Download Chapter 17 ELECTRIC POTENTIAL

Chapter 17
ELECTRIC POTENTIAL
Conceptual Questions
1. (a) The electric field does positive work on –q as it moves closer to +Q.
(b) The potential increases as –q moves closer to +Q.
(c) The potential energy of –q decreases.
(d) If the fixed charge instead has a value –Q, the electric field does negative work, the potential decreases, and
the potential energy increases.
2. Such a capacitor can be built by replacing the air between the capacitor plates with a dielectric material. This
change not only increases the maximum possible voltage across the capacitor but also increases the amount of
charge on the capacitor plates for a given potential difference.
3. While standing on a high voltage wire, the magnitude of a bird’s electric potential varies between –100 kV and
+100 kV. Important for the bird is the fact that although its body is at a non-zero potential, the potential difference
across its body is small. If a large potential difference existed across its body, the bird would be electrocuted.
4. A positive charge in an electric field moves toward a position of lower potential. A negative charge in this
situation moves toward a position of higher potential.
5. Zero work is required to move a charge between two points at the same potential. An external force may need to
be applied to move the charge but the work done to start the charge in motion will be negated by the work done to
stop it.
6. If the charge of a point particle is negative, its electric potential energy decreases as it is moved towards a region
of higher electric potential.
7. If all parts of a conductor in electrostatic equilibrium were not at the same potential, electric fields would exist
within the conductor and charges would not remain stationary. The assumption of electrostatic equilibrium would
therefore be invalid.
8. There is no physical significance to zero potential—only potential differences have physical consequences. The
potential of the earth is often taken to be zero and therefore an object that is grounded has zero potential. This is
only a reference value however and the potential of the earth could be taken to be any other quantity as long as
other values were appropriately offset by the same potential.
9. If the electric field is zero throughout a region of space, the electric potential must be constant throughout that
region.
10. The woman’s head has acquired a net charge. The microscopic charges (electrons or ions) distribute themselves so
as to maximize their separation from one another, as a result of their mutual repulsion. This is why the charges
move out onto the woman’s hair, which then spreads out in response to the repulsive electrical forces. The
charged strands of hair orient themselves parallel to the electric field lines, which emanate radially outward from
the woman’s head, as though it were a charged conducting sphere.
11. If the potential is constant throughout a region of space, the electric field must be zero throughout that region.
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College Physics
Chapter 17: Electric Potential
12. If a uniform electric field exists throughout a region of space, the potential must be linearly changing in the
direction parallel to the field and unchanging in the directions perpendicular to the field.
13. It doesn’t matter which points we choose because the potential on each plate is constant over the whole plate.
14. The factor of 1/2 appears because the average height of the water in the pool is (1/2)h. The work required to fill
the pool is exactly equal to the potential energy of the water in the pool, (1/2)Mgh. This is analogous to the
charging of a capacitor. The charge Q on the capacitor is like the mass M of water in the pool, and the electrical
potential energy QV of the charge is like the gravitational potential energy Mgh. Thus, by analogy, the total
potential energy of a charged capacitor is (1/2)QV, which is correct.
15. Nothing happens to the capacitance, which depends only on the geometry and electrical properties of the materials
in the capacitor. Since C = Q/V, if the charge doubles then the voltage doubles as well.
16. Cow A is more likely to be killed because the potential difference between its front and hind legs will be greater
than that for cow B.
17. We can’t say anything about the electric field if all we know is the potential at a single point. The electric field
tells us how the potential changes if we move from one point to another.
18. As long as the person touching the dome is isolated from the ground, there is no complete circuit for current to
flow through, so she is safe.
19. The electric field points from regions of higher potential to regions of lower potential. Therefore, the upper
atmosphere is at a higher potential than the Earth.
20. Since the capacitor is connected to a battery the whole time, its voltage V remains constant. The capacitance is
proportional to , so when the dielectric is removed the capacitance decreases by a factor of 3. Since Q = CV,
and V remains constant, the charge on the capacitor Q decreases by a factor of 3 as well. The electric field
remains constant, since V is constant. The energy stored decreases by a factor of 3.
21. In this case the capacitor plates are isolated, so it is the charge Q that remains constant. Again the capacitance
decreases by a factor of 3, so V must increase by a factor of 3 to keep Q = CV the same. Thus, the electric field
increases by a factor of 3 and the energy stored in the capacitor increases by a factor of 3 as well.
22. The plates are isolated so the charge remains constant. The capacitance is given by C   e0 A / d . As the plates are
moved closer, all these quantities remain constant except d, which decreases. Therefore, the capacitance increases.
Since Q = CV, and Q is constant, V must decrease. The electric field remains constant. The energy stored in the
capacitor decreases as well.
23. The potential close to the positive charge must be positive, while close to the negative charge it must be negative.
Therefore, there is a point in region B where the potential is zero. If we move very far away from the two charges,
they will look like a single point of charge –3 C, so the potential very far away must be negative. Thus, there
must be a point in region A with a potential of zero as well. The electric field can only be zero in region A, and
this does not occur at the same point where the potential is zero.
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Chapter 17: Electric Potential
College Physics
Problems
1. Strategy Use Eq. (17-1).
Solution Compute the electric potential energy.
qq
(8.988  109 N  m 2 C2 )(5.0  106 C)(  2.0  106 C)
UE  k 1 2 
 18 mJ
r
5.0 m
2. (a) Strategy Use Eq. (17-1).
Solution Compute the electric potential energy.
(8.988  109 N  m 2 C2 )(1.602  1019 C)(  1.602  1019 C)
qq
  4.36  1018 J
UE  k 1 2 
r
0.0529  109 m
(b) Strategy and Solution The negative sign signifies that the force between the two charges is attractive; the
potential energy is lower than if the two were separated by a larger distance.
3. Strategy The work done by the applied force is positive, since the direction of the applied force was in the
direction of motion. (The force between the two charges is repulsive.) The potential energy of the charges is
positive, so the work done on the charges is equal to their potential energy. Use Eq. (17-1).
Solution Compute the work done on the charges.
qq
(8.988  109 N  m 2 C2 )(6.5  106 C) 2
W  UE  k 1 2 
 8.4 J
r
0.045 m
4. Strategy The work done by the external agent is positive since the potential energy increases. Use Eq. (17-1).
Solution Find the work done by the external agent.
ke2
(8.988  109 N  m 2 C 2 )(1.602  1019 C)2
W  U  U f  U i 
 U 
 0  2.3  1013 J
rf
1.0  1015 m
5. Strategy The work done on the charges is equal to their potential energy. Let the upper charge by 1, the lower
left-hand charge be 2, and the right-hand charge be 3. Also, let a  0.16 m and b  0.12 m. Use Eq. (17-2).
Solution Compute the work done on the charges.
q q
q1q3
q q 
 2 3
W  UE  k  1 2 
 b
a 
a 2  b2

 (5.5  106 C)(6.5  106 C) (5.5  106 C)(2.5 106 C)
 (8.988  109 N  m 2 C2 ) 

0.12 m

(0.16 m) 2  (0.12 m) 2

(6.5  106 C)(2.5  106 C) 

