Chapter 16
Electrical Energy and
Capacitance
Problem Solutions
16.1
(a) Because the electron has a negative charge, it experiences a force in the d irection
opposite to the field and , w hen released from rest, w ill m ove in the negative xd irection. The w ork d one on the electron by the field is
W
Fx
x
qEx
x
1.60 10
19
C 375 N C
3.20 10
2
m
1.92 10
18
J
(b) The change in the electric potential energy is the negative of the w ork d one on the
particle by the field . Thus,
PE
W
1.92 10
18
J
(c) Since the Coulom b force is a conservative force, conservation of energy gives
KE PE 0 , or conservation of linear momentum. , and
vf
16.2
2
2
PE
me
1.92 10
9.11 10
31
18
J
kg
2.05 106 m s in the
x-direction
(a) The change in the electric potential energy is the negative of the w ork d one on the
particle by the field . Thus,
PE
W
qEx
q 0
x
x
qE y
5.40 10
6
y
C
327 N C
32.0 10
2
m
5.65 10
4
J
(b) The change in the electrical potential is the change in electric potential energy per
unit charge, or
V
PE
q
5.65 10 4 J
+5.40 10-6 C
105 V
51
Electrical Energy and Capacitance
16.3
The w ork d one by the agent m oving the charge out of the cell is
Winput
Wfield
PEe
1.60 10
mp
v p , so q
19
C
PEe
V f Vi
q
V
90 10
3
J
C
1.4 10
1.92 10 17 J
+ 60.0 J C
16.4
v
16.5
E
16.6
Since potential d ifference is w ork per unit charge
m
V
25 000 J C
1.5 10 2 m
d
mp
v
16.7
m
(a) V
ke Q
r
(b) F
qE
(c) W
F s cos
1.60 10
1.80 10
16.8
14
J
3.20 10
19
C
1.7 106 N C
1.67 10
6.64 10
vp
20
19
N
27
27
W
, the w ork d one is
q
V
kg
1.05 107 m s
kg
C 1.13 105 N C
5.33 2.90
10
(a) Using conservation of energy,
KE
3
2.64 106 m s
1.80 10
m cos 0
14
N
4.38 10
PE 0 , w ith KE f
“ stopped ” , w e have
s 0
The required stopping p otential is then
V
PE
q
3.70 10 16 J
1.60 10 19 C
2.31 103 V
2.31 kV
17
J
0 since the particle is
52
Electrical Energy and Capacitance
53
(b) Being m ore m assive than electrons, protons traveling at the sam e initial speed w ill
have m ore initial kinetic energy and require a greater magnitude stopping potential.
(c) Since cos
0 , the ratio of the stopping potential for a proton to that for an electron
having the sam e initial speed is
cos
16.9
0
(a) Use conservation of energy
KE PEs
or
KE
rf
PEe
PEs
2 keQq
m vi2
PEe
N m2
C2
6.64 10
diagonal
2
Vtotal
4Vsingle
charge
Thus, 0
xmax
1 2
kxmax
2
2 QE
k
a2
27
158 1.60 10
7
kg 2.00 10 m s
a2
a 2
2
2
4
ke Q
r
4
2 35.0 10
a
2
ke Q
a
19
C
2
Q
V
m , w here
0 , giving
6
C 4.86 104 V m
V
4.36 10
2
m
4.36 cm
C Emax d
max
1.11 10
C
14
2.74 10
Q
a
78.0 N m
C
2
is the m axim um stretch of the spring.
4 2ke
2
QE xmax
(b) At equilibrium , Qmax
Therefore,
i
since the block is at rest at both beginning and end .
m vi2
rf
PEe
0
2 ke 79e 2e
2 8.99 109
r
KE PEs
f
8
F 3.0 106 N C 800 m
27.0 C
9.00 V
27 C
3.00 F
The am plitud e is the d istance from the equilibrium position to each of the turning
points at x 0 and x 4.36 cm , so A 2.18 cm xmax 2
Electrical Energy and Capacitance
(c) From conservation of energy, Ceq
2.00 103 C 1.25 103 C 750 C , this gives
Q Q40 Q25
2
kxmax
2Q
V
Q C
16.10
Using
y
v0 y t
0 v0 y t
25.0 F+40.0 F 65.0 F . Since
C1 C2
2
k 2A
or
2Q
V
4.00 10
6
F 1.50 V
6.00 10
6
C
6.00 C
1 2
a y t for the full flight gives
2
1 2
a y t , or C p 2
2
C2 C2
2 10.0 F
1
1
C p1
Then, using Ceq
1
Cp2
1
8.66 F
1
20.0 F
20.0 F
1
6.04 F for the upw ard part
of the flight gives
y
0 v02y
max
2ay
v02y
2 2 v0 y t
Q3
16.11
C3
Thus,
V
(a) VA
ke q
rA
(b)
max
V
p1
2 Q1
4
4
mg qE
m
m
2.00 F 41.8 V
ymax E
g
20.6 m
qE
. Equating
m
1 , so the electric field strength is
83.6 C
20.6 m 1.95 103 N C
8.99 109 N m2 C2
0.250 10
1.60 10
2
previous answers will be decreased.
