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Physics II
Homework IX
Chapter 29: 3, 6, 10, 12, 20, 24, 30, 37, 58, 71
CJ
29.3. Model: The mechanical energy of the proton is conserved. A parallel plate
capacitor has a uniform electric field.
Visualize:
The figure shows the before-and-after pictorial representation.
Solve: The proton loses potential energy and gains kinetic energy as it moves toward
the negative plate. The potential energy is defined as U  U0  qEx, where x is the
distance from the negative plate and U0 is the potential energy at the negative plate (at x 
0 m). Thus, the change in the potential energy of the proton is
Up  Uf  Ui  U0  0 J   U0  qEd   qEd
The change in the kinetic energy of the proton is
K  K f  K i  21 mvf2  21 mvi2  21 mvf2
Applying the law of conservation of energy K  Up  0 J, we have
1
2
mvf2   qEd   0 J  vf2 
2qEd
m
When the amount of charge on each plate is doubled, then the final velocity of the proton
is
vf2 
2qEd
m
Dividing these equations,
vf2 E
 vf 

E
vf2
For a parallel-plate capacitor
vf 
E    0  Q A 0 .
E
vf
E
Therefore,
Q
vf  2  50,000 m/s   7.07  10 4 m/s
Q
Assess: The proton’s velocity is expected to increase because an increased charge on
the capacitor plates leads to a higher electric field between the plates and hence to an
increased force on the proton.
29.6. Model:
The charges are point charges.
Visualize: Please refer to Figure Ex29.6.
Solve: For a system of point charges, the potential energy is the sum of the
potential energies due to all pairs of charges:
Uelec  
i,j
Kqi q j
rij
 U12  U13  U23  K
qq
qq
q1q2
K 1 3 K 2 3
r12
r13
r23
  2 nC  1.0 nC   2 nC  1.0 nC   1 nC  1.0 nC  
 9.0  10 9 N m 2 /C2 



0.03 m
0.03 m
0.03 m


6
6
6
6
 0.60  10 J  0.60  10 J  0.30  10 J=  0.90  10 J


Assess: Note that U12  U21, U13  U31, and U23  U32.
29.10.
Model: An external electric field supplies energy to a dipole.
Visualize: On an energy diagram, the oscillation occurs between the points
where the potential-energy curve crosses the total energy line.
Solve: (a) The potential energy of an electric dipole moment in a uniform
electric field is given by Equation 29.23: U   p  E   pE cos . This means
U 0   pE  2  J
U 60   pE cos60   21 pE   21  2J   1 J
The mechanical energy Emech  U  K. We know that at   60°, K  60  0 J. So,
Emech  U 60  K 60  1  J  0 J   1  J
(b) Conservation of mechanical energy gives
U 60  K 60  U 0  K 0  1  J  0 J  2  J  K 0  K 0  1  J
29.12. Model: Energy is conserved. The potential energy is determined by the
electric potential.
Visualize:
The figure shows a before-and-after pictorial representation of a proton moving through a
potential difference of 1000 V. A positive charge speeds up as it moves into a region of
lower potential (U  K).
Solve: The potential energy of charge q is U  qV. Conservation of energy,
expressed in terms of the electric potential V, is
Kf  qVf  Ki  qVi,  Kf  Ki  q(Vi  Vf)  0 J qV

1
2
mv  q  1000 V  vf 
2
f


2 1.60  10 19 C 1000 V 
1.67  10
27
kg
 4.38  105 m/s
Assess: Note that the proton of our problem has nothing to do with creating the
potential difference of 1000 V.
29.20. Model: The electric potential between the plates of a parallel plate capacitor is
determined by the uniform electric field between the plates.
Solve: (a) The potential difference across the plates of a capacitor is



0.708  10 9 C 1.0  10 3 m
 
 Q A d  Qd 
VC  Ed    d 
 200 V
0
A 0
4.0  10 4 m 2 8.85  10 12 C2 /N m 2
 0 



(b) For d  2.0 mm, VC  400 V.
Assess: Note that the units in part (a) are N m/C. But Exercise 29.17 showed that
1 N/C  1 V/m, so 1 N m/C  1 V. We also see that the potential difference across a
parallel-plate capacitor is directly proportional to the plate separation.
29.24. Solve: Outside a charged sphere of radius R, the electric potential is identical
to that of a point charge Q at the center. That is,
V
1
Q
4 0 r
rR
The potential of the ball bearing is the potential right on the surface of the ball bearing.
Thus,
V
 9.0  10
9