  3.0 J
0.16 m

6. Strategy Let q1  q2  q  10.0 nC and d = 4.00 cm. Use Eq. (17-2).
Solution Find the total electric potential energy for the two charges.
kq q
kq 2
(8.988  109 N  m 2 C2 )(10.0  109 C) 2
UE  1 2  

 11.2 J
r
2d
2(0.0400 m)
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College Physics
Chapter 17: Electric Potential
7. Strategy Let q1  q2  q  10.0 nC, d = 4.00 cm, and q3   4.2 nC. Use Eq. (17-2).
Solution Find the total electric potential energy of the three charges at point a.
 q 2 qq3 qq3  kq  q
qq
qq
q q 
 1 


U E  k  1 2  1 3  2 3   k 

   q3 1   
2
3
2
r
r
r
d
d
d
d
 3 

13
23 
 12


(8.988  109 N  m 2 C2 )(10.0  109 C)  10.0  109 C 2(4.2  109 C) 



  17.5 J
0.0400 m
2
3


8. Strategy Let q1  q2  q  10.0 nC, d = 4.00 cm, and q3   4.2 nC. Use Eq. (17-2).
Solution Find the total electric potential energy of the three charges at point b.
 q 2 qq3 qq3 
qq
qq
q q 
kq 2


UE  k  1 2  1 3  2 3   k  

 2d
2d
r13
r23 
d
d 
 r12

(8.988  109 N  m 2 C2 )(10.0  109 C) 2

 11.2 J
2(0.0400 m)
9. Strategy Let q1  q2  q  10.0 nC, d = 4.00 cm, and q3   4.2 nC. Use Eq. (17-2).
Solution Find the total electric potential energy of the three charges at point c.
 q 2 qq3 qq3 
qq
qq
q q 
kq 2


UE  k  1 2  1 3  2 3   k  

 2d 2d
2d 
2d
r13
r23 
 r12

(8.988  109 N  m 2 C2 )(10.0  109 C) 2

 11.2 J
2(0.0400 m)
10. Strategy Use Eq. (17-2).
Solution Find the electric potential energy.
qq
qq
q q 
UE  k  1 2  1 3  2 3 
r13
r23 
 r12


3.0 106 C
1.0  106 C 
 (8.988  109 N  m 2 C2 )  (4.0  106 C) 


 (4.0 m)2  (3.0 m)2

3.0 m



6
6

(3.0 10 C)(  1.0  10 C)

  2.8 mJ
4.0 m

y (m)
4.0
3
2
2.0
0.0
1
0.0
2.0
11. Strategy Use Eqs. (6-8) and (17-1).
Solution Compute the work done by the electric field.
Wfield  U  U i  U f  U12  (U12  U13  U 23 )  U13  U 23  
kq1q3 kq2q3

r13
r23
 8.00  109 C

8.00  109 C
  (8.988  109 N  m 2 C2 )(2.00  109 C) 

  2.70 J
 0.0400 m
0.0400 m  0.1200 m 

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4.0
x (m)
Chapter 17: Electric Potential
College Physics
12. Strategy Use Eqs. (6-8) and (17-1).
Solution Compute the work done by the electric field.
kq1q3 kq2q3

r13
r23
1
1


C) 

  1.80 J
 0.0800 m 0.0400 m 
Wfield  U  U i  U f  U12  (U12  U13  U 23 )  U13  U 23  
  (8.988  109 N  m 2 C2 )(2.00 109 C)(8.00  109
13. Strategy Use Eqs. (6-8) and (17-1).
Solution Compute the work done by the electric field.
Wfield  U  U i  U f  U12  U13i  U 23i  (U12  U13f  U 23f )  U13i  U13f  U 23i  U 23f

 1
 1
1 
1 
 k  q1q3 


  q2q3 

 r13i r13f 
 r23i r23f  


1
1


 (8.988  109 N  m 2 C2 ) (8.00  109 C)(2.00  109 C) 


 0.0400 m 0.0800 m 

1
1


 (8.00  109 C)(2.00  109 C) 

   4.49 J
 0.0400 m  0.1200 m 0.0400 m  
14. Strategy Use Eqs. (6-8) and (17-1).
Solution Compute the work done by the electric field.
Wfield  U  U i  U f  U12  U13i  U 23i  (U12  U13f  U 23f )  U13i  U13f  U 23i  U 23f

 1
 1
1 
1 
 k  q1q3 


  q2q3 


 r13i r13f 
 r23i r23f  
1
1
1
1


 (8.988  109 N  m 2 C2 )(8.00  109 C)(2.00  109 C) 




0.0800
m
0.120
m
0.0400
m
0.120
m


 1.80 J
15. Strategy The potential difference is the change in electric potential energy per unit charge. Use U E  qV .
Solution Find the change in the electric potential energy.
U E  qV  (3.0 nC)(25 V)  75 nJ
16. Strategy The potential difference is the change in electric potential energy per unit charge. Use U E  qV .
Solution Find the change in the electric potential energy.
U E  qV  q(Vf  Vi )  e(VB  VA )  (1.602  1019 C)[360 V  (240 V)]  1.92  1017 J
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College Physics
Chapter 17: Electric Potential
17. Strategy Use the principle of superposition and Eq. (17-9).
Solution Sum the electric fields at the center due to each charge.
 







E  E a  Eb  Ec  Ed  Ea  Eb  Ea  Eb  0
Do the same for the potential at the center.
kQ
4kQ 4(8.988  109 N  m 2 C2 )(9.0  106 C)

 2.3  107 V
V  i 
ri
r
(0.020 m)2  (0.020 m) 2
2
18. Strategy Use the principle of superposition and Eq. (17-9).
Solution Sum the electric fields at the center due to each charge.
 









k q
kq 
E  Ea  Eb  Ec  Ed  Ea  Eb  Ec  Eb  Ea  Ec   2a  2c  toward c
 r
r 

8.988  109 N  m 2 C2
(9.0  106 C  3.0  106 C) toward c  5.4  108 N C toward c

2
 (0.020 m)2  (0.020 m)2 


2


Do the same for the potential at the center.
kQ
k
V   i  (3q  q  q  q)  0
ri
r
19. Strategy Use Eqs. (17-7), (17-8), (17-9), and (6-8).
Solution Find the potentials, potential difference, change in electric potential energy, and work done by the
electric field.
(a) V 
kQ (8.988  109 N  m 2 C2 )(50.0  109 C)

 1.5 kV
r
0.30 m
(b) V 
(8.988  109 N  m 2 C2 )(50.0  109 C)
 900 V
0.50 m
 1
1 
1 
 1

 600 V
(c) V  kQ     (8.988  109 N  m 2 C2 )(50.0  109 C) 
r
r
0.50
m
0.30
m 

A
 B
V  0, so the potential increases .
(d) U E  qV  (1.0  109 C)(6.0 102 V)  6.0  107 J
U E  0, so the potential energy decreases .
(e) Wfield  U E  6.0  107 J
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Chapter 17: Electric Potential
College Physics
20. Strategy Use Eq. (17-9).
Solution Find the electric potential at point P due to the charges.
 2.0  103 C
kQ
 4.0  103 C
V   i  (8.988  109 N  m 2 C2 ) 

