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Unformatted text preview: increased by a factor of 5 then the force
will decrease by a factor of 25. The new force is, then, 3.5 N
= 0.14 N
25
______________________________________________________________________________
F= 17. REASONING The electrons transferred increase the magnitudes of the positive and negative
charges from 2.00 µC to a greater value. We can calculate the number N of electrons by
dividing the change in the magnitude of the charges by the magnitude e of the charge on an
electron. The greater charge that exists after the transfer can be obtained from Coulomb’s law
and the value given for the magnitude of the electrostatic force.
SOLUTION The number N of electrons transferred is N= qafter − qbefore
e where qafter and qbefore are the magnitudes of the charges after and before the transfer of
electrons occurs. To obtain qafter , we apply Coulomb’s law with a value of 68.0 N for the
electrostatic force:
F =k qafter
r2 2 or qafter = Fr 2
k Using this result in the expression for N, we find that N= Fr 2
− qbefore
k
=
e ( 68.0 N )( 0.0300 m )2 − 2.00 ×10−6 C
8.99 ×109 N ⋅ m 2 / C2
= 3.8 × 1012
−19
1.60 × 10
C 956 ELECTRIC FORCES AND ELECTRIC FIELDS 18. REASONING AND SOLUTION Calculate the magnitude of each force acting on the
center charge. Using Coulomb’s law, we can write F43 = k q4 q3
2
r43 (8.99 × 109 N ⋅ m2 / C2 ) ( 4.00 × 10−6 C ) ( 3.00 × 10−6 C )
=
( 0.100 m )2 = 10.8 N (toward the south)
F53 = k q5 q3
2
r53 = (8.99 × 109 N ⋅ m2 / C2 ) ( 5.00 × 10−6 C ) ( 3.00 × 10−6 C )
( 0.100 m )2 = 13.5 N (toward the east)
Adding F43 and F53 as vectors, we have
2
2
F = F43 + F53 = (10.8 N )2 + (13.5 N )2 = 17.3 N F43 –1 10.8 N = tan = 38.7° S of E 13.5 N F53 ______________________________________________________________________________ θ = tan –1 19. REASONING According to Newton’s second law, the centripetal acceleration experienced
by the orbiting electron is equal to the centripetal force divided by the electron’s mass.
Recall from Section 5.3 that the centripetal force Fc is the name given to the net force
required to keep an object on a circular path of radius r. For an electron orbiting about two
protons, the centripetal force is provided almost exclusively by the electrostatic force of
attraction between the electron and the protons. This force points toward the center of the
circle and its magnitude is given by Coulomb’s law. SOLUTION The magnitude ac of the centripetal acceleration is equal to the magnitude Fc
of the centripetal force divided by the electron’s mass: ac = Fc / m (Equation 5.3). The
centripetal force is provided almost entirely by the electrostatic force, so Fc = F, where F is
the magnitude of the electrostatic force of attraction between the electron and the two
protons, Thus, ac = F / m . The magnitude of the electrostatic force is given by Coulomb’s law, F = k q1 q2 / r 2 (Equation 18.1), where q1 = −e and q2 = +2e are the magnitudes
of the charges, r is the radius of the orbit, and k = 8.99 ×109 N ⋅ m 2 / C2 . Substituting this
expression for F into ac = F / m , and using m = 9.11 × 10
we find that −31 kg for the mass of the electron, Chapter 18 Problems ac =
= F
=
m k 957 −e +2e
k −e +2e
r2
=
m
m r2 ( 8.99 ×109 N ⋅ m 2 / C2 ) −1.60 ( 9.11×10 −31 × 10−19 C +2 × 1.60 × 10−19 C kg )( 2.65 ×10 −11 m) 2 = 7.19 × 1023 m/s 2 ____________________________________________________________________________________________ 20. REASONING
The drawing at the right shows the forces that act
on the charges at each corner. For example, FAB is
the force exerted on the charge at corner A by the
charge at corner B. The directions of the forces
are consistent with the fact that like charges repel
and unlike charges attract.
Coulomb’s law
indicates that all of the forces shown have the
2 same magnitude, namely, F = k q / L , where q
is the magnitude of each of the charges and L is the
length of each side of the equilateral triangle. The
magnitude is the same for each force, because q
and L are the same for each pair of charges. FBC
+
B
FBA
FAB 2 +
A − C
FAC FCA
FCB The net force acting at each corner is the sum of the two force vectors shown in the drawing,
and the net force is greatest at corner A. This is because the angle between the two vectors at
A is 60º. With the angle less than 90º, the two vectors partially reinforce one another. In
comparison, the angles between the vectors at corners B and C are both 120º, which means
that the vectors at those corners partially offset one another.
The net forces acting at corners B and C have the same magnitude, since the magnitudes of
the individual vectors are the same and the angles between the vectors at both B and C are
the same (120º). Thus, vector addition by either the tailtohead method (see Section 1.6) or
the component method (see Section 1.8) will give resultant vectors that have different
directions but the same magnitude. The magnitude of the net force is the smallest at these
two co...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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