 × 10−23 J K) (20 K) = 2.76 × 10−22 J.
kT
= 3.60.
∆E
Therefore, although T is quite small, there is still plenty of energy to excite CO molecules into the first rotational
level. This allows astronomers to detect the 2.59 mm wavelength radiation from such molecular clouds.
IDENTIFY and SET UP: El = l (l + 1)% 2 / 2 I , so El and the transition energy ∆ E depend on I. Different isotopic
molecules have different I.
mNa mCl
(3.8176 × 10−26 kg)(5.8068 × 10−26 kg)
EXECUTE: (a) Calculate I for Na 35Cl: mr =
=
= 2.303 × 10 −26 kg
mNa + mCl 3.8176 × 10−26 kg + 5.8068 × 10−26 kg
∆E = 7.67 × 10−23 J (from part (a)). So 42.43. I = mr r 2 = (2.303 × 10−26 kg)(0.2361× 10−9 m)2 = 1.284 × 10−45 kg ⋅ m2
l = 2 → l = 1 transition
" % 2 # 2% 2 2(1.055 × 10−34 J ⋅ s) 2
∆ E = E2 − E1 = (6 − 2) $ % =
=
= 1.734 × 10−23 J
2I '
I
1.284 × 10−45 kg ⋅ m 2
&
hc
hc (6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s)
∆E =
so λ =
=
= 1.146 × 10 −2 m = 1.146 cm
λ
∆E
1.734 × 10 −23 J
l = 1 →= 0 transition
l
" %2 # %2 1
∆ E = E1 − E0 = (2 − 0) $ % =
= (1.734 × 10−23 J) = 8.67 × 10−24 J
& 2I ' I 2
hc (6.626 × 10 −34 J ⋅ s)(2.998 × 108 m/s)
=
= 2.291 cm
λ=
∆E
8.67 × 10−24 J
mNa mCl
(3.8176 × 10−26 kg)(6.1384 × 10−26 kg)
(b) Calculate I for Na 37Cl: mr =
=
= 2.354 × 10−26 kg
mNa + mCl 3.8176 × 10 −26 kg + 6.1384 × 10−26 kg I = mr r 2 = (2.354 × 10−26 kg)(0.2361 × 10−9 m)2 = 1.312 × 10−45 kg ⋅ m2 Molecules and Condensed Matter 4211 l = 2 → l = 1 transition
2% 2 2(1.055 × 10−34 J ⋅ s) 2
=
= 1.697 × 10 −23 J
I
1.312 × 10 −45 kg ⋅ m 2
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 1.171 × 10−2 m = 1.171 cm
λ=
∆E
1.697 × 10−23 J
∆E = l = 1 →= 0 transition
l %2 1
= (1.697 × 10−23 J) = 8.485 × 10−24 J
I2
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
λ=
=
= 2.341 cm
∆E
8.485 × 10−24 J
The differences in the wavelengths for the two isotopes are:
l = 2 → l = 1 transition: 1.171 cm − 1.146 cm = 0.025 cm
l = 1 → l = 0 transition: 2.341 cm − 2.291 cm = 0.050 cm
EVALUATE: Replacing 35 Cl by 37 Cl increases I, decreases ∆ E and increases λ . The effect on λ is small but
measurable.
∆E
The vibration frequency is, from Eq.(42.8), f =
= 1.12 × 1014 Hz. The force constant is
h
k ′ = (2"f ) 2 mr = 777 N m.
∆E = 42.44. 42.45. 1#
k′
1 2k ′
"
En = $ n + % %
! E0 = %
2'
mr
2 mH
&
1
2(576 N m )
! E0 = (1.054 × 10−34 J ⋅ s)
= 4.38 × 10−20 J = 0.274 eV.
2
1.67 × 10−27 kg 42.46. This is much less than the H 2 bond energy.
(a) The frequency is proportional to the reciprocal of the square root of the reduced mass, and in terms of the
atomic masses, the frequency of the isotope with the deuterium atom is
12 " m m ( mH + mF ) #
f = f0 $ F H
%
& mF mD ( mD + mF ) '
42.47. 12 " 1 + ( mF mD ) #
= f0 $
%.
& 1 + (mF mH ) ' Using f 0 from Exercise 42.13 and the given masses, f = 8.99 × 1013 Hz.
IDENTIFY and SET UP: Use Eq.(42.6) to calculate I. The energy levels are given by Eq.(42.9). The transition
energy ∆ E is related to the photon wavelength by ∆ E = hc / λ .
EXECUTE: (a) mr = mH mI
(1.67 × 10−27 kg)(2.11 × 10−25 kg)
=
= 1.657 × 10−27 kg
mH + mI 1.67 × 10−27 kg + 2.11 × 10−25 kg I = mr r 2 = (1.657 × 10−27 kg)(0.160 × 10−9 m)2 = 4.24 × 10−47 kg ⋅ m 2 " %2 #
k′
(b) The energy levels are Enl = l (l + 1) $ % + ( n + 1 )%
(Eq.(42.9))
2
2I '
mr
&
" %2 #
k′
= ω = 2π f so Enl = l (l + 1) $ % + ( n + 1 )hf
2
m
& 2I ' (i) transition n = 1 → n = 0, l = 1 → l = 0
" %2 #
%2
∆ E = (2 − 0) $ % + (1 + 1 − 1 )hf = + hf
2
2
I
& 2I '
hc
hc
hc
c
∆E =
so λ =
=
=
λ
∆ E (% 2 / I ) + hf (% / 2π I ) + f
% 2π I λ= = 1.055 × 10−34 J ⋅ s
= 3.960 × 1011 Hz
2π (4.24 × 10−47 kg ⋅ m 2 ) c
2.998 × 108 m/s
=
= 4.30 µ m
(% / 2π I ) + f 3.960 × 1011 Hz + 6.93 × 1013 Hz 4212 Chapter 42 (ii) transition n = 1 → n = 0, l = 2 → l = 1
" %2 #
2% 2
∆ E = (6 − 2) $ % + hf =
+ hf
I
& 2I '
c
2.998 × 108 m/s
λ=
=
= 4.28 µ m
2(% / 2π I ) + f 2(3.960 × 1011 Hz) + 6.93 × 1013 Hz
(iii) transition n = 2 → n = 1, l = 2 → l = 3 42.48.
42.49. " %2 #
3% 2
∆ E = (6 − 12) $ % + hf = −
+ hf
I
& 2I '
c
2.998 × 108 m/s
=
= 4.40 µ m
λ=
−3(% / 2π I ) + f −3(3.960 × 1011 Hz) + 6.93 × 1013 Hz
EVALUATE: The vibrational energy change for the n = 1 → n = 0 transition is the same as for the n = 2 → n = 1
transition. The rotational energies are much smaller than the vibrational energies, so the wavelengths for all three
transitions don’t differ much.
1
1
1
e −∆E / kT
The sum of the probabilities is f ( EF + ∆E ) + f ( EF − ∆E ) = −∆E kT
+ ∆E kT
= −∆E kT
+
= 1.
e
+1 e
+1 e
+ 1 1 + e −∆E kT
32 3 " 4 3% 2n 2 3
Since potassium is a metal we approximate EF = EF0 . ! EF =
.
2m
/
851 kg m3
= 1.31× 1028 electron m3
But the electron concentration n = ! n =
m
6.49 × 10−26 kg 32 3 " 4 3 (1.054 × 10−34 J ⋅ s)2 (1.31× 1028 /m3 )2 3
= 3.24 × 10−19 J = 2.03 eV.
2(9.11× 10−31 kg)
IDENTIFY: The only difference between the two isotopes is their mass, which will affect their reduced mass and
hence their moment of inertia.
%2
SET UP: The rotational energy states are given by E = l (l + 1)
and the reduced mass is given by mr =
2I
m1m2/(m1 + m2).
EXECUTE: (a) If we call m the mass of the Hatom, the mass of the deuterium atom is 2m and the reduced masses
of the molecules are
H2 (hydrogen): mr(H) = mm/(m + m) = m/2
D2 (deuterium): mr(D) = (2m)(2m)/(2m + 2m) = m
Using I = mr r02, the moments of inertia are IH = mr02/2 and ID = mr02. The ratio of the rotational energies is then
! EF = 42.50. 2
EH l (l + 1) ( % / 2 I H ) I D mr02
=
=
=
= 2.
ED l (l + 1) ( % 2 / 2 I D ) I H m r 2
0
2 1#
k′
"
$ n + %%
2 ' mr (H)
E
mr (D)
m
&
=
=
= 2.
(b) The ratio of the vibrational energies is H =
ED "
mr (H)
m/2
k′
1#
$ n + %%
2 ' mr (D)
& 41.51. EVALUATE: The electrical force is the same for both molecules since both H and D have the same charge, so it is
reasonable that the force constant would be the same for both of them.
IDENTIFY and SET UP: Use the description of the bcc lattice in Fig.42.12c in the textbook to calculate the
number of atoms per unit cell and then the number of atoms per unit volume.
EXECUTE: (a) Each unit cell has one atom at its center and 8 atoms at its corners that are each shared by 8 other
unit cells. So there are 1 + 8/ 8 = 2 atoms per unit cell.
2
n
=
= 4.66 × 1028 atoms/m3
V (0.35 × 10−9 m)3
(b) EF0 = 32 / 3 π 4 / 3% 2 " N #
$%
2m
&V ' 2/3 In this equation N/V is the number of free electrons per m3 . But the problem says to assume one free electron per
atom, so this is the same as n/V calculated in part (a).
m = 9.109 × 10 −31 kg (the electron mass), so EF0 = 7.563 × 10 −19 J = 4.7 eV
EVALUATE: Our result for metallic lithium is similar to that calculated for copper in Example 42.8. Molecules and Condensed Matter 42.52. (a) d
%e 2 1
1
8 A4π P0
U tot =
− 8 A 9 . Setting this equal to zero when r = r0 gives r07 =
dr
4"P0 r 2
r
%e 2 and so U tot = 42.53. 4213 %e 2 " 1 r07 #
7 %e 2
= −1.26 × 10−18 J = −7.85 eV.
$ − + 8 % . At r = r0 , U tot = −
4"P0 & r 8r '
32"P0r0 (b) To remove a Na + Cl− ion pair from the crystal requires 7.85 eV. When neutral Na and Cl atoms are formed
from the Na + and Cl− atoms there is a net release of energy −5.14 eV + 3.61 eV = −1.53 eV, so the net energy
required to remove a neutral Na, Cl pair from the crystal is 7.85 eV − 1.53 eV = 6.32 eV.
dE
(a) IDENTIFY and SET UP: p = − tot . Relate E tot to EF0 and evaluate the derivative.
dV
3N
3 " 32 / 3 π 4 / 3% 2 # 5 / 3 −2 / 3
EF0 = $
EXECUTE: Etot = NEav =
%N V
5
5&
2m
' " 32 / 3 π 4 / 3% 2 # " N #
dEtot 3 " 32 / 3 π 4 / 3% 2 # 5 / 3 " 2 −5 / 3 #
=$
%N $− V
%$ %
% so p = $
5&
2m
dV
5m
&3
'
'
&
'& V ' 5/3 , as was to be shown. (b) N / V = 8.45 × 10 28 m −3
" 32 / 3 π 4 / 3 (1.055 × 10−34 J ⋅ s) 2 #
28
10
5
−3 5 / 3
p=$
% (8.45 × 10 m ) = 3.81× 10 Pa = 3.76 × 10 atm.
5(9.109 × 10−31 kg)
&
' 42.54. (c) EVALUATE: Normal atmospheric pressure is about 105 Pa, so these pressures are extremely large. The
electrons are held in the metal by the attractive force exerted on them by the copper ions.
53
/ 5 32 3 " 4 3% 2 " N # 2 3 " − N # 0 5
dp
32 3 " 4 3% 2 " N #
= −V 1 ⋅
⋅ $ % $ 2 % 2 = p.
(a) From Problem 42.53, p =
$ % . B = −V
dV
5m
5m & V '
& V ' & V '2 3
13
3
4 N
5 3 2 3 π 4 3% 2
= 8.45 × 1028 m −3 . B = ⋅
(8.45 × 10 28 m −3 )5 3 = 6.33 × 1010 Pa.
V
3
5m
6.33 × 1010 Pa
(c)
= 0.45. The copper ions themselves make up the remaining fraction.
1.4 × 1011 Pa
(b) 42.55. (a) EF0 = 32 3 " 4 3% 2
2m 23 1
"N#
mc2 .
$ % . Let EF0 =
100
V'
&
32 0
2m 2c 2
23 2 m3c3
23 2 m3c 3
"N# /
=1
=
=
= 1.67 × 1033 m −3 .
$%
23 43 22
1003 23" 2% 3 3000" 2% 3
& V ' 3 (100)3 " % 4
8.45 × 10 28 m −3
(b)
= 5.06 × 10 −5. Since the real concentration of electrons in copper is less than one part in 10 −4 of the
1.67 × 1033 m −3
concentration where relativistic effects are important, it is safe to ignore relativistic effects for most applications.
6(2 × 1030 kg)
(c) The number of electrons is N e =
= 6.03 × 1056. The concentration is
1.99 × 10−26 kg
Ne
6.03 × 1056
=4
= 6.66 × 1035 m −3 .
V
" (6.00 × 106 m)3
3
6.66 × 1035 m −3
≅ 400 so relativistic effects will be very important.
1.67 × 1033 m −3
IDENTIFY: The current through the diode is related to the voltage across it.
SET UP: The current through the diode is given by I = IS (eeV/kT – 1).
EXECUTE: (a) The current through the resistor is (35.0 V)/(125 ) = 0.280 A = 280 mA, which is also the
current through the diode. This current is given by I = IS (eeV/kT – 1), giving 280 mA = 0.625 mA(eeV/kT – 1) and 1 +
−23
kT ln 449 (1.38 × 10 J/K ) (293 K)ln 449
(280/0.625) = 449 = eeV/kT. Solving for V at T = 293 K gives V =
=
=
e
1.60 × 10−19 C
0.154 V
(b) R = V/I = (0.154 V)/(0.280 A) = 0.551
EVALUATE: At a different voltage, the diode would have different resistance. (d) Comparing this to the result from part (a) 42.56. 4214 Chapter 42 42.57. (a) U = qi q
q 2 " −1 1
1
1
1
1 1 # q2 " 2 2
1
1#
6j r j = 4"P $ d + r − r + d − r − d + r − d % = 4"P $ r − d − r + d − r − d %.
4"P0 i < ij
'
'
0&
0& "
#
d d 2 # 2 2d 2
1
1
1$ 1
1 % 1" d d2
But
+
=$
+
% ≈ $ 1 − + 2 + ⋅ ⋅⋅ + 1 + + 2 % ≈ + 3
r + d r − d r $1+ d 1− d % r &
rr
r r' r
r
&
r
r'
−2q 2 " 1 d 2 # −2 p 2
2 p2
!U =
−
.