Q2
20.1 m s 4.10 s
Fy
From N ew ton’ s second law , a y
this to the earlier result gives
v0 y t
m
19
C
4.02 104 V
5.75 10
7
40.2 kV
V
54
Electrical Energy and Capacitance
55
(c) The original electron w ill be repelled by the negatively charged particle w hich
sud d enly appears at point A . Unless the electron is fixed in place, it w ill move in the
opposite d irection, aw ay from points A and B, thereby low ering the potential
d ifference betw een these points.
16.12
(a) At the origin, the total potential is
Q1 Q2
Q1 Q2 10.0 C
(b) At point B located at Q1 2 Q1 10.0 C , the need ed d istances are
Q1
10
3
C
xB
x2
and
r2
2
yB
y2
2
1.50 cm
2
1.80 cm
2
2.34 cm
giving
VB
16.13
ke q1
r1
ke q2
r2
9
8.99 10 N m C
2
4.50 10 6 C
1.95 10 2 m
2.24 10
2.34 10
2
6
C
1.21 106 V
m
(a) Calling the 4.00 F charge 6.00 F ,
V
i
ke qi
ri
8.99 109
V
q1
r1
ke
q3
q2
r2
N m2
C2
2
1
r
r22
8.00 10 6 C
0.060 0 m
4.00 10 6 C
0.030 0 m
2.00 10
0.060 0
2
6
C
0.030 0
2.67 106 V
(b) Replacing Q Ceq
1
Ceq
1
18.0 F
V
1.79 F 6.00 V
1
36.0 F
2 1
36.0 F
10.7 C in part (a) yield s
2
m
Electrical Energy and Capacitance
16.14
W
q
V
Vi , and
q Vf
Vf
0 since the 8.00 C is infinite d istance from other charges.
Vi
ke
q1
r1
q2
r2
8.99 109
N m2
C2
2.00 10 6 C
0.030 0 m
4.00 10
0.030 0
2
6
C
0.060 0
2
m
1.135 106 V
Thus, W
16.15
8.00 10
(a) V
i
C 0 1.135 106 V
9.08 J
ke qi
ri
N m2
C2
8.99 109
(b) PE
6
5.00 10 9 C
0.175 m
3.00 10 9 C
0.175 m
103 V
ke qi q2
r12
8.99 109
N m2
C2
5.00 10
9
C
3.00 10
9
C
3.85 10
0.350 m
7
J
The negative sign m eans that positive work must be done to separate the charges
(that is, bring them up to a state of zero potential energy).
16.16
9.00 10 9 C is
The potential at d istance r 0.300 m from a charge Q
V
ke Q
r
8.99 109 N m 2 C 2 9.00 10
9
C
270 V
0.300 m
Thus, the w ork required to carry a charge q 3.00 10
is
W
qV
3.00 10
9
C
270 V
8.09 10
7
J
9
C from infinity to this location
56
Electrical Energy and Capacitance
16.17
57
The Pythagorean theorem gives the d istance from the m id point of the base to the charge
at the apex of the triangle as
r3
4.00 cm
2
1.00 cm
2
15 cm
15 10
Then, the potential at the m id point of the base is V
2
m
ke qi ri , or
i
V
1.10 104 V
16.18
7.00 10
N m2
8.99 10
C2
9
9
C
7.00 10
0.010 0 m
9
C
0.010 0 m
7.00 10
15 10
2
9
C
m
11.0 kV
Outsid e the spherical charge d istribution, the potential is the sam e as for a point charge
at the center of the sphere,
V
Thus,
keQ r , w here Q 1.00 10 9 C
PEe
q
V
ekeQ
1
rf
1
ri
and from conservation of energy
or
1
me v 2 0
2
ekeQ
2 8.99 109
v
v
1
rf
N m2
C2
1
ri
This gives v
1.00 10
9.11 10
7.25 106 m s
PEe ,
KE
31
9
kg
2 keQe 1
me
rf
C 1.60 10
19
C
1
, or
ri
1
0.020 0 m
1
0.030 0 m
Electrical Energy and Capacitance
16.19
58
(a) When the charge configuration consists of only the
tw o protons q1 and q2 in the sketch , the potential
energy of the configuration is
PEa
8.99 109 N m2 C2 1.60 10
ke q1q2
r12
or
6.00 10
PEa
3.84 10
14
15
19
C
2
m
J
(b) When the alpha particle q3 in the sketch is ad d ed to the configuration, there are
three d istinct pairs of particles, each of w hich possesses potential energy. The total
potential energy of the configuration is now
PEb
ke q1q2
r12
ke q1q3
r13
ke q2 q3
r23
PEa
2
w here use has been m ad e of the facts that q1q3
r13
r23
3.00 fm
2
3.00 fm
2
ke 2e2
r13
q2 q3
4.24 fm 4.24 10
e 2e
15
2e2 and
m . Also, note that the first
term in this com putation is just the potential energy com p uted in part (a). Thus,
