Chapter 16
Electrical Energy and
Capacitance
Problem Solutions
16.1
(a) The work done is W = F ⋅ s cos θ = ( qE ) ⋅ s cos θ , or
W = ( 1.60 × 10 −19 C ) ( 200 N C ) ( 2.00 × 10 −2 m ) cos0° = 6.40 × 10 −19 J
(b) The change in the electrical potential energy is
∆PEe = − W = − 6.40 × 10 −19 J
(c) The change in the electrical potential is
∆V =
16.3
∆ PEe − 6.40 × 10−19 J
=
= − 4.00 V
q
1.60 × 10-19 C
The work done by the agent moving the charge out of the cell is
Winput = −Wfield = − ( −∆PEe ) = +q ( ∆V )
J 

−20
= ( 1.60 × 10−19 C)  + 90 × 10−3
 = 1.4× 10 J
C

25 000 J C
∆V
=
= 1.7 × 106 N C
d
1.5 × 10−2 m
16.5
E=
16.8
From conservation of energy, 1 mv 2f − 0 = q ( ∆V )
2
(a) For the proton,
vf =
(
or v f =
)
2 1.60 × 10−19 C ( − 120 V)
1.67 × 10
−27
kg
2 q( ∆V )
m
= 1.52 × 105 m s
35
36 CHAPTER 16
(b) For the electron,
16.12
vf =
(
)
2 − 1.60 × 10−19 C ( + 120 V)
9.11× 10
− 31
kg
= 6.49 × 106 m s
q
q 
V = V1 + V2 = ke  1 + 2  where r1 = 0.60 m − 0 = 0.60 m , and
 r1 r2 
r2 = 0.60 m − 0.30 m = 0.30 m. Thus,
N ⋅ m 2  3.0 × 10−9 C 6.0× 10−9 C

2
V =  8.99 × 109
+

 = 2.2 × 10 V
2
C
0.60
m
0.30
m



V=∑
i
16.15
(a)
N ⋅ m 2  5.00 × 10−9 C 3.00× 10−9 C

=  8.99 × 109
−

 = 103 V
C2  0.175 m
0.175 m 

PE =
(b)
k e qi
ri
k e qi q2
r12
2
5.00 × 10−9 C) ( − 3.00× 10−9 C)

9 N⋅m (
=  8.99 × 10
= − 3.85× 10−7 J

2
C 
0.350 m

The negative sign means that positive work must be done to separate the charges
(that is, bring them up to a state of zero potential energy).
16.19
From conservation of energy, ( KE + PEe ) f = ( KE + PEe ) i , which gives
0+
ke Qq 1
2 k Qq 2 k ( 79e ) ( 2 e )
= mα v 2i + 0 or rf = e 2 = e
rf
2
mαv i
m αv i2
2
N ⋅ m2 

2  8.99 × 109
( 158) ( 1.60× 10−19 C)

2
C 
rf = 
= 2 .74 × 10−14 m
2
7
−27
( 6.64× 10 kg) ( 2.00× 10 m s)
16.22
(a)
Q = C ( ∆V ) = ( 4.00 × 10−6 F ) ( 12.0 V ) = 48.0 × 10 −6 C = 48.0 µ C
Electrical Energy and Capacitance
(b)
16.23
37
Q = C ( ∆V ) = ( 4.00 × 10 −6 F ) ( 1.50 V ) = 6.00 × 10 −6 C = 6.00 µ C
(a) C = ∈0
6
2
A 
C 2  ( 1.0 × 10 m )
=  8.85 × 10−12
= 1.1 × 10 −8 F
2 
d 
N ⋅ m  ( 800 m)
Qmax = C ( ∆ V ) max = C ( Emax d )
(b)
16.25
= ( 1.11 × 10−8 F) ( 3.0 × 106 N C) ( 800 m) = 27 C
∆V
20.0 V
=
= 1.11 × 10 4 V m = 11.1 kV m directed toward the negative
d
1.80 × 10 -3 m
plate
(a) E =
(b) C = ∈0 A =
d
( 8.85× 10
−12
C2 N ⋅ m2 ) ( 7.60 × 10−4 m2 )
1.80 × 10-3 m
= 3.74 × 10−12 F = 3.74 pF
(c)
Q = C ( ∆V ) = ( 3.74 × 10 −12 F ) ( 20.0 V ) = 7.47 × 10 −11 C = 74.7 pC on one plate and
− 74.7 pC on the other plate.
16.29
(a) For series connection,
1
1
1
CC
=
+
⇒ Ceq = 1 2
C eq C 1 C 2
C1 +C2
 CC 
Q=C eq ( ∆ V ) =  1 2  ∆ V
 C1 + C2 
 ( 0.050 µ F ) ( 0.100 µ F) 
=
 ( 400 V) = 13.3 µ C on each
 0.050 µ F + 0.100 µ F 
(b) Q1 =C1 ( ∆V ) = ( 0.050 µ F ) ( 400 V ) = 20.0 µ C
Q2 =C2 ( ∆V ) = ( 0.100 µ F ) ( 400 V ) = 40.0 µ C
38 CHAPTER 16
16.31
(a) Using the rules for combining capacitors in series and in parallel, the circuit is
reduced in steps as shown below. The equivalent capacitor is shown to be a
2.00 µ F capacitor.
(b) From Figure 3:
From Figure 2:
Qac =Cac ( ∆V ) ac = ( 2.00 µ F ) ( 12.0 V ) = 24.0 µ C
Qab =Qbc = Qac = 24.0 µ C
Thus, the charge on the 3.00 µ F capacitor is Q3 = 24.0 µ C
Continuing to use Figure 2, ( ∆ V ) ab =
and
( ∆ V ) 3 = ( ∆ V ) bc =
Qab 24.0 µ C
=
= 4.00 V
Cab 6.00 µ F
Qbc 24.0 µ C
=
= 8.00 V
Cbc 3.00 µF
From Figure 1, ( ∆V ) 4 = ( ∆V ) 2 = ( ∆V ) ab = 4.00 V
and
Q4 =C4 ( ∆V ) 4 = ( 4.00 µ F ) ( 4.00 V ) = 16.0 µ C
Q2 =C2 ( ∆V ) 2 = ( 2.00 µ F ) ( 4.00 V ) = 8.00 µ C
Electrical Energy and Capacitance
39
16.33
(a) The equivalent capacitance of the upper branch between points a and c in Figure 1
is
Cs =
( 15.0 µF ) ( 3.00 µF )
15.0 µ F + 3.00 µ F
= 2.50 µF
Then, using Figure 2, the total capacitance between points a and c is
Cac = 2.50 µ F+6.00 µ F=8.50 µ F
From Figure 3, the total capacitance is