 4.0 m
ri
(4.0 m)2  (3.0 m) 2

 2.7 MV
y (m)




4.0
− 4.0 mC
2.0
0.0
2.0 mC
0.0
2.0
P
4.0
x (m)
21. Strategy and Solution
(a) Since V is positive, q is positive .
1
(b) V  , so since the potential is doubled, the distance is halved or 10.0 cm .
r
22. Strategy Just outside the surface of the sphere, the electric potential is given by V  Er , where r is the radius of
the sphere.
Solution Find the electric potential.
V  Er  (8.40  105 V m)(0.750 m)  6.30 105 V
kqi
, the minimum or most negative value of the potential is the case
ri
where the two negative charges are closer to x  0 than the two positive charges.
23. (a) Strategy and Solution Since V  
y
+
_
_
+
x
(b) Strategy Let d = 1.0 m and q = 1.0 C. Use Eq. (17-9).
Solution Find the potential at the origin.
 q
kQ
2  4(8.988  109 N  m 2 C2 )(1.0  106 C)
q q
q  kq  2
2
2
V   i  k 3  d  d  3  





 d
3 
1.0 m
ri
d  d 3
2
2
2 
2
 36 kV
24. (a) Strategy Use Eq. (16-5).
Solution Find RE .
E0 
kQ0
R02
E
k (3Q0 )
RE2
, so RE 
3 R0 .
(b) Strategy Use Eq. (17-8).
Solution Find RV .
V0 
kQ0
k (3Q0 )
V 
, so RV  3R0 .
R0
RV
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College Physics
Chapter 17: Electric Potential
25. Strategy Use Eq. (17-9).
Solution Find the electric potential at the third corner, B.
kQ
k
8.988  109 N  m 2 C2
V   i  (QA  QB ) 
(2.0  109 C  1.0  109 C)  9.0 V
ri
r
1.0 m
26. (a) Strategy Use Eq. (17-9).
Solution Find the electric potential at each point.
 4.2  109 C  6.4  109 C 
kq
kq

Va      (8.988  109 N  m 2 C2 ) 
  270 V
 0.060 m
r
r
0.159 m 

 4.2  109 C  6.4  109 C 
Vb  (8.988  109 N  m 2 C2 ) 

  160 V
 0.120 m
0.120 m 

(b) Strategy Use Eq. (17-6).
Solution Compute the potential difference for the trip from a to b.
V  Vb  Va  160 V  270 V   430 V
(c) Strategy The work done by an external agent is equal to the change in electric potential energy of the point
charge when moved from a to b. Use Eq. (17-7).
Solution Compute the work done.
W  qV  (1.50  109 C)(  430 V)   6.5  107 J
27. (a) Strategy Use Eq. (17-9).
Solution Find the potential at the points.
 2.50  109 C 2.50  109 C 
kQ1 kQ2

 (8.988  109 N  m 2 C2 ) 

  300 V
 0.050 m

r1
r2
0.150 m


kQ
kQ
kQ
kQ
If Q  2.50  109 C, Vb  1  2 

 0 .
r1
r2
r
r
Va 
(b) Strategy The work done by an external agent is equal to the change in electric potential energy of the point
charge when moved from infinity to b. Use Eq. (17-7).
Solution Compute the work done.
W  qV  q(Vb  V )  q(0  0)  0
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Chapter 17: Electric Potential
College Physics
28. (a) Strategy Let d  4.00 cm, r  12.0 cm, and q  8.00 nC. Use Eq. (17-9). Let V  0 at infinity.
Solution Find the potentials at points a and b.
kq kq
kq k (q) 3kq 3(8.988  109 N  m 2 C2 )(8.00  109 C)
Va  1  2 



 1350 V
r1
r2
d
4d
4d
4(0.0400 m)
Vb 
kq k (q)
kq
(8.988  109 N  m 2 C2 )(8.00  109 C)



 899 V
2d
d
2d
2(0.0400 m)
(b) Strategy The potential difference is the change in electric potential energy per unit charge. Use U E  qV .
Solution Compute the change in electric potential energy.
U E  qV  (2.00  109 C)(  899 V  1348 V)   4.49 J
29. (a) Strategy Let d  4.00 cm, r  12.0 cm, and q  8.00 nC. Use Eq. (17-9). Let V  0 at infinity.
Solution Find the potentials at points b and c.
kq kq
kq k (q)
kq
(8.988  109 N  m 2 C2 )(8.00  109 C)
Vb  1  2 



 899 V
r1
r2
2d
d
2d
2(0.0400 m)
kq k (q)
Vc 

 0
r
r
(b) Strategy The potential difference is the change in electric potential energy per unit charge. Use U E  q V .
Solution Compute the change in electric potential energy.
U E  qV  (2.00  109 C)(0  899 V)  1.80 J
30. Strategy Rewrite each unit in terms of kg, m, s, and C.
Solution Show that 1 N C  1 V m.
1 N C
kg  m s 2
J
kg  m 2 s 2 kg  m s 2
and 1 V m 


, therefore 1 N C  1 V m.
C
mC
mC
C
31. (a) Strategy Use Eq. (16-4b).
Solution Find the electric force that acts on the particle.


F  qE  (4.2  109 C)(240 N C to the right)  1.0 to the right
(b) Strategy The work done on the particle is equal to the electric force times the displacement of the particle.
Solution Compute the work done on the particle.
W  Fd  (1.0 N)(0.25 m)  0.25 J
(c) Strategy The electric field points in the direction of decreasing potential, so Va  Vb and Va  Vb  0. Use
Eq. (17-10).
Solution Compute the potential difference.
Va  Vb  Ed  (240 N C)(0.25 m)  60 V
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College Physics
Chapter 17: Electric Potential
32. (a) Strategy and Solution Positive work is required to move an electron (negative charge) from low potential to
high potential, so Y is at the higher potential.
(b) Strategy Use Eqs. (6-8) and (17-5).
Solution Find the potential difference.
qV  eV  U E  Wfield , so V  VY  VX 
Wfield
8.0  1019 J

 5.0 V .
e
1.602  1019 J
33. Strategy V  Ed for a uniform electric field.
Solution Find the distance between the equipotential surfaces.
V
1.0 V
d

 1.0 cm
E
100.0 N C
34. Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces
drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer together
where the field is stronger. The electric field always points in the direction of maximum potential decrease.

Solution Outside the sphere, E is radially directed (toward the sphere), and V  r 1. The equipotential surfaces


are perpendicular to E at any point, so they are spheres . Inside the sphere, E  0 and V is constant.
35. Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces
drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer together
where the field is stronger. The electric field always points in the direction of maximum potential decrease.

