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Solutions to Problems
1.
The work done by the electric field can be found from Eq. 17-2b.
W
Vba   ba  Wba  qVba   7.7  106 C  55 V   4.2  10 4 J
q

2.

The work done by the electric field can be found from Eq. 17-2b.
W
Vba   ba  Wba  qVba   1.60  1019 C 1.80  102 V  2.88  1017 J
q



  1 e  180 V   1.80  102 eV
3.
The
The kinetic energy gained is equal to the work done on the electron by the electric field.
potential difference must be positive for the electron to gain potential energy. Use Eq. 17-
2b.
Vba  
Wba
q



 Wba   qVba   1.60  10 19 C 2.3  10 4 V  3.7  10 15 J


   1 e  2.3  104 V  2.3  10 4 eV
4.
The kinetic energy gained by the electron is the work done by the electric force. Use Eq. 172b to calculate the potential difference.
Wba
7.45 1017 J
Vba  

 466 V
q
1.60 1019 C


The electron moves from low potential to high potential, so plate B is at the higher
potential.
5.
The magnitude of the electric field can be found from Eq. 17-4b.
V
220 V
E  ba 
 3.8  104 V m
3
d
5.8  10 m
6. The magnitude of the voltage can be found from Eq. 17-4b.
V
E  ba  Vba  Ed   640 V m  11.0  10 3 m  7.0 V
d

7.

The distance between the plates can be found from Eq. 17-4b.
V
V
45 V
E  ba  d  ba 
 3.0  102 m
d
E 1500 V m
8.
The gain of kinetic energy comes from a loss of potential energy due to conservation of
energy, and
the magnitude of the potential difference is the energy per unit charge.
PE
KE 65.0 103 eV
V 


 3.25  104 V
q
q
2e
The negative sign indicates that the helium nucleus had to go from a higher potential to a
lower potential.
9.
Find the distance corresponding to the maximum electric field, using Eq. 17-4b.
V
V
200 V
E  ba  d  ba 
 6.67  105 m  7  105 m
d
E 3  106 V m
10. By the work energy theorem, the total work done, by the external force and the electric field
together, is the change in kinetic energy. The work done by the electric field is given by Eq.
17-2b.
Wexternal  Welectric  KE final  KE initial  Wexternal  q Vb  Va   KEfinal 
Vb  Va  
Wexternal  KE final
q

15.0  104 J  4.82  10 4 J
6
8.50  10 C
 1.20  102 V
11. The kinetic energy of the electron is given in each case. Use the kinetic energy to find the
speed.
(a) mv  KE  v 
2KE
(b) 12 mv 2  KE  v 
2KE
2
1
2
m
m



2  750 eV  1.60 1019 J eV
31
9.1110 kg

  1.6 10

2 3.2 103 eV 1.60 1019 J eV
31
9.1110 kg
7
ms
  3.4 10
7
ms
12. The kinetic energy of the proton is given. Use the kinetic energy to find the speed.
1
2
mv  KE  v 
2
2KE
m



2 3.2 103 eV 1.60 1019 J eV
27
1.67  10 kg
  7.8 10
5
ms
13. The kinetic energy of the alpha particle is given. Use the kinetic energy to find the speed.
1
2
mv  KE  v 
2
2KE
m



2 5.53 106 eV 1.60 1019 J eV
27
6.64 10 kg
  1.63 10
7
ms
14. Use Eq. 17-5 to find the potential.
V
1 Q
4 0 r

 8.99  109 N m 2 C 2
4.00  106 C
 1.50 10
1
m
 2.40 105 V
15. Use Eq. 17-5 to find the charge.
1 Q

1

V
 Q   4 0  rV  
 0.15 m 125 V   2.1109 C
9
2
2 
4 0 r
 8.99 10 N m C 
16. The work required is the difference in potential energy between the two locations. The test
charge has potential energy due to each of the other charges, given in Conceptual Example
17-7 as PE  k
Q1Q2
. So to find the work, calculate the difference in potential energy
r
between the two locations. Let Q represent the 35 C charge, let q represent the 0.50 C
test charge, and let d represent the 32 cm distance.
kQq kQq
kQq
kQq
PEinitial 

PE final 

d 2 d 2
 d 2  0.12 m d 2  0.12 m
Work  PE final  PE initial 

d
2  0.12 m 
1
 kQq 

kQq
  0.16 m  0.12 m 
 8.99  109 N m 2 C2


 kQq 

d 2
kQq
d
 2
2  0.12 m 
1
0.16 m  0.12 m 



0.16 m 
2
 35 10 C  0.50 10 C 16.07 m   2.5 J
6
6
1
An external force needs to do positive work to move the charge.
17.
18. (a) The electric potential is given by Eq. 17-5.
1 Q
1.60  1019 C
V
 8.99  109 N m 2 C2
 5.754  105 V  5.8 105 V
4 0 r
2.5  1015 m
(b) The potential energy of a pair of charges is derived in Conceptual Example 17.7.