$ + 3 %=
3
4"P0 & d r ' 4"P0 r 4"P0 d 3 (b) U = qi q
q 2 " −1 1
1
1
1
1 1 # q 2 " −2 2 2 2d 2 #
6 r j = 4"P $ d − r + r + d + r − d − r − d % = 4"P $ d − r + r + r 3 % =
4"P0 i < j ij
'
'
0&
0& −2 p 2
−2q 2 " 1 d 2 #
2 p2
+
.
$ − 3 % !U =
4"P0 d 3 4"P0 r 3
4"P0 & d r '
If we ignore the potential energy involved in forming each individual molecule, which just involves a different
choice for the zero of potential energy, then the answers are:
−2 p 2
. The interaction is attractive.
(a) U =
4"P0 r 3
(b) U =
42.58. +2 p 2
. The interaction is repulsive.
4"P0 r 3 " 1 e2 #
1 e2
1 e2
(a) Following the hint, k ′dr = − d $
dr and %) = % 2k ′ m = %
=
%=
2
3
"P0 mr03
& 4"P0 r ' r = r0 2"P0 r0
1.23 × 10−19 J = 0.77 eV, where ( m 2) has been used for the reduced mass. (b) The reduced mass is doubled, and the energy is reduced by a factor of 2 to 0.54 eV. NUCLEAR PHYSICS 28
14 Si has 14 protons and 14 neutrons. 85
37 Rb has 37 protons and 48 neutrons. (c)
43.2. (a)
(b) 43.1. 205
81 43 Tl has 81 protons and 124 neutrons. (a) Using R = (1.2 fm)A1 3 , the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm.
(b) Using 4π R 2 for each of the radii in part (a), the areas are 163 fm 2 , 353 fm 2 and 633 fm 2 . 43
π R gives 195 fm3 , 624 fm 3 and 1499 fm3 .
3
(d) The density is the same, since the volume and the mass are both proportional to A: 2.3 × 1017 kg m3 (see
Example 43.1).
(c) 3 43.3. (e) Dividing the result of part (d) by the mass of a nucleon, the number density is 0.14 fm3 = 1.40 × 1044 m .
IDENTIFY: Calculate the spin magnetic energy shift for each spin state of the 1s level. Calculate the energy
splitting between these states and relate this to the frequency of the photons.
SET UP: When the spin component is parallel to the field the interaction energy is U = − µ z B. When the spin component is antiparallel to the field the interaction energy is U = + µ z B. The transition energy for a transition
between these two states is ∆ E = 2µ z B, where µ z = 2.7928µ n . The transition energy is related to the photon
frequency by ∆ E = hf , so 2 µ z B = hf . hf
(6.626 × 10−34 J ⋅ s)(22.7 × 106 Hz)
=
= 0.533 T
2µ z
2(2.7928)(5.051× 10−27 J/T)
EVALUATE: This magnetic field is easily achievable. Photons of this frequency have wavelength
λ = c/f = 13.2 m. These are radio waves.
#
#
(a) As in Example 43.2, ∆E = 2(1.9130)(3.15245 × 10−8 eV T)(2.30 T) = 2.77 × 10−7 eV. Since µ and S are in
opposite directions for a neutron, the antiparallel configuration is lower energy. This result is smaller than but
comparable to that found in the example for protons.
∆E
c
(b) f =
= 66.9 MHz, = = 4.48 m.
h
f
IDENTIFY: Calculate the spin magnetic energy shift for each spin component. Calculate the energy splitting
between these states and relate this to the frequency of the photons.
#
#
#
(a) SET UP: From Example 43.2, when the zcomponent of S (and µ ) is parallel to B, U = −  µ z  B =
#
#
#
−2.7928µ n B. When the zcomponent of S (and µ ) is antiparallel to B , U = − µ z  B = +2.7928µ n B. The state
B= EXECUTE: 43.4. 43.5. with the proton spin component parallel to the field lies lower in energy. The energy difference between these two
states is ∆ E = 2(2.7928µ n B ).
∆ E 2(2.7928µ n ) B 2(2.7928)(5.051× 10−27 J/T)(1.65 T)
=
=
h
h
6.626 × 10−34 J ⋅ s
f = 7.03 × 107 Hz = 7.03 MHz EXECUTE: ∆ E = hf so f = And then λ =
EVALUATE: c 2.998 × 108 m/s
=
= 4.26 m
f
7.03 × 107 Hz
From Figure 32.4 in the textbook, these are radio waves. 431 432 Chapter 43 (b) SET UP: From Eqs. (27.27) and (41.22) and Fig.41.14 in the textbook, the state with the zcomponent of
#
#
µ parallel to B has lower energy. But, since the charge of the electron is negative, this is the state with the
#
electron spin component antiparallel to B. That is, for the ms = − 1 state lies lower in energy.
2
" e # " %#
" e% #
1
1
For the ms = + 1 state, U = +(2.00232) $
% $ + % B = + 2 (2.00232) $
% B = + 2 (2.00232)µ B B.
2
2m ' & 2 '
&
& 2m '
For the ms = − 1 state, U = − 1 (2.00232) µB B. The energy difference between these two states is
2
2
∆ E = (2.00232) µ B B. EXECUTE: ∆ E 2.00232 µ B B (2.00232)(9.274 × 10 −24 J/T)(1.65 T)
=
=
= 4.62 × 1010 Hz = 46.2 GHz. And
h
h
6.626 × 10−34 J ⋅ s
c 2.998 × 108 m/s
λ= =
= 6.49 × 10−3 m = 6.49 mm.
f
4.62 × 1010 Hz
EVALUATE: From Figure 32.4 in the textbook, these are microwaves. The interaction energy with the magnetic
field is inversely proportional to the mass of the particle, so it is less for the proton than for the electron. The
smaller transition energy for the proton produces a larger wavelength.
(a) 146mn + 92mH − mU = 1.93 u
∆ E = hf so f = 43.6. 43.7. 43.8. 43.9. (b) 1.80 × 103 MeV
(c) 7.56 MeV per nucleon (using 931.5 MeV/u and 238 nucleons).
IDENTIFY and SET UP: The text calculates that the binding energy of the deuteron is 2.224 MeV. A photon that
breaks the deuteron up into a proton and a neutron must have at least this much energy.
hc
hc
E=
so λ =
λ
E
(4.136 × 10 −15 eV ⋅ s)(2.998 × 108 m/s)
EXECUTE: λ =
= 5.575 × 10−13 m = 0.5575 pm.
2.224 × 106 eV
EVALUATE: This photon has gammaray wavelength.
IDENTIFY: The binding energy of the nucleus is the energy of its constituent particles minus the energy of the
carbon12 nucleus.
SET UP: In terms of the masses of the particles involved, the binding energy is
EB = (6mH + 6mn – mC12)c2.
EXECUTE: (a) Using the values from Table 43.2, we get
EB = [6(1.007825 u) + 6(1.008665 u) – 12.000000 u)](931.5 MeV/u) = 92.16 MeV
(b) The binding energy per nucleon is (92.16 MeV)/(12 nucleons) = 7.680 MeV/nucleon
(c) The energy of the C12 nucleus is (12.0000 u)(931.5 MeV/u) = 11178 MeV. Therefore the percent of the mass
92.16 MeV
= 0.8245% .
that is binding energy is
11178 MeV
EVALUATE: The binding energy of 92.16 MeV binds 12 nucleons. The binding energy per nucleon, rather than
just the total binding energy, is a better indicator of the strength with which a nucleus is bound.
IDENTIFY: Conservation of energy tells us that the initial energy (photon plus deuteron) is equal to the energy
after the split (kinetic energy plus energy of the proton and neutron). Therefore the kinetic energy released is equal
to the energy of the photon minus the binding energy of the deuteron.
SET UP: The binding energy of a deuteron is 2.224 MeV and the energy of the photon is E = hc/λ. Kinetic
energy is K = ½mv2.
EXECUTE: (a) The energy of the photon is Eph = hc λ = ( 6.626 ×10 −34 J ⋅ s )( 3.00 × 108 m/s ) 3.50 × 10−13 m = 5.68 × 10−13 J . The binding of the deuteron is EB = 2.224 MeV = 3.56 × 10 −13 J . Therefore the kinetic energy
is K = (5.68 − 3.56) × 10−13 J = 2.12 × 10−13 J = 1.32 MeV.
(b) The particles share the energy equally, so each gets half. Solving the kinetic energy for v gives
v= 43.10. 2K
2(1.06 × 10−13 J)
=
= 1.13 × 107 m/s
m
1.6605 × 10 −27 kg EVALUATE: Considerable energy has been released, because the particle speeds are in the vicinity of the speed of light.
(a) 7( mn + mH ) − mN = 0.112 u, which is 105 MeV, or 7.48 MeV per nucleon.
(b) Similarly, 2( mH + mn ) − mHe = 0.03038 u = 28.3 MeV, or 7.07 MeV per nucleon, slightly lower (compare to
Figure 43.2 in the textbook). Nuclear Physics 43.11. 433 (a) IDENTIFY: Find the energy equivalent of the mass defect.
1
SET UP: A 15 B atom has 5 protons, 11 − 5 = 6 neutrons, and 5 electrons. The mass defect therefore is
1
∆ M = 5mp + 6mn + 5me − M ( 15 B). EXECUTE: ∆ M = 5(1.0072765 u) + 6(1.0086649 u) + 5(0.0005485799 u) − 11.009305 u = 0.08181 u. The energy equivalent is EB = (0.08181 u)(931.5 MeV/u) = 76.21 MeV.
(b) IDENTIFY and SET UP: Eq.(43.11): EB = C1 A − C2 A2 /3 − C3Z ( Z − 1) / A1/3 − C4 ( A − 2 Z ) 2 / A
The fifth term is zero since Z is odd but N is even. A = 11 and Z = 5.
EXECUTE: EB = (15.75 MeV)(11) − (17.80 MeV)(11)2/3 − (0.7100 MeV)5(4)/111/3 − (23.69 MeV)(11 − 10)2 /11. EB = +173.25 MeV − 88.04 MeV − 6.38 MeV − 2.15 MeV = 76.68 MeV
76.68 MeV − 76.21 MeV
= 0.6%
76.21 MeV
EVALUATE: Eq.(43.11) has a greater percentage accuracy for 62 Ni. The semiempirical mass formula is more
accurate for heavier nuclei.
(a) 34mn + 29mH − mCu = 34(1.008665) u + 29(1.007825) u − 62.929601 u = 0.592 u, which is 551 MeV, The percentage difference between the calculated and measured EB is 43.12. or 8.75 MeV per nucleon (using 931.5 MeV/u and 63 nucleons).
(b) In Eq.(43.11), Z = 29 and N = 34, so the fifth term is zero. The predicted binding energy is
(29)(28)
(5)2
2
EB = (15.75 MeV)(63) − (17.80 MeV)(63) 3 − (0.7100 MeV)
− (23.69 MeV)
.
1
3
(63)
(63) 43.13. EB = 556 MeV . The fifth term is zero since the number of neutrons is even while the number of protons is odd,
making the pairing term zero. This result differs from the binding energy found from the mass deficit by 0.86%, a
very good agreement comparable to that found in Example 43.4.
IDENTIFY In each case determine how the decay changes A and Z of the nucleus. The β + and β − particles have
charge but their nucleon number is A = 0.
(a) SET UP: α decay: Z increases by 2, A = N + Z decreases by 4 (an α particle is a 4 He nucleus)
2
EXECUTE: 239
94 β − decay: Z increases by 1, A = N + Z remains the same (a β − particle is an electron, (b) SET UP:
EXECUTE: 4
Pu → 2 He + 235 U
92 24
11 e) Na → e + Mg
0
−1 24
12 (c) SET UP β + decay: Z decreases by 1, A = N + Z remains the same (a β + particle is a positron, 0
+1 e) EXECUTE:
O→ e+ N
EVALUATE: In each case the total charge and total number of nucleons for the decay products equals the charge
and number of nucleons for the parent nucleus; these two quantities are conserved in the decay.
(a) The energy released is the energy equivalent of mn − mp − me = 8.40 × 10−4 u, or 783 keV.
15
8 43.14. 0
−1 0
+1 15
7 (b) mn > mp , and the decay is not possible.
43.15. IDENTIFY: The energy of the photon must be equal to the difference in energy of the two nuclear energy levels.
SET UP: The energy difference is ∆E = hc/λ. 43.16. hc ( 6.626 ×10 −34 J ⋅ s )( 3.00 × 108 m/s ) = 8.015 × 10−15 J = 0.0501 MeV
0.0248 × 10−9 J
EVALUATE: Since the wavelength of this photon is much shorter than the wavelengths of visible light, its energy
is much greater than visiblelight photons which are frequently emitted during electron transitions in atoms. This
tells us that the energy difference between the nuclear shells is much greater than the energy difference between
electron shells in atoms, meaning that nuclear energies are much greater than the energies of orbiting electrons.
IDENTIFY: The energy released is equal to the mass defect of the initial and final nuclei.
SET UP: The mass defect is equal to the difference between the initial and final masses of the constituent particles.
EXECUTE: (a) The mass defect is 238.050788 u – 234.043601 u – 4.002603 u = 0.004584 u. The energy released
is (0.004584 u)(931.5 MeV/u) = 4.270 MeV.
(b) Take the ratio of the two kinetic energies, using the fact that K = p2/2m:
EXECUTE: ∆E = λ = 2
pTh
K Th 2mTh mα
4
=
=
=
.
2
pα
Kα
mTh 234
2mα 434 Chapter 43 The kinetic energy of the Th is
4
4
K Total =
(4.270 MeV) = 0.07176 MeV = 1.148 × 10−14 J
234 + 4
238
Solving for v in the kinetic energy gives K Th = v= 43.17. 2K
2(1.148 × 10−14 J)
=
= 2.431× 105 m/s
−27
m
(234.043601) (1.6605 × 10 kg ) EVALUATE: As we can see by the ratio of kinetic energies in part (b) , the alpha particle will have a much higher
kinetic energy than the thorium.
If β − decay of 14 C is possible, then we are considering the decay 14 C → 14 N + β − .
6
7 ∆m = M ( 14 C) − M ( 14 N) − me
6
7
∆m = (14.003242 u − 6(0.000549 u)) − (14.003074 u − 7(0.000549 u)) − 0.0005491 u
∆m = +1.68 × 10−4 u. So E = (1.68 × 10−4 u)(931.5 MeV u ) = 0.156 MeV = 156 keV
43.18. (a) A proton changes to a neutron, so the emitted particle is a positron ( β + ).