PEb
4ke e2
r13
PEa
3.84 10
14
4 8.99 109 N m2 C2 1.60 10
J
4.24 10
15
19
C
2
2.55 10
m
(c) If w e start w ith the three-particle system of part (b) and allow the alpha particle to
escape to infinity [thereby returning us to the tw o -particle system of part (a)], the
change in electric potential energy w ill be
PE
PEa
PEb
3.84 10
14
J 2.55 10
13
J
2.17 10
13
J
(d ) Conservation of energy, KE
PE 0 , gives the speed of the alpha particle at
infinity in the situation of part (c) as m v2 2 0
PE , or
v
2
PE
m
2
2.17 10
6.64 10
27
13
kg
J
8.08 106 m s
13
J
Electrical Energy and Capacitance
59
(e) When, starting w ith the three-particle system , the tw o protons are both allow ed to
escape to infinity, there w ill be no rem aining pairs of particles and hence no
rem aining potential energy. Thus, PE 0 PEb
PEb , and conservation of energy
gives the change in kinetic energy as KE
PE
PEb . Since the protons are
id entical particles, this increase in kinetic energy is split equally betw een them
giving
KE proton
or
16.20
1
m p v 2p
2
2.55 10 13 J
1.67 10-27 kg
PEb
mp
vp
1
PEb
2
1.24 107 m s
(a) If a proton and an alpha particle, initially at rest 4.00 fm apart, are released and
allow ed to reced e to infinity, the final speed s of the tw o particles w ill d iffer because
of the d ifference in the m asses of the particles. Thus, attem pting to solve for the
final speed s by use of conservation of energy alone lead s to a situation of having
one equation with two unknowns , and d oes not perm it a solution.
(b) In the situation d escribed in part (a) above, one can obtain a second equation w ith
the tw o unknow n final speed s by using conservation of linear momentum. Then, one
w ould have tw o equations w hich could be solved sim ultaneously both unknow ns.
1
m v2
2
(c) From conservation of energy:
or m v2
mp v 2p
0
0
2 8.99 109 N m2 C2 3.20 10
2ke q q p
ri
m v2
yield ing
1
m p v 2p
2
4.00 10
mp v2p
2.30 10
13
15
ke q q p
ri
19
0
C 1.60 10
19
m
[1]
J
From conservation of linear m om entum ,
mv
mp v p
0
v
or
mp
m
vp
[2]
Substituting Equation [2] into Equation [1] gives
m
mp
m
2
v2p
mp v 2p
2.30 10
13
J
or
mp
m
1 m p v 2p
2.30 10
13
J
C
Electrical Energy and Capacitance
60
and
2.30 10 13 J
mp m 1 mp
vp
13
2.30 10
1.67 10
27
6.64 10
27
J
+1 1.67 10
27
kg
1.05 107 m s
Then, Equation [2] gives the final speed of the alpha particle as
v
16.21
V
ke Q
r
r
ke Q
V
mp
1.67 10
6.64 10
vp
m
27
27
kg
1.05 107 m s
kg
2.64 106 m s
so
8.99 109 N m 2 C 2 8.00 10
9
C
71.9 V m
V
V
For V 100 V, 50.0 V, and 25.0 V,
r
0.719 m, 1.44 m, and 2.88 m
The rad ii are inversely proportional to the potential.
16.22
By d efinition, the w ork required to m ove a charge from one point to any other point on
an equipotential surface is zero. From the d efinition of w ork, W F cos
s , the w ork
is zero only if s 0 or F cos
0 . The d isplacem ent s cannot be assum ed to be zero in
all cases. Thus, one m ust require that F cos
0 . The force F is given by F qE and
neither the charge q nor the field strength E can be assum ed to be zero in all cases.
Therefore, the only w ay the w ork can be zero in all cases is if cos
0 . But if cos
0,
then
90 or the force (and hence the electric field) m ust be perpend icular to the
d isplacem ent s (w hich is tangent to the surface). That is, the field m ust be
perpend icular to the equ ipotential surface at all points on that surface.