PE  k
Q1Q2
r


 8.99  10 N m C
9
2
2

1.60 10
19
C

2.5  1015 m
2
 9.2  10 14 J
19. The potential at the corner is the sum of the potentials due to each of the charges, using Eq.
17-5.
V
k  3Q 
L

kQ
2L

k  2Q 
L

kQ 
1 
1 

L 
2
2kQ
2L


2 1
20. By energy conservation, all of the initial potential energy will change to kinetic energy of the
electron when the electron is far away. The other charge is fixed, and so has no kinetic
energy. When the electron is far away, there is no potential energy.
Einitial  Efinal  PE initial  KE final 
v
2k  e  Q 
mr


k  e  Q 
r
 12 mv 2 
 1.60 10 C  1.25 10 C 
 9.1110 kg   0.325 m 
19
2 8.99  109 N m 2 C 2
7
31
 3.49  107 m s
21. By energy conservation, all of the initial potential energy of the charges will change to
kinetic energy when the charges are very far away from each other. By momentum
conservation, since the initial momentum is zero and the charges have identical masses, the
charges will have equal speeds in opposite directions from each other as they move. Thus
each charge will have the same kinetic energy.
kQ 2
Einitial  Efinal  PE initial  KE final 
 2 12 mv 2 
r

v
kQ 2
mr
8.99 10 N m C  9.5 10 C 
1.0 10 kg   0.035 m 
9

2
6
2
6

2
 4.8  103 m s
22.
(a)
Because of the inverse square nature of the
x
d
electric
q2  0
field, any location where the field is zero must be
q1  0
closer to the weaker charge  q2  . Also, in between the
two charges, the fields due to the two charges are parallel to each other (both to the left)
and cannot cancel. Thus the only places where the field can be zero are closer to the
weaker charge, but not between them. In the diagram, this is the point labeled as “x”.
Take to the right as the positive direction.
q
q1
2
E  k 22  k
 0  q2  d  x   q1 x 2 
2
x
d  x
q2
2.0  106 C
 5.0 cm   22 cm left of q2
3.0  106 C  2.0  106 C
(b) The potential due to the positive charge is positive
d
x1
x2
everywhere, and the potential due to the negative
charge is negative everywhere. Since the negative
q2  0
q1  0
charge is smaller in magnitude than the positive charge,
any point where the potential is zero must be closer to the negative charge. So consider
locations between the charges (position x1 ) and to the left of the negative charge
x
q1 
d
q2
(position x2 ) as shown in the diagram.
Vlocation 1 
Vlocation 2 
kq1
 d  x1 
kq1
 d  x2 


kq2
x1
kq2
x2
 2.0 10 C   5.0 cm   2.0 cm
q  q 
 5.0 10 C 
 2.0 10 C   5.0 cm   10.0 cm
qd


q  q 
1.0 10 C 
 0  x1 
6
q2 d
2

6
1
6
 0  x2
2
6
1
2
So the two locations where the potential is zero are 2.0 cm from the negative charge
towards the positive charge, and 10.0 cm from the negative charge away from the
positive charge.
23. Let the side length of the equilateral triangle be L. Imagine bringing the
e
L
electrons in from infinity one at a time. It takes no work to bring the first
electron to its final location, because there are no other charges present.
L
L
Thus W1  0 . The work done in bringing in the second electron to its
final location is equal to the charge on the electron times the potential
e
(due to the first electron) at the final location of the second electron.
2
 ke  ke
Thus W2   e     
. The work done in bringing the third electron to its final
 L L
location is equal to the charge on the electron times the potential (due to the first two
2
 ke ke  2ke
electrons). Thus W3   e      
. The total work done is the sum
L
 L L
W1  W2  W3 .
W  W1  W2  W3  0 
ke 2
L

2ke 2
L

3ke 2
L



3 8.99  109 N m 2 C 2 1.60  10 19 C
1.0 10
10
m


e
2
 6.9  1018 J
24. (a) The potential due to a point charge is given by Eq. 17.5
1 1
kq kq
Vba  Vb  Va    kq   
rb ra
 rb ra 

 8.99  109 N m 2 C2
 3.8 10 C   0.881 m  0.721 m   8.6 10 V
6
3

(b) The magnitude of the electric field due to a point
charge is given by Eq. 16-4a. The direction of the
electric field due to a negative charge is towards the
charge, so the field at point a will point downward, and
the field at point b will point to the right. See the
vector diagram.

