Home Work Solutions 4/5
1. Figure 24-42 shows a thin plastic rod of length L = 12.0 cm and uniform positive charge Q = 56.1 fC lying on an
x axis. With V = 0 at infinity, find the electric potential at point P1 on the axis, at distance d = 2.50 cm from one
end of the rod.
Figure 24-42
Sol
Consider an infinitesimal segment of the rod, located between x and x + dx. It has length dx and
contains charge dq =  dx, where  = Q/L is the linear charge density of the rod. Its distance from
P1 is d + x and the potential it creates at P1 is
dV 
dq
1  dx

.
4 0 d  x 4 0 d  x
1
To find the total potential at P1, we integrate over the length of the rod and obtain:
V

4 0

L
0
L
dx

Q
 L

ln(d  x) 
ln 1  
d  x 4 0
4 0 L  d 
0
(8.99 109 N  m 2 C2 )(56.11015 C) 
0.12 m 

ln 1 
103 10
V.-3 V
6.31
  7.39
0.12 m
 0.025 m 
2. A plastic disk of radius R = 64.0 cm is charged on one side with a uniform surface charge density σ = 7.73 fC/m2,
and then three quadrants of the disk are removed. The remaining quadrant is shown in Fig. 24-45. With V = 0 at
infinity, what is the potential due to the remaining quadrant at point P, which is on the central axis of the original
disk at distance D = 25.9 cm from the original center?
Sol
The disk is uniformly charged. This means that when the full disk is present each quadrant
contributes equally to the electric potential at P, so the potential at P due to a single quadrant is
one-fourth the potential due to the entire disk. First find an expression for the potential at P due to
the entire disk. We consider a ring of charge with radius r and (infinitesimal) width dr. Its area is
2r dr and it contains charge dq = 2r dr. All the charge in it is a distance r 2  D2 from P, so the
potential it produces at P is
dV 
1 2p  rdr
 rdr

.
4p  0 r 2  D 2 2 0 r 2  D 2
The total potential at P is
V

2 0

rdr
R
0
r 2  D2


r 2  D2
2 0
R
0

  2
R  D2  D  .

2 0 
The potential Vsq at P due to a single quadrant is
Vsq 
V   2
(7.731015 C/m2 ) 

R  D2  D 
(0.640 m)2  (0.259 m) 2  0.259 m 

12
2
2
 8(8.85 10 C /N  m ) 

4 8 0 
 4.71105 V.
Note: Consider the limit D ? R. The potential becomes
Vsq 
  2

R  D2  D  


8 0
8 0
  1 R2

 D 1 
2
2
D


q
  R 2  R 2 / 4


D

 sq



 8 0 2 D 4 0 D 4 0 D
where qsq   R 2 / 4 is the charge on the quadrant. In this limit, we see that the potential
resembles that due to a point charge qsq .
3. The thin plastic rod of length L = 10.0 cm in Fig. 24-42 has a nonuniform linear charge density λ = cx, where c =
49.9 pC/m2. (a) With V = 0 at infinity, find the electric potential at point P2 on the y axis at y = D = 3.56 cm. (b)
Find the electric field component Ey at P2. (c) Why cannot the field component Ex at P2 be found using the result
of (a)?
Sol
(a) Consider an infinitesimal segment of the rod from x to x + dx. Its contribution to the potential
at point P2 is
dV 
1
 (x)dx
4 0
x 2  y2

1
cx
4 0
x 2  y2
dx.
Thus,
V
rod
dVP 
 L  y  y
C/m )  (0.120 m)  (0.0356 m)  0.0356 m 
c L
x
c
dx 