(b) The number of nucleons in the nucleus decreases by 4 and the number of protons by 2, so the emitted particle is
an alphaparticle.
(c) A neutron changes to a proton, so the emitted particle is an electron ( β − ). 43.19. (a) As in the example, (0.000898 u)(931.5 MeV u) = 0.836 MeV.
(b) 0.836 MeV − 0.122 MeV − 0.014 MeV = 0.700 MeV. 43.20. (a) 90
39 Sr → β − + 90 X . X has 39 protons and 90 protons plus neutrons, so it must be
39 (b) Use base 2 because we know the half life. A = A0 2 t=−
43.21. T1 2 log 0.01
log 2 =− −t T 1 2 and 0.01A0 = A0 2 −t T1 2 90 Y. . (28 yr)log 0.01
= 190 yr .
log 2 IDENTIFY and SET UP: T1 / 2 = ln 2 λ The mass of a single nucleus is 124mp = 2.07 × 10−25 kg . ∆N / ∆t = 0.350 Ci = 1.30 × 1010 Bq ; ∆N / ∆t = λ N
EXECUTE: T1/ 2 =
43.22. ln 2 λ N= 6.13 × 10−3 kg
∆N / ∆t 1.30 × 1010 Bq
= 2.96 × 1022 ; λ =
=
= 4.39 × 10−13 s −1
−25
2.07 × 10 kg
N
2.96 × 1022 = 1.58 × 1012 s = 5.01× 104 yr Note that Eq.(43.17) can be written as follows: N = N 0 2 − t / T1 2 . The amount of elapsed time since the source was created is roughly 2.5 years. Thus, we expect the current activity to be N = (5000 Ci)2− ( 2.5 yr)/(5.271 yr) = 3600 Ci. The
source is barely usable. Alternatively, we could calculate λ =
43.23. ln(2)
= 0.132(years)−1 and use the Eq. 43.17 directly
T1 2 to obtain the same answer.
IDENTIFY and SET UP: As discussed in Section 43.4, the activity A = dN / dt obeys the same decay equation as
Eq. (43.17): A = A0e − λ t . For 14C, T1/2 = 5730 y and λ = ln2 / T1/ 2 so A = A0e− (ln 2)t / T1/ 2 ; Calculate A at each t;
A0 = 180.0 decays/min.
EXECUTE: (a) t = 1000 y, A = 159 decays/min
(b) t = 50,000 y, A = 0.43 decays/min 43.24. EVALUATE: The time in part (b) is 8.73 halflives, so the decay rate has decreased by a factor or ( 1 )8.73.
2
IDENTIFY and SET UP: The decay rate decreases by a factor of 2 in a time of one halflife.
EXECUTE: (a) 24 d is 3T1/2 so the activity is (375 Bq) /(23 ) = 46.9 Bq
(b) The activity is proportional to the number of radioactive nuclei, so the percent is 17.0 Bq
= 36.2%
46.9 Bq (c) 131 I →01 e + 131 Xe The nucleus 131 Xe is produced.
−
53
54
54
EVALUATE: Both the activity and the number of radioactive nuclei present decrease by a factor of 2 in one halflife. Nuclear Physics 43.25. 3
(a) 3 H → −0 e + 2 He
1
1 (b) N = N 0e − t , N = 0.100 N 0 and = (ln 2) T1
0.100 = e 43.26. 435 − t (ln 2) T1 2 2 ; −t (ln 2) T1 2 = ln(0.100); t = − ln(0.100)T1 2
ln 2 = 40.9 y dN
= 500 µ Ci = (500 × 10−6 )(3.70 × 1010 s −1 ) = 1.85 × 107 decays s
dt
ln 2
ln 2
ln 2
T1 2 =
→=
=
= 6.69 × 10−7 s .
T1 2 12 d(86,400s d) (a) dN
dN dt 1.85 × 107 decays / s
= N!N=
=
= 2.77 × 1013 nuclei . The mass of this many 131 Ba nuclei is
dt
6.69 × 10−7 s −1
m = 2.77 × 1013 nuclei × (131× 1.66 × 10−27 kg nucleus ) = 6.0 × 10−12 kg = 6.0 × 10−9 g = 6.0 ng
(b) A = A0e − λ t . 1 * Ci = (500 * Ci) e − λ t . ln(1/500) = −λt . " 1d #
ln(1 500)
= 9.29 × 106 s $
% = 108 days
−7 −1
λ
6.69 × 10 s
& 86, 400 s '
(ln 2)t
− t (ln 2) / T1 / 2
= ln( A A0 ) .
A = A0e − λ t = A0e
.−
T1 2 t=−
43.27. ln(1 500) T1 2 = −
43.28. =− (ln 2)t
(ln 2)(4.00 days)
=−
= 2.80 days
ln( A A0 )
ln(3091 8318) dN
ln 2
ln 2
= λN . λ =
=
= 1.36 × 10−11 s −1 .
dt
T1 2 1620 yr ( 3.15 × 107 s/yr )
" 6.022 × 1023 atoms #
25
N =1g$
% = 2.665 × 10 atoms .
226 g
&
' dN
= λ N = (2.665 × 1025 )(1.36 × 10−11 s −1 ) = 3.62 × 1010 decays/s = 3.62 × 1010 Bq
dt
"
#
1 Ci
Convert to Ci: 3.62 × 1010 Bq $
% = 0.98 Ci
3.70 × 1010 Bq '
&
43.29. IDENTIFY and SET UP: Calculate the number N of 14 C atoms in the sample and then use Eq. (43.17) to find the
decay constant λ . Eq. (43.18) then gives T1 / 2 .
EXECUTE: Find the total number of carbon atoms in the sample.
n = m/M;
N tot = nN A = mN A / M = (12.0 × 10 −3 kg)(6.022 × 1023 atoms/mol)/(12.011× 10−3 kg/mol)
N tot = 6.016 × 1023 atoms, so (1.3 × 10−12 )(6.016 × 1023 ) = 7.82 × 1011 carbon14 atoms
∆N / ∆ t = −180 decays/min = −3.00 decays/s −∆N / ∆t
= 3.836 × 10−12 s −1
N
T1/ 2 = (ln 2) / λ = 1.807 × 1011 s = 5730 y
EVALUATE: The value we calculated agrees with the value given in Section 43.4.
360 × 106 decays
= 4.17 × 103 Bq = 1.13 × 10−7 Ci = 0.113 *Ci.
86,400 s
∆ N / ∆ t = −λ N ; 43.30.
43.31. (a) λ= dN
0.693
0.693
=
= 3.75 × 10−4 s −1.
= 7.56 × 1011 Bq = 7.56 × 1011 decays s . λ =
dt
T1 2
(30.8 min)(60 s min) N0 = 1 dN 7.56 × 1011 decays s
=
= 2.02 × 1015 nuclei.
3.75 × 10−4 s −1
λ dt 436 Chapter 43 (b) The number of nuclei left after one halflife is N0
= 1.01 × 1015 nuclei, and the activity is half:
2 dN
= 3.78 × 1011 decays s.
dt
(c) After three half lives (92.4 minutes) there is an eighth of the original amount, so N = 2.53 × 1014 nuclei, and an
" dN #
10
eighth of the activity: $
% = 9.45 × 10 decays s.
dt '
&
43.32. The activity of the sample is 3070 decays min
= 102 Bq kg, while the activity of atmospheric carbon is
(60 sec min) (0.500 kg) ln (102 255)
= 7573 y.
1.21 × 10−4 y
IDENTIFY and SET UP: Find λ from the halflife and the number N of nuclei from the mass of one nucleus and
the mass of the sample. Then use Eq.(43.16) to calculate  dN / dt , the number of decays per second.
EXECUTE: (a)  dN / dt = λ N
0.693
0.693
λ=
=
= 1.715 × 10−17 s −1
9
T1/ 2
(1.28 × 10 y)(3.156 × 107 s/1 y) 255 Bq kg (see Example 43.9). The age of the sample is then t = −
43.33. The mass of 40 K atom is approximately 40 u, so the number of
1.63 × 10−9 kg
1.63 × 10−9 kg
N=
=
= 2.454 × 1016.
40 u
40(1.66054 × 10−27 kg) 40 ln (102 255) λ =− K nuclei in the sample is Then  dN / dt = λ N = (1.715 × 10−17 s −1 )(2.454 × 1016 ) = 0.421 decays/s 43.34. (b)  dN / dt = (0.421 decays/s)(1 Ci/(3.70 × 1010 decays/s)) = 1.14 × 10−11 Ci
EVALUATE: The very small sample still contains a very large number of nuclei. But the half life is very large, so
the decay rate is small.
(a) rem = rad × RBE. 200 = x(10) and x = 20 rad.
(b) 1 rad deposits 0.010 J kg , so 20 rad deposit 0.20 J kg . This radiation affects 25 g (0.025 kg) of tissue, so the total energy is (0.025 kg)(0.20 J kg ) = 5.0 × 10−3 J = 5.0 mJ
(c) Since RBE = 1 for β rays, so rem = rad. Therefore 20 rad = 20 rem.
43.35.
43.36. 1 rad = 10 −2 Gy, so 1 Gy = 100 rad and the dose was 500 rad.
rem = (rad)(RBE) = (500 rad)(4.0) = 2000 rem. 1 Gy = 1 J kg, so 5.0 J kg .
IDENTIFY and SET UP: For x rays RBE = 1 so the equivalent dose in Sv is the same as the absorbed dose in J/kg.
EXECUTE: One wholebody scan delivers (75 kg)(12 × 10−3 J/kg) = 0.90 J . One chest x ray delivers 0.90 J
= 900 chest x rays to deliver the same total energy.
1.0 × 10−3 J
IDENTIFY and SET UP: For x rays RBE = 1 and the equivalent dose equals the absorbed dose.
EXECUTE: (a) 175 krad = 175 krem = 1.75 kGy = 1.75 kSv
(5.0 kg)(0.20 × 10−3 J/kg) = 1.0 × 10−3 J . It takes 43.37. (1.75 × 103 J/kg)(0.150 kg) = 2.62 × 102 J
(b) 175 krad = 1.75 kGy ; (1.50)(175 krad) = 262 krem = 2.62 kSv 43.38. 43.39. The energy deposited would be 2.62 × 102 J , the same as in (a).
EVALUATE: The energy required to raise the temperature of 0.150 kg of water 1 C° is 628 J, and 2.62 × 102 J is
less than this. The energy deposited corresponds to a very small amount of heating.
(a) 5.4 Sv (100 rem Sv) = 540 rem.
(b) The RBE of 1 gives an absorbed dose of 540 rad.
(c) The absorbed dose is 5.4 Gy, so the total energy absorbed is (5.4 Gy) (65 kg) = 351 J. The energy required to
raise the temperature of 65 kg by 0.010° C is (65 kg) (4190 J kg ⋅ K) (0.01C°) = 3 kJ.
(a) We need to know how many decays per second occur. λ= 0.693
0.693
=
= 1.79 × 10−9 s −1.
(12.3 y) (3.156 × 107 s y)
T1 2 The number of tritium atoms is N 0 = 1 dN (0.35 Ci) (3.70 × 1010 Bq Ci )
=
= 7.2540 × 1018 nuclei .
1.79 × 10−9 s −1
λ dt Nuclear Physics 437 The number of remaining nuclei after one week is
−9 −1
N = N 0e− λt = (7.25 × 1018 )e − (1.79 ×10 s ) (7) (24) (3600s) = 7.2462 × 1018 nuclei. ∆ N = N 0 − N = 7.8 × 1015 decays. So the
energy absorbed is Etotal = ∆ N Eγ = (7.8 × 1015 ) (5000 eV) (1.60 × 10−19 J eV) = 6.24 J. 43.40. 43.41. The absorbed dose is (6.24 J)
= 0.125 J kg = 12.5 rad. Since RBE = 1, then the equivalent dose is 12.5 rem.
(50 kg)
(b) In the decay, antinetrinos are also emitted. These are not absorbed by the body, and so some of the energy of
the decay is lost (about 12 keV ).
(0.72 × 10 −6 Ci) (3.7 × 1010 Bq Ci ) (3.156 × 107 s) = 8.41 × 1011 % particles. The absorbed dose is
(8.41 × 1011 ) (4.0 × 106 eV) (1.602 × 10−19 J eV )
= 1.08 Gy = 108 rad. The equivalent dose is (20) (108 rad) = 2160 rem.
(0.50 kg)
(a) IDENTIFY and SET UP: Determine X by balancing the charge and nucleon number on the two sides of the
reaction equation.
EXECUTE: X must have A = 2 + 14 − 10 = 6 and Z = 1 + 7 − 5 = 3. Thus X is 6 Li and the reaction is
3
H + 14 N → 6 Li + 10 B
7
3
5
(b) IDENTIFY and SET UP: Calculate the mass decrease and find its energy equivalent.
EXECUTE: The neutral atoms on each side of the reaction equation have a total of 8 electrons, so the electron
masses cancel when neutral atom masses are used. The neutral atom masses are found in Table 43.2.
mass of 2 H + 14 N is 2.014102 u + 14.003074 u = 16.017176 u
1
7
2
1 mass of 6 Li + 10 B is 6.015121 u + 10.012937 u = 16.028058 u
3
5
The mass increases, so energy is absorbed by the reaction. The Q value is
(16.017176 u − 16.028058 u)(931.5 MeV/u) = −10.14 MeV
(c) IDENTIFY and SET UP: The available energy in the collision, the kinetic energy K cm in the center of mass
reference frame, is related to the kinetic energy K of the bombarding particle by Eq. (43.24).
EXECUTE: The kinetic energy that must be available to cause the reaction is 10.14 MeV. Thus
4
K cm = 10.14 MeV. The mass M of the stationary target ( 17 N) is M = 14 u. The mass m of the colliding particle
( 2 H) is 2 u. Then by Eq. (43.24) the minimum kinetic energy K that the 2 H must have is
1
1
"M +m#
" 14 u + 2 u #
K =$
% K cm = $
% (10.14 MeV) = 11.59 MeV
&M'
& 14 u ' EVALUATE: The projectile 43.42.
43.43. ( H ) is much lighter than the target (
2
1 14
7 N ) so K is not much larger than K cm . The K we have calculated is what is required to allow the mass increase. We would also need to check to see if at this
energy the projectile can overcome the Coulomb repulsion to get sufficiently close to the target nucleus for the
reaction to occur.
m3 He + m 2 H − m 4 He − m1 H = 1.97 × 10−2 u, so the energy released is 18.4 MeV.
2 1 2 1 IDENTIFY and SET UP: Determine X by balancing the charge and the nucleon number on the two sides of the
reaction equation.