Eb
Ea
E b  Ea

Eb
Ea
Eb 
kq
Ea 
kq
2
b
r
2
a
r
right 
8.99 10
9
Eb
9
N m 2 C2
 tan 1
6
2
65899
44114
4
4
V m , right
 3.8 10 C  down  6.5899 10
 0.72 m 
 4.4114 10
E b  Ea  Ea2  Eb2 
Ea
 3.8 10 C  right  4.4114  10
 0.88 m 
8.99 10
down 
  tan 
N m 2 C2
6
2
 
2
V m  6.5899  104 V m

2
4
V m , down
 7.9  104 V m
 56o
25. We assume that all of the energy the proton gains in being accelerated by the
voltage is changed to potential energy just as the proton’s outer edge reaches the
outer radius of the silicon nucleus.
e 14e 
PEinitial  PEfinal  eVinitial  k

r
14  1.60 1019 C
14e
9
2
2
Vinitial  k
 8.99  10 N m C
 4.2  106 V
15
r
1.2  3.6  10 m



r

26. Use Eq. 17-2b and Eq. 17-5.
k   q    kq k   q  
1 1
1 
 kq
 1
VBA  VB  VA  

 
 kq 
  



b   b
d b 
 d b b b d b 
 d b
1  2kq  2b  d 
 1
 
b d  b
 d b b 
 2kq 
27. (a) The electric potential is found from Eq. 17-5.
Vinitial  k
qp

 8.99 10 N m C
9
2
2

1.60 10
19
10
C
  27 V
ratom
0.53  10 m
(b) The kinetic energy can be found from the fact that the magnitude of the net force on the
electron, which is the attraction by the proton, is causing circular motion.
qp qe
m v2
e2
e2
Fnet  e  k 2
 me v 2  k
 KE  12 me v 2  12 k
ratom
ratom
ratom
ratom

1
2
8.99 10
1.60 10 C   2.17110
C 
 0.53 10 m 
19
9
N m2
 2.171 1018 J 
2
10
1eV
2
18
J  2.2  10 18 J
 13.57 eV  14 eV
1.60  1019 J
(c) The total energy of the electron is the sum of its KE and PE. The PE is found from Eq.
17-2a,
and is negative since the electron’s charge is negative.
e2
Etotal  PE  KE  eV  me v   k
2
1
2
ratom
 k
1
2
e2
ratom
 k
1
2
e2
ratom
 2.171 1018 J  2.2  1018 J  14 eV
(d) If the electron is taken to infinity at rest, both its PE and KE would be 0. The amount of
energy
needed by the electron to have a total energy of 0 is just the opposite of the answer to
part (c), 2.2  1018 J or 14 eV .
28. The dipole moment is the product of the magnitude of one of the charges times the
separation distance.



p  Ql  1.60 1019 C 0.53 1010 m  8.5 1030 C m
29. The potential due to the dipole is given by Eq. 17-6b.
8.99  109 N m 2 C 2 4.8  10 30 C m cos 0
kp cos 

(a) V 
2
r2
1.1  109 m


 3.6  102 V
(b) V 
kp cos 
r
2

8.99 10
9
N m C
kp cos 
r
2


2
9
8.99 10
9
30

C m cos 45
Q
 4.8 10
1.110 m 
Q
r
o
2
N m 2 C2
9
r

 4.8 10
1.110 m 
2
 2.5  102 V
(c) V 


Q
30
Q

C m cos135o
2
r

2
 2.5  10 V
Q
Q
30. We assume that p1 and p 2 are equal in magnitude, and that each makes a 52o angle with p .
The magnitude of p1 is also given by p1  qd , where q is the charge on the hydrogen atom,
and d is the distance between the H and the O.
p
p  2 p1 cos 52o  p1 
 qd 
2 cos 52o
q
p
2d cos 52
o


6.1  1030 C m
2 0.96  10
10

m cos 52
o
 5.2  1020 C
This is about 0.32 times the charge on an electron.
31. The capacitance is found from Eq. 17-7.
Q 2.5  103 C
Q  CV  C  
 2.9  106 F
V
850 V
32. The voltage is found from Eq. 17-7.
Q 16.5  108 C
Q  CV  V  
 17.4 V
C
9.5  109 F
33. The capacitance is found from Eq. 17-7.
Q 95  1012 C
Q  CV  C  
 7.9  1013 F
V
120 V
34. We assume the capacitor is fully charged, according to Eq. 17-7.


Q  CV  7.00  106 F 12.0 V   8.4  105 C
35. The area can be found from Eq. 17-8.
 0.20 F  2.2  103 m
0 A
Cd
C
 A

 5.0  107 m 2
12
2
2
d
0
8.85  10 C N  m




36. Let Q1 and V1 be the initial charge and voltage on the capacitor, and let Q2 and V2 be the
final charge and voltage on the capacitor. Use Eq. 17-7 to relate the charges and voltages to
the capacitance.
Q1  CV1 Q2  CV2 Q2  Q1  CV2  CV1  C V2  V1  
C
Q2  Q1
V2  V1

18  106 C
24 V
 7.5  107 F
37. The desired electric field is the value of V d for the capacitor. Combine Eq. 17-7 and Eq.
17-8 to find the charge.
 AV
Q  CV  0
  0 AE  8.85  1012 C2 N  m 2 35.0  104 m 2 8.50  105 V m
d




 2.63  108 C
38.
Combine Eq. 17-7 and Eq. 17-8 to find the area.
Q  CV 
 0 AV
d
  0 AE 
 5.2 10 C 
A

 0.29 m
 E
 2000 V 
8.85

10
C
N

m

  10 m 
6
Q
12
0
2
2
2
3
39. From Eq. 17-4a, the voltage across the capacitor is the magnitude of the electric field times
the separation distance of the plates. Use that with Eq. 17-7.
72  106 C
Q
Q  CV  CEd  E 