Solution Outside the cylinder, E is radially directed away from the axis of the cylinder. The equipotential

surfaces are perpendicular to E at any point, so they are cylinders .
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Chapter 17: Electric Potential
College Physics
36. Strategy The rate at which work is done by the electric organs is equal to the rate of change of the electric
potential energy. Use Eq. (17-7). The total amount of work done in one pulse is equal to the rate times the
duration of the pulse.
Solution
(a) Compute the rate at which work is done.
W q V
 q 

 V    (0.20  103 V)(18 C s)  3.6 kW
t
t
 t 
(b) Compute the total amount of work done.
W
W
t  (3.6  103 W)(0.0015 s)  5.4 J
t
37. (a) Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential
surfaces drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer
together where the field is stronger. The electric field always points in the direction of maximum potential
decrease.
Solution The electric field lines are radial. They begin on the point charge and end on the inner surface of
the shell. Then they begin again on the surface and extend to infinity.
+
(b) Strategy Use Eqs. (16-5) and (17-8), and the principle of superposition.
Solution For r  r1, E is that due to the point charge, E  kq r 2 . For r1  r  r2 , E  0, since this is inside a
conductor. For r  r2 , E once again is that due to the point charge, kq r 2 . For r  r1, V  kq r (point
charge). For r1  r  r2 , V  kq r1 , since V is continuous, and it is constant in a conductor. For r  r2 ,
V
kq  kq kq 
    (to preserve continuity). The graphs of the electric field magnitude and potential:
r1  r
r2 
E
V
kq
r 12
kq
r1
kq
r22
0
r1
r2
r
0
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College Physics
Chapter 17: Electric Potential
38. Strategy Since the electric field is uniform, we can use Eq. (17-10).
Solution Find the magnitude of the charge on the drop in terms of the elementary charge e.
F
Fe
Fe
dFe
(0.16 m)(9.6  1016 N)
F  qE and V   Ed , so q 
e  2e .




E
Ee
(V d )e eV (1.602  1019 C)(480 V)
39. Strategy Since the electric field is uniform, we can use Eq. (17-10). Use Newton’s second law.
Solution Find the magnitude of the charge on the drop.
F  qE  mg  0, so
q 
mg
mg
mgd (1.0  1015 kg)(9.80 m s 2 )(0.16 m)



 1.6  1019 C  e .
E
V d
V
9.76  103 V
qE
mg
40. Strategy Use conservation of energy and Eq. (17-7).
Solution Find the change in kinetic energy.
K  U  qV  2(1.602 1019 C)(200.0 103 V  500.0 103 V)  9.612 1014 J
41. Strategy Use conservation of energy and Eq. (17-7).
Solution Find the potential difference.
1
mv 2 (9.109  1031 kg)(7.26  106 m s) 2
U  eV  K   mv 2 , so V 

 150 V .
2
2e
2(1.602  1019 C)
42. (a) Strategy The electric field always points in the direction of maximum potential decrease. Electrons, being
negatively charged, move in the direction opposite the direction of the electric field; that is, in the direction of
potential increase.
Solution Since the speed of the electron decreased, it must have traveled in the direction of the electric field,
so it moved in the direction of potential decrease; that is, to a lower potential .
(b) Strategy The kinetic energy of the electron decreased, so its potential energy increased. Use conservation of
energy and Eq. (17-7).
Solution Compute the potential difference the electron moved through.
V 
U K m(vf 2  vi 2 ) (9.109  1031 kg)[(2.50  106 m s) 2  (8.50 106 m s)2 ]



 188 V
q
e
2e
2(1.602  1019 C)
43. Strategy According to Example 17.8, the speed of the electrons at the anode is proportional to the square root of
the potential difference. Use a proportion.
Solution Find the speed of the electrons.
v
V2
V2
6.0 kV
v  V , so 2 
and v2  v1
 (6.5  107 m s)
 4.6  107 m s .
v1
V1
V1
12 kV
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Chapter 17: Electric Potential
College Physics
44. Strategy According to Example 17.8, the speed of the electrons at the anode is proportional to the square root of
the potential difference. Use a proportion.
Solution Find the potential difference.
2
2
2
 3.0  107 m s 
 v2 
V2  v2 
v  V , so
   and V2  V1    (12 kV) 
  2.6 kV .
7


V1  v1 
 v1 
 6.5  10 m s 
45. Strategy and Solution

(a) Electrons travel opposite the direction of the electric field, so E is directed upward .
(b) For a uniform electric field, Fy  eE 
vy
v y md
e V
eV
 ma y , so a y 
. Thus, t 
.

d
ay
e V
md
(c) Since the electron gains kinetic energy, its potential energy decreases .
46. Strategy The field is uniform. Use conservation of energy and Eq. (17-10).
Solution Find the kinetic energy increase.
K  U  qV  eEd  (1.602 1019 C)(500.0 N C)(0.0030 m)  2.4  1019 J .
47. Strategy Use conservation of energy and Eq. (17-7).
Solution Find the final kinetic energy.
K  K f  Ki  U  q V  2eV , so
K f  Ki  2eV  1.20  1016 J  2(1.602  1019 C)(  0.50  103 V)  2.8  1016 J .
48. Strategy The force between the nuclei is repulsive, since they both have positive charge. Use conservation of
energy and Eq. (17-1).
Solution Find the closest distance that a helium nucleus approaches the gold nucleus.
kq q
1
U f  1 2  Ki  mHe vi 2 , so
r
2
2kqAu qHe 2k (79e)(2e) 316(8.988  109 N  m 2 C2 )(1.602  1019 C)2


 4.85  1014 m .
r
mHe vi 2
mHe vi 2
(6.68  1027 kg)(1.50  107 m s)2
Note that the radius of the gold nucleus is about 7  1015 m and the radius of the gold atom is about 1 1010 m.
49. Strategy The electron must have enough kinetic energy at point A to overcome the potential decrease between A
and C. Use conservation of energy and Eq. (17-7).
Solution Find the required kinetic energy.
K A  U  eV  (1.602  1019 C)(  60.0 V  100.0 V)  2.56  1017 J
50. Strategy and Solution Since positive charges move through decreases in potential, and since the potential and
potential energy are greatest at A, the proton will spontaneously travel from point A to point E. So, K A  0 .
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College Physics
Chapter 17: Electric Potential
51. Strategy Use the definition of capacitance, Eq. (17-14).
Solution Find the magnitude of the charge on each plate.
Q  C V  (2.0 μF)(9.0 V)  18 μC
52. (a) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Find the potential difference between the plates.
Q 0.75 μC
Q  C V , so V  
 50 mV .
C 15.0 μF
(b) Strategy and Solution The plate with the positive charge is at the higher potential, so the 0.75-C plate.
53. Strategy Use the definition of capacitance, Eq. (17-14).
Solution
Q  C V  (10.2  106 F)(  60.0 V)   6.12  104 C
612 C of charge must be removed from each plate.
54. (a) Strategy Use Eq. (17-10).
Solution Compute the maximum potential difference across the capacitor.
Vmax  Emax d  (3  106 V m)(0.0010 m)  3 kV
(b) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the magnitude of the greatest charge.
Q  C V  (2.0  106 F)(3  103 V)  6 mC
55. Strategy and Solution

