Homework Solutions
Physics 8B – Spring 2012
Chpt. 30 – 1,6,10,12,20,21,26,34,44,45,62,69
30.1. Solve: The potential difference ∆V between two points in space is
xf
∆V =
V ( xf ) − V ( xi ) =
− ∫ Ex dx
xi
where x is the position along a line from point i to point f. When the electric field is uniform,
xf
∆V =− Ex ∫ dx =− Ex ∆x =− (1000 V/m )( 0.30 m − 0.10 m ) =−200 V
xi
30.6. Solve: The Van de Graaff generator or the motor that runs the belt does work in lifting a positive ion
(q = e) against the downward force on the positive charge that is moving up the belt. The work done is
W=
∆U =∆
q V=
q (1.0 ×106 V − 0 V ) =
(1.60 ×10−19 C )(1.0 ×106 V ) =1.6 ×10−13 J
Assess: The work done by the generator in lifting the charge is stored as electric potential energy of the charge.
30.10. Model: The electric field is perpendicular to the equipotential lines and points “downhill.”
Visualize: Please refer to Figure EX30.10. Three equipotential surfaces at potentials of −200 V, 0 V, and 200
V are shown.
Solve: The electric field component perpendicular to the equipotential surface is
∆V
200 V
−
=
−
=
−20 kV/m
E=
∆s
0.01 m
The electric field vector is in the third quadrant, 45° below the negative x-axis. That is,

=
E ( 20 kV/m, 45° below − x -axis )
30.12. Model: The electric field is the negative of the slope of the graph of the potential function.
Visualize: Please refer to Figure EX30.12.
Solve: There are three regions of different slope. For 0 cm < x < 1 cm,
∆V
50 V − 0 V
= = 5000 V/m ⇒ E x = −5000 V/m
∆ x 0.01 m − 0 m
For 1 cm < x < 2 cm,
∆V −50 V − ( 50 V )
=
= −10,000 V/m ⇒ E x = 10,000 V/m
dx 0.02 m − 0.01 m
For 2 cm < x < 3 cm,
0 V − ( −50 V )
∆V
= = 5000 V/m ⇒ E x = −5000 V/m
∆ x 0.03 m − 0.02 m
1
Assess:
∆V ∆ x and E x are the negative of each other.
30.20. Model: Assume that the battery is ideal and the wires connecting the battery with the capacitor have
zero resistance.
Solve: Using Equation 30.19,
Q
48 ×10−6 C
=
∆Vbat =
∆VC = =
=
24 V
C
2.0 ×10−6 F
30.21. Model: Assume that the battery is ideal.
Solve: Equation 30.19 is
∆VC =
Q
Q
45 ×10−9 C
= ⇒ =
=
= 5.0 ×10−9 F
C
C
∆VC
9.0 V
30.26. Visualize: Please refer to Figure EX30.26.
Solve: Any two electrodes, regardless of their shape, form a capacitor whose capacitance is defined as
C
= Q ∆VC . The capacitance is
20 ×10−9 C
2.0 ×10−10 F =
200 pF
C=
=
100 V
30.34. Model: The electrodes form a parallel-plate capacitor.
Solve: (a) The capacitance of the equivalent vacuum-filled capacitor is
=
C0
ε0 A
=
d
(8.85 ×10
C 2 /Nm 2 )( 5.0 ×10−3 m )
= 2.21×10−12 F
−3
0.10
×
10
m
(
)
2
−12
From Table 30.1, the dielectric constant of mylar is κ = 3.1. With the battery attached, the potential difference
across the plates is ∆VC =
∆Vbatt =
9.0 V . The charge on the plates is
Q = C ∆V = κ C0 ∆V = ( 3.1) ( 2.21 × 10−12 F ) ( 9.0 V ) = 6.17 × 10−11 C = 62 pC
The electric field inside the capacitor is
=
E
E0 1 ∆VC  1 
9.0 V

=
= 
=

 20 kV/m
−3
×
κ κ d
3.1
0.10
10
m



2
(b) With the battery connected, ∆VC =
9.0 V. The capacitor is now vacuum-insulated.
=
E E=
0
∆VC
9.0 V
=
= 90 kV/m
d
0.10 ×10−3 m
Q = C0 ∆VC = ( 2.21×10−12 F ) ( 9.0 V ) = 20 pC
Assess: Since the battery remains connected as the mylar is withdrawn, the potential difference across the
plates does not change.
30.44. Model: Assume the charged rod is a line of charge of length L.
Visualize: Please refer to Figure P30.44.
Solve: (a) Divide the charged rod into N small segments, each of length ∆x and with charge ∆q. The segment i
located at position x i contributes a small amount of potential V i at point P:
=
Vi
∆q
∆q
( Q∆ x L )
=
=
4πε 0 ri 4πε 0 ( x0 − xi ) 4πε 0 ( x0 − xi )
Point P is at a distance x 0 from the origin. This is done to avoid confusion with x i . The V i are now summed and
the sum is converted to an integral giving
V=
 x +L 2
L/2
dx
Q
Q
=
ln  0
 − ln ( x0 − x )  − L / 2=