 4.5  104 V m
6
3
Cd
0.80  10 F 2.0  10 m





40. After the first capacitor is disconnected from the battery, the total charge must remain
constant. The voltage across each capacitor must be the same when they are connected
together, since each capacitor plate is connected to a corresponding plate on the other
capacitor by a constant-potential connecting wire. Use the total charge and the final
potential difference to find the value of the second capacitor.
QTotal  C1V1
Q1  C1Vfinal Q2  C2Vfinal
initial
QTotal  Q1
final
  C1  C2  Vfinal  C1V1
 Q2
final
final
final
  C1  C2 Vfinal 
initial
 V1

 125 V 
initial

C2  C1
 1    7.7  106 F  
 1   5.6  10 5 F
 Vfinal

15
V




41. The total charge on the combination of capacitors is the sum of the charges on the two
individual capacitors, since there is no battery connected to them to supply additional charge.
The voltage across each capacitor must be the same after they are connected, since each
capacitor plate is connected to a corresponding plate on the other capacitor by a constantpotential connecting wire. Use the total charge and the fact of equal potentials to find the
charge on each capacitor and the common potential difference.
Q1  C1V1
Q2  C2V2
Q1  C1Vfinal Q2  C2Vfinal
initial
initial
QTotal  Q1
 Q2
initial
initial
 Q1
initial
 C2V2
C1V1
Vfinal 
initial
initial
initial
C1  C2
 Q2
final
final
final
 C1V1
final
 C2V2
initial
 C1Vfinal  C2Vfinal 
initial
 2.50 10 F  875 V    6.80 10 F   652 V 

 9.30 10 F 
6
6
6
 711.95 V  712 V  V1  V2
Q1




 C1Vfinal  2.50  106 F  711.95   1.78  103 C
final
Q2
 C2Vfinal  6.80  106 F  711.95   4.84  103 C
final
42. Use Eq. 17-9 to calculate the capacitance with a dielectric.
C  K 0
A
d

  2.2  8.85  10
 5.5 10 m 
 1.8 10 m


2
12
C Nm
2
2
2
3
 3.3  1011 F
43. Use Eq. 17-9 to calculate the capacitance with a dielectric.
C  K 0
A
d

  7  8.85  10
12
C Nm
2
2

  5.0  102 m 
 3.2 10 m 
3
2
 1.5  1010 F
44. The initial charge on the capacitor is Qinitial  CinitialV . When the mica is inserted, the
capacitance changes to Cfinal  KCinitial , and the voltage is unchanged since the capacitor is
connected to the same battery. The final charge on the capacitor is Qfinal  CfinalV .


Q  Qfinal  Qinitial  CfinalV  CinitialV   K  1 CinitialV   7  1 3.5  109 F  22 V 
 4.6  107 C
45. The capacitance is found from Eq. 17-7, with the voltage given by Eq. 17-4 (ignoring the
sign).
Q
0.775 106 C
Q  CV  C  Ed   C 

 4.82  109 F
4
3
Ed
8.24 10 V m 1.95 10 m



The plate area is found from Eq. 17-9.
4.82  109 F 1.95  10 3 m
A
Cd
C  K 0
 A

 0.283 m 2
d
K  0  3.75  8.85  1012 C 2 N  m 2





46. The stored energy is given by Eq. 17-10.
PE  12 CV 2 
1
2
 2.2 10 F  650 V 
9
2
 4.6 104 J
47. The capacitance can be found from the stored energy using Eq. 17-10.
2  PE 
2 1200 J 
PE  12 CV 2  C 

 9.6  10 5 F
2
2
3
V
5.0  10 V


48. The two charged plates form a capacitor. Use Eq. 17-8 to calculate the capacitance, and Eq.
17-10 for the energy stored in the capacitor.
C
0 A
d
PE 
1
2
Q2
C

1
2
Q2d
0 A
 4.2 10 C  1.5 10 m 
8.85 10 C N  m 8.0 10 m 
4

1
2
12
2
2
3
2
2
2
 2.3  103 J
49. (a) Use Eq. 17-8 to estimate the capacitance.
C
0 A
d
8.85 10

12

C2 N  m2   4.5in  .0254 m in 
 5.0 10 m 
2
2
 7.265  1012 F  7 1012 F
(b) Use Eq. 17-7 to estimate the charge on each plate.