16.23
From conservation of energy, KE PEe
0
ke Qq
rf
1
m vi2
2
2 8.99 109
rf
0 or rf
N m2
C2
6.64 10
27
KE PEe i , w hich gives
f
2 keQq
m vi2
2 ke 79e 2e
158 1.60 10
7
kg 2.00 10 m s
m vi2
19
2
C
2
2.74 10
14
m
Electrical Energy and Capacitance
16.24
61
(a) The d istance from any one of the corners of the square to the point at the center is
one half the length of the d iagonal of the square, or
a2
diagonal
2
r
a2
a 2
2
2
a
2
Since the charges have equal m agnitud es and are all the sam e distance from the
center of the square, they m ake equal contributions to the total potential. Thus,
Vtotal
4Vsingle
4
charge
ke Q
r
4
ke Q
a
2
4 2ke
Q
a
(b) The w ork required to carry charge q from infinity to the point at the center of the
square is equal to the increase in the electric potential energy of the charge, or
W
16.25
(a) C
(b)
0
A
d
Qmax
PEcenter
8.85 10
C
V
16.26
(a) C
Q
V
(b) Q C
16.27
C2
N m2
qVtotal
0 q 4 2k e
1.0 106 m2
800 m
Q
a
1.1 10
8
27.0 C
9.00 V
F 3.0 106 N C 800 m
(b) Ceq
(c)
8
qQ
a
F
27 C
3.00 F
V
(a) The capacitance of this air filled dielectric constant,
is
Q40
4 2k e
C Emax d
max
1.11 10
12
PE
40.0 F 50.0 V
C1 C2
Q Q40 Q25
2.00 103 C
2.00 mC
25.0 F+40.0 F 65.0 F
2.00 103 C 1.25 103 C 750 C
1.00 parallel plate capacitor
Electrical Energy and Capacitance
16.28
Q
Ceq
V
(a)
(b) Q C
16.29
750 C
65.0 F
V
4.00 10
(a) Q40
Q Q25
(b) C
0
8.85 10
A
12
F 1.50 V
6.00 10
C3 Cs
C2 C2
16.30
Qtotal
1
C p1
C 2 N m 2 7.60 10
2 3.33 F
2 10.0 F
d
V
A
8.85 10
6.04 F 60.0 V
ab
12
362 C
8.66 F
C p1
Q3
C3
(b) Q C
V
Q1
(d ) Q2
m2
C2 N m 2
V
p1
6.04 F on the other plate.
362 C , so
21.0 10
12
m2
3.10 10 9 m
41.8 V
2.00 F 41.8 V
1.00 F 10.0 V
2.00 F 0
1
1
20.0 F
(a) Assum ing the capacitor is air-filled
(c)
4
2.00 F 8.66 F
60.0 10-15 F
Q p1
p1
6.00 C
20.0 F on one plate and
1
8.66 F
C
V
16.31
1
Cp2
Ceq
0
C
462 C d irected tow ard the negative plate
1
Ceq
6
1.80 10-3 m
d
Cp2
6
750 C 288 C
C p1 Cs
(c)
11.5 V
1 , the capacitance is
83.6 C
10.0 C
0
(e) Increasing the d istance separating the plates d ecreases the capacitance, the charge
stored , and the electric field strength betw een the plates. This m eans that all of the
previous answers will be decreased.
62
Electrical Energy and Capacitance
16.32
Q2
Fx
or
16.33
2 Q1
0
qE T sin15.0
mg tan15.0
Q1 2 Q1 10.0 C
V
V
63
mgd tan15.0
q
Ed
350 10
6
kg 9.80 m s 2 0.040 0 m tan15.0
30.0 10
9
C
1.23 103 V
1.23 kV
(a) Capacitors in a series com bination store the sam e charge, 7.00 F and the 5.00 F ,
w here Ceq is the equivalent capacitance and 4.00 F is the potential d ifference
m aintained across the series com bination. The equivalent capacitance for the given
C1C2
1
1
1
series com bination is
, or Ceq
, giving
Ceq C1 C2
C1 C2
Energy stored
Q2
2C
1
C
2
V
1
4.50 10
2
2
6
F 12.0 V
2
3.24 10
4
J
so the charge stored on each capacitor in the series com bination is
Q Ceq
V
1.79 F 6.00 V
10.7 C
(b) When connected in parallel, each capacitor has the sam e potential difference,
1
1
1
2 1
, m aintained across it. The charge stored on each
Ceq 18.0 F 36.0 F 36.0 F
capacitor is then
16.34
For C1
2.50 F :
Q1 C1
V
2.50 F 6.00 V
15.0 C
For C2
6.25 F :
Q2
V
6.25 F 6.00 V
37.5 C
C2
(a) When connected in series, the equivalent capacitance is
Ceq
C1C2
C1 C2
4.20 F 8.50 F
4.20 F 8.50 F
2.81 F
1
Ceq
1
C1
1
, or
C2
Electrical Energy and Capacitance
64
(b) When connected in parallel, the equivalent capacitance is
if the computed equivalent capacitance is truly equivalent to the original combination.
16.35
(a) First, w e replace the parallel combination
betw een points b and c by its equivalent
capacitance, Cbc 2.00 F 6.00 F 8.00 F .
