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Unformatted text preview: e 7 ' i 1
g 56 of the text, note that tne Pythagorean theorem tells us that the 4.00 {JG 7 F the charge of the l' ht ' ‘ ‘
exerts on the 3.00 #0 Change of the ﬁght gigatélrégZde. The magnitude of the force that the 4.00 ,uC charge _ 1 lQl 9 «
FH_.__ I. 2 0
1 wig; (5.00 ><10—2 m )2 143.2 N. rE‘he force is repulsive since both charges have the same sign.
If we denote by 4') the angle hetween the maxis and the line connecting the 4.00 M) charge with the origin, we can express the force F1 in component form as
E 2 43.2 N (— cos qii— sin 95;) = 43.2 N (—0.800? — 0.6003) = —34.5 N’i— 25.9 N}.
 The force of the —3.00 n0 charge on the charge of the light brigade has magnitude 1 long! 2 (9.00 x109 Nm2/02 )(3.00 x 106 execs x 105 C) F = — z T
2 4m r2 (4.00x109m)2 5051‘ This force is attractive since the two charges have opposite signs. Write the force on the charge of the light
brigade in Cartesian form as t —p 1. F2 = 50.6 Ni. The total force is the vector sum: from 2 F. + E = —34.6 Ni— 25.9 Nj+ 50.6 Ni : 16,0 Ni— 25.9 N}. b) The magnitude of the total force is
Ftoga} 3 N)2 4* Z c) The acute angle 9 that the total force makes with the xdirection is found from Fy total _ —25.9 N Fm total _ 16.0N : "‘1'62 => 9 x —58.3 . tané‘ z Therefore, the total force points along a direction 58.3“ below the x—axis. 16.27
a) Find the electric ﬁeld produced by each charge at point P from the general expression
E 2: "L912:
are) 7‘2 Where f‘ is 3 unit vector pointing radially away from the charge. In general, the magnitude of E is E: _1_i¢il_
Limo 7‘2 Therefore: the magnitude of the electric ﬁeid produced at P by the 2.00 ,uC charge is = (9.00 x :09 NrnZ/CE )(200 x 106 C) 2.1 13N .'
(4.00111? 13x0 /C E1 (Q7750 {7/ The charge called Billy produces a ﬁeld Where lql is located of magnitude E (9.00 x 109 Nm2/02 )} — 2.00 x 103 C
1 I l_ 5 Since Bill}r is negative, this ﬁeld is directed toward Billy, parallel to (Hi). Therefore, 733: —(4.50 X 105 N/C)i. Using the Pythagorean theorem to ﬁnd the distance of Milly from the point Where a" = «(1.00 m)2 + (2.00 m}2 x 2.24 m. Thus, the charge called Milly produces a ﬁeld where I 19} is located, we have ql is located of magnitude 9 _2 2 M3
E2 : (9.00x10 Ng2/4Cm))l:.00x10 C] 21.79x106N/C. E2 =1.79 x 106 N/o(cos9i+ 51119;) 2.00113  —1.00rn a
: . 1 6\T ' '
179x 0 i/C[(2.24711)1+(2.24m l3]
2 (1.60 x 106 N/C)§— (0.799 x 10“ N/C)j. I The total electric ﬁeld at the place Where M is to be piawd is the vectm sum 0f the ﬁe}
Milly and Billy at that location, ds produced by —a E=E1+E2 2 (4.50 x 106 N/C)l+ {1.60 X 196 N/oﬁ— (0.799 x106
= (“2.90 x 106 N/Cﬁ— (0.799 x 106 N/Cﬁ. gative. The force on this N/CJE C) Your charge is He charge is F 2 9E : (—3.00 x 1073 C)((—2.90 ><106 N/eﬁ— (0.799 x 105 N/C)j]
= (8.70 x 103 N)l+ (2.40 x 103 IN}. 598 CHAPTER 16. ELECTRICAL CHARGES, ELECTRICAL FORCES, AND .. . d) The magnitude of the force is F 2 (8.70 x103 N)? + (—2.40 x 103 N)2 = 9.03 x103 N. 16.32
a) The charges shown in Figure 13.32 on page 758 of the text are point charges, so the magnitude of the electric ﬁeid each creates is 1 lQi _ three 4"? ' E Referring to Figure 13.32, the charge on the :raxis produces a field at the origin of magnitude 9W. 2 2 ~ MS ..
zwwzaiw 108 5/0. E1 (0.500 m)? Since the charge producing this eiectric ﬁeld is negative, the ﬁeld is directed toward the charge creating it,
parailel to i, so the ﬁeld vector is E; 2 (5.40 x 105 N/C)i.
The charge on the y—axis produces a ﬁeld at the origin of magnitude (9.00 X 109 Nm2/02)1(}.0 X 10—6 CI = . 5N1 .
(0.2mm? 244x10 i/C E2: Since the charge producing this electric ﬁeld is positive, the ﬁeld is directed away from the charge creating
it, parallel to ~j, so the ﬁeid vector is a. m (44.4 x 105 51/0);
The total electric ﬁeld at the origin is the vector sum of the ﬁelds produced by both charges at that location,
1?: = E1 + a m (5.40 x 105 won — (14.4 x 105 N/C )3.
b) The magnitude of the field at the origin is
E = (5.40 x105 N/o)2 4 (—14.4 x 105 we 2 = 15.4 x 105 N/C.