Q  CV  7.265 1012 F  9 V   6.54 1011 C  7 1011 C
(c) Use Eq. 17-4b to estimate the (assumed uniform) electric field between the plates. The
actual
location of the field measurement is not critical in this approximation.
V
9V
E 
 180 V m  200 V m
d 5.0  102 m
(d) By energy conservation, the work done by the battery to charge the plates is the potential
energy
stored in the capacitor, given by Eq. 17-10.
PE  12 QV 
1
2
 6.54 10
11

C  9 V   2.94 1010 J  3 1010 J
(e) If a dielectric is inserted, the capacitance changes, and so the charge on the capacitor
and the
energy stored also change. The electric field does not change because it only depends
on the battery voltage and the plate separation.
50. (a) From Eq. 17-8 for the capacitance, C   0
A
d
, since the separation of the plates is
doubled, the
capacitance is reduced to half its original value. Then from Eq. 17-10 for the energy,
Q2
1
PE  2
, since the charge is constant and the capacitance is halved, the energy is
C
doubled. So the energy stored changes by a factor of 2 .
(b) The work done to separate the plates is the source of the increase of stored energy. So
the work
is the change in the stored energy.
PE final  PEinitial  2PEinitial  PEinitial  PEinitial 
1
2
Q2
C

1
2
Q2
Q2 d

A
2 0 A
0
d
51. (a) The energy stored in the capacitor is given by Eq. 17-10, PE  12 CV 2 . Assuming the
capacitance is constant, then if the potential difference is doubled, the stored energy is
multiplied by 4 .
(b) Now we assume the potential difference is constant, since the capacitor remains
connected to a
battery. Then the energy stored in the capacitor is given by PE  12 QV , and so the
stored energy is multiplied by 2 .
52. (a) Before the two capacitors are connected, all of the stored energy is in the first capacitor.
Use
Eq. 17-10.
PE  12 CV 2 
1
2
 2.70 10 F 12.0 V 
6
2
 1.944 104 J  1.94 10 4 J
(b) After the first capacitor is disconnected from the battery, the total charge must remain
constant,
and the voltage across each capacitor must be the same, since each capacitor plate is
connected to a corresponding plate on the other capacitor by a connecting wire which
always has a constant potential. Use the total charge and the fact of equal potentials to
find the charge on each capacitor, and then calculate the stored energy.


Qtotal  C1Vinitial  2.70  10 6 F 12.0 V   3.24  10 5 C
Q1  C1V
Q2  C2V
Q1  Qtotal
C1
C1  C2
Q1

C1
Q2
C2

Qtotal  Q1
C2

 2.70 10 F   1.3057 10
C
 6.70 10 F 
6

 3.24  10
5
5
6
C
Q 2  Qtotal  Q1  3.24  10 5 C  1.3057  10 5 C  1.9343  10 5 C
 1.3057  105 C 2 1.9343  105 C 2 

PE total  PE1  PE 2  12
 12
 12 

C1
C2
4.00  106 F  
  2.70  106 F 



2
2
 1.3057  105 C  1.9343  105 C  
1 
  7.83  105 J
2

6
6
  2.70  10 F 
 4.00 10 F  

Q12
Q22
(c) Subtract the two energies to find the change.
PE  PE final  PE initial  7.83  105 J  1.944  104 J  1.16  10 4 J
53. Consider three parts to the electron’s motion. First,
during the horizontal acceleration phase, energy will be
conserved and so the horizontal speed of the electron vx
can be found from the accelerating potential V .
Secondly, during the deflection phase, a vertical force
will be applied by the uniform electric field which gives
the electron an upward velocity, v y . We assume that
22 cm
E
vx
30o
xscreen
xfield
there is very little upward displacement during this time.
Finally, after the electron leaves the region of electric
field, it travels in a straight line to the top of the screen, moving at an angle of approximately
30o.
Acceleration:
PEinitial  KE final  eV  12 mvx2  vx 
2eV
m
Deflection:
time in field: xfield  vx tfield  tfield 
Fy  eE  ma y  a y 
eE
xfield
vx
v y  v0  a y tfield  0 
m
eE xfield
mvx
Screen:
xscreen  vx tscreen  tscreen 
xscreen
vx
yscreen  v y tscreen  v y
eE xfield
yscreen
xscreen

vy
vx

mvx
vx

eE xfield
mvx2

xscreen
vx
11cm
E
yscreen m
yscreen mvx2
xscreen exfield
2eV


2 7.0  103 V  0.11m 
2V yscreen
m



xscreen exfield xscreen xfield  0.22 m  cos 30o   0.028 m 
 2.89  105 V m  2.9  105 V m
As a check on our assumptions, we calculate the upward distance that the electron would
move while in the electric field.
E  xfield 
 eE   xfield  eE  xfield 
 0 


 
4V
 2eV 
 m   vx 
2m 

 m 
2
y  v0tfield  a t
2
y field
1
2
 2.89 10

5
2
2
1
2

V m 2.8  102 m

2
 8  103 m
4  7000 V 
This is about 7% of the total 11 cm vertical deflection, and so for an estimation, our
approximation is acceptable.
54. If there were no deflecting field, the electrons would hit
the center of the screen. If an electric field of a certain
direction moves the electrons towards one extreme of the
yscreen  15cm
E
screen, then the opposite field will move the electrons to
the opposite extreme of the screen. So we solve for the
vx
field to move the electrons to one extreme of the screen.
Consider three parts to the electron’s motion, and see the
xscreen  34 cm
diagram, which is a top view. First, during the horizontal
xfield
acceleration phase, energy will be conserved and so the
horizontal speed of the electron vx can be found from the accelerating potential V .
Secondly, during the deflection phase, a vertical force will be applied by the uniform electric
field which gives the electron a leftward velocity, v y . We assume that there is very little
leftward displacement during this time. Finally, after the electron leaves the region of
electric field, it travels in a straight line to the left edge of the screen.
Acceleration:
PEinitial  KE final  eV  12 mvx2  vx 
2eV
m
Deflection:
time in field: xfield  vx tfield  tfield 
Fy  eE  ma y  a y 
eE
m
xfield
vx
v y  v0  a y tfield  0 
eE xfield
mvx
Screen:
xscreen  vx tscreen  tscreen 
xscreen
vx
yscreen  v y tscreen  v y
xscreen
vx
eE xfield
yscreen
xscreen
E