1
1 
Ceq = 
+

 8.50 µ F 20.0 µ F 
(b)
−1
= 5.96 µ F
Qab = Qac = Qcb = ( ∆V ) ab Ceq
= ( 15.0 V) ( 5.96 µ F ) = 89.5 µ C
Thus, the charge on the 20.0 µ C is
Q20 = Qcb = 89.5 µ C
 89.5 µ C 
 = 10.53 V
 20.0 µ F 
( ∆ V ) ac = ( ∆ V ) ab − ( ∆ V ) bc = 15.0 V − 
Then, Q6 = ( ∆V ) ac ( 6.00 µ F ) = 63.2 µ C and
Q15 = Q3 = ( ∆V ) ac ( 2.50 µ F ) = 26.3 µ C
40 CHAPTER 16
16.35
The circuit may be reduced in steps as shown above.
Using the Figure 3,
Qac = ( 4.00 µ F ) ( 24.0 V ) = 96.0 µ C
Then, in Figure 2, ( ∆ V ) ab =
and
Qac 96.0 µ C
=
= 16.0 V
C ab 6.00 µ F
( ∆V ) bc = ( ∆V ) ac − ( ∆V ) ab = 24.0 V − 16.0 V = 8.00 V
Q1 = C1 ( ∆V ) ab = ( 1.00 µ F ) ( 16.0 V ) = 16.0 µ C
Finally, using Figure 1,
Q5 = ( 5.00 µ F ) ( ∆V ) ab = 80.0 µ C ,
and
16.43
Q8 = ( 8.00 µ F ) ( ∆V ) bc = 64.0 µ C
Q4 = ( 4.00 µ F ) ( ∆V ) bc = 32.0 µ C
The capacitance is
C=
∈0 A
d
( 8.85 × 10
=
−12
C2 N ⋅ m2
) ( 2 .00× 10
-3
5.00 × 10 m
−4
m2 )
= 3.54 × 10−13 F
and the stored energy is
1
1
2
2
W = C ( ∆V ) = ( 3.54 × 10 −13 F ) ( 12.0 V ) = 2.55 × 10 −11 J
2
2
Electrical Energy and Capacitance
16.45
41
The capacitance of this parallel plate capacitor is
6
2
A 
C 2  ( 1.0 × 10 m )
−12
C = ∈0 =  8.85 × 10
= 1.1 × 10 −8 F

d 
N ⋅ m 2  ( 800 m)
With an electric field strength of E = 3.0 × 10 6 N C and a plate separation of d = 800 m ,
the potential difference between plates is
∆V = Ed = ( 3.0 × 106 V m ) ( 800 m ) = 2.4 × 109 V
Thus, the energy available for release in a lightning strike is
2
1
1
2
W = C ( ∆V ) = ( 1.1 × 10 −8 F ) ( 2.4 × 109 V ) = 3.2 × 1010 J
2
2
16.47
The initial capacitance (with air between the plates) is Ci = Q ( ∆V ) i , and the final
capacitance (with the dielectric inserted) is C f = Q ( ∆V ) f where Q is the constant
quantity of charge stored on the plates.
Thus, the dielectric constant is κ =
16.49
Cf
Ci
=
( ∆V ) i
( ∆V ) f
=
100 V
= 4.0
25 V
(a) The dielectric constant for Teflon® is κ = 2.1 , so the capacitance is
−12
2
2
−4
2
κ ∈0 A ( 2 .1 ) ( 8.85 × 10 C N ⋅ m ) ( 175× 10 m )
C=
=
d
0.040 0 × 10-3 m
C = 8.13 × 10 −9 F = 8.13 nF
(b) For Teflon®, the dielectric strength is Emax = 60.0 × 10 6 V m , so the maximum
voltage is
Vmax = Emax d = ( 60.0 × 106 V m ) ( 0.040 0 × 10 -3 m
Vmax = 2.40 × 10 3 V = 2.40 kV
)
42 CHAPTER 16
16.60
From Q = C ( ∆V ) , the capacitance of the capacitor with air between the plates is
C0 =
Q0 150 µ C
=
∆V
∆V
After the dielectric is inserted, the potential difference is held to the original value, but
the charge changes to Q = Q0 + 200 µ C=350 µ C . Thus, the capacitance with the dielectric
slab in place is
C=
Q
350 µ C
=
∆V
∆V
The dielectric constant of the dielectric slab is therefore
κ=
C  350 µ C   ∆V  350
=
= 2 .33

=
C0  ∆ V   150 µ C  150