EXECUTE: X must have A = +2 + 9 − 4 = 7 and Z = +1 + 4 − 2 = 3. Thus X is 7 Li and the reaction is
3
7
4
H + 9 Be = 3 Li + 2He
4
(b) IDENTIFY and SET UP: Calculate the mass decrease and find its energy equivalent.
EXECUTE: If we use the neutral atom masses then there are the same number of electrons (five) in the reactants
as in the products. Their masses cancel, so we get the same mass defect whether we use nuclear masses or neutral
atom masses. The neutral atoms masses are given in Table 43.2.
2
9
1 H + 4 Be has mass 2.014102 u + 9.012182 u = 11.26284 u
2
1 Li + 4He has mass 7.016003 u + 4.002603 u = 11.018606 u
2
The mass decrease is 11.026284 u − 11.018606 u = 0.007678 u.
This corresponds to an energy release of 0.007678 u(931.5 MeV/1 u) = 7.152 MeV.
(c) IDENTIFY and SET UP: Estimate the threshold energy by calculating the Coulomb potential energy when the
2
9
1 H and 4 Be nuclei just touch. Obtain the nuclear radii from Eq. (43.1).
7
3 EXECUTE: The radius RBe of the 9 Be nucleus is RBe = (1.2 × 10−15 m)(9)1/3 = 2.5 × 10−15 m.
4
2
The radius RH of the 1 H nucleus is RH = (1.2 × 10−15 m)(2)1/3 = 1.5 × 10−15 m.
The nuclei touch when their centertocenter separation is
R = RBe + RH = 4.0 × 10−15 m. 438 Chapter 43 The Coulomb potential energy of the two reactant nuclei at this separation is
1 q1q2
1 e(4e)
U=
=
4π P0 r
4π P0 r
4(1.602 × 10−19 C) 2
= 1.4 MeV
(4.0 × 10−15 m)(1.602 × 10−19 J/eV)
This is an estimate of the threshold energy for this reaction.
EVALUATE: The reaction releases energy but the total initial kinetic energy of the reactants must be 1.4 MeV in
order for the reacting nuclei to get close enough to each other for the reaction to occur. The nuclear force is strong
but is very shortrange.
IDENTIFY and SET UP: 0.7% of naturally occurring uranium is the isotope 235 U . The mass of one 235 U nucleus
is about 235mp.
1.0 × 1019 J
= 3.13 × 1029 . The mass of
EXECUTE: (a) The number of fissions needed is
6
(200 × 10 eV)(1.60 × 10−19 J/eV)
U = (8.988 × 109 N ⋅ m 2 / C2 ) 43.44. 235 U required is (3.13 × 1029 )(235mp ) = 1.23 × 105 kg . 1.23 × 105 kg
= 1.76 × 107 kg
0.7 × 10−2
EVALUATE: The calculation assumes 100% conversion of fission energy to electrical energy.
IDENTIFY and SET UP: The energy released is the energy equivalent of the mass decrease. 1 u is equivalent to
931.5 MeV. The mass of one 235 U nucleus is 235mp.
1
1
EXECUTE: (a) 235 U + 0 n → 144 Ba + 89 Kr + 3 0 n
92
56
36
We can use atomic masses since the same number of electrons are included on each side of the reaction equation
and the electron masses cancel. The mass decrease is
1
89
1
∆M = m ( 235 U ) + m ( 0 n ) − / m ( 144 Ba ) + m ( 36 Kr ) + 3m ( 0 n ) 0
92
56
3
4
∆M = 235.043930 u + 1.0086649 u − 143.922953 u − 88.917630 u − 3(1.0086649 u)
∆M = 0.1860 u . The energy released is (0.1860 u)(931.5 MeV/u) = 173.3 MeV .
(b) 43.45. (b) The number of 235 U nuclei in 1.00 g is 1.00 × 10−3 kg
= 2.55 × 1021 . The energy released per gram is
235mp (173.3 MeV/nucleus)(2.55 × 1021 nuclei/g) = 4.42 × 1023 MeV/g . 43.46. (a) 28
14 24
A
Si + γ ! 12 Mg + Z X. A + 24 = 28 so A = 4. Z + 12 = 14 so Z = 2. X is an α particle. (b) E' = −∆mc 2 = (23.985042 u + 4.002603 u − 27.976927 u) (931.5 MeV u ) = 9.984 MeV
43.47.
43.48. The energy liberated will be
4
M (3 He) + M ( 2 He) − M (7 Be) = (3.016029 u + 4.002603 u − 7.016929 u)(931.5 MeV u) = 1.586 MeV.
2
4
(a) Z = 3 + 2 − 0 = 5 and A = 4 + 7 − 1 = 10.
(b) The nuclide is a boron nucleus, and mHe + mLi − mn − mB = −3.00 × 10−3 u, and so 2.79 MeV of energy is absorbed. 43.49. A
Nuclei: Z X Z + → A−4
Z −2 Y ( Z − 2) + + 4 He2 + . Add the mass of Z electrons to each side and we find:
2 43.50. A
A
4
∆m = M ( Z X) − M ( Z − 4 Y) − M ( 2He), where now we have the mass of the neutral atoms. So as long as the mass of
−2
the original neutral atom is greater than the sum of the neutral products masses, the decay can happen.
A
Denote the reaction as Z X → Z +1A Y + e − . The mass defect is related to the change in the neutral atomic masses by [ mX − Zme ] − [mY − ( Z + 1) me ] − me = ( mX − mY ),
where mX and mY are the masses as tabulated in, for instance, Table (43.2).
43.51. A
Z XZ + → A
Z −1 A
Y ( Z −1) + + β + . Adding (Z –1) electrons to both sides yields Z X + → ( ) A
Z −1 Y + β + . So in terms of masses: A
A
A
∆m = M ( Z X + ) − M ( Z −1A Y ) − me = M ( Z X ) − me − M ( Z −1A Y ) − me = M ( Z X ) − M ( Z −1A Y ) − 2me . So the decay will 43.52. occur as long as the original neutral mass is greater than the sum of the neutral product mass and two electron masses.
IDENTIFY and SET UP: m = ρV . 1 gal = 3.788 L = 3.788 × 10−3 m 3 . The mass of a 235 U nucleus is 235mp.
1 MeV = 1.60 × 10−13 J
EXECUTE: (a) For 1 gallon, m = ρV = (737 kg/m3 )(3.788 × 10−3 m3 ) = 2.79 kg = 2.79 × 103 g 1.3 × 108 J/gal
= 4.7 × 104 J/g
2.79 × 103 g/gal Nuclear Physics (b) 1 g contains 439 1.00 × 10−3 kg
= 2.55 × 1021 nuclei
235mp (200 MeV/nucleus)(1.60 × 10−13 J/MeV)(2.55 × 1021 nuclei) = 8.2 × 1010 J/g
(c) A mass of 6mp produces 26.7 MeV.
(26.7 MeV)(1.60 × 10−13 J/MeV)
= 4.26 × 1014 J/kg = 4.26 × 1011 J/g
6mp (d) The total energy available would be (1.99 × 1030 kg)(4.7 × 107 J/kg) = 9.4 × 1037 J energy
energy
9.4 × 1037 J
so t =
=
= 2.4 × 1011 s = 7600 yr
t
power 3.86 × 1026 W
EVALUATE: If the mass of the sun were all proton fuel, it would contain enough fuel to last
" 4.3 × 1011 J/g #
10
(7600 yr) $
% = 7.0 × 10 yr .
4
& 4.7 × 10 J/g '
power = 43.53. A
24
Using Eq: (43.12): Z M = ZM H + Nmn − EB c 2 ! M (11 Na) = 11M H +13mn − EB / c 2 . But EB = (15.75 MeV)(24) − (17.80 MeV)(24)2 3 − (0.7100 MeV)
(23.69 MeV) (11)(10)
−
(24)1 3 (24 − 2(11))2
− (39 MeV)(24)−4 3 = 198.31 MeV.
24
24
! M (11 Na) = 11(1.007825 u) + 13(1 .008665 u) − % error = (198.31 MeV)
= 23.9858 u
931.5 MeV u 23.990963 − 23.9858
× 100 = 0.022%.
23.990963 24
If the binding energy term is neglected, M (11 Na) = 24.1987 u and the percentage error would be 43.54.
43.55. 24.1987 − 23.990963
× 100 = 0.87%.
23.990963
226
226
The α particle will have
of the released energy (see Example 43.5).
( mTh − mRa − mα ) =
230
230
5.032 × 10−3 u or 4.69 MeV.
(a) IDENTIFY and SET UP: The heavier nucleus will decay into the lighter one.
25
25
EXECUTE: 13 Al will decay into 12 Mg.
(b) IDENTIFY and SET UP: Determine the emitted particle by balancing A and Z in the decay reaction.
25
25
0
EXECUTE: This gives 13 Al →12 Mg + +1 e. The emitted particle must have charge +e and its nucleon number
must be zero. Therefore, it is a β + particle, a positron.
(c) IDENTIFY and SET UP: Calculate the energy defect ∆M for the reaction and find the energy equivalent of
25
25
∆M . Use the nuclear masses for 13 Al and 12 Mg, to avoid confusion in including the correct number of electrons if
neutral atom masses are used.
25
25
EXECUTE: The nuclear mass for 13 Al is M nuc (13 Al) = 24.990429 u − 13(0.000548580 u) = 24.983297 u.
25
25
The nuclear mass for 12 Mg is M nuc (12 Mg) = 24.985837 u − 12(0.000548580 u) = 24.979254 u.
The mass defect for the reaction is
25
25
∆ M = M nuc (13 Al) − M nuc (12 Mg) − M (0 1 e) = 24.983297 u − 24.979254 u − 0.00054858 u = 0.003494 u
+ Q = (∆ M )c 2 = 0.003494 u(931.5 MeV/1 u) = 3.255 MeV
EVALUATE: The mass decreases in the decay and energy is released. Note:
25
12
25
13 25
13 Al can also decay into Mg by the electron capture.
0
25
Al + −1 e →12 Mg 0
25
The −1 electron in the reaction is an orbital electron in the neutral 13 Al atom. The mass defect can be calculated
using the nuclear masses:
25
25
∆ M = M nuc ( 13 Al ) + M (0 1 e) − M nuc ( 12 Mg ) = 24.983287 u + 0.00054858 u − 24.979254 u = 0.004592 u.
− Q = ( ∆ M ) c 2 = (0.004592 u)(931.5 MeV/1 u) = 4.277 MeV
2 The mass decreases in the decay and energy is released. 4310 Chapter 43 43.56. (a) m210 Po − m 206 Pb − m4 He = 5.81 × 10−3 u, or Q = 5.41 MeV. The energy of the alpha particle is ( 206 210) times this,
84 82 2 or 5.30 MeV (see Example 43.5).
(b) m210 Po − m209 Bi − m1 H = −5.35 × 10−3 u < 0, so the decay is not possible.
84 83 1 (c) m210 Po − m 209 Po − mn = −8.22 × 10−3 u < 0, so the decay is not possible.
84 84 (d) m 210 At > m 210 Po , so the decay is not possible (see Problem (43.50)).
85 84 (e) m 210 Bi + 2me > m 210 Po , so the decay is not possible (see Problem (43.51)).
83 43.57. 84 IDENTIFY and SET UP: The amount of kinetic energy released is the energy equivalent of the mass change in the
4
decay. me = 0.0005486 u and the atomic mass of 17 N is 14.003074 u. The energy equivalent of 1 u is 931.5 MeV. 14 C has a halflife of T1/ 2 = 5730 yr = 1.81 × 1011 s . The RBE for an electron is 1.0. EXECUTE: (a) 14
6 C → e − + 14 N + υe
7 (b) The mass decrease is ∆M = m ( 14 C ) − / me + m ( 14 N ) 0 . Use nuclear masses, to avoid difficulty in accounting for
6
7
3
4 atomic electrons. The nuclear mass of
The nuclear mass of 14
7 14
6 C is 14.003242 u − 6me = 13.999950 u . N is 14.003074 u − 7me = 13.999234 u . ∆ M = 13.999950 u − 13.999234 u − 0.000549 u = 1.67 × 10−4 u . The energy equivalent of ∆ M is 0.156 MeV.
(c) The mass of carbon is (0.18)(75 kg) = 13.5 kg . From Example 43.9, the activity due to 1 g of carbon in a living organism is 0.255 Bq. The number of decay/s due to 13.5 kg of carbon is (13.5 × 103 )(0.255 Bq/g) =
3.4 × 103 decays/s . (d) Each decay releases 0.156 MeV so 3.4 × 103 decays/s releases 530 MeV/s = 8.5 × 10−11 J/s .
(e) The total energy absorbed in 1 yr is (8.5 × 10−11 J/s)(3.156 × 107 s) = 2.7 × 10 −3 J . The absorbed dose is
2.7 × 10−3 J
= 3.6 × 10 −5 J/kg = 36 µ Gy = 3.6 mrad . With RBE = 1.0 , the equivalent dose is 36 µSv = 3.6 mrem .
75 kg 43.58. IDENTIFY and SET UP: mπ = 264me = 2.40 × 10−28 kg . The total energy of the two photons equals the rest mass energy mπ c2 of the pion.
EXECUTE: (a) Eph = 1 mπc 2 = 1 (2.40 × 10−28 kg)(3.00 × 108 m/s) 2 = 1.08 × 10−11 J = 67.5 MeV
2
2 Eph = hc λ so λ = hc 1.24 × 10−6 eV ⋅ m
=
= 1.84 × 10−14 m = 18.4 fm
Eph
67.5 × 106 eV These are gamma ray photons, so they have RBE = 1.0 .
(b) Each pion delivers 2(1.08 × 10−11 J) = 2.16 × 10−11 J .
The absorbed dose is 200 rad = 2.00 Gy = 2.00 J/kg . 43.59. The energy deposited is (25 × 10−3 kg)(2.00 J/kg) = 0.050 J .
0.050 J
= 2.3 × 109 mesons .
The number of π 0 mesons needed is
2.16 × 10−11 J/meson
EVALUATE: Note that charge is conserved in the decay since the pion is neutral. If the pion is initially at rest the
photons must have equal momenta in opposite directions so the two photons have the same λ and are emitted in
opposite directions. The photons also have equal energies since they have the same momentum and E = pc .
IDENTIFY and SET UP: Find the energy equivalent of the mass decrease. Part of the released energy appears as
the emitted photon and the rest as kinetic energy of the electron.