vy
vx
mvx


vx
yscreen mv
eE xfield
yscreen m
2
x
xscreen exfield

mvx2
2eV


2 6.0  103 V  0.15 m 
2V yscreen
m



xscreen exfield xscreen xfield
 0.34 m  0.026 m 
 2.04  105 V m  2.0  105 V m
As a check on our assumptions, we calculate the upward distance that the electron would
move while in the electric field.
E  xfield 
 eE   xfield  eE  xfield 
 0 


 
4V
 2eV 
 m   vx 
2m 

 m 
2
y  v0tfield  a t
2
y field
1
2
 2.04 10

5
2
2
1
2

V m 2.6  102 m

2
 6  103 m
4  6000 V 
This is about 4% of the total 15 cm vertical deflection, and so for an estimation, our
approximation is acceptable. And so the field must vary from
2.0  105 V m to  2.0  105 V m
55. (a) The electron was accelerated through a potential difference –6.3 kV in gaining 6.3 keV of
KE.
The proton is accelerated through the opposite potential difference as the electron, and
has the exact opposite charge. Thus the proton gains the same KE, 6.3 keV.
(b) Both the proton and the electron have the same KE. Use that to find the ratio of the
speeds.
1
2
mp vp2  12 me ve2 
ve
vp

mp

me
1.67  1027 kg
9.11 1031 kg
 42.8
56. (a) The energy is related to the charge and the potential difference by Eq. 17-3.
PE 4.2  106 J
PE  qV  V 

 1.05  106 V  1.1 106 V
q
4.0 C
(b) The energy (as heat energy) is used to raise the temperature of the water and boil it.
Assume
that room temperature is 20oC.
Q  mcT  mLf 
m
Q
cT  Lf

4.2  106 J
J 


o
5 J 
 4186 kg o C   80 C    22.6  10 kg 




57. The energy density is given by Eq. 17-11.
Energy density  12  0 E 2 
1
2
8.85 10
12

 1.6 kg
C2 N  m2 150 V m   1.0 107 J m3
2
58. The electric force on the electron must be the same magnitude as the weight of the electron.
The magnitude of the electric force is the charge on the electron times the magnitude of the
electric field. The electric field is the potential difference per meter: E  V d .
FE  mg  eV d  mg 
 9.1110

mgd
31

kg 9.80 m s 2
3.0 10 m   1.7 10
2
12
V
e
1.60  1019 C
Since it takes such a tiny voltage to balance gravity, the thousands of volts in a television set
are more than enough (by many orders of magnitude) to move electrons upward against the
force of gravity.
V
59. The energy in the capacitor, given by Eq. 17-10, is the heat energy absorbed by the water,
given by Eq. 14-2.
PE  Qheat  12 CV 2  mcT 

V
2mcT

C
2  2.5 kg   4186

 o
o
  95 C  21C 
kg C 
 622 V  620 V
J
o
4.0 F
60. Since the capacitor is disconnected from the battery, the charge on it cannot change. The
capacitance of the capacitor is increased by a factor of K, the dielectric constant.
C
C
1
Q  CinitialVinitial  CfinalVfinal  Vfinal  Vinitial initial  Vinitial initial   24.0 V 
 11V
Cfinal
KCinitial
2.2
61. Combine Eq. 17-7 with Eq. 17-8 and Eq. 17-4.
Q  CV 
0 A
d
V  0 A
V
d

  0 AE  8.85  1012 C2 N  m 2
 56 10
4
m2
 3.0 10
6
V m

 1.5  107 C
62. Use Eq. 17-5 to find the potential due to each charge. Since the triangle is
equilateral, the 30-60-90 triangle relationship says that the distance from a
corner to the midpoint of the opposite side is 3L 2 .
VA 
VB 
VC 
k  Q 
L 2
k  Q 
L 2
k Q
L2