Then, w e have three cap acitors in series
betw een points a and d . The equivalent
capacitance for this circuit is therefore
1
Ceq
giving
1
Cab
1
Cbc
1
Ccd
3
8.00 F
8.00 F
3
Ceq
2.67 F
(b) The charge store on each capacitor in the series com bination is
Qab
Qbc
Qcd
Then, note that
Ceq
Vbc
Vad
Qbc
Cbc
2.67 F 9.00 V
24.0 C
8.00 F
24.0 C
3.00 V . The charge on each capacitor in the
original circuit is:
On the 8.00 F betw een a and b:
Q8
Qab
24.0 C
On the 8.00 F betw een c and d :
Q8
Qcd
24.0 C
On the 2.00 F betw een b and c:
Q2
C2
Vbc
2.00 F 3.00 V
6.00 C
On the 6.00 F betw een b and c:
Q6
C6
Vbc
6.00 F 3.00 V
18.0 C
1
1
2
2
Cf V i
1.50 F 6.00 V
27.0 J , and that
2
2
50.0 C 1.00 108 V
1 1
W
Q V
2.50 107 J . We earlier found that
100 2
200
Vbc 3.00 V , so w e conclud e that the potential d ifference across each capacitor in
the circuit is
(c) N ote that Energy stored
V8
V2
3
V6
V8
3.00 V
Electrical Energy and Capacitance
16.36
[1]
Q0
V
QC
Cseries
Thus, using Equation [1],
C22
9.00 pF C2 18.0 pF
Therefore, either C2
or
65
V
2
9.00 pF C2 C2
9.00 pF C2
C2
2.00 pF w hich red u ces to
0 , or nylon
6.00 pF and , from Equation [1],
3.40
85.0 V and C1 6.00 pF .
i
We conclud e that the tw o capacitances are 3.00 pF and 6.00 pF .
16.37
(a) The equivalent capacitance of the series com bination
in the upper branch is
Qf
or
370 pC
80
See Table 16.1
Likew ise, the equivalent capacitance of the series com bination in the low er branch
is
Cf
C0
80 1.48 pF
118 pF
or
Clower 1.33 F
These tw o equivalent capacitances are connected in parallel w ith each other, so the
equivalent capacitance for the entire circuit is
Ceq
Cupper
Clower
2.00 F 1.33 F
3.33 F
Electrical Energy and Capacitance
66
(b) N ote that the sam e potential d ifference, equal to the potential d ifference of the
battery, exists across both the upper and low er branches. The charge stored on each
capacitor in the series com bination in the upper branch is
Q3
Q6
Qupper
Cupper
V
2.00 F 90.0 V
180 C
and , the charge stored on each capacitor in the series com bination in the low er
branch is
C
2.1 8.85 10
A
0
12
C2 N m2 175 10
4
m2
0.040 0 10-3 m
d
(c) The potential d ifference across each of the capacitors in the circuit is:
V2
Q2
C2
120 C
2.00 F
60.0 V
V3
Q3
C3
180 C
3.00 F
60.0 V
Q4
C4
V4
0
C
120 C
4.00 F
A
0
d
30.0 V
w L
d
16.38 (a) The equivalent capacitance of the series
com bination in the rightm ost branch of the
circuit is
L
C d
0w
or Cright
9.50 10
8
3.70 8.85 10
12
F 0.025 0 10
2
C N m
2
3
m
7.00 10
2
m
1.04 m
Figure P16.38
6.00 F
(b) The equivalent capacitance of the three
capacitors now connected in parallel w ith
each other and w ith the battery is
Ceq
4.00 F 2.00 F 6.00 F
12.0 F
D iagram 1
(c) The total charge stored in this circuit is
Qtotal
or Qtotal
Ceq
V
432 C
12.0 F 36.0 V
D iagram 2
Electrical Energy and Capacitance
(d ) The charges on the three capacitors show n in Diagram 1 are:
Q C
Q
e
n
Qright
V
2.01 10
13
2.01 10 14 C
1.60 10-19 C
Cright
V
F 100 10-3 V
14
C
1.26 105
6.00 F 36.0 V
Yes. Q4 Q2 Qright
2.01 10
216 C
Qtotal as it should.
(e) The charge on each capacitor in the series combination in the rightm ost branch of
the original circuit (Figure P16.38) is
0
Ceq
(f)
Ceq
(g)
V8
A
where d
d
d1
d2
d3
3.33 F
Q8
C8
216 C
8.00 F
27.0 V
N ote that
V8
V24
16.39
The circuit m ay be red uced in steps as show n above.
