BW7 Ch 21: P. 10-19
P21.10. Prepare: Energy is conserved. The potential energy is determined by the electric potential. The figure shows
a before-and-after pictorial representation of a proton moving through a potential difference.
Solve: (a) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region
of lower potential to a region of higher potential.
(b) Using the conservation of energy equation,
1
1 1 2

(Ki  K f ) 
mv i  0 J 

q
( e)  2

mv 2i (1.67  1027 kg)(8.0 105 m/s) 2
 V 

 3340 V  3000 V
2e
2(1.60  1019 C)
K f  U f  Ki  U i  K f  qVf  Ki  qVi  Vf  Vi 
(c)
Ki 
1 2
mv i  qV  (e)(3340V)  3000 eV
2
Assess: A positive V confirms that the proton moves into a higher potential region.
P21.11. Prepare: Let the distance between the two charges be called 2r. Then the distance from the observation point
midway between the charges to either charge is r. The potential near a charged particle is given by Equation 21.11:
V  KQ / r.
Solve: The total potential midway between the charges is equal to the sum of the potential of either charge by itself. Thus
we have:
300 V 
KQ KQ 2KQ
KQ



 150 V
r
r
r
r
We don’t know Q or r but this ratio of KQ and r is all we need to solve the problem. Since the total distance between
the charges is 2r , a point 25% of the way from one particle to the other is a distance r / 2 from the closer particle and a
distance 3r / 2 from the farther particle. Thus the potential at such a point is:
V
KQ
KQ 8KQ  8 


   (150 V)  400 V
r / 2 3r / 2
3r
3
Assess: Going from the midpoint of the two charges to a point closer to one of them increases the potential. This means
that we would have to do work to move a positive test charge from the midpoint toward one of the charges. This makes
sense considering that very close to either charge, the field is strong and a positive test charge placed there would
experience a strong repulsive force.
P21.12. Prepare: The electric potential difference between the plates is determined by the uniform electric field in
the parallel-plate capacitor and is given by Equation 21.6.
Solve: (a) The potential difference VC across a capacitor of spacing d is related to the electric field inside by Equation
21.6:
VC
E
 VC  Ed  (1.0  105 V/m)(0.002 m)  200 V
d
(b) The electric field of a capacitor is related to the charge by Equation 20.7:
Q   0 AE  (8.85 1012 C2/(N  m2 ))(4.0 104 m2 )(1.0 105 V/m)  3.5 10 10 C
Assess: A charge of 0.35 nC on the positive plate and an equal negative charge on the negative plate create a significant
potential difference across the parallel.
P21.13. Prepare: The electric potential between the plates of a parallel plate capacitor is determined by the uniform
electric field between the plates by Equation 21.6.
Solve: (a) Using Equations 21.6 and 20.7, the potential difference across the plates of a capacitor is
VC  Ed 
(Q/A)
0
d
Qd
(0.708 109 C)(1.0 103 m)

 200 V
A 0 (4.0 104 m 2 )(8.85 10 12 C 2/(N  m 2 ))
(b) For d  2.0 mm, VC  400 V.
Assess: Note that the units in part (a) are N  m/C. But 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.
P21.14. Prepare: Please refer to Figure P21.12. The electric field inside a parallel-place capacitor is determined by the
potential difference between the plates given by Equation 21.7. The proton’s potential energy inside the capacitor is also
determined by the capacitor’s potential difference.
Solve: (a) Because the right plate is at a higher potential compared with the left plate, the positive plate is on the right
and has a potential of 300 V.
(b) The electric field strength inside the capacitor is
VC 300 V  0 V
E

 1.0  105 V/m
d
3.0 103 m
(c) The potential energy of a charge q is U  qV. A proton on the left plate will have zero potential energy. A proton at
the midpoint of the capacitor is at a potential of 150 V. Thus, its potential energy is
U  (1.6 1019 C)(150 V)  2.4 1017 J
Assess: Because the right plate is at a higher potential compared with the left plate, the proton’s potential energy at
midpoint was expected to be positive.
P21.15. Prepare: The charge is a point charge. The electric potential of a charge q is given by Equation 21.10.
Solve: (a)
V
q
1 q
25 109 C 225 N  m 2/C
r 
 (9.0  109 N  m 2 /C2 )

4 0 r
4 0 V
V
V
1
For V  1000 V,
r1000 
225 N  m 2 /C
 0.225 m
1000 V
For V  2000 V, r2000  0.113 m; for V  3000 V, r3000  0.075 m; for V  4000 V, r4000  0.056 m.
(b)
Assess: The radius of an equipotential surface increases with the decrease in potential. This is what we would expect
because V  1/r.
P21.16. Prepare: Please refer to Figure P21.16. The charge is a point charge. We will use Equations 21.10 and 21.1
to calculate the potential and the potential energy of the charge.
Solve: (a) The electric potential of the point charge q is
V
 2.0 109 C  18.0 N  m 2 /C
q
 (9.0  109 N  m2 /C2 ) 

4 0 r
r
r


1
For points A and B, r  0.01 m. Thus,
VA  VB 
18.0 N  m2 /C
Nm
V
 1800
 1800   m  1800 V
0.01 m
C
m
For point C, r  0.02 m and VC  900 V.
(b) The potential energy of a charge q at a point where the electric potential is V is U  qV. The expression for the
potential in part (a) assumes that we have chosen V  0 V to be the potential at r  . So, we are obtaining
potential/potential energy relative to a zero of potential/potential energy at infinity. Thus,
U A  U B  (q)V  (e)(V )   (1.60 1019 C)(1800 V)  2.9 1016 J
UC  (1.60 1019 C)(900 V)  1.4 1016 J
(c) The potential differences are
VAB  VB  VA  1800 V 1800 V  0 V
VBC  VC  VB  900 V  1800 V  900 V
Assess: Clearly VA  VB and VC and VA, so, as expected, VAB  0 and VBC is negative.
P21.17. Prepare: 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
Solve:
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  109 N  m2 /C2 )(2.0 109 )(1.60 1019 C)
 5800 V
0.5  103 m
P21.18. Prepare: Please refer to Figure P21.18. The net potential is the sum of the potentials due to each charge given
by Equation 21.10.
Solve: The potential at the dot is
V
 2.0 109 C 2.0 109 C 2.0 109 C 
q1
1 q2
1 q3


 (9.0  109 N  m2 /C2 ) 


  1400 V
4 0 r1 4 0 r2 4 0 r3
0.050 m
0.030 m 
 0.040 m
1
Assess: Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.
P21.19. Prepare: The electric potential difference between the plates is determined by the uniform electric field in
the parallel-plate capacitor as given by Equation 21.6.
Solve: (a) The potential of an ordinary AA or AAA battery is 1.5 V. Actually, this is the potential difference between the
two terminals of the battery. If the electric potential of the negative terminal is taken to be zero, then the positive terminal
is at a potential of 1.5 V.
(b) If a battery with a potential difference of 1.5 V is connected to a parallel-plate capacitor, the potential difference
between the two capacitor plates is also 1.5 V. Thus, VC  1.5 V  V  V  Ed
where d is the separation between the two plates. The electric field inside a parallel-plate capacitor is
E
 Q 
Q
(1.5 V)( A 0 ) (1.5 V) (2.0 102 m) 2 (8.85 1012 C 2/N  m 2 )
 1.5 V  

 8.3  1012 C
d  Q 
A 0
d
2.0  103 m
 A 0 
Thus, the battery moves 8.3  1012 C of electron charge from the positive to the negative plate of the capacitor.
Assess: This is the charge on the positive plate. The other plate has a charge of –8.3  10–12 C.