98
EXECUTE: 179 Au →198 Hg + −01 e
80
The mass change is 197.968225 u − 197.966752 u = 1.473 × 10 −3 u
(The neutral atom masses include 79 electrons before the decay and 80 electrons after the decay. This one
additional electron in the products accounts correctly for the electron emitted by the nucleus.) The total energy
released in the decay is (1.473 × 10 −3 u)(931.5 MeV/u) = 1.372 MeV. This energy is divided between the energy of
the emitted photon and the kinetic energy of the β − particle. Thus the β − particle has kinetic energy equal to
1.372 MeV − 0.412 MeV = 0.960 MeV.
98
EVALUATE: The emitted electron is much lighter than the 180 Hg nucleus, so the electron has almost all the final
kinetic energy. The final kinetic energy of the 198 Hg nucleus is very small. Nuclear Physics 4311 43.60. (See Problem (43.51)) m11 C − m11 B − 2me = 1.03 × 10−3 u. Decay is energetically possible. 43.61. IDENTIFY and SET UP: The decay is energetically possible if the total mass decreases. Determine the nucleus
3
0
produced by the decay by balancing A and Z on both sides of the equation. 17 N → +1 e + 13C. To avoid confusion in
6
including the correct number of electrons with neutral atom masses, use nuclear masses, obtained by subtracting
the mass of the atomic electrons from the neutral atom masses. 6 5 13
7 EXECUTE: The nuclear mass for The nuclear mass for 13
6 N is M nuc ( 13
7 N ) = 13.005739 u − 7(0.00054858 u) = 13.001899 u. ( C ) = 13.003355 u − 6(0.00054858 u) = 13.000064 u.
13
6 C is M nuc The mass defect for the reaction is
0
∆ M = M nuc ( 13 N ) − M nuc ( 13 C ) − M ( +1 e ). ∆ M = 13.001899 u − 13.000064 u − 0.00054858 u = 0.001286 u.
7
6
43.62. EVALUATE: The mass decreases in the decay, so energy is released. This decay is energetically possible.
ln 2
(a) A leastsquares fit to log of the activity vs. time gives a slope of λ = 0.5995 h −1 , for a halflife of
= 1.16 h. λ (2.00 × 104 Bq)
(b) The initial activity is N 0λ , and this gives N 0 =
= 1.20 × 108.
(0.5995 hr −1 )(1 hr 3600 s)
(c) N 0e− t = 1.81 × 106.
43.63. 43.64. dN (t )
dN (t )
but
= − N (t ) so − N 0 = A0 . Taking the derivative of
dt
dt
dN (t )
N (t ) = N 0e − t !
= − N 0e − t = A0e − t , or A(t ) = A0e− t .
dt
− (ln 2) ( t / T1 2 )
From Eq.43.17 N (t ) = N 0e − t but N 0e − t = N 0e The activity A(t ) ≡ = N 0 / e− (ln 2) 0
3
4 43.65. (t / T1 2 ) = N 0 / eln ( 1/ 2) 0
3
4 ( t / T1 2 ) n t
"1#
. So N (t ) = N 0 $ % where n =
.
T1 2
& 2' (We have used that a ln x = ln( x a ), eax = (e x )a , and eln x = x. )
IDENTIFY and SET UP: Onehalf of the sample decays in a time of T1/2.
10 × 109 yr
= 5.0 × 104
EXECUTE: (a)
200,000 yr
4 (b) ( 1 )5.0×10 . This exponent is too large for most handheld calculators. But ( 1 ) = 10−0.301 so
2
2
4 4 ( 1 )5.0×10 = (10−0.301 )5.0×10 = 10−15,000
2
43.66. IDENTIFY and SET UP: N= EXECUTE: T1 / 2 = ln 2 λ . The mass of a single nucleus is 149mp = 2.49 × 10−25 kg . ∆N / ∆t = −λ N . 12.0 × 10−3 kg
= 4.82 × 1022 . ∆ N / ∆ t = −2.65 decays/s
2.49 × 10−25 kg ∆ N / ∆ t 2.65 decays/s
ln 2
=
= 5.50 × 10−23 s −1 ; T1/ 2 =
= 1.26 × 1022 s = 3.99 × 1014 yr
4.82 × 1022
λ
N
IDENTIFY: Use Eq. (43.17) to relate the initial number of radioactive nuclei, N 0 , to the number , N, left after time t. λ =− 43.67. SET UP: We have to be careful; after 87 Rb has undergone radioactive decay it is no longer a rubidium atom. Let
N85 be the number of 85 Rb atoms; this number doesn’t change. Let N 0 be the number of 87 Rb atoms on earth when the solar system was formed. Let N be the present number of 87 Rb atoms.
EXECUTE: The present measurements say that 0.2783 = N /( N + N85 ).
( N + N85 )(0.2783) = N , so N = 0.3856 N85 . The percentage we are asked to calculate is N 0 /( N 0 + N85 ).
N and N 0 are related by N = N 0e − λ t so N 0 = e + λ t N . Thus N0
Neλ t
(0.3855eλ t ) N85
0.3856eλ t
.
= λt
=
=
λt
N 0 + N85 Ne + N85 (0.3856e ) N85 + N85 0.3856eλ t + 1 t = 4.6 × 109 y; λ =
−11 eλ t = e(1.459×10 0.693
0.693
=
= 1.459 × 10−11 y −1
T1/2
4.75 × 1010 y y−1 )(4.6×109 y) = e 0.16711 = 1.0694 4312 Chapter 43 Thus 43.68. N0
(0.3856)(1.0694)
=
= 29.2%.
N 0 + N85 (0.3856)(1.0694) + 1 EVALUATE: The halflife for 87 Rb is a factor of 10 larger than the age of the solar system, so only a small
fraction of the 87 Rb nuclei initially present have decayed; the percentage of rubidium atoms that are radioactive is
only a bit less now than it was when the solar system was formed.
(a) (6.25 × 1012 )(4.77 × 106 MeV)(1.602 × 10−19 J eV) (70.0 kg) = 0.0682 Gy = 0.682 rad
(b) (20)(6.82 rad ) = 136 rem
m ln(2)
(c) N λ =
= 1.17 × 109 Bq = 31.6 mCi .
Amp T1 2 6.25 × 1012
= 5.34 × 103 s, about an hour and a half. Note that this time is so small in comparison with the
1.17 × 109 Bq
halflife that the decrease in activity of the source may be neglected.
IDENTIFY and SET UP: Find the energy emitted and the energy absorbed each second. Convert the absorbed
energy to absorbed dose and to equivalent dose.
EXECUTE: (a) First find the number of decays each second:
" 3.70 × 1010 decays/s #
6
2.6 × 10−4 Ci $
% = 9.6 × 10 decays/s
1 Ci
&
'
The average energy per decay is 1.25 MeV, and onehalf of this energy is deposited in the tumor. The energy
delivered to the tumor per second then is
1
(9.6 × 106 decays/s)(1.25 × 106 eV/decay)(1.602 × 10−19 J/eV) = 9.6 × 10−7 J/s.
2
(b) The absorbed dose is the energy absorbed divided by the mass of the tissue:
9.6 × 10−7 J/s
= (1.9 × 10−6 J/kg ⋅ s)(1 rad/(0.01 J/kg)) = 1.9 × 10−4 rad/s
0.500 kg
(c) equivalent dose (REM) = RBE × absorbed dose (rad)
In one second the equivalent dose is 0.70(1.9 × 10−4 rad) = 1.3 × 10−4 rem.
(d) 43.69. (d) (200 rem/1.3 × 10−4 rem/s) = 1/ 5 × 106 s(1 h/3600 s) = 420 h = 17 days.
EVALUATE: The activity of the source is small so that absorbed energy per second is small and it takes several
days for an equivalent dose of 200 rem to be absorbed by the tumor. A 200 rem dose equals 2.00 Sv and this is
large enough to damage the tissue of the tumor.
"1 1 # −
(240) $
%
2−240 122.2
= 2 & 26.9 122.2 ' = 124.
−240 26.9
2 43.70. (a) After 4.0 min = 240 s, the ratio of the number of nuclei is 43.71. (b) After 15.0 min = 900 s, the ratio is 7.15 × 107.
IDENTIFY and SET UP: The number of radioactive nuclei left after time t is given by N = N 0e − λt . The problem says N / N 0 = 0.21; solve for t.
EXECUTE: 0.21 = e − λ t so ln(0.21) = −λt and t = −ln(0.21)/λ Example 43.9 gives λ = 1.209 × 10−4 y −1 for 14C. Thus t =
EVALUATE: The halflife of remaining is less than
43.72. () 12
2 14 − ln(0.21)
= 1.3 × 104 y.
1.209 × 10−4 y C is 5730 y, so our calculated t is more than two halflives, so the fraction = 1.
4 IDENTIFY: The tritium (H3) decays to He3. The ratio of the number of He3 atoms to H3 atoms allows us to
calculate the time since the decay began, which is when the H3 was formed by the nuclear explosion. The H3
decay is exponential.
SET UP: The number of tritium (H3) nuclei decreases exponentially as N H = N 0,H e− λt , with a halflife of 12.3 years. The amount of He3 present after a time t is equal to the original amount of tritium minus the number
of tritium nuclei that are still undecayed after time t.
EXECUTE: The number of He3 nuclei after time t is
N He = N 0,H − N H = N 0,H − N 0,H e− λt = N 0,H (1 − e − λt ) . Taking the ratio of the number of He3 atoms to the number of tritium (H3) atoms gives
− λt
N He N 0,H (1 − e ) 1 − e − λ t
=
= − λt = eλt − 1.
NH
N 0,H e − λ t
e Nuclear Physics 4313 ln (1 + N He / N H )
ln 2
, we have
. Using the given numbers and T1 / 2 =
λ
λ
ln (1 + 4.3)
ln 2
ln 2
λ=
=
= 0.0563/ y and t =
= 30 years.
0.0563/ y
T1 / 2 12.3 y
EVALUATE: One limitation on this method would be that after many years the ratio of H to He would be too
small to measure accurately.
(a) IDENTIFY and SET UP: Use Eq.(43.1) to calculate the radius R of a 2 H nucleus. Calculate the Coulomb
1
potential energy (Eq.23.9) of the two nuclei when they just touch.
EXECUTE: The radius of 2 H is R = (1.2 × 10−15 m)(2)1/3 = 1.51 × 10−15 m. The barrier energy is the Coulomb
1
Solving for t gives t = 43.73. potential energy of two 2 H nuclei with their centers separated by twice this distance:
1
1 e2
(1.602 × 10−19 C)2
= (8.988 × 109 N ⋅ m 2 / C2 )
= 7.64 × 10−14 J = 0.48 MeV
4π P0 r
2(1.51 × 10−15 m)
(b) IDENTIFY and SET UP: Find the energy equivalent of the mass decrease.
2
2
EXECUTE: 1 H + 1 H → 3 He + 1 n
2
0
If we use neutral atom masses there are two electrons on each side of the reaction equation, so their masses cancel.
The neutral atom masses are given in Table 43.2.
2
2
1 H + 1 H has mass 2(2.014102 u) = 4.028204 u
U= 3
2 He + 1 n has mass 3.016029 u + 1.008665 u = 4.024694 u
0 The mass decrease is 4.028204 u − 4.024694 u = 3.510 × 10−3 u. This corresponds to a liberated energy of
(3.510 × 10 −3 u)(931.5 MeV/u) = 3.270 MeV, or (3.270 × 106 eV)(1.602 × 10 −19 J/eV) = 5.239 × 10−13 J.
2
(c) IDENTIFY and SET UP: We know the energy released when two 1 H nuclei fuse. Find the number of reactions
2
obtained with one mole of 1 H.
2
EXECUTE: Each reaction takes two 1 H nuclei. Each mole of D 2 has 6.022 × 1023 molecules, so 6.022 × 10 23 pairs of atoms. The energy liberated when one mole of deuterium undergoes fusion is (6.022 × 10 23 )(5.239 × 10 −13 J) = 43.74. 3.155 × 1011 J/mol.
EVALUATE: The energy liberated per mole is more than a million times larger than from chemical combustion of
one mole of hydrogen gas.
In terms of the number N of cesium atoms that decay in one week and the mass
m = 1.0 kg, the equivalent dose is N
N
N
((RBE) ' E ' + (RBE)e E e ) = ((1)(0.66 MeV) + (1.5)(0.51 MeV)) = (2.283 × 10−13 J), so
m
m
m
(1.0 kg)(3.5 Sv)
N=
= 1.535 × 1013 . The number N 0 of atoms present is related to
(2.283 × 10−13 J) 3.5 Sv = N by N 0 = Ne t . λ = ln 2
0.693
= 7.30 × 10−10 sec −1 .
Ty2 (30.07 yr)(3.156 × 107 sec yr) Then N 0 = Ne t = (1.535 × 1013 )e(7.30×10
43.75. (a) vcm = v −10 s −1 ) (7 days) (8.64×104 s day) = 1.536 × 1013. m
m
vm
"M#
=$
. v′ = v − v
.
m
M
% v . v′ =
m+M &m+M '
m+M
m+M 1
1
1 mM 2
1 Mm 2
1 M " mM
m2
′
K ′ = mvm2 + Mv′2 =
v2 +
v2 =
+
$
M
2
2
2 (m + M ) 2
2 (m + M ) 2
2 (m + M ) & m + M m + M
K′ = #2
%v .
' M "1 2#
M
K ≡ K cm .
$ mv % ! K ′ =
m+M &2
m+M
' (b) For an endoergic reaction K cm = −Q ( Q < 0 ) at threshold. Putting this into part (a) gives −Q = −( M + m)
M
K th ! K th =
Q
M +m
M 4314 43.76. 43.77. Chapter 43 Mα
K ∞ , where K ∞ is the energy that the α particle would have if the nucleus were infinitely massive.
Mα + m
186
Then, M = M Os − M α − K ∞ = M Os − M α −
( 2.76 MeV c2 ) = 181.94821 u .
182
235
∆ m = M ( 92 U ) − M ( 140 Xe ) − M ( 94 Sr ) − mn
54
38
∆ m = 235.043923 u − 139.921636 u − 93.915360 u − 1.008665 u = 0.1983 u
K= ! E = ( ∆ m ) c 2 = ( 0.1983 u ) ( 931.5 MeV u ) = 185 MeV.
43.78. (a) A leastsquares fit of the log of the activity vs. time for the times later than 4.0 h gives a fit with correlation
− (1 − 2 × 10−6 ) and decay constant of 0.361 h −1, corresponding to a halflife of 1.92 h. Extrapolating this back to time 0 gives a contribution to the rate of about 2500/s for this longerlived species. A leastsquares fit of the log of the activity
vs. time for times earlier than 2.0 h gives a fit with correlation = 0.994, indicating the presence of only two species.
− t 1.733 h )
− t 0.361 h )
(b) By trial and error, the data is fit by a decay rate modeled by R = ( 5000 Bq ) e (
+ ( 2500 Bq ) e (
. This 43.79. would correspond to halflives of 0.400 h and 1.92 h.
(c) In this model, there are 1.04 × 10 7 of the shorterlived species and 2.49 × 10 7 of the longerlived species.