k  3Q 

L 2
k Q 
L 2

3L 2

k  3Q 
L 2
k Q 
k  3Q 

3L 2

k  Q 
3L 2
2kQ 
1 
kQ
4 
 6.85


L 
L
3
6kQ
 3.46
3L

2kQ 
kQ
L
1 
kQ
2
 5.15


L 
L
3
B
Q
A
Q
C
3Q
63.
(a)
Because of the inverse square nature of the
x
d
electric
field, any location where the field is zero must be
q2  0
q1  0
closer to the weaker charge  q2  . Also, in between the
two charges, the fields due to the two charges are parallel to each other (both to the left)
and cannot cancel. Thus the only places where the field can be zero are closer to the
weaker charge, but not between them. In the diagram, this is the point labeled as “x”.
Take to the right as the positive direction.
q
q1
2
E  k 22  k
 0  q2  d  x   q1 x 2 
2
x
d  x
q2
2.6  106 C
1.6 cm   11.1cm left of q2
3.4  106 C  2.6  106 C
(b) The potential due to the positive charge is positive
d
x1
x2
everywhere, and the potential due to the negative
charge is negative everywhere. Since the negative
q2  0
q1  0
charge is smaller in magnitude than the positive charge,
any point where the potential is zero must be closer to the negative charge. So consider
locations between the charges (position x1 ) and to the left of the negative charge
x
d
q1 
q2
(position x2 ) as shown in the diagram.
Vlocation 1 
Vlocation 2 
kq1
 d  x1 
kq1
 d  x2 


kq2
x1
kq2
x2
 2.6 10 C  1.6 cm   0.69 cm
q  q 
 6.0 10 C 
 2.6 10 C  1.6 cm   5.2 cm
qd


q  q 
 0.8 10 C 
 0  x1 
6
q2 d
2

6
1
6
 0  x2
2
6
1
2
So the two locations where the potential is zero are 0.7 cm from the negative charge
towards the positive charge, and 5.2 cm from the negative charge away from the
positive charge.
64. The voltage across the capacitor stays constant since the capacitor remains connected to the
battery. The capacitance changes according to Eq. 17-9.
 A
 A
Cinitial  0  2600  1012 F Cfinal  K 0  KCinitial  5Cinitial
d
d


Qfinal  Qinitial  CfinalV  CinitialV   5Cintial  Cinitial  V  4CinitialV  4 2600  1012 F  9.0 V 
 9.4  108 C
65. To find the angle, the horizontal and vertical components of the velocity are needed. The
horizontal component can be found using conservation of energy for the initial acceleration
of the electron. That component is not changed as the electron passes through the plates.
The vertical component can be found using the vertical acceleration due to the potential
difference of the plates, and the time the electron spends between the plates.
Horizontal:
PE inital  KE final  qV  12 mvx2
Vertical:
FE  qE y  ma  m
v
y
 vy 0 
t
t
 vy 
x
vx
qE y t
m

qE y x
mvx
Combined:
qE y x
tan  
vy
vx

mvx
vx
 250 V 
 0.065 m 
qE y x qE y x E y x  0.013 m 




 0.1136
2
mvx
2qV
2  5,500 V 
2V
  tan 1 0.1136  6.5o
66. There is no other source of charge except for the original capacitor. Thus the total charge
must remain at Q0 . Also, since the plates of the one capacitor are connected via
equipotential wires to the plate of the other capacitor, the two capacitors must have the same
voltage across their plates. Use the total charge and fact of equal potentials to find the
charge on each capacitor and the common potential difference.
Q0  Q1  Q2
Q1  C1V  Q0
Q1  C1V
Q2  C2V
C1
Q0  C1V  C2V   C1  C2 V  V 
Q2  C2V  Q0
C1  C2
Q0
C1  C2
C2
C1  C2
67. Use Eq. 17-8 for the capacitance.
12
2
2
4
2
0 A
 0 A 8.85  10 C N  m 1.0  10 m
C
 d

 9  1016 m
d
C
1F 



No , this is not practically achievable. The gap would have to be smaller than the radius of
a proton.
68. Since the E-field points downward, the surface of the Earth is a lower potential than points
above the surface. Call the surface of the Earth 0 volts. Then a height of 2.00 m has a
potential of 300 V. We also call the surface of the Earth the 0 location for gravitational PE.
Write conservation of energy relating the charged spheres at 2.00 m (where their speed is 0)
and at ground level (where their electrical and gravitational potential energies are 0).
Einitial  Efinal  mgh  qV  12 mv 2  v 


2  gh 
qV 

m 

6.5  10 4 C   300 V  

2
v  2  9.80 m s   2.00 m  
  6.3184 m s
 0.540 kg 



6.5  104 C   300 V  

2
v  2  9.80 m s   2.00 m  
  6.2030 m s
 0.540 kg 


v  v  6.3184 m s  6.2030 m s  0.12 m s
69. (a) Use Eq. 17-10 to calculate the stored energy.
PE  12 CV 2 
1
2
 5.0 10 F3.0 10 V 
8
4
2
 22.5 J
23J
(b) The power is the energy converted per unit time.
Energy 0.12  22.5 J 
P

 3.4  105 W
6
time
8.0 10 s
70. (a) Use Eq. 17-8.
C
0 A
d
8.85 10

(b) Use Eq. 17-7.
12

C2 N  m 2 110  106 m 2
1500 m 


  6.49 10
7
F  6.5  107 F

Q  CV  6.49  107 F 3.5  107 V  22.715 C  23C
(c) Use Eq. 17-10.
PE  12 QV 
 22.715C   3.5 107 V  
1
2
4.0 108 J
71. The kinetic energy of the electrons (provided by the UV light) is converted completely to
potential energy at the plate since they are stopped. Use energy conservation to find the
emitted speed, taking the 0 of PE to be at the surface of the barium.
KEinitial  PE final  12 mv 2  qV 
v
2qV
m