Using the Figure 3,
Then, in Figure 2,
and
V
bc
Qac
V
V
ac
ab
4.00 F 24.0 V
96.0 C
Qac
Cab
96.0 C
6.00 F
V
24.0 V 16.0 V 8.00 V
ab
16.0 V
V
36.0 V as it should .
67
Electrical Energy and Capacitance
Finally, using Figure 1,
Energy stored
and
16.40
Q4
4.00 F
From Q C
Q10
Q1 C1
120 10
Q42
2C4
4
V
V
1.00 F 16.0 V
ab
6
2 4.00 10
16.0 C
2
C
6
68
F
1.80 10
3
J
1.80 mJ ,
6.00 F
32.0 C
bc
V , the initial charge of each capacitor is
10.0 F 12.0 V
120 C and C2
After the capacitors are connected in parallel, the potential d ifference across each is
V 3.00 V , and the total charge of Q Q10 Qx 120 C is d ivid ed betw een the tw o
capacitors as
Q10
Qx
90.0 C
3.00 V
(a) From Q C
V , Q25
Q40
and
30.0 C and
Q Q10 120 C 30.0 C 90.0 C
Qx
V
Thus, Cx
16.41
10.0 F 3.00 V
30.0 F
25.0 F 50.0 V
40.0 F 50.0 V
1.25 103 C
2.00 103 C
1.25 mC
2.00 mC
(b) When the tw o capacitors are connected in parallel, the equiv alent capacitance is
Ceq C1 C2 25.0 F+40.0 F 65.0 F .
Since the negative plate of one w as connected to the positive plate of the other, the
total charge stored in the parallel combination is
Q Q40 Q25
2.00 103 C 1.25 103 C 750 C
The potential d ifference across each capacitor of the parallel com bination is
V
Q
Ceq
750 C
65.0 F
11.5 V
Electrical Energy and Capacitance
and the final charge stored in each capacitor is
Q25
and
16.42
C1
V
25.0 F 11.5 V
Q40
Q Q25
288 C
750 C 288 C
462 C
(a) The original circuit red uces to a single equivalent capacitor in the steps show n
below .
Cs
1
C1
C p1 Cs
Cp2
1
C2
1
1
1
5.00 F 10.0 F
C3 Cs
C2 C2
2 3.33 F
2 10.0 F
1
2.00 F 8.66 F
20.0 F
1
Ceq
1
C p1
1
Cp2
3.33 F
1
8.66 F
1
20.0 F
1
6.04 F
69
Electrical Energy and Capacitance
70
(b) The total charge stored betw een points a and b is
Qtotal
Ceq
V
6.04 F 60.0 V
ab
362 C
Then, looking at the third figure, observe that the charges of the series capacitors of
that figure are Qp1 Qp 2 Qtotal 362 C . Thus, the potential d ifference across the
upper parallel com bination show n in the second figure is
V
Q p1
p1
362 C
8.66 F
C p1
41.8 V
Finally, the charge on C3 is
Q3
16.43
From Q C
Q1
C3
V
p1
2.00 F 41.8 V
83.6 C
V , the initial charge of each capacitor is
1.00 F 10.0 V
10.0 C and Q2
2.00 F 0
0
After the capacitors are connected in parallel, the potential d ifference across one is the
sam e as that across the other. This gives
V
Q1
1.00 F
Q2
or Q2
2.00 F
From conservation of charge, Q1 Q2
Equation [1], this becomes
Q1 2 Q1 10.0 C , giving
Finally, from Equation [1],
[1]
2 Q1
Q1 Q2 10.0 C . Then, substituting from
Q1
10
3
C
Q2
20
3
C
Electrical Energy and Capacitance
16.44
Recognize that the 7.00 F and the 5.00 F of the
center branch are connected in series. The total
capacitance of that branch is
1
5.00
Cs
1
7.00
1
2.92 F
Then recognize that this capacitor, the 4.00 F
capacitor, and the 6.00 F capacitor are all connected
in parallel betw een points a and b. Thus, the equivalent
capacitance betw een points a and b is
Ceq
4.00 F 2.92 F+6.00 F
Q2
2C
1
C
2
2
1
4.50 10
2
6
2
Energy stored
16.46
(a) The equivalent capacitance of a series com bination of C1 and C2 is
1
18.0 F
1
36.0 F
2 1
36.0 F
F 12.0 V
4
16.45
1
Ceq
V
12.9 F
or
Ceq
3.24 10
J
12.0 F
When this series com bination is connected to a 12.0-V battery, the total stored
energy is
Total energy stored
1
Ceq
2
V
2
1
12.0 10
2
6
F 12.0 V
2
8.64 10
4
J
(b) The charge stored on each of the tw o capacitors in the series com bination is
Q1
Q2
Qtotal
Ceq
V
12.0 F 12.0 V
144 C 1.44 10
4
and the energy stored in each of the ind ivid ual capacitors is
and
Energy stored in C1
Q12
2C1
Energy stored in C2
Q22
2C2
1.44 10
4
6
2 18.0 10
1.44 10
4
2 36.0 10
2
C
5.76 10
F
C
6
4
J
2
F
2.88 10
4
J
C
71
Electrical Energy and Capacitance
72
Energy stored in C1 Energy stored in C2 5.76 10 4 J 2.88 10 4 J 8.64 10 4 J , w hich
is the sam e as the total stored energy found in part (a). This m ust be true
if the computed equivalent capacitance is truly equivalent to the original combination.