c) If a proton is placed at the origin, the force on it is
f = qﬁ
: (1.602 x 10’1g C) [(5.40 X105 N/C)i m (24.4 x 105 N/C )3] = (8.65 x 10"14 Nﬁ (23.1 ><10"14 N);
d) The magnitude of the force on the proton is
F a (8.65 x 1014 N)2 + (—23.1 x 1014 N)2 = 24.7 x 10"” N. The acceleration of the proton is found from Newton’s second law. Using the magnitudes of the vectors, we have F 24.7 x 10w14 N Fzmoz—ﬁ: a:— _ "~————— = 14 2‘
m T 1.57 x 10—27 kg 1.48 x 10 mg 16.40. 601 16.40 Since the torque on an electric dipole is 1'" 2 if x E,
to ﬁnd the torque on the water molecule, we need to know its dipole moment ff and the electric ﬁeid I3”) in
which it ﬁnds itseif. Referring to Figure P40 on page 759 of the text, we see that the magnitude of the torque on the dipole is .— = {13‘ x E: = pEsinl? a (6.0 x 10—3“ cm)(250 N/C) sin 120° 21.3 x 102? Nm. The vector product righthand rule indicates that the direction of the torque vector is perpendicular to and
out of the page. This problem illustrates the principle behind the microwave oven. Water makes up a high percentage
of most food. When micrOWaves pass through the food from one direction, they exert a torque on the
water molecules which causes them to rotate until 6 3 0° and the torque they experience vanishes. The
microwaves then reﬂect off the inside of the oven and come though the food from a different direction,
causing the water molecules to rotate so they align with the new direction. As this process continues, the
water molecules rotate quite a bit and this collective motion contributes to the thermodynamic internal
energy. An advantage of microwave heating is that it doesn’t need to start at the surface and work its way
in, like other forms of heating, such as conduction. Rather, microwave heating occurs uniformly throughout all of the food.
Although some people refer to microwaving their food as “nuking” it, as you see, there are no nuciear processes involved. 16.50. 605 16.50
In Example 16.17 on page 738 of the text, we saw that two oppositely charged parallel plates produce a.
a; electric field between them of magnitude
E = i7
50
' 47FEQE X 3.06 m5 2
Z 2 r M = 2.7 x 10 C m .
“1* U EOE 471' 4mm) x 109 NEH/c?) / h) Since 103 mm = 1 m, we have (103)2 mm2 2 106 rim12 : 1 m2. Hence the number of fundamental units of charge per square millimeter is 2.7 x 104 cm? :17 X 108 mm—z
(106 Hung/r112) (1.602 x10“19 C) ' 16.58
a) The force on the particle of mass m and charge 9‘ is the electric force, Whose magnitude is Feiectric : To find the particle’s acceleration, apply Newton’s second law using vector magnitudes, f q E Felectric : mo ::> quE : me :> o = The acceleration of a charged particle in a uniforrn E ﬁeld is constant, so we can use the kinematic equations
for motion with a constant acceleration, letting i be in the direction of motion and choosing the origin at
the initial position of the proton: ' 39(t) = $0 + vmgt + ::> tﬁ 2—ma 293m
Vex IQFE‘ For a proton to travel 1.00 In, the time is 20.00 m)(1.6? x 10“27 kg) M : ms
gm“ ‘ (1.502 X 1019 oxsoo N/C) 8'34 X10 3' “Mawmsmmmmwmmmmemm b) For an electron, the time is t _ 2(1.00 m)(9.11 x 18—31 kg) _ 1 9, 10M?
elem“ (1.502 x 1049 C)(300 N/C) * ‘ D X S‘ a) The ratio of the two times is: tproton __ X 10%; S
telectron X 10"“? s :: 42.8. Now use the equation for t at the end of part a) to evaluate the above ratio Without any numbers: 2 x mproton
lfproton __ / mproton 42 8
tElectron 2 1: melectron melectron qE So the ratio of times depends only upon the ratio of the masses (as long as both particles have the same
chare‘el. \A.....i u VJ\\J\JK)A1/\_l" 16.66 Referring to Figure P. 56 on page 7'61, choose a coordinate system with ipointing right, pointing
up, and origin at the launch point of the electron. With these choices, we make the important observation
that, since We are projecting an electron into the given uniform electric field, the motion of this electron is
equivalent to the motion of a massive projectile in a gravitational ﬁeld. In other words, we can treat this
problem exactly like we treated projectile motion problems in Chapter 4, with the only difference being that
(my = —eE/m rather than —g. The magnitude of the electrical force on the electron is F = fglE : 6E. Apply Newton’s second law to the electron to ﬁnd the magnitude of its acceleration. Using vector magni—
tudes, we have Plate! BE Rota} : ma => 3 z
m m In the chosen coordinate system the acceleration is down, and ye =0m _ . $0 — 0 m
oyo : o0 s1ni9 Umo : B0 6059
BE 2
ay = —m— cm 2 0 m/s .
m
Therefore, the equations for the velocity and position components are
. 3E
Dy“) _ b0 8mg — v$(t) 2 U9 c086
eE t9 t = 9 t
310?) = (1:0 sin l9)t — 3 ml ) 00(008 ) Impact occurs Where y 2 0 in, so E t2
0 In 3 (nosinﬂhf ~— (L) 3. m .4 Use the quadratic formula to solve this for t, and thus, to ﬁnd the time the electron is in the air. The two
solutions are t : l} s, which corresponds to the launch time (i.e., the time at which 3; = 0 m l» and 2 Zmo
tze—Eﬂsintlm 0 try eE sin 6. Finally, use this time in the equation for :c to ﬁnd the distance the electron travels: x = (no cos 6%
= (no Cost?) (2*mZOEsmg)
_ 2mo§
eE
: sin 501]" cos 600°
(1.602 x 1019 C}{250 N/G) = 0.492 In . sin 9 cos 8 ...
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This note was uploaded on 10/01/2009 for the course PHYS 102 taught by Professor Henry during the Spring '09 term at Johns Hopkins.
 Spring '09
 HENRY
 Physics

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