2 1.60  1019 C  3.02 V 
31
9.11 10 kg
 1.03  106 m s
72. Let d1 represent the distance from the left charge to point b, and let d 2 represent the
distance from the right charge to point b. Let Q represent the positive charges, and let q
represent the negative charge that moves. The change in potential energy is given by Eq. 173, PE b  PE a  qVba  q Vb  Va  .
d1  122  142 cm  18.44 cm

PE b  PE a  q Vb  Va   q 
d 2  142  242 cm  27.78 cm
kQ
 0.1844 m

 kQq 

1
 0.1844 m
 8.99  109


  kQ
kQ  




0.2778 m   0.12 m 0.24 m  
kQ
  1
1 




0.2778 m   0.12 m 0.24 m  
1
 1.5 10 C  33 10 C  3.477 m   1.547 J  1.5 J
6
6
1
73.
Use Eq. 17-7 with Eq. 17-9 to find the charge.
Q  CV  K  0
A
d

V   3.7  8.85  10
12
C Nm
2
2

  0.55  102 m 
 0.15 10 m 
3
2
12 V  
2.5  10 10 C
74.
(a)
Use Eq. 17-5 to calculate the potential due to the charges. Let the distance
between the charges
be d.
Vmid 
kQ1
kQ2

 d 2  d 2

2k
d
 Q1  Q2  

2 8.99  109 N m 2 C2
 0.23m 

 4.5 10
6
C  8.2  106 C

 2.9  105 V
(b) Use Eq. 16-4b to calculate the electric field. Note that the field due to each of the two
charges
will point to the left, away from the positive charge and towards the negative charge.
Find the magnitude of the field using the absolute value of the charges.
k Q2
kQ1
4k
Emid 

 2  Q1  Q2 
2
2
 d 2  d 2 d

4 8.99  109 N m 2 C 2

 0.23 m 
2

 4.5 10
6

C  8.2  106 C  8.6  106 V m , left
75. (a) Use Eq. 17-7 and Eq. 17-8 to calculate the charge.
Q  CV 
0 A
d
V
8.85 10
12
C2 N  m 2
 5.0 10
4
 2.0 10
m
4
m2
 12 V   4.248 10
11
C
 4.2  1011C
(b) The charge does not change. Since the capacitor is not connected to a battery, no charge
can
flow to it or from it. Thus Q  4.248 1011 C  4.2 1011C .
(c) The separation distance was multiplied by a factor of 1.5. The capacitance is inversely
proportional to the separation distance, so the capacitance was divided by a factor of
1.5. Since Q  CV , if the charge does not change and the capacitance is divided by a
factor of 1.5, then the voltage must have increased by a factor of 1.5.
Vfinal  1.5Vinitial  1.5 12 V   18 V
(d) The work done is the change in energy stored in the capacitor.
W  PEfinal  PEinitial  12 QVfinal  12 QVinitial  12 Q Vfinal  Vinitial 

1
2
 4.248 10
11

C 18 V  12 V   1.3 10 10 J
76. The energy stored in the capacitor is given by Eq. 17-10. The final energy is half the initial
energy. Find the final charge, and then subtract the final charge from the initial charge to
find the charge lost.
Q2
Q2
1
Efinal  12 Einitial  12 final  12 12 initial  Qfinal 
Qinitial
C
C
2


Qlost  Qinitial  Qfinal  Qinitial  1 
1 
1 

6
  CV  1 
   2.5  10 F   6.0 V  0.2929 
2
2

 4.4  106 C
77. (a) We assume that Q2 is held at rest. The energy of the system will then be conserved, with
the
initial potential energy of the system all being changed to kinetic energy after a very
long time.
kQ1Q2 1
Einitial  Efinal  PE initial  KE final 
 2 m1v12 
r
v1 
2kQ1Q2
m1r

 5.0 10 C 
kg  4.0  10 m 
6
2 8.99  109 N m 2 C2

1.5 10
6
2
2
 2.7  103 m s
(b) In this case, both the energy and the momentum of the system will be conserved. Since
the
initial momentum is zero, the magnitudes of the momenta of the two charges will be
equal.
m
pinitial  pfinal  0  m1v1  m2 v2  v2  v1 1
m2
Einitial  Efinal  PE initial  KE final 
kQ1Q2
r
v1 
2
 2

m1   1 m1
 m v  m v   m1v1  m2  v1
 m1  m2  v12
 2
m2  
m2



1
2
2
1 1
1
2
2kQ1Q2 m2
m1  m1  m2  r
2
2 2

1
2

 5.0 10 C   2.5 10
kg  4.0  10 m  4.0  10 kg 
6
2 8.99  109 N m 2 C 2
1.5 10
6
2
 2.2  103 m s
78. Calculate VAB  VA  VB . Represent the 0.10 m distance by the variable d.
2
6
6
kg