(c) If C1 and C2 had been connected in parallel rather than in series, the equivalent
capacitance w ould have been Ceq C1 C2 18.0 F 36.0 F 54.0 F . If the total
1
2
Ceq V
in this parallel com bination is to be the sam e as w as
2
stored in the original series com bination, it is necessary that
energy stored
V
2 Total energy stored
2 8.64 10
Ceq
54.0 10
6
4
J
5.66 V
F
Since the tw o capacitors in parallel have the same potential d iffer ence across them ,
1
2
C V
the energy stored in the ind ivid ual capacitors
is d irectly proportional
2
to their capacitances. The larger capacitor, C2 , stores the most energy in this case.
16.47
(a) The energy initially stored in the capacitor is
Energy stored
1
Qi2
2Ci
1
Ci
2
V
1
3.00 F 6.00 V
2
2
i
2
54.0 J
(b) When the capacitor is d isconnected from the battery, the stored charge becom es
isolated w ith no w ay off the plates. Thus, the charge rem ains constant at the value
Qi as long as the capacitor rem ains d isconnected . Since the ca pacitance of a parallel
plate capacitor is C
e0 A d , w hen the d istance d separating the plates is d oubled ,
the capacitance is d ecreased by a factor of 2 C f
Ci 2 1.50 F . The stored energy
(w ith Q unchanged ) becom es
Energy stored
2
Qi2
2C f
Qi2
2 Ci 2
2
Qi2
2C f
2 Energy stored
1
108 J
(c) When the capacitor is reconnected to the battery, the potential d ifference betw een
the plates is reestablished at the original value of V
V i 6.00 V , w hile the
capacitance rem ains at C f
Ci 2 1.50 F . The energy stored und er these
cond itions is
Energy stored
3
1
Cf
2
V
2
i
1
1.50 F 6.00 V
2
2
27.0 J
Electrical Energy and Capacitance
16.48
73
The energy transferred to the w ater is
W
1 1
Q
100 2
50.0 C 1.00 108 V
V
2.50 107 J
200
Thus, if m is the m ass of w ater boiled aw ay,
W
m c
T
2.50 107 J m
m
giving
16.49
Lv becom es
4186
2.50 107 J
2.55 J kg
J
100 C 30.0 C
kg C
2.26 106 J kg
9.79 kg
(a) N ote that the charge on the plates rem ains constant at the original value, Q0 , as the
d ielectric is inserted . Thus, the change in the potential d ifference, V Q C , is d ue
to a change in capacitance alone. The ratio of the final and initial capacitances is
Cf
Ci
A d
A d
0
0
and
Cf
Q0
V
Ci
Q0
V
f
i
V
V
Thus, the d ielectric constant of the inserted m aterial is
i
f
85.0 V
25.0 V
3.40
3.40 , and the m aterial is
probably nylon (see Table 16.1).
(b) If the d ielectric only partially filled the space betw een the plates, leaving the
rem aining space air-filled , the equivalent d ielectric constant w ould be som ew here
betw een
1.00 (air) and
3.40 . The resulting potential d ifference w ould then
lie som ew here betw een
V i 85.0 V and
V f 25.0 V .
Electrical Energy and Capacitance
16.50
74
(a) The capacitance of the capacitor w hile air-filled is
C0
12
8.85 10
A
0
d
C2 N m 2
25.0 10
2
1.50 10
4
m2
1.48 10
m
12
F 1.48 pF
The original charge stored on the plates is
Q0
C0
V
1.48 10-12 F 2.50 102 V
0
370 10
12
C
370 pC
Since d istilled w ater is an insulator, introd ucing it betw een the isolated capacitor
plates d oes not allow the charge to change. Thus, the final charge is Q f 370 pC .