(a) Since E does not depend upon the separation of the plates ( E   e0 ), it stays the same .
(b) Since V  d , V increases if d increases.
56. Strategy Use the definition of capacitance, Eq. (17-14), and the fact that the potential difference depends upon
the plate separation.
Solution
(a) Compute the potential difference between the plates.
Q 0.800  106 C
V  
 667 V
C
1.20  109 F
(b) Since V  d , V doubles if d doubles.
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Chapter 17: Electric Potential
College Physics
57. Strategy and Solution
(a) The battery maintains a constant potential difference between the plates: ΔV stays the same.
(b) The electric field magnitude increases because the same potential difference occurs over a shorter distance (E
= ΔV/d for a uniform field).
(c) To maintain a constant potential difference while the plate spacing changes, the battery must change the
charge on the plates. A larger electric field means that the charge increases. Check: Q = CΔV, C increases
and ΔV doesn’t change, so Q increases.
58. Strategy Use the definition of capacitance, Eq. (17-14), and the fact that the potential difference depends upon
the plate separation.
Solution
(a) Find the magnitude of the charge on each plate.
Q  C V  (1.20 nF)(12 V)  14 nC
(b) The capacitor remains connected to the battery so the potential difference stays the same. From E = ΔV/d,
increasing d means that the electric field decreases. The electric field is proportional to the charge per unit
area on the plates, so the charge decreases.
59. Strategy The capacitance of a parallel plate capacitor is directly proportional to its area. Form a proportion.
Solution Find the capacitance for each situation.
(a)
C2
(b)
C2
C1
C1

A2

A2
A1
A1
, so C2 
A2
, so C2 
A2
A1
A1
C1 
C1 
1A
2 1C
A1 1
2A
3 1
A1

1
(0.694 pF)  0.347 pF .
2
C1 
2
(0.694 pF)  0.463 pF .
3
60. Strategy Use Eq. (17-10).
Solution Compute the plate separation.
V
1.5 V
d

 1500 km
E
1.0  106 V m
61. Strategy Use the definition of capacitance, Eq. (17-14).
Solution Find the capacitance of the spheres.
Q
3.2  1014 C
Q  C V , so C 

 8.0 pF .
V
0.0040 V
62. Strategy Use the definition of capacitance and the definition of potential for a spherical conductor.
Solution Find the capacitance of the Moon if wrapped in aluminum foil.
Q
Q r
1.737  106 m
Q  C V , so C 

 
 1.933  104 F .
kQ
V
k 8.988 109 N  m 2 C2
r
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College Physics
Chapter 17: Electric Potential
63. Strategy The electrons are accelerated by the electric field between the plates of the capacitor. When they emerge
from the positive plate, their speed will be greater than their speed when they entered the capacitor.
Solution Find the acceleration of the electrons while they are inside the capacitor.
F
ma
eV
V  Ed  d 
d , so a 
.
e
e
md
Find the speed of the electrons as they emerge from the capacitor.
2eV
 eV 
vfx 2  vix 2  2a x x  2ad  2 
, so
d 
m
 md 
vfx  vix 2 
2eV
2(1.602  1019 C)(40.0 V)
 (2.50  106 m s) 2 
 4.51 106 m s .
m
9.109  1031 kg
64. Strategy Use the definition of electric flux and Gauss’s law.
Solution
(a) The Gaussian surface is a cylinder whose axis is parallel to a radius vector (of the sphere) through it. One end
is just within the conductor and the other is just outside it. The ends have area A, with a radius much smaller
than that of the sphere, so the electric field is approximately uniform. Find E just outside the conductor.
Q
 E  EA cos  
e0
Cylindrical surface:  E  EA cos 90  EA(0)  0
End inside conductor:  E  EA cos  (0) A cos   0, since the electric field is zero inside a conductor.

Q
Q
End just outside the conductor:  E  EA cos 0  EA  , so E 
 .
Ae0 e0
e0
(b) Consider an area A of the surface of an arbitrary conductor. If A is small enough such that its surface is
approximately flat, then the electric field will be nearly uniform just outside the surface. Comparing an area
of the same size on a spherical conductor with the same charge density to that of the arbitrary conductor, we
see that the electric field just outside either conductor should be  e0 ; as long as A is small enough that it is
approximately flat, then this holds for any conductor.
65. (a) Strategy Use Eq. (16-6).
Solution The electric field between the plates is
Q
4.0  1011 C
E

 3.3  103 V m .
e0 A [8.854 1012 C2 (N  m 2 ) ](0.062 m)(0.022 m)
(b) Strategy Use the definition of the dielectric constant, Eq. (17-17).
Solution Find the electric field between the plates of the capacitor with the dielectric.
E
E
3.3  103 V m
 6.0  102 V m .
  0 , so E  0 
E
5.5

66. Strategy Assume the field is uniform. Use Eq. (17-10).
Solution Find the maximum possible height for the bottom of the thundercloud.
V
1.00  108 V

 300 m .
d
E
3.33  105 V m
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Chapter 17: Electric Potential
College Physics
67. (a) Strategy Use Eq. (17-10).
Solution Compute the magnitude of the average electric field between the cow’s front and hind legs.
V (400  200)  103 V
E

  105 V m
d
1.8 m
Since the electric field always points in the direction of decreasing potential, the average electric field is
 105 V m toward the hind legs .
(b) Strategy and Solution The front and hind legs of Cow B are nearly at the same potential, whereas those for
Cow A span a potential difference of approximately 200 kV, thus Cow A is more likely to be killed.
68. Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the capacitance of the capacitor.
Q
0.020  106 C
C

 83 pF .
V
240 V
69. Strategy The spark flies between the spheres when the electric field between them exceeds the dielectric
strength. The magnitude of the electric field is given by V d , where d is the distance between the spheres.
Solution Find d.
V
V
900 V

 0.30 mm .
E
, so d 
d
E
3.0  106 V m
70. Strategy Use Eq. (17-16).
Solution
(a) Find the greatest  d , since A and e0 are constant.
3.5
7.0
2.0
 35,
 3.5, and
 0.2. Since 35 > 3.5 > 0.2, the paper is the best choice.
0.10
2.0
10.0
e A 3.5[8.854  1012 C2 (N  m 2 )](120 10 4 m 2 )
C  0 
 3.7 nF
d
0.10  103 m
(b) Compute the smallest capacitance.
2.0[8.854  1012 C2 (N  m 2 )](120  104 m 2 )
C
 21 pF
10.0  103 m
71. Strategy Use Eq. (17-16).
Solution Compute the capacitance of the capacitor.
e A 2.5[8.854  1012 C2 (N  m 2 )](0.30 m)(0.40 m)
C  0 
 89 nF
d
0.030  103 m
72. Strategy Use Eq. (17-16).
Solution Find the average dielectric constant of the tissue in the limb.
eA
dC
(0.030 m)(0.59  1012 F)
C   0 , so  

 5.0 .
d
e0 A [8.854 1012 C2 (N  m 2 )](4.0 104 m 2 )
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College Physics
Chapter 17: Electric Potential
73. (a) Strategy Use Eq. (17-18c).
Solution Compute the capacitance.
Q2
Q 2 (8.0  102 C)2
U
, so C 

 7.1 F .
2C
2U
2(450 J)
(b) Strategy Use Eq. (17-18a).
Solution Compute the potential difference.
1
2U
2(450 J)
U  QV , so V 