N m 2 /C2 2.0  10 9 1.60  10 19 C
0.5  10
3
m
  5760 V
29.30. Model: The net potential is the sum of the potentials due to each charge.
Visualize: Please refer to Figure Ex29.30.
Solve: (a) Let q1  5 nC, q2  10 nC, and q3  10 nC. Also, r1  2 cm, r2  4
2
2
cm, and r3   2 cm    4 cm   4.47 cm . Each individual potential is simply that of a point
charge, so
VB  V1  V2  V3 
q1
4 0 r1

q2
4 0 r2

q3
4 0 r3

1  q1 q2 q3 
 
 
4 0  r1 r2 r3 
 5  10 C 10  10 9 C 10  10 9 C 
 9.0  109 N m2 /C2 


  2010 V
0.04 m
0.0447 m 
 0.02 m


9
(b) The potential energy of an electron at point B is
Uelectron  qelectronVB  eVB  (1.6  1019 C)(2010 V)  3.22  1016 J
Assess: Note that the units in part (a) are N m/C. But Problem 29.17 showed that
1 N/C  1 V/m, so 1 N m/C  1 V.
29.37. Solve: Let the two unknown, positive charges be Q1 and Q2. They are
separated by a distance r12  5.0  102 m . Their electric potential energy is



72  10 6 J 5.0  10 2 m
1 Q1Q2
6
Q
Q

 4.0  10 16 C2

U
 72  10 J
1 2
9.0  109 N m 2 /C2
4 0 r12
The electric force between these two charges is
F
1 Q1Q2
4.0  1016 C2
9
2
2

9.0

10
N
m
/C
 1.44  10 3 N
2
2
4 0 r122
5.0  10 m




Assess: As far as the magnitudes are concerned,
F
U
72  106 J

 1.44  103 N
r 5.0  102 m
29.58. Model: The electric field inside a capacitor is uniform.
Solve: (a) While the capacitor is attached to the battery, the plates are at the
same potentials as the terminals of the battery. Thus, the potential difference
across the capacitor is VC  15 V. The electric field strength inside the capacitor
is
E
Because
E    0  Q A 0 ,
VC
15 V

 3000 V/m
d
0.5  102 m
the charge on each plate is
 (3000 V/m)  (0.05 m)2 (8.85  1012 C2/N m2)  2.09  1010 C
Q  EA 0
(b) After the electrodes are pulled away to a new separation of d  1.0
cm, the potential difference across the capacitor stays the same as before.
That is, VC  VC  15 V . The electric field strength inside the capacitor is
E 
V 
15 V

 1500 V

d
0.01 m
The charge on each plate is
Q  EA 0 
(1500 V) (0.05 m)2(8.85  1012 C2/N m2)  1.04  1010 C
(c) The potential difference VC  VC  15 V is unchanged when the
electrodes are expanded to area A  4A . The electric field between the plates is
E  E 
VC
15 V

 3000 V/m
d
0.5  102 m
The new charge is
Q  EA 0

4EA 0  4Q
 8.34  1012 C
29.71. Model: Because the rod is thin, assume the charge lies along the semicircle of
radius R.
Visualize: Please refer to Figure P29.71. The bent rod lies in the xy-plane with
point P as the center of the semicircle. Divide the semicircle into N small segments
of length s and of charge Q  Q  R s , each of which can be modeled as a point
charge. The potential V at P is the sum of the potentials due to each segment of
charge.
Solve: The total potential is
V  Vi  
i
i
1 Q
1
1 Q
1 Q
 Q 

s 

 R  4  R i  .
4 0 R
4 0 R i   R 
4 0  R2 i
0
All of the terms come to the front of the summation because these quantities did not
change as far as the summation is concerned. The summation does not have to convert to
an integral because the sum of all the  around the semicircle is . Hence, the potential
at the center of a charged semicircle is
Vcenter 
1 Q
1 Q

4 0  R 4 0 R