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Chapter 19: Electric Charges, Forces, and Fields
14. How much equal charge should be placed on the Earth and the Moon so that the electrical repulsion balances the gravitational force of 1.98×1020 N?
Treat the Earth and Moon as point charges a distance 3.84×108 m apart.
14. Picture the Problem: Electric charges are placed on the Moon and the Earth that exert electrostatic force on each other.
Strategy: Solve Coulomb’s law (Equation 19-5) for the charge q.
q q
q2
Solution: Solve Eq. 19-5 for r: F  k 1 2 2  k 2
r
r
r2F
q 

k
2
3.84 10 m  1.98 10 N   5.70 10
8.99 10 N  m /C 
8
2
20
13
9
2
2
C
Insight: This is a large amount of charge, but the Earth and Moon are very large bodies. It would require the removal of
only one electron from every 1036 atoms in the Earth to accumulate this charge. However, both the Earth and Moon are
almost electrically neutral, and the gravitational force dominates the interaction between the two.
96. How many electrons must be transferred away from a bee to produce a charge of +93.0 pC?
96. Picture the Problem: A bumblebee acquires an electrostatic charge in active flight.
Strategy: Use the magnitude of an electron’s charge e to find the number of electrons that correspond to the 93.0-pC
total charge.
Q
93.0 1012 C

 5.81108 electrons
e 1.60 1019 C/electron
Insight: The 93.0-pC charge may be small, but it amounts to 581 million electrons!
N
Solution: Find the number of electrons:
97. Suppose two bees, each with a charge of 93.0 pC, are separated by a distance of 1.20 cm. Treating the bees as point charges, what is the magnitude of the
electrostatic force experienced by the bees? (In comparison, the weight of a 0.140-g bee is 1.37×10−3 N.)
97. Picture the Problem: Two bumblebees each acquire electrostatic charges in active flight.
Strategy: Use Coulomb’s law to find the magnitude of the force between the two charged bees.
 93.0 1012 C 
kq q
F  12 2   8.99  109 N  m 2 /C2 
2
r
 0.0120 m 
Solution: Use Coulomb’s law to find
the force between two charged bees:
2
 5.40 107 N
Insight: The electrical force between the bees is a tiny fraction (about 1 2540 ) of their weight because the amount of
electrical charge is quite small. It would require a charge of 4,690 pC on each bee for the electrical force to equal the weight!
98. The force required to detach a grain of pollen from an avocado stamen is approximately 4.0×10−8 N. What is the maximum distance at which the
electrostatic force between a bee and a grain of pollen is sufficient to detach the pollen? Treat the bee and pollen as point charges, and assume the pollen has
a charge opposite in sign and equal in magnitude to the bee.
98. Picture the Problem: A charged bee attracts an oppositely charged pollen grain.
Strategy: Because the bee and the pollen grain have the same magnitude charge, the magnitude of the attractive force
between them, as given by Coulomb’s law (Equation 19-5), is F  kq 2 r 2 . Solve this expression for r in order to find
the distance between the bee and the pollen grain when the attractive force is sufficient to detach the pollen grain.
k
8.99 109 N  m2 /C2
  93 1012 C 
 0.044 m  4.4 cm
F
4.0 108 N
Insight: This is a substantial distance, over 2 body lengths for a 2.0-cm bumblebee!
Solution: Solve Equation 19-5 for r:
rq
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19 – 1
James S. Walker, Physics, 5th Edition
Chapter 19: Electric Charges, Forces, and Fields
45. Predict/Calculate Figure 19-42 shows a system consisting of three charges, q1 = +5.00 µC, q2 = +5.00 µC, and q3 = −5.00 µC, at the vertices of an
equilateral triangle of side d = 2.95 cm. (a) Find the magnitude of the electric field at a point halfway between the charges q1 and q2. (b) Is the magnitude of
the electric field halfway between the charges q2 and q3 greater than, less than, or the same as the electric field found in part (a)? Explain. (c) Find the
magnitude of the electric field at the point specified in part (b).
45. Picture the Problem: Three charges are positioned as shown at right.
Strategy: Each of the three charges produces its own electric field that surrounds it.
The total electric field at any point is the vector sum of the fields from each charge.
Use Equation 19-10 and the component method of vector addition to find the
magnitude electric field at the points indicated in the problem statement. Let q1 be
at the origin and q3 be on the positive x-axis.
Solution: 1. (a) At a point halfway between
charges q1 and q2 , the vectors E1 and E 2
cancel one another. The remaining contribution comes from q3 . Find the distance r
r
r 2   d 2  d 2
2
r  3d 2 4
 3  0.0295 m  4
2
from q3 to the midpoint of the opposite side:
2. Apply Equation 19-10 to find E3 :
r  0.02555 m
E3 
9
2
2
6
k q3  8.99 10 N  m /C  5.00 10 C 