 1.1 104 V .
Q 8.0  102 C
2
74. Strategy Use Eq. (17-19). The dielectric strength of air is 3 kV mm , which is equal to the maximum electric
field.
Solution Compute the maximum electric energy density in dry air.
1
1
u   e0 E 2  (1.00054)[8.854  1012 C2 (N  m 2 )](3  106 V m) 2  40 J m3
2
2
75. Strategy The capacitance of a capacitor is inversely proportional to the distance between the plates and
U  Q 2 (2C ).
Solution Form a proportion to find the ratio of the new capacitance to the old.
C2 d1
d1
1



C1 d 2 1.50d1 1.50
Form a proportion to find the energy stored in the capacitor in terms of the old.
U 2 Q 2 (2C2 ) C1


 1.50, so U 2  1.50U1.
U1 Q 2 (2C1 ) C2
Thus, the energy increases by 50% .
76. Strategy The energy stored in the capacitor is given by U  12 C (V ) 2 . When the plate separation is increased,
the capacitance changes but the potential difference stays the same, so the energy in the capacitor changes as well.
The work done on the capacitor in separating the plates is negative the change in energy.
Solution Form a proportion. The capacitance is inversely proportional to the plate separation.
U f Cf (V )2 Cf
d
1.00 cm


 i 
 0.500. So, the energy is reduced by half.
2
U i Ci (V )
Ci df 2.00 cm
Find the work done on the capacitor.
1
0.500 e0 A
W  U i  U f  U i  0.500U i  0.500U i  0.500 Ci ( V )2 
( V )2
di
2
2

0.500[8.854 1012 C2 (N  m 2 )](314  10 4 m 2 )
(20.0 V) 2  2.78 nJ
2(0.0100 m)
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Chapter 17: Electric Potential
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77. (a) Strategy Use Eq. (17-15).
Solution Find the capacitance for the thundercloud.
e A [8.854  1012 C2 (N  m 2 )](4500 m)(2500 m)
C 0 
 0.18 F
d
550 m
(b) Strategy Use Eq. (17-18c).
Solution Find the energy stored in the capacitor.
Q2
(18 C)2
U

  108 J
2C 2(0.1811 106 F)
78. (a) Strategy The capacitance after the slab is removed is equal to the capacitance with the slab divided by the
dielectric constant.
Solution Compute the capacitance.
C 6.0 F
C0  
 2.0 F
3.0

(b) Strategy Use Eqs. (17-10) and (17-17).
Solution Find the potential difference across the capacitor.
E0   E
E0d   Ed
V0   V  3.0(1.5 V)  4.5 V
(c) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the charge on the plates.
Q  C V  (2.0 F)(4.5 V)  9.0 C
(d) Strategy Use Eq. (17-18c).
Solution Compute the energy stored in the capacitor.
Q 2 (9.0  106 C) 2
U

 20 J
2C 2(2.0  106 F)
79. (a) Strategy Use the definition of capacitance, Eq. (17-14), and Eq. (17-15).
Solution Find the charge on the capacitor.
eA
[8.854  1012 C2 (N  m 2 )](0.100 m) 2 (150 V)
Q  C V  0 V 
 18 nC
d
0.75  103 m
(b) Strategy Use Eq. (17-18a).
Solution Compute the energy stored in the capacitor.
1
1
U  QV  (17.7  109 C)(150 V)  1.3 J
2
2
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College Physics
Chapter 17: Electric Potential
80. (a) Strategy Use Eq. (17-15).
Solution Compute the capacitance.
e A [8.854  1012 C2 (N  m 2 )](0.100 m)2
C 0 
 59.0 pF
d
0.75  103 m  0.750  103 m
(b) Strategy From Problem 63, Q  18 nC. Use Eq. (17-18c) and conservation of energy.
Solution Compute the new energy stored in the capacitor.
Q2
(18  109 C)2
U

 2.7 J
2C 2(59.0  1012 F)
Work was done on the capacitor when the plates were separated; that work has been stored in the capacitor
as potential energy.
81. (a) Strategy U  Pt where P  10.0 kW and t  2.0 ms. Use Eq. (17-18b).
Solution Find the initial potential difference.
1
2U
2(10.0 kW)(2.0 ms)
U  C (V ) 2 , so V 

 630 V .
2
C
100.0  106 F
(b) Strategy Use Eq. (17-18c).
Solution Find the initial charge.
Q2
U
, so Q  2CU  2(100.0  106 F)(10.0 kW)(2.0 ms)  0.063 C .
2C
82. Strategy Use Eq. (17-18a).
Solution Compute the energy stored in the capacitor.
1
1
U  QV  (0.020  106 C)(240 V)  2.4 J
2
2
83. Strategy The work done by the external agent is equal to the change in potential energy of the capacitor. Use
Eq. (17-18c) and the fact that the capacitance is inversely proportional to the plate separation.
Solution Find the work required to double the plate separation.
 Q 2  df
 (0.80  106 C) 2
Q2 Q2
Q 2  Ci
W  U 
(2  1)  0.27 mJ


 1 

  1 
9
2Cf 2Ci 2Ci  Cf
 2Ci  di
 2(1.20  10 F)
84. Strategy Use Eq. (17-18b).
Solution Find the required potential difference.
1
2U
2(300 J)
U  C (V )2 , so V 

 8 kV .
2
C
9  106 F
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Chapter 17: Electric Potential
College Physics
85. (a) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the charge that passes through the body tissues.
Q  C V  (15  106 F)(9.0  103 V)  0.14 C
(b) Strategy Use Eq. (17-18b) and the definition of average power.
Solution Find the average power delivered to the tissues.
E U C (V ) 2 (15  106 F)(9.0  103 V)2
Pav 



 0.30 MW
t t
2t
2(2.0  103 s)
86. Strategy Assume the that thundercloud and Earth system behaves like a capacitor. Use Eq. (17-18a).
Solution Find the electric potential energy released by the lightning strike.
1
1
U  QV  (25.0 C)(1.00  108 V)  1.25 GJ .
2
2
87. (a) Strategy Assume that the thundercloud and Earth system acts like a capacitor. Use Eq. (17-18a).
Solution Find the electric potential energy released by the lightning strike.
1
1
U  QV  (20.0 C)(1.00  109 V)  10.0 GJ
2
2
(b) Strategy Use the definition of latent heat.
Solution Find the mass of sap that is vaporized.
Q  energy absorbed  mLV  0.100(10.0  109 J), so m 
1.00  109 J
 443 kg .
2,256,000 J kg
(c) Strategy Divide 10.0% of the total energy released from the lightning strike by the homeowner’s monthly
energy use.
Solution
t 
0.100(10.0  109 J)
(400.0  103 W  h month)(3600 s h)
 0.694 month
88. Strategy Use Eq. (17-9).
Solution Find the potential midway between the charges.
kq kq
k
8.988  109 N  m 2 C2
V  1  2  (q1  q2 ) 
(12.0  109 C  22.0 109 C)  873 V
0.700 m
r
r
r
2
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College Physics
Chapter 17: Electric Potential
89. (a) Strategy Let qL  qR  10.0 nC. Use Eqs. (17-5) and (17-9).
Solution Find the potential energy of the point charge at each location.
 kq
 1
kq 
1 
U a  qVa  q  L  R   kqqL   
rR 
 rL
 rL rR 
1
1