(d) After 5.0 h, there would be 1.80 × 103 of the shorterlived species and 4.10 × 10 6 of the longerlived species.
(a) There are two processes occurring: the creation of 128 I by the neutron irradiation, and the decay of the newly
dN
= K − λ N where K is the rate of production by the neutron irradiation. Then
produced 128 I . So
dt
K (1 − e− t )
N
t
N
dN ′
= 5 dt. /ln ( K − λ N ′ ) 0 0 = −λt . ln ( K − λ N ) = ln K − λt . N ( t ) =
. The graph is given in
50 K − λ N ′ 0 3
4
λ
Figure 43.79. " 0.693 #
"
−$
%t #
25 min '
% . So the activity is
= (1.5 × 106 decays s ) × $ 1 − e &
)
%
$
&
'
′
(1.5 × 106 decays s ) (1 − e−0.02772 t ) , with t in minutes. So the activity " −dN # at various times is:
$
%
dt '
& (b) The activity of the sample is λ N ( t ) = K (1 − e − t − dN ′
− dN ′
(t = 1 min) = 4.1 × 104 Bq;
(t = 10 min) = 3.6 × 105 Bq;
dt
dt
− dN ′
− dN ′
(t = 25 min) = 7.5 × 105 Bq;
(t = 50 min) = 1.1× 106 Bq;
dt
dt
− dN ′
− dN ′
(t = 75 min) = 1.3 × 106 Bq;
(t = 180 min) = 1.5 × 106 Bq;
dt
dt
(1.5 × 106 ) (60)
= 3.2 × 109 atoms .
λ
( 0.02772 )
(d) The maximum activity is at saturation, when the rate being produced equals that decaying and so it equals
1.5 × 106 decays s.
(c) N max = K = Figure 43.79
43.80. The activity of the original iron, after 1000 hours of operation, would be
(9.4 × 10−6 Ci) (3.7 × 1010 Bq Ci)2−(1000 h) (45 d×24 h d) = 1.8306 × 105 Bq . The activity of the oil is 84 Bq, or
4.5886 × 10−4 of the total iron activity, and this must be the fraction of the mass worn, or mass of 4.59 × 10−2 g . The rate at which the piston rings lost their mass is then 4.59 × 10−5 g h . 44 PARTICLE PHYSICS AND COSMOLOGY 44.1. 44.2. (a) IDENTIFY and SET UP: Use Eq.(37.36) to calculate the kinetic energy K.
"
#
1
EXECUTE: K = mc 2 $
− 1% = 0.1547 mc 2
2
2
& 1− v / c
'
m = 9.109 × 10−31 kg, so K = 1.27 × 10−14 J
(b) IDENTIFY and SET UP: The total energy of the particles equals the sum of the energies of the two photons.
Linear momentum must also be conserved.
EXECUTE: The total energy of each electron or positron is E = K + mc 2 = 1.1547 mc 2 = 9.46 × 10−13 J. The total
energy of the electron and positron is converted into the total energy of the two photons. The initial momentum of
the system in the lab frame is zero (since the equalmass particles have equal speeds in opposite directions), so the
final momentum must also be zero. The photons must have equal wavelengths and must be traveling in opposite
directions. Equal λ means equal energy, so each photon has energy 9.46 × 10 −14 J.
(c) IDENTIFY and SET UP: Use Eq. (38.2) to relate the photon energy to the photon wavelength.
EXECUTE: E = hc / λ so λ = hc / E = hc /(9.46 × 10−14 J) = 2.10 pm
EVALUATE: The wavelength calculated in Example 44.1 is 2.43 pm. When the particles also have kinetic energy,
the energy of each photon is greater, so its wavelength is less.
The total energy of the positron is
E = K + mc 2 = 5.00 MeV + 0.511 MeV = 5.51 MeV. We can calculate the speed of the positron from Eq.(37.38):
E= mc 2
1− 44.3. v2
c2 2 ! 2 " mc 2 #
v
" 0.511 MeV #
= 1− $
% = 1− $
% = 0.996.
c
E'
& 5.51 MeV '
& IDENTIFY and SET UP: By momentum conservation the two photons must have equal and opposite momenta.
Then E = pc says the photons must have equal energies. Their total energy must equal the rest mass energy
E = mc 2 of the pion. Once we have found the photon energy we can use E = hf to calculate the photon frequency
and use λ = c / f to calculate the wavelength. EXECUTE: The mass of the pion is 270me , so the rest energy of the pion is 270(0.511 MeV) = 138 MeV. Each photon has half this energy, or 69 MeV. E = hf so f = E (69 × 106 eV)(1.602 × 10−19 J/eV)
=
= 1.7 × 1022 Hz
h
6.626 × 10−34 J ⋅ s c 2.998 × 108 m/s
=
= 1.8 × 10−14 m = 18 fm.
f
1.7 × 1022 Hz
EVALUATE: These photons are in the gamma ray part of the electromagnetic spectrum.
(a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of 2.27 × 1023 Hz and a λ= 44.4. wavelength of 1.32 × 10 −15 m.
(b) The energy of each photon will be 938.3 MeV + 830 MeV = 1768 MeV, with frequency 42.8 × 1022 Hz and wavelength 7.02 × 10 −16 m.
44.5. (a) ∆ m = mπ + − mµ + = 270 me − 207 me = 63 me ! E = 63(0.511 MeV) = 32 MeV.
(b) A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you
could always find a frame where energy was not conserved. This cannot occur.
441 442 Chapter 44 hc
hc
h
(6.626 × 10−34 J ⋅ s)
=
=
=
= 1.17 × 10−14 m = 0.0117 pm
E m* c 2 m* c (207)(9.11 × 10−31 kg)(3.00 × 108 m s) 44.6. (a) λ = 44.7. In this case, the muons are created at rest (no kinetic energy).
(b) Shorter wavelengths would mean higher photon energy, and the muons would be created with nonzero kinetic energy.
IDENTIFY: The energy released comes from the mass difference.
SET UP: The mass difference is the initial mass minus the final mass.
∆ m = mµ − − me− − me+
EXECUTE: Using the masses from Table 44.2, we have
∆ m = mµ − − me− − me+ = (105.7 MeV/c 2 ) − (0.511 MeV/c 2 ) − (0.511 MeV/c 2 ) = 105 MeV/c 2 44.8. Multiplying these masses by c2 gives E = 105 MeV.
EVALUATE: This energy is observed as kinetic energy of the electron and positron.
IDENTIFY and SET UP: Calculate the mass change in each reaction, using the atomic masses in Table 44.2. A
mass change of 1 u is equivalent to an energy of 931.5 MeV.
1
EXECUTE: (a) and (b) Eq.(44.1): 4 He + 9 Be → 12 C + 0 n
2
4
6
∆ M = m ( 4 He ) + m ( 9 Be ) − / m ( 12 C ) + m ( 1 n ) 0
3
4 ∆ M = 4.00260 u + 9.01218 u − 12.00000 u − 1.00866 u = 0.00612 u
The mass decreases and the energy liberated is 5.70 MeV. The reaction is exoergic.
1
4
Eq.(44.2): 0 n + 10 B → 7 Li + 2 He
5
3 ∆ M = m ( 1 n ) + m ( 10 B ) − / m ( 7 Li ) + m ( 4 He ) 0
3
4 44.9. ∆ M = 1.00866 u + 10.01294 u − 7.01600 u − 4.00260 u = 0.00300 u
The mass decreases and the energy liberated is 2.79 MeV. The reaction is exoergic.
(c) The reactants in the reactions of Eq.(44.1) have positive nuclear charges and a threshold kinetic energy is
required for the reactants to overcome their Coulomb repulsion and get close enough for the reaction to occur. The
neutron in Eq.(44.2) is neutral so there is no Coulomb repulsion and no threshold energy for this reaction.
IDENTIFY: The antimatter annihilates with an equal amount of matter.
SET UP: The energy of the matter is E = (∆ m)c 2 .
EXECUTE: Putting in the numbers gives E = (∆ m)c 2 = (400 kg + 400 kg)(3.00 × 108 m s)2 = 7.2 × 1019 J. 44.10. This is about 70% of the annual energy use in the U.S.
EVALUATE: If this huge amount of energy were released suddenly, it would blow up the Enterprise! Getting
useable energy from matterantimatter annihiliation is not so easy to do!
IDENTIFY: With a stationary target, only part of the initial kinetic energy of the moving electron is available.
Momentum conservation tells us that there must be nonzero momentum after the collision, which means that there
must also be left over kinetic energy. Therefore not all of the initial energy is available.
SET UP: The available energy is given by Ea2 = 2mc 2 ( Em + mc 2 ) for two particles of equal mass when one is
initially stationary. In this case, the initial kinetic energy (20.0 GeV = 20,000 MeV) is much more than the rest
energy of the electron (0.511 MeV), so the formula for available energy reduces to Ea = 2mc 2 Em .
EXECUTE: (a) Using the formula for available energy gives
Ea = 2mc 2 Em = 2(0.511 MeV)(20.0 GeV) = 143 MeV 44.11. (b) For colliding beams of equal mass, each particle has half the available energy, so each has 71.5 MeV. The total
energy is twice this, or 143 MeV.
EVALUATE: Colliding beams provide considerably more available energy to do experiments than do beams
hitting a stationary target. With a stationary electron target in part (a), we had to give the moving electron
20,000 MeV of energy to get the same available energy that we got with only 143 MeV of energy with the
colliding beams.
(a) IDENTIFY and SET UP: Eq. (44.7) says ω = q B / m so B = mω / q . And since ω = 2π f , this becomes B = 2π mf / q .
EXECUTE: A deuteron is a deuterium nucleus ( H ). Its charge is q = +e. Its mass is the mass of the neutral
2
1 atom (Table 43.2) minus the mass of the one atomic electron:
m = 2.014102 u − 0.0005486 u = 2.013553 u (1.66054 × 10−27 kg /1 u) = 3.344 × 10−27 kg
B= 2π mf 2π (3.344 × 10−27 kg)(9.00 × 106 Hz)
=
= 1.18 T
q
1.602 × 10−19 C 2
1 H Particle Physics and Cosmology (b) Eq.(44.8): K = 443 q 2 B 2 R 2 [(1.602 × 10−19 C)(1.18 T)(0.320 m)]2
=
.
2m
2(3.344 × 10−27 kg) K = 5.471 × 10 −13 J = (5.471 × 10 −13 J)(1 eV/1.602 × 10 −19 J) = 3.42 MeV
2K
=
m K = 1 mv 2 so v =
2 2(5.471 × 10−13 J)
= 1.81 × 107 m/s
3.344 × 10−27 kg v / c = 0.06, so it is ok to use the nonrelativistic expression for kinetic energy.
ω eB
(a) 2 f = =
= 3.97 × 107 s.
π mπ
eBR
(b) ω R =
= 3.12 × 107 m/s
m
(c) For threefigure precision, the relativistic form of the kinetic energy must be used,
( γ − 1 )mc 2
= 5.11× 106 V.
eV = ( γ − 1 )mc 2 , so eV = ( γ − 1 )mc 2 , so V =
e
(a) IDENTIFY and SET UP: The masses of the target and projectile particles are equal, so Eq. (44.10) can be used.
Ea2 = 2mc 2 ( Em + mc 2 ). Ea is specified; solve for the energy Em of the beam particles. EVALUATE:
44.12. 44.13. Ea2
− mc 2
2mc 2
The mass for the alpha particle can be calculated by subtracting two electron masses from the 4 He atomic mass:
2 EXECUTE: Em = m = mα = 4.002603 u − 2(0.0005486 u) = 4.001506 u
Then mc 2 = (4.001506 u)(931.5 MeV/u) = 3.727 GeV.
Em = 44.14. Ea2
(16.0 GeV)2
− mc 2 =
− 3.727 GeV = 30.6 GeV.
2
2mc
2(3.727 GeV) (b) Each beam must have 1 Ea = 8.0 GeV.
2
EVALUATE: For a stationary target the beam energy is nearly twice the available energy. In a colliding beam
experiment all the energy is available and each beam needs to have just half the required available energy.
1000 × 103 MeV
= 1065.8, so v = 0.999999559c.
(a) γ =
938.3 MeV
(b) Nonrelativistic: ω = Relativistic: ω =
44.15. eB
= 3.83 × 108 rad s.
m eB 1
= 3.59 × 105 rad s.
m' (a) IDENTIFY and SET UP: For a proton beam on a stationary proton target and since Ea is much larger than the proton rest energy we can use Eq.(44.11): Ea2 = 2mc 2 Em .
EXECUTE: 44.16. Em = Ea2
(77.4 GeV) 2
=
= 3200 GeV
2
2mc
2(0.938 GeV) (b) IDENTIFY and SET UP: For colliding beams the total momentum is zero and the available energy Ea is the
total energy for the two colliding particles.
EXECUTE: For protonproton collisions the colliding beams each have the same energy, so the total energy of
each beam is 1 Ea = 38.7 GeV.
2
EVALUATE: For a stationary target less than 3% of the beam energy is available for conversion into mass. The
beam energy for a colliding beam experiment is a factor of (1/83) times smaller than the required energy for a
stationary target experiment.
IDENTIFY: Only part of the initial kinetic energy of the moving electron is available. Momentum conservation
tells us that there must be nonzero momentum after the collision, which means that there must also be left over
kinetic energy.
SET UP: To create the η0, the minimum available energy must be equal to the rest mass energy of the products,
which in this case is the η0 plus two protons. In a collider, all of the initial energy is available, so the beam energy
is the available energy.
EXECUTE: The minimum amount of available energy must be rest mass energy
Ea = 2mp + mη = 2(938.3 MeV) + 547.3 MeV = 2420 MeV 444 Chapter 44 44.17. Each incident proton has half of the rest mass energy, or 1210 MeV = 1.21 GeV.
EVALUATE: As we saw in problem 44.10, we would need much more initial energy if one of the initial protons
were stationary. The result here (1.21 GeV) is the minimum amount of energy needed; the original protons could
have more energy and still trigger this reaction.
Section 44.3 says m(Z0 ) = 91.2 GeV c 2 .
E = 91.2 × 109 eV = 1.461 × 10−8 J; m = E c 2 = 1.63 × 10−25 kg; m (Z0 ) m (p) = 97.2 44.18. (a) We shall assume that the kinetic energy of the Λ 0 is negligible. In that case we
can set the value of the photon’s energy equal to Q:
Q = (1193 − 1116) MeV = 77 MeV = Ephoton . (b) The momentum of this photon is p= Ephoton
c = (77 × 106 eV)(1.60 × 10−18 J eV)
= 4.1× 10−20 kg ⋅ m s
(3.00 × 108 m s ) To justify our original assumption, we can calculate the kinetic energy of a Λ 0 that has this value of momentum
K Λ0 =
44.19. 44.20. p2
E2
(77 MeV) 2
=
=
= 2.7 MeV << Q = 77 MeV.