4 0 0 x 2  y 2
4 0
 (8.99 109 N  m 2 C2 )(49.9 1012
 4.02 102 V.
(b) The y component of the field there is
2
2
2
2
2
Ey  
VP
c d

y
4 0 dy


L2  y 2  y 
c 
1 
4 0 


L2  y 2 
y


0.0356 m
 (8.99  109 N  m 2 C2 )(49.9  1012 C/m 2 )  1 


(0.120 m)2  (0.0356 m)2 
 0.321 N/C.
(c) We obtained above the value of the potential at any point P strictly on the y-axis. In order to
obtain Ex(x, y) we need to first calculate V(x, y). That is, we must find the potential for an arbitrary
point located at (x, y). Then Ex(x, y) can be obtained from Ex ( x, y)  V ( x, y) / x .
4. Figure 25-43 displays a 12.0 V battery and 3 uncharged capacitors of capacitances C1 = 4.00 μF, C2 = 6.00 μF,
and C3 = 3.00 μF. The switch is thrown to the left side until capacitor 1 is fully charged. Then the switch is
thrown to the right. What is the final charge on (a) capacitor 1, (b) capacitor 2, and (c) capacitor 3?
Sol
The charges on capacitors 2 and 3 are the same, so these capacitors may be replaced by an
equivalent capacitance determined from
1
1
1 C2  C3



.
Ceq C2 C3
C2 C3
Thus, Ceq = C2C3/(C2 + C3). The charge on the equivalent capacitor is the same as the charge on
either of the two capacitors in the combination, and the potential difference across the equivalent
capacitor is given by q2/Ceq. The potential difference across capacitor 1 is q1/C1, where q1 is the
charge on this capacitor. The potential difference across the combination of capacitors 2 and 3
must be the same as the potential difference across capacitor 1, so q1/C1 = q2/Ceq. Now some of
the charge originally on capacitor 1 flows to the combination of 2 and 3. If q0 is the original charge,
conservation of charge yields q1 + q2 = q0 = C1 V0, where V0 is the original potential difference
across capacitor 1.
(a) Solving the two equations
q1 q2

C1 Ceq
q1  q2  C1V0
for q1 and q2, we obtain
C12  C2  C3 V0
C12V0
C12V0
q1 


.
C2C3
Ceq  C1
C
C

C
C

C
C
1
2
1
3
2
3
 C1
C2  C3
With V0 = 16.0 V, C1= 4.00 F, C2= 6.00 F and C3 =3.00 F, we find Ceq = 2.00 F and q1 = 42.7 C.
(b) The charge on capacitors 2 is
q2  C1V0  q1  (4.00 F)(16.0 V)  42.7 C  21.3 C .
(c) The charge on capacitor 3 is the same as that on capacitor 2:
q3  C1V0  q1  (4.00  F)(16.0 V)  42.7 C  21.3 C .
5. The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are
charged by a 6.00 V battery. They are then disconnected from the battery and pulled apart (without discharge) to
a separation of 8.00 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial
stored energy, (c) the final stored energy, and (d) the work required to separate the plates.
Sol
(a) Let q be the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is
given by  0 A d i , the charge is q  CV   0 AVi d i . After the plates are pulled apart, their
separation is d f and the potential difference is Vf. Then q   0 AV f 2d f and
Vf 
df
0 A
q
d f 0 A
d
Vi  f Vi .
 0 A di
di
With di  3.00 103 m , Vi  6.00 V, and d f  8.00 103 m , we have V f  16.0 V .
(b) The initial energy stored in the capacitor is
 AV 2 (8.85 1012 C2 /N  m2 )(8.50 104 m 2 )(6.00 V) 2
1
U i  CVi 2  0 i 
 4.511011 J.
3
2
2d i
2(3.00 10 m)
(c) The final energy stored is
2
1  0 A 2 1  0 A  d f  d f   0 AVi 2  d f
Uf 
Vf 
 Vi   
  Ui .
2 df
2 d f  di 
di  di  di
With d f / di  8.00 / 3.00 , we have U f  1.20 1010 J.
(d) The work done to pull the plates apart is the difference in the energy:
W = Uf – Ui = 7.52 1011 J.