 (8.988  109 N  m 2 C2 )( 4.2  109 C)(10.0  109 C) 

  6.3 J
0.0400
m
0.1200
m


1
1


9
2
2
9
9
U b  qVb  (8.988 10 N  m C )( 4.2  10 C)(10.0  10 C) 

 0
0.0400
m
0.0400
m


1
1


U c  qVc  (8.988  109 N  m 2 C2 )( 4.2  109 C)(10.0  109 C) 

 0
 0.0800 m 0.0800 m 
(b) Strategy The work done by the external force is negative the work done by the field. Use Eq. (6-8).
Solution Find the work required to move the point charge.
W  Wfield  U  U a  U b   6.3 J  0   6.3 J
90. Strategy Use Eqs. (6-8) and (17-7).
Solution Compute the work done by the electric field.
Wfield  U   ( e)V  (1.602  1019 C)[100.0 V  (100.0 V)]  3.204  1017 J
91. Strategy The potential at the surface of a conducting sphere is equal to the magnitude of the electric field times
the radius of the sphere.
Solution Compute the potential.
V  Er  (3.0  106 N C)(0.15 m)  450 kV
92. Strategy Let q  2.0  1021 C and r = 1.0 nm. Use Eq. (17-9).
Solution Find the potential at the sodium ions due to the other three ions.
kq
kq kq
 q 2q q  kq (8.988  109 N  m 2 C2 )(2.0  1021 C)
V  1  2  3  k 
 

 9.0 mV
r1
r2
r3
2(1.0  109 m)
 r r 2r  2r
93. (a) Strategy Electric field lines begin on positive charges and end on negative charges. The same number of
field lines begins on the plate and ends on the negative point charge. Field lines never cross. Use the
principles of superposition and symmetry.
Solution The electric field lines for the cylinder and sheet:
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Chapter 17: Electric Potential
College Physics
(b) Strategy Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential
surfaces drawn such that the potential difference between adjacent surfaces is constant, the surfaces are closer
together where the field is stronger. The electric field always points in the direction of maximum potential
decrease.
Solution The equipotential surfaces for the cylinder and sheet:
94. Strategy Use Newton’s second law and Eq. (4-9).
Solution Find the acceleration.
eE
Fy  eE  ma y , so a y 
.
m
Find the time to reach the lower plate.
y 
1
a y (t )2 , so t 
2
2(9.109  1031 kg)(0.040 m)
2my

eE
(1.602  1019 C)(5.0  104 N C)
 3.0 ns .
95. Strategy Assume the field is uniform. Use Eq. (17-10).
Solution Compute the magnitude of the electric field in the membrane.
V 90  103 V
E

 9  106 V m .
d
10  109 m
96. Strategy Use Newton’s second law, x  vx t , and Eqs. (4-7) and (4-9).
Solution

(a) Since E points downward, the negatively charged electron’s change in velocity is directed upward.
Find the acceleration.
eE
Fy  eE  ma y , so a y 
.
m
Find the time interval.
x
t 
vx
Find the change in velocity.
 eE   x  (1.602  1019 C)(2.0  104 N C)(0.060 m)
v y  a y t  
 7.0  106 m s


(9.109  1031 kg)(3.0  107 m s)
 m   vx 

So, v  7.0 106 m s upward .
(b) Find the deflection of the electrons.
2
y 
1
1  eE   x 
(1.602  1019 C)(2.0  104 N C)(0.060 m) 2
a y (t )2  
 7.0 mm

 

2
2  m   vx 
2(9.109  1031 kg)(3.0  107 m s) 2
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College Physics
Chapter 17: Electric Potential
97. Strategy The negatively charged particle will accelerate toward the positively charged plate while it is between
the plates of the capacitor.
Solution The particle is between the plates for a time given by t  x vx . During this time, the particle travels a
vertical distance y  0.00100 m. Find the acceleration of the particle.
2
2v 2 y
1
1  x 
a(x)2
y  a (t ) 2  a   
, so a  x
.
2
2  vx 
2vx 2
(x) 2
V
, where d is the plate separation and N is the
d
number of excess electrons on the particle. According to Newton’s second law, the acceleration of the particle is
Ne dV NeV

a
. We set the two expressions for the acceleration of the particle equal and solve for N.
m
md
The magnitude of the electrical force on the particle is NeE  Ne
2md vx 2 y 2(5.00  1019 kg)(0.00200 m)(35.0 m s)2 (0.00100 m)
NeV 2vx 2 y

, so N 

 51 .
md
(x)2
eV (x)2
(1.602  1019 C)(3.00 V)(0.0100 m) 2
98. (a) Strategy Compute the electrical and gravitational forces on the particle and compare. Refer to Problem 83.
Solution The gravitational force on the particle is mg  (5.00  1019 kg)(9.80 m s 2 )  4.90  1018 N .
The electrical force on the particle is
NeV 51(1.602  1019 C)(3.00 V)

 1.23  1014 N .
d
0.00200 m
Compare the forces.
1.226  1014 N
 2.50  103
4.90  1018 N
The electrical force is 2.50  103 times larger than the gravitational force.
(b) Strategy Use the results from Problem 83. The horizontal component of the velocity doesn’t change.
Solution The horizontal component of the velocity is vx  35.0 m s .
Compute the y-component of the particle’s velocity.
2v 2 y  x  2vx y 2(35.0 m s)(0.00100 m)
v y  at  x

 7.00 m s
 
x
0.0100 m
(x) 2  vx 
99. Strategy Find the charge on the capacitor. Use Eqs. (17-14) and (17-15).
Solution The charge on the capacitor is Q  Ne, where N is the number of excess electrons.
eA
e AV [8.854  1012 C2 (N  m 2 )](0.0100 m)2 (3.00 V)
Q  Ne  C V  0 V , so N  0

 8.29  106 .
d
de
(0.00200 m)(1.602  1019 C)
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Chapter 17: Electric Potential
College Physics
100. (a) Strategy Use the definition of capacitance, Eq. (17-14).
Solution Compute the capacitance.
Q
0.020  106 C
C

 83 pF
V
240 V
(b) Strategy Use Eq. (17-15).
Solution Find the area of a single plate.
eA
dC (0.40  103 m)(8.33  1011 F)
C  0 , so A 

 3.8  103 m 2 .
2
12 2
e0
d
8.854  10
C (N  m )
(c) Strategy Use Eq. (17-10).
Solution Compute the voltage required to ionize the air between the plates.
V  Ed  (3.0  103 V mm)(0.40 mm)  1.2 kV
101. Strategy The energy in the capacitor is converted into heat in the water. Use Eqs. (14-4) and (17-18b).
Solution Find the temperature change of the water.
1
C (V ) 2
(200.0  106 F)(12.0 V)2
Q  mcT  U  C (V ) 2 , so T 