2m 2mc 2 2(1116 MeV) Thus, we can ignore the momentum of the Λ 0 without introducing a large error.
IDENTIFY and SET UP: Find the energy equivalent of the mass decrease.
EXECUTE: The mass decrease is m( 6 + ) − m(p) − m(π 0 ) and the energy released is
mc 2 ( 6 + ) − mc 2 (p) − mc 2 (π 0 ) = 1189 MeV − 938.3 MeV − 135.0 MeV = 116 MeV. (The mc 2 values for each
particle were taken from Table 44.3.)
EVALUATE: The mass of the decay products is less than the mass of the original particle, so the decay is
energetically allowed and energy is released.
IDENTIFY: If the initial and final rest mass energies were equal, there would be no left over energy for kinetic
energy. Therefore the kinetic energy of the products is the difference between the mass energy of the initial
particles and the final particles.
SET UP: The difference in mass is ∆m = M Ω− − mΛ 0 − mK − . EXECUTE: Using Table 44.3, the energy difference is
E = ( ∆ m)c 2 = 1672 MeV − 1116 MeV − 494 MeV = 62 MeV 44.21. EVALUATE: There is less rest mass energy after the reaction than before because 62 MeV of the initial energy
was converted to kinetic energy of the products.
Conservation of lepton number.
(a) µ − → e − + ve + vµ ! Lµ : +1 ≠ −1, Le : 0 ≠ +1 + 1 , so lepton numbers are not conserved.
(b) 0 − → e − + ve + v0 ! Le : 0 = +1 − 1 ; L0 : +1 = +1 , so lepton numbers are conserved.
(c) π + → e + + γ . Lepton numbers are not conserved since just one lepton is produced from zero original leptons.
(d) n → p + e − + 44.22. 44.23. e ! Le : 0 = +1 − 1, so the lepton numbers are conserved. IDENTIFY and SET UP: p and n have baryon number +1 and p has baryon number −1 . e+, e − , υe and γ all
have baryon number zero. Baryon number is conserved if the total baryon number of the products equals the total
baryon number of the reactants.
EXECUTE: (a) reactants: B = 1 + 1 = 2 . Products: B = 1 + 0 = 1 . Not conserved.
(b) reactants: B = 1 + 1 = 2 . Products: B = 0 + 0 = 0 . Not conserved.
(c) reactants: B = +1 . Products: B = 1 + 0 + 0 = +1 . Conserved.
(d) reactants: B = 1 −= 0 . Products: B = 0 . Conserved.
1
IDENTIFY and SET UP: Compare the sum of the strangeness quantum numbers for the particles on each side of
the decay equation. The strangeness quantum numbers for each particle are given Table 44.3.
EXECUTE: (a) K + → µ + + vµ ; SK + = +1, S µ + = 0, Svµ = 0 S = 1 initially; S = 0 for the products; S is not conserved
(b) n + K + → p + π 0 ; Sn = 0, S K + = +1, S p = 0, Sπ 0 = 0
S = 1 initially; S = 0 for the products; S is not conserved
(c) K + + K − → π 0 + π 0 ; SK + = +1; SK − = −1; Sπ 0 = 0 S = +1 − 1 = 0 initially; S = 0 for the products; S is conserved Particle Physics and Cosmology 445 (d) p + K − → Λ 0 + π 0 ; Sp = 0, SK − = −1, SΛ0 = −1, Sπ 0 = 0. 44.24. S = −1 initially; S = −1 for the products; S is conserved
EVALUATE: Strangeness is not a conserved quantity in weak interactions and strangeness nonconserving
reactions or decays can occur.
(a) Using the values of the constants from Appendix F,
e2
1
, or 1 137 to three figures.
= 7.29660475 × 10−3 =
4π P0 %c
137.050044
(b) From Section 38.5, v1 = 44.25.
44.26. " e2 #
e2
. But notice this is just $
% c , as claimed.
2P0 h
& 4π P0 %c ' / f20
(J ⋅ m)
f2
= 1 and thus
is dimensionless. (Recall f 2 has units of energy times distance.)
1 2=
−1
%c 4 (J ⋅ s)(m ⋅ s )
%c
3
(a) The diagram is given in Figure 44.26. The Ω − particle has Q = −1 (as its label suggests) and S = −3. Its
appears as a “hole”in an otherwise regular lattice in the S − Q plane. The mass difference between each S row is around 145 MeV (or so). This puts the Ω − mass at about the right spot. As it turns out, all the other particles on
this lattice had been discovered already and it was this “hole” and mass regularity that led to an accurate prediction
of the properties of the Ω !
2
−1
−1
(b) See diagram. Use quark charges u = + , d = , and s =
as a guide.
3
3
3 Figure 44.26
44.27. IDENTIFY and SET UP: Each value for the combination is the sum of the values for each quark. Use Table 44.4.
EXECUTE: (a) uds
2
Q = 3e− 1e− 1e =0
3
3 B = 1 + 1 + 1 =1
3
3
3
S = 0 + 0 − 1 = −1
C = 0+0+0 =0
(b) cu
The values for u are the negative for those for u.
2
2
Q = 3e− 3e=0
1
B = 1 −= 0
3
3
S =0+0=0
C = +1 + 0 = +1
(c) ddd
Q = − 1 e − 1 e − 1 e = −e
3
3
3 B = 1 + 1 + 1 = +1
3
3
3
S =0+0+0 =0
C = 0+0+0 =0
(d) d c
2
Q = − 1 e − 3 e = −e
3
1
B = 1 −= 0
3
3
S =0+0=0
C = 0 − 1 = −1
EVALUATE: The charge, baryon number, strangeness and charm quantum numbers of a particle are determined
by the particle's quark composition. 446 Chapter 44 44.28. ( mγ − 2mτ )c 2 = (9460 MeV − 2(1777 MeV)) = 5906 MeV (see Sections 44.3 and 44.4 for masses). 44.29. (a) The antiparticle must consist of the antiquarks so n = udd .
(b) So n = udd is not its own antiparticle.
(c) ( = cc so ( = cc = ( so the ( is its own antiparticle.
(a) S = 1 indicates the presence of one s antiquark and no s quark. To have baryon number 0 there can be only
one other quark, and to have net charge +e that quark must be a u, and the quark content is us .
(b) The particle has an s antiquark, and for a baryon number of –1 the particle must consist of three antiquarks. 44.30. 44.31. 44.32. For a net charge of –e, the quark content must be dd s .
(c) S = −2 means that there are two s quarks, and for baryon number 1 there must be one more quark. For a charge
of 0 the third quark must be a u quark and the quark content is uss.
IDENTIFY: A proton is made up of uud quarks and a neutron consists of udd quarks.
SET UP: If a proton decays by β + decay, we have p → e + + n + ve (both charge and lepton number are
conserved).
EVALUATE: Since a proton consists of uud quarks and a neutron is udd quarks, it follows that in β + decay a u
quark changes to a d quark.
(a) Using the definition of z from Example 44.9 we have that
1+ z =1+
Now we use Eq.(44.13) to obtain 1 + z = (λ0 − λs ) λ0 = λ0
.
λs c+v
1+ v / c
1+ β
=
=
.
c−v
1− v/c
1− β (b) Solving the above equation for β we obtain β = (1 + z ) 2 − 1 1.52 − 1
=
= 0.3846.
(1 + z ) 2 + 1 1.52 + 1 Thus, v = 0.3846 c = 1.15 × 108 m s.
(c) We can use Eq.(44.15) to find the distance to the given galaxy,
r=
44.33. v
(1.15 × 108 m s)
=
= 1.6 × 103 Mpc
H 0 (7.1 × 104 ( m s ) Mpc) (a) IDENTIFY and SET UP: Use Eq.(44.14) to calculate v.
/ (λ / λ ) 2 − 1 0
/ (658.5 nm/590 nm) 2 − 1 0
EXECUTE: v = 1 0 s 2 2 c = 1
2 c = 0.1094c
2
3 (658.5 nm/590 nm) + 1 4
3 (λ0 / λs ) + 1 4
v = (0.1094)(2.998 × 108 m/s) = 3.28 × 107 m/s
(b) IDENTIFY and SET UP: Use Eq.(44.15) to calculate r.
v
3.28 × 104 km/s
EXECUTE: r =
=
= 1510 Mly
H 0 (71 (km/s)/Mpc)(1 Mpc/3.26 Mly) EVALUATE: The red shift λ0 / λ S − 1 for this galaxy is 0.116. It is therefore about twice as far from earth as the
44.34.
44.35. galaxy in Examples 44.9 and 44.10, that had a red shift of 0.053.
c
3.00 × 108 m s
From Eq.(44.15), r =
=
= 1.5 × 104 Mly.
H 0 20(km s) Mly
(b) This distance represents looking back in time so far that the light has not been able to reach us.
(a) IDENTIFY and SET UP: Hubble's law is Eq.(44.15), with H 0 = 71 (km/s)/(Mpc). 1 Mpc = 3.26 Mly.
EXECUTE: r = 5210 Mly so v = H 0 r = ((71 km/s)/Mpc)(1 Mpc/3.26 Mly)(5210 Mly) = 1.1× 105 km/s
(b) IDENTIFY and SET UP: Use v from part (a) in Eq. (44.13).
EXECUTE: λ0
c+v
1+ v/c
=
=
λS
c−v
1− v / c v
1.1 × 108 m/s
λ
1 + 0.367
=
= 0.367 so 0 =
= 1.5
8
λS
c 2.9980 × 10 m/s
1 − 0.367 44.36. EVALUATE: The galaxy in Examples 44.9 and 44.10 is 710 Mly away so has a smaller recession speed and
redshift than the galaxy in this problem.
IDENTIFY and SET UP: mH = 1.67 × 10 −27 kg . The ideal gas law says pV = nRT . Normal pressure is
1.013 × 105 Pa and normal temperature is about 27 °C = 300 K . 1 mole is 6.02 × 10 23 atoms . Particle Physics and Cosmology EXECUTE: (a) 447 6.3 × 10−27 kg/m3
= 3.8 atoms/m3
1.67 × 10−27 kg/atom (b) V = (4 m)(7 m)(3 m) = 84 m3 and (3.8 atoms/m3 )(84 m3 ) = 320 atoms
(c) With p = 1.013 × 105 Pa , V = 84 m3 , T = 300 K the ideal gas law gives the number of moles to be n= 44.37. pV
(1.013 × 105 Pa)(84 m3 )
=
= 3.4 × 103 moles
RT (8.3145 J/mol ⋅ K)(300 K) (3.4 × 103 moles)(6.02 × 1023 atoms/mol) = 2.0 × 1027 atoms
EVALUATE: The average density of the universe is very small. Interstellar space contains a very small number of
atoms per cubic meter, compared to the number of atoms per cubit meter in ordinary material on the earth, such as air.
IDENTIFY and SET UP: Find the energy equivalent of the mass decrease.
3
3
EXECUTE: (a) p + 2 H → 2 He or can write as 1 H + 2 H → 2 He
1
1
1
If neutral atom masses are used then the masses of the two atomic electrons on each side of the reaction will
cancel.
3
Taking the atomic masses from Table 43.2, the mass decrease is m ( 1 H ) + m ( 2 H ) − m ( 2 He ) = 1.007825 u +
1
1 2.014102 u − 3.016029 u = 0.005898 u. The energy released is the energy equivalent of this mass decrease:
(0.005898 u)(931.5 MeV/u) = 5.494 MeV
1
3
(b) 0 n + 2 He → 4 He
2
If neutral helium masses are used then the masses of the two atomic electrons on each side of the reaction equation
1
3
will cancel. The mass decrease is m ( 0 n ) + m ( 2 He ) − m ( 4 He ) = 1.008665 u + 3.016029 u − 4.002603 u =
2 44.38. 0.022091 u. The energy released is the energy equivalent of this mass decrease:
(0.022091 u)(931.l5 MeV/u) = 20.58 MeV
EVALUATE: These are important nucleosynthesis reactions, discussed in Section 44.7.
3m( 4 He) − m(12 C) = 7.80 × 10−3 u, or 7.27 MeV. 44.39. ∆ m = me + mp − mn − mve so assuming mve ≈ 0, ∆ m = 0.0005486 u + 1.007276 u − 1.008665 u = −8.40 × 10−4 u
! E = (∆ m)c 2 = ( −8.40 × 10−4 u)(931.5 Me V u) = −0.783 MeV and is endoergic.
44.40. m12 C + m4 He − m16 O = 7.69 × 10−3 u, or 7.16 MeV, an exoergic reaction. 44.41. IDENTIFY and SET UP: The Wien displacement law (Eq.38.30) sys λmT equals a constant. Use this to relate 6 2 8 λm,1 at T1 to λm,2 at T2 .
EXECUTE: λm,1T1 = λm,2T2 " T2 #
" 2.728 K #
−3
% = 1.062 × 10 m $
% = 966 nm
& 3000 K '
& T1 '
EVALUATE: The peak wavelength was much less when the temperature was much higher.
(a) The dimensions of % are energy times time, the dimensions of G are energy times time per mass squared, and λm,1 = λm,2 $ 44.42. so the dimensions of %G / c 3 are / (E ⋅ T)(E ⋅ L M 2 ) 0
1
2
( L T)3
3
4
1/ 2 44.43. 1/2 / E 0 /T
=1 21
3M4 3 L 2 2 0 / L 0 / T2 0
2 = 1 2 1 2 = L.
4 3T4 3 L 4 12 " (6.626 × 10 −34 J ⋅ s)(6.673 × 10−11 N ⋅ m 2 kg 2 ) #
" %G #
−35
(b) $ 3 % = $
% = 1.616 × 10 m.
2" (3.00 × 108 m s)3
&c '
&
'
IDENTIFY and SET UP: For colliding beams the available energy is twice the beam energy. For a fixedtarget
experiment only a portion of the beam energy is available energy (Eqs.44.9 and 44.10).
EXECUTE: (a) Ea = 2(7.0 TeV) = 14.0 TeV
(b) Need Ea = 14.0 TeV = 14.0 × 106 MeV. Since the target and projectile particles are both protons Eq. (44.10) can be used: Ea2 = 2mc 2 ( Em + mc 2 )
Ea2
(14.0 × 106 MeV)2
− mc 2 =
− 938.3 MeV = 1.0 × 1011 MeV = 1.0 × 105 TeV.
2mc 2
2(938.3 MeV)
EVALUATE: This shows the great advantage of colliding beams at relativistic energies.