 6.89 107 N/C
2
r2
0.02555
m


3. (b) At this location, the electric fields from q2 and q3 add, and the resulting field points toward q3 . The field due to
q1 will have the same magnitude as found in part (a) and will be perpendicular to the combined fields of q2 and q3 .
The vector sum of the electric fields from all three charges will have a magnitude greater than that found in part (a).
k q1
 cos30 xˆ  sin 30 yˆ  
4. (c) Find the components of E1 :
E1 
5. Find the components of E2 :
E2 
6. Find the components of E3 :
E3 
7. Let q1  q2  q3  q
and find the vector sum:
Enet  E1  E2  E3
2
3d 4
k q2
 cos 60 xˆ  sin 60 yˆ  
k q2
2
 cos 60 xˆ  sin 60 yˆ  
k q3
2
 d 2
k q3
 d 2
Enet 
k q1  2 3
2 
xˆ  yˆ 
2 

3 
d  3
d2
d2
 2 xˆ  2
3 yˆ
 2 xˆ  2
3 yˆ


kq 
2 3
2
 
4
 xˆ    4 3  yˆ 
2 

3 
d 
3
 
2
8. Determine the magnitude of Enet :
Enet
kq
 2
d
2

2 3 2
kq

 4 
    4 3   2 65.78
3  3
d


8.99 10

9
N  m 2 /C2  5.00  10 6 C 
 0.0295 m 
2
8.110 
Enet  4.19  108 N/C  419 MN/C
Insight: As expected, the field is larger at the point midway between q2 and q3 , about six times larger in magnitude
than at the point midway between q1 and q2 .
Copyright © 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19 – 2
James S. Walker, Physics, 5th Edition
Chapter 20: Electric Potential and Electric Potential Energy
Chapter 20: Electric Potential and Electric Potential Energy
35. A charge of 4.07 µC is held fixed at the origin. A second charge of 3.45 µC is released from rest at the position (1.25 m, 0.570 m). (a) If the mass of the
second charge is 2.36 g, what is its speed when it moves infinitely far from the origin? (b) At what distance from the origin does the second charge attain
half the speed it will have at infinity?
35. Picture the Problem: Two charged particles are separated by a certain distance, with one held fixed at the origin and
the other allowed to move freely.
Strategy: The second charge will accelerate away from the origin due to the electrical repulsion between the two
charges. Use conservation of energy to set the loss of potential energy of the second charge equal to the gain of kinetic
energy and solve for the final speed vf . When the particle’s speed is half the speed it will attain at infinity, its kinetic
energy is only one-fourth of its final value. That means the potential energy at that point is three-fourths of its initial
value. Use this fact to find the distance from the origin at which the speed is half what it will attain at infinity.
Solution: 1. (a) Set Ei  Ef
and solve for vf :
U i  Ki  U f  K f
qVi  0i  0  12 m vf 2
vf 

2. (b) Set U  34 U i and solve for r:
2qVi

m
2q k Q

m ri
2k q Q
m ri
2  8.99  109 N  m 2 /C2  3.45  10 6 C  4.07  106 C 
 0.00236 kg 
kQ 3 kQ

 r  34 ri 
r
4 ri
4
3
1.25 m    0.570 m 
2
1.25 m    0.570 m 
2
2
2
 8.82 m/s
 1.83 m
Insight: The location at which K  12 Kf is a bit farther away, at r  2 ri  2.75 m.
(Continued next page)
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
20 – 1
Chapter 20: Electric Potential and Electric Potential Energy
James S. Walker, Physics, 5th Edition
38. Predict/Calculate Figure 20-32 shows three charges at the corners of a rectangle. (a) How much work must be done to move the +2.7-µC charge to
infinity? (b) Suppose, instead, that we move the −6.1-µC charge to infinity. Is the work required in this case greater than, less than, or the same as when we
moved the +2.7-µC charge to infinity? Explain. (c) Calculate the work needed to move the −6.1-µC charge to infinity.
38. Picture the Problem: Three charges are arranged at the corners of a
rectangle as indicated in the diagram at right.
Strategy: The work required to move the 2.7-µC charge to infinity is equal
to the change in its electric potential energy due to the other two charges.
Sum the electric potential energies due to the other two charges to find the
electric potential energy at the corner of the rectangle. At infinity the
electric potential energies are zero. Use these facts to find the work
required to move the 2.7-µC charge to infinity, and then repeat the
procedure to find the work required to move the − 6.1-µC charge.
Solution: 1. (a) Find
U  U f  Ui  0  U1  U 2  :
  k q q k q q 
q
q 
W  U  0   2 1  2 3    k q2  1  3 
r
r
r
r
23
 
 21 23 
  21
  6.1 10 6 C

N  m2 
3.3  10 6 C
6
   8.99  109

  2.7  10 C  
2
C 
 0.25 m
(0.25 m) 2  (0.16 m) 2




W  0.86 J
2. (b) The  6.1- C charge is repelled by the 3.3- C charge more than it is attracted by the 2.7- C charge. The work
required will be negative, which is less than the work required in part (a).
3. (c) Repeat step 1 for
the − 6.1-µC charge:
  k q q k q q 
q
q 
W  U  0   1 2  1 3    k q1  2  3 
r
r
r
r
13
 
 12 13 
  12

 2.7  10 6 C 3.3 10 6 C 
N  m2 
6
   8.99  109

6.1

10
C






0.16 m 
C2 

 0.25 m
W   0.54 J
Insight: Notice that the work done by the electric field is W  U (Equation 20-1), but the work you must do to move
the charge against the field is W  U . The negative work of part (c) means the charge would fly off on its own.
Copyright © 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
20 – 2