 3.44 mK .
2
2mc
2(1.00 cm3 )(1.00 g cm3 )[4.186 J (g  K)]
102. (a) Strategy Use Eqs. (17-16) and (17-18b).
Solution Find the energy stored in the capacitor.
1
1  e A 
5.2[8.854  1012 C2 (N  m 2 )](1.0  107 m 2 )(90.0  103 V)2
U  C (V ) 2   0  (V )2 
2
2 d 
2(7.5  109 m)
 2.5  1012 J
(b) Strategy Divide the total charge by the charge of one ion. Use the definition of capacitance, Eq. (17-14), and
Eq. (17-16).
Solution Find the number of ions outside of the membrane.
Q C V  e0 AV 5.2[8.854  1012 C2 (N  m 2 )](1.0  107 m 2 )(90.0  103 V)



 3.4  108 ions
e
e
ed
(1.602  1019 C ion)(7.5  109 m)
103. Strategy Use Eq. (17-16).
Solution Find the capacitance of the axon.
e A 5[8.854 1012 C2 (N  m 2 )](5  1012 m 2 )
C  0 
 5  1014 F
d
4.4  109 m
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College Physics
Chapter 17: Electric Potential
104. Strategy Use conservation of energy, Wfield  U , and the fact that the field is uniform.
Solution Find the kinetic energy of each electron when it leaves the space between the plates.
 V 
K  K f  Ki  U  Wfield  eE y  e 
 y, so
 d 
(1.602  1019 C)(100.0  103 V)(0.0030 m)
 V 
15
2.0
10
J
K f  Ki  e 
y




 6.0  1015 J .

0.0120 m
 d 
105. (a) Strategy For a parallel plate capacitor, E   e0 and V  Ed .
Solution Find the potential difference between the plates.
 d (4.0 106 C m 2 )(0.0060 m)
V  Ed 

 2.7 kV
e0
8.854  1012 C 2 (N  m 2 )
(b) Strategy Use conservation of energy and the fact that U  q V .
Solution Find the kinetic energy of each point charge just before it hits the positive plate.
K  K f  Ki  U  qV , so K f  Ki  q V  0  (2.5 109 C)(2711 V)  6.8 J .
106. Strategy Use conservation of energy, Wfield  U , and the fact that the field is uniform.
Solution Find the final kinetic energy of the alpha particle.
K  K f  Ki  U  Wfield  2eEd , so
K f  Ki  2eEd  0  2(1.602  1019 C)(10.0  103 V m)(0.010 m)  3.2  1017 J .
107. Strategy Use Eqs. (6-8) and (17-7).
Solution Find the work done by the electric field.
Wfield  U   qV   eV  (1.602  1019 C)(  90.0  103 V)  1.44  1020 J
108. Strategy Electric field lines begin on positive charges and end on negative charges. The same number of field
lines begins on the plate and ends on the negative point charge. Field lines never cross. Use the principles of
superposition and symmetry. The electric field always points in the direction of maximum potential decrease.
Equipotential surfaces are perpendicular to electric field lines at all points. For equipotential surfaces drawn such
that the potential difference between adjacent surfaces is constant, the surfaces are closer together where the field
is stronger.
Solution The electric field lines and equipotential surfaces for the cube:
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Chapter 17: Electric Potential
College Physics
109. Strategy The energy stored in the capacitor is directly proportional to the capacitance. Use Eq. (17-16) and the
fact that V is constant. Form a proportion.
Solution Determine what happens to the energy stored in the capacitor.


1
U C  e0 A df  di
d


 i 1 
 1   0.200, so the energy is reduced by 20.0%.
 e0 A
1.25
U0
C0
df
d
1
1
i
110. Strategy Use Eq. (17-18b) and form a proportion. V is constant.
Solution Find the energy stored in the capacitor after the dielectric is inserted.
1 C (V ) 2
U
C  C0
 2


  , so U   U 0  3.0U 0 .
2
1
U0
C0
C0
C0 (V )
2
111. (a) Strategy Treat the axon as a parallel plate capacitor. Use Eq. (17-16) and the fact that the area of the curved
surface of a cylinder is (2 r ) L, where r and L are the radius and length of a cylinder, respectively.
Solution Calculate the capacitance per unit length of the axon.
 e A  e [(2 r ) L]
, so
C 0  0
d
d
C 2 e0r 2 (7.0)[8.854 1012 C2 (N  m 2 )](5.0 106 m)


 3.2  107 F m .
9
L
d
6.0  10 m
(b) Strategy Use Eq. (17-10) and the fact that the magnitude of the electric field inside a parallel plate capacitor
is given by  e0 .
Solution The outside of the membrane has the positive charge, since the potential is higher outside than
inside.
E
 e V 7.0[8.854 1012 C 2 (N  m 2 )](0.085 V)

d , so   0
V  Ed  0 d 

 8.8  104 C m 2 .

 e0
d
6.0  109 m
112. (a) Strategy Use Eq. (17-18b), the definition of capacitance, Eq. (17-14), and the relationships between the
quantities (energy, potential difference, capacitance) before and after the dielectric is inserted.
Solution Calculate the initial energy stored in the capacitor (without the dielectric).
1
1
U i  Ci (Vi ) 2  (4.00 106 F)(100.0 V)2  20.0 mJ
2
2
C
Vi
Cf   Ci and Q  Ci Vi  Cf Vf . So, Vf  i Vi 
.

Cf
Calculate the final energy.
2
Uf 
1
1
1 1
20.0 mJ
 V 
 1
Cf (Vf )2   Ci  i    Ci (Vi )2   U i 
 3.3 mJ
2
2
2
6.0







(b) Strategy and Solution Since the energy of the capacitor increases when the dielectric is removed, an
external agent has to do positive work to remove the dielectric .
708
College Physics
Chapter 17: Electric Potential
113. (a) Strategy Use Eq. (17-10).
Solution Compute the minimum thickness of the titanium dioxide.
V
5.00 V
d

 1.25 m
E
 106 V m
(b) Strategy Use Eq. (17-16).
Solution Find the area of the plates.
eA
dC
(1.25  106 m)(1.0 F)
C   0 , so A 

 1600 m 2 .
 e0 90.0[8.854  1012 C2 (N  m2 )]
d
114. Strategy Use the definition of capacitance, Eq. (17-14), to find the charge that moves through the membrane.
Then divide the charge by e.
Solution
C 
(a) Q  C V    AV  (1 106 F cm 2 )(0.05 cm 2 )[20  103 V  (90  103 V)]  6 nC
 A
(b)
Q
5.5  109 C

 3  1010 ions
e 1.602  1019 C ion
115. Strategy Let E0  20.0 V m , E1 be the field outside of the dielectric after it is inserted, and E2 be the field
inside the dielectric. Use the principle of superposition for the potential after the dielectric is inserted and the fact
that E2  E1  .
Solution Find the electric field inside the dielectric.
Initially: V  E0d
d
d
d E d Ed  1
 E2  E1   1   1 1  
2
2
2  2
2  
Solve for E1 in terms of E0.
Finally: V  E1
E1d
2
2 E0
 1
.
1     E0d , so E1 
1  1


Calculate E2.
E2 
E1


2E0 2(20.0 V m)

 8.0 V m
 1
4.0  1
709