4πε 0 L − L∫/ 2 x0 − x 4πε 0 L 
4πε 0 L  x0 − L 2 
Q
L/2
Replacing x 0 with x, the potential due to a line charge of length L at a distance x from the center is
V=
Q
 x + L/2 
ln 

4πε 0 L  x − L/2 
(b) Because Ex = − dV dx ,

Q d
Q 
1
1
Q
1
Ex =
−
−
−
ln ( x + L/2 ) − ln ( x − L/2 )  =

= 2

4πε 0 L dx
4πε 0 L  ( x + L/2 ) ( x − L/2 )  4πε 0 x − L2 /4
Assess: When L = 0 m, Ex = Q 4πε 0 x 2. This is the electric field of a point charge Q a distance x away from a
point charge, as expected.
30.45. Model: The electric field is the negative of the slope of the potential graph.
Visualize: Please refer to Figure P30.45.
Solve: Since the contours are uniformly spaced along the y-axis above and below the origin, the slope method
is the easiest to apply. Point 1 is in the center of a 75 V change (25 V to 100 V) over a distance of 2 cm, so the
slope ∆V/∆ s is 37.5 V/cm or 3750 V/m. Point 2 has the same potential difference in half the distance. Thus the slope at
point 2 is 7500 V/m. The magnitudes of the electric fields at points 1 and 2 are 3750 V/m and 7500 V/m. The
directions of the electric fields are downward at point 1 and upward at point 2, that is, from the higher potential
to the lower potential. That is,


=
E1 (=
3750 V m, down )
E2 ( 7000 V m, up )
30.62. Model: Assume the battery is an ideal battery.
Visualize:
3
The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the
figure.
Solve: Because C 2 and C 3 are in series,
1
1
1
1
1
10
−1
=
+
=
+
= ( µ F ) ⇒ C=
eq 23
Ceq 23 C2 C3 4 µ F 6 µ F 24
=
µF
24
10
2.4 µ F
C eq 23 and C 1 are in parallel, so
C eq = C eq 23 + C 1 = 2.4 µF + 5 µF = 7.4 µF
A potential difference of ∆V C = 9 V across a capacitor of equivalent capacitance 7.4 µF produces a charge
Q = C eq ∆V C = (7.4 µF)(9 V) = 66.6 µC
Because C eq is a parallel combination of C 1 and C eq 23 , these capacitors have ∆V1 =
∆Veq 23 =
∆VC =
9 V. Thus the
charges on these two capacitors are
Q 1 = (5 µF)(9 V) = 45 µC
Q eq 23 = (2.4 µF)(9 V) = 21.6 µC
Because Q eq 23 is due to a series combination of C 2 and C 3 , Q 2 = Q 3 = 21.6 µC. This means
∆V2 =
Q2 21.6 µ C
=
= 5.4 V
C2
4 µF
∆V3 =
Q3 21.6 µ C
=
= 3.6 V
C3
6 µF
In summary, Q 1 = 45 µC, V 1 = 9 V; Q 2 = 21.6 µC, V 2 = 5.4 V; and Q 3 = 21.6 µC, V 3 = 3.6 V.
30.69. Model: Assume the battery is ideal.
Visualize:
Solve:
When the capacitors are individually charged, their charges are
Q1 C=
=
1V
V)
(10 µ F )(10 =
100 µ C
Q2 C=
=
2V
V)
( 20 µ F )(10 =
200 µ C
These two capacitors are then connected with the positive plate of C 1 connected with the negative plate of C 2 as
shown in the figure. Let the new charges on C 1 and C 2 be Q1′ and Q2′ . Then,
Q1′ +=
Q2′ 200 µ C −
=
100µ
C
100µ C
The voltages across parallel capacitors are the same, so
Q′ Q′
C
∆V1′ =
∆V2′ = 1 = 2 ⇒ Q2′ = 2 Q1′
C1 C2
C1
Substituting this expression for Q2′ into the previous equation,
Q1′ +
 20 µ F 
C2
′ 100 µ C ⇒ Q1′ 1 +
Q=
1
= 100 µC
C1
 10 µ F 
Solving these equations,
we get Q1′ 33
=
=
µ C nd
a Q2′ 67 µ C. Finally,
Q′ 33 µ C
3.3 V =
∆V1′ = 1 =
=
∆V2′
C1 10 µ F
4