Em = 448 Chapter 44 44.44. K + mpc 2 = 44.45. IDENTIFY and SET UP: Section 44.3 says the strong interaction is 100 times as strong as the electromagnetic
interaction and that the weak interaction is 10 −9 times as strong as the strong interaction. The Coulomb force is
kq q
mm
Fe = 12 2 and the gravitational force is Fg = G 1 2 2 .
r
r
(9.0 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C)2
= 200 N
EXECUTE: (a) Fe =
(1 × 10−15 m) 2 Fg = hc λ ,K= hc λ − mpc 2 = 652 MeV. (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.67 × 10−27 kg)2
= 2 × 10−34 N
(1 × 10−15 m) 2 (b) Fstr ≈ 100 Fe ≈ 2 × 104 N . Fweak ≈ 10 −9 Fstr ≈ 2 × 10−5 N
(c) Fstr > Fe > Fweak > Fg
(d) Fe ≈ 1 × 1036 Fg . Fstr ≈ 100 Fe ≈ 1× 1038 Fg . Fweak ≈ 10−9 Fstr ≈ 1× 1029 Fg 44.46. EVALUATE: The gravity force is much weaker than any of the other three forces. Gravity is important only when
one very massive object is involved.
In Eq.(44.9), Ea = ( mΣ0 + mK0 )c 2 , and with M = mp , m = m" − and Em = (m" − )c 2 + K , K= Ea2 − ( mπ − c 2 ) 2 − ( mp c 2 ) 2
2mpc 2 − ( mπ − )c 2 (1193 MeV + 497.7 MeV) 2 − (139.6 MeV)2 − (938.3 MeV)2
− 139.6 MeV = 904 MeV.
2(938.3 MeV)
IDENTIFY: With a stationary target, only part of the initial kinetic energy of the moving proton is available.
Momentum conservation tells us that there must be nonzero momentum after the collision, which means that there
must also be left over kinetic energy. Therefore not all of the initial energy is available.
K= 44.47. SET UP: The available energy is given by Ea2 = 2mc 2 ( Em + mc 2 ) for two particles of equal mass when one is initially stationary. The minimum available energy must be equal to the rest mass energies of the products, which in
this case is two protons, a K+ and a K − . The available energy must be at least the sum of the final rest masses.
EXECUTE: The minimum amount of available energy must be
Ea = 2mp + mK + + mK − = 2(938.3 MeV) + 493.7 MeV + 493.7 MeV = 2864 MeV = 2.864 GeV
Solving the available energy formula for Em gives Ea2 = 2mc 2 ( Em + mc 2 ) and
Em = 44.48. Ea2
(2864 MeV)2
− mc 2 =
− 938.3 MeV = 3432.6 MeV
2
2mc
2(938.3 MeV) Recalling that Em is the total energy of the proton, including its rest mass energy (RME), we have
K = Em – RME = 3432.6 MeV – 938.3 MeV = 2494 MeV = 2.494 GeV
Therefore the threshold kinetic energy is K = 2494 MeV = 2.494 GeV.
EVALUATE: Considerably less energy would be needed if the experiment were done using colliding beams of
protons.
(a) The decay products must be neutral, so the only possible combinations are π 0π 0π 0 or π 0π +π −
(b) m10 − 3mπ 0 = 142.3 Me V c 2 , so the kinetic energy of the π 0 mesons is 142.3 MeV. For the other reaction, K = (mη0 − mπ 0 − mπ + − mπ − )c 2 = 133.1 MeV.
44.49. IDENTIFY and SET UP: Apply conservation of linear momentum to the collision. A photon has momentum
hc
. All the mass of the electron and
p = h / λ , in the direction it is traveling. The energy of a photon is E = pc = λ positron is converted to the total energy of the two photons, according to E = mc 2 . The mass of an electron and of
a positron is me = 9.11 × 10 −31 kg
EXECUTE: (a) In the lab frame the initial momentum of the system is zero, since the electron and positron have
equal speeds in opposite directions. According to momentum conservation, the final momentum of the system
must also be zero. A photon has momentum, so the momentum of a single photon is not zero.
(b) For the two photons to have zero total momentum they must have the same magnitude of momentum and move
in opposite directions. Since E = pc , equal p means equal E. Particle Physics and Cosmology 449 (c) 2 Eph = 2mec 2 so Eph = me c 2 Eph = 44.50. 44.51. 44.52. hc λ so hc λ = mec 2 and λ = h
6.63 × 10−34 J ⋅ s
=
= 2.43 pm
me c (9.11 × 10−31 kg)(3.00 × 108 m/s) These are gamma ray photons.
EVALUATE: The total charge of the electron/positron system is zero and the photons have no charge, so charge is
conserved in the particleantiparticle annihilation.
(a) If the π − decays, it must end in an electron and neutrinos. The rest energy of π − (139.6 MeV) is shared
between the electron rest energy (0.511 MeV) and kinetic energy (assuming the neutrino masses are negligible). So
the energy released is 139.6 MeV – 0.511 MeV = 139.1 MeV.
(b) Conservation of momentum leads to the neutrinos carrying away most of the energy.
(a) The baryon number is 0, the charge is +e , the strangeness is 1, all lepton numbers are zero, and the particle is K + .
(b) The baryon number is 0, the charge is −e , the strangeness is 0, all lepton numbers are zero, and the particle is π − .
(c) The baryon numbers is –1, the charge is 0, the strangeness is zero, all lepton numbers are 0, and the particle is
an antineutron.
(d) The baryon number is 0, the charge is +e , the strangeness is 0, the muonic lepton number is –1, all other
lepton numbers are 0, and the particle is µ + .
∆ t = 7.6 × 10−21 s ! ∆ E = % 1.054 × 10−34 J ⋅ s
=
= 1.39 × 10−14 J = 87 keV .
7.6 × 10−21 s
∆t ∆ E 0.087 MeV
=
= 2.8 × 10−5.
m( c 2 3097 MeV
44.53.
44.54. %
(1.054 × 10−34 J ⋅ s)
=
= 1.5 × 10−22 s.
∆ E (4.4 × 106 eV)(1.6 × 10−19 J/eV) φ → K + + K − . The total energy released is the energy equivalent of the mass decrease.
(a) EXECUTE: The mass decrease is m(φ ) − m(K + ) − m(K − ). The energy equivalent of the mass decrease is
IDENTIFY and SET UP: mc 2 (φ ) − mc 2 (K + ) − mc 2 (K − ). The rest mass energy mc 2 for the φ meson is given in Problem 44.53, and the values for K +and K − are given in Table 44.3. The energy released then is 1019.4 MeV − 2(493.7 MeV) =
32.0 MeV. The K + gets half this, 16.0 MeV.
EVALUATE: (b) Does the decay φ → K + + K − + π 0 occur? The energy equivalent of the K + + K − + π 0 mass is
493.7 MeV + 493.7 MeV + 135.0 MeV = 1122 MeV. This is greater than the energy equivalent of the φ mass.
The mass of the decay products would be greater than the mass of the parent particle; the decay is energetically
forbidden.
(c) Does the decay φ → K + + π − occur? The reaction φ → K + + K − is observed. K + has strangeness +1 and K −
has strangeness −1, so the total strangeness of the decay products is zero. If strangeness must be conserved we
deduce that the φ particle has strangeness zero. π − has strangeness 0, so the product K + + π − has strangeness 44.55. −1. The decay φ → K + + π − violates conservation of strangeness. Does the decay φ → K + + µ − occur? µ − has
strangeness 0, so this decay would also violate conservation of strangeness.
(a) The number of protons in a kilogram is " 6.023 × 1023 molecules mol #
25
(1.00 kg) $
% (2 protons molecule) = 6.7 × 10 .
18.0 × 10−3 kg mol
&
'
Note that only the protons in the hydrogen atoms are considered as possible sources of proton decay. The energy
per decay is mp c 2 = 938.3 MeV = 1.503 × 10−10 J, and so the energy deposited in a year, per kilogram, is 44.56. " ln(2) #
−10
−3
(6.7 × 1025 ) $
% (1 y) (1.50 × 10 J) = 7.0 × 10 Gy = 0.70 rad
18
& 1.0 × 10 y '
(b) For an RBE of unity, the equivalent dose is (1) (0.70 rad) = 0.70 rem.
IDENTIFY and SET UP: The total released energy is the equivalent of the mass decrease. Use conservation of
linear momentum to relate the kinetic energies of the decay particles.
EXECUTE: (a) The energy equivalent of the mass decrease is
mc 2 (Ξ − ) − mc 2 ( Λ 0 ) − mc 2 (π − ) = 1321 MeV − 1116 MeV − 139.6 MeV = 65 MeV 4410 Chapter 44 (b) The Ξ − is at rest means that the linear momentum is zero. Conservation of linear momentum then says that the
Λ 0 and π − must have equal and opposite momenta:
mΛ 0 vΛ 0 = mπ − vπ − "m 0 #
vπ − = $ Λ % vΛ 0
$m − %
& π'
Also, the sum of the kinetic energies of the Λ 0 and π − must equal the total kinetic energy K tot = 65 MeV
calculated in part (a):
K tot = K Λ0 + Kπ −
2
K Λ0 + 1 mπ − vπ − = K tot
2 Use the momentum conservation result:
2 "m 0 # 2
K Λ0 + mπ − $ Λ % vΛ 0 = K tot
$m − %
&π'
" mΛ 0 # 1
K Λ0 + $
% m v 2 = K tot
$ m − % ( 2 Λ0 Λ0 )
&π'
" mΛ0 #
K Λ0 $ 1 +
% = K tot
$ m−%
π'
&
K tot
65 MeV
=
= 7.2 MeV
K Λ0 =
1+mΛ0 / mπ − 1 + (1116 MeV)/(139.6 MeV)
1
2 K Λ0 + Kπ − = K tot so Kπ − = K tot − K Λ 0 = 65 MeV − 7.2 MeV = 57.8 MeV 7.2 MeV
= 11%.
65 MeV
57.8 MeV
The fraction for the π − is
= 89%.
65 MeV
EVALUATE: The lighter particle carries off more of the kinetic energy that is released in the decay than the
heavier particle does.
dR
dR dt HR
(a) For this model,
= HR, so
=
= H , presumed to be the same for all points on the surface.
dt
R
R
dr dR
(b) For constant θ ,
=
θ = HRθ = Hr.
dt dt
dR dt
(c) See part (a), H 0 =
.
R
dR
(d) The equation
= H 0 R is a differential equation, the solution to which, for constant H 0 , is R(t ) =
dt
R0e H 0 t , where R0 is the value of R at t = 0 . This equation may be solved by separation of variables, as The fraction for the Λ 0 is 44.57. dR dt d
= ln ( R) = H 0 and integrating both sides with respect to time.
R
dt
(e) A constant H 0 would mean a constant critical density, which is inconsistent with uniform expansion.
44.58. r
dR 1 dr r d& 1 dr
d&
since
=
−
=
= 0.
From Problem 44.57, r = Rθ ! R = . So
θ
dt & dt & 2 dt & dt
dt
1 dR
1 dr 1 dr
dr " 1 dR #
dv
d " r dR # d " dR #
So
=
=
!v=
=$
=0=
% r = H 0 r. Now
$
%=
$&
%
dt & R dt '
d&
d& & R dt ' d& & dt '
R dt R& dt r dt
dR
dR K
dθ
1 dR θ K 1
"K#
!θ
= K where K is a constant. !
= ! R = $ % t since
= 0 ! H0 =
=
= . So the
dt θ
dt
R dt Kt θ t
dt
&θ '
1
where T is the present age of the universe.
T
v0 − vcm
(a) For mass m, in Eq. (37.23) u = −vcm , v′ = v0 , and so vm =
. For mass
1 − v0 vcm c 2
M , u = −vcm , v′ = 0, so vM = −vcm . current value of the Hubble constant is
44.59. Particle Physics and Cosmology 4411 (b) The condition for no net momentum in the center of mass frame is mγ m vm + M γ M vM = 0, where γ m and γ M correspond to the velocities found in part (a). The algebra reduces to β mγ m = ( β 0 − β ′)γ 0γ M , where
v0
v
, β ′ = cm , and the condition for no net momentum becomes m( β 0 − β ′)γ 0γ M = M β ′γ M , or
c
c
β0
mv0
m
β′ =
.
= β0
. vcm =
2
M
m + M 1 − (v0 / c) 2
m + M 1 − β0
1+
mγ 0
(c) Substitution of the above expression into the expressions for the velocities found in part (a) gives the relatively
M
m
. After some more algebra,
,vM = −v0γ 0
simple forms vm = v0γ 0
m + Mγ 0
mγ 0 + M β0 = γm = m + Mγ 0
m + M + 2mM γ 0
2 2 ,γM = M + mγ 0
m + M 2 + 2mM γ 0
2 , from which mγ m + M γ M = m 2 + M 2 + 2mM γ 0 . This last expression, multiplied by c 2 , is the available energy Ea in the center of mass frame, so that
Ea2 = (m 2 + M 2 + 2mM'0 )c 4 = (mc 2 ) 2 + (Mc 2 ) 2 + (2Mc 2 )(m'0c 2 ) = ( mc 2 ) 2 + ( Mc 2 ) 2 + 2 Mc 2 Em , which is Eq.(44.9). 44.60. Λ0 → n + π 0
(a) E = ( ∆ m)c 2 = (mΛ0 )c 2 − (mn )c 2 ( mπ 0 )c 2 = 1116 MeV − 939.6 MeV − 135.0 MeV = 41.4 MeV (b) Using conservation of momentum and kinetic energy; we know that the momentum of the neutron and pion
must have the same magnitude, pn = pπ . K n = En − mn c 2 = ( mn c 2 ) 2 + ( pn c) 2 − mn c 2 = ( mn c 2 ) 2 + ( pn c) 2 − mn c 2
K n = (mn c 2 ) 2 + Kπ2 + 2mπ c 2 Kπ − mn c 2 = Kπ + K n = Kπ + ( mn c 2 ) 2 + Kπ2 + 2mπ c 2 Kπ − mn c 2 = E.
( mn c 2 ) 2 + Kπ2 + 2mπ c 2 Kπ = E 2 + ( mn c 2 ) 2 + Kπ2 + 2 Emπ c 2 − 2 EKπ − 2mn c 2 Kπ . Collecting terms we find:
Kπ (2mπ c 2 + 2 E + 2mn c 2 ) = E 2 + 2 Emn c 2 ! Kπ = (41.4 MeV) 2 + 2(41.4 MeV)(939.6 MeV)
= 35.62 MeV.
2(135.0 MeV) + 2(41.4 MeV) + 2(939.6 MeV) So the fractional energy carried by the pion is 35.62
= 0.86, and that of the neutron is 0.14.
41.4 ...
View
Full
Document
This note was uploaded on 11/02/2010 for the course PHYS 0030 taught by Professor Cutts during the Fall '07 term at Brown.
 Fall '07
 Cutts
 Physics

Click to edit the document details