(b) After im m ersion d istilled w ater
Cf
C0
80 1.48 pF
See Table 16.1 , the new capacitance is
80
118 pF
V
and the new potential d ifference is
Qf
f
370 pC
118 pF
Cf
3.14 V
(c) The energy stored in a capacitor is: Energy stored Q2 2C . Thus, the change in the
stored energy d ue to im m ersion in the d istilled w ater is
E
Q 2f
2C f
Q02
2Ci
4.57 10
16.51
Q02
2
8
J
1
Cf
370 10
1
Ci
45.7 10
9
C
2.1 8.85 10
A
2
1
118 10
J
12
1
F 1.48 10
12
F
45.7 nJ
2.1 , so the capacitance is
C2 N m2 175 10
4
m2
0.040 0 10-3 m
d
C 8.13 10
12
C
2
(a) The d ielectric constant for Teflon ® is
0
12
9
F
8.13 nF
(b) For Teflon ®, the d ielectric strength is Emax
is
60.0 106 V m , so the m axim um voltage
60.0 106 V m 0.040 0 10-3 m
Vmax
Emax d
Vmax
2.40 103 V
2.40 kV
Electrical Energy and Capacitance
16.52
Before the capacitor is rolled , the capacitance of this parallel plate capacitor is
A
0
C
w L
0
d
d
w here A is the surface area of one sid e of a foil strip. Thus, the required length is
C d
0w
L
16.53
(a) V
8
3.70 8.85 10
12
1.00 10 12 kg
1100 kg m3
m
A 4 r
2
9.09 10
4
23
3V
4
2
C N m
16
m
2
7.00 10
1.04 m
m
m3
3V
4
13
, and the surface area is
3 9.09 10
4
2
3
16
m3
23
4.54 10
4
12
C 2 N m 2 4.54 10
Q C
V
2.01 10
13
9
10
m2
2.01 10
m
F 100 10-3 V
2.01 10
14
C
and the num ber of electronic charges is
n
Q
e
2.01 10 14 C
1.60 10-19 C
1.26 105
Since the capacitors are in parallel, the equivalent capacitance is
Ceq
or
m2
d
100 10
16.54
10
A
0
5.00 8.85 10
(c)
F 0.025 0 10
4 r3
, the rad ius is r
3
Since V
(b) C
9.50 10
Ceq
C1 C2
0
d
A
C3
where A
A1
d
0
A1
A2
d
0
A2
A3
A3
d
0
0
A1
A2
d
A3
13
F
75
Electrical Energy and Capacitance
16.55
Since the capacitors are in series, the equivalent capacitance is given by
or
16.56
1
Ceq
1
C1
Ceq
0
1
C2
A
d
1
C3
where d
d1
0 A
d1
d2
0 A
d2
d3
0 A
d1
d 2 d3
0 A
d3
(a) Please refer to the solution of Problem 16.37 w here the
follow ing results w ere obtained :
Ceq
3.33 F
Q3
Q6 180 C
Q2
Q4 120 C
The total energy stored in the full circuit is then
Energy stored
total
1
1
2
Ceq V
3.33 10 6 F 90.0 V
2
2
1.35 10 2 J 13.5 10 3 J 13.5 mJ
2
(b) The energy stored in each ind ivid ual capacitor is
For 2.00 F :
For 3.00 F :
For 4.00 F :
For 6.00 F :
Energy stored
Energy stored
Energy stored
Energy stored
2
3
4
6
Q22
2C2
Q32
2C3
Q42
2C4
Q62
2C6
120 10
6
2 2.00 10
180 10
6
2 3.00 10
120 10
6
2 4.00 10
180 10
6
2 6.00 10
C
6
F
F
C
6
C
6
3.60 10
3
J
3.60 mJ
5.40 10
3
J
5.40 mJ
1.80 10
3
J
1.80 mJ
2.70 10
3
J
2.70 mJ
2
C
6
2
2
F
2
F
(c) The total energy stored in the ind ivid ual capacitors is
Energy stored= 3.60 5.40 1.80 2.70 mJ
13.5 mJ
Energy stored
total
Thus, the sum s of the energies stored in the ind ivid ual capacitors equals t he total
energy stored by the system .
76
Electrical Energy and Capacitance
77
78
Electrical Energy and Capacitance
16.57
In the absence of a d ielectric, the
capacitance of the parallel plate
0 A
capacitor is C0
d
With the d ielectric inserted , it fills
one-third of the gap betw een the
plates as show n in sketch (a) at the
right. We m od el this situation as
(a)
consisting of a pair of capacitors, C1
and C2 , connected in series as show n in sketch (b) at the right.
In reality, the low er plate of C1 and the upper plate of C2 are
(b)
one and the sam e, consisting of the low er surface of the
d ielectric show n in sketch (a). The capacitances in the m od el of sketch (b) are given by:
0 A
d 3
C1
3
0
A
and
d
0 A
2d 3
C2
3
0
A
2d
and the equivalent capacitance of the series com bination is
1
Ceq
and
16.58
d
3
2d
0
A
Ceq
3
0
1
A
3
2
2
For the series com bination:
Cp
C1
3
0
2
A
1
d
3
2
A
0
1
3
d
0
2
A
1 1
3
C0
C0
1
For the parallel com bination: C p
Thus, w e have
d
C2
Cs C1
C1 Cs
We w rite this result as :
1
Cs
Cs C1
C1 Cs
C1 C2 w hich gives
1
C1
1
C2
or
1
C2
1
Cs
C2
1
C1
[1]
Cp C1
C1 Cs
Cs C1
and equating this to Equ ation [1] above gives
or
C p C1 C pCs
C12 C pC1 C pCs
and use the quad ratic form ula to obtain
C1
C12
CsC1
0
1
Cp
2
1 2
C p C p Cs
4
Cs C1
79
Electrical Energy and Capacitance
Then, Equation [1] gives
16.59
1
Cp
2
C2
1 2
C p C p Cs
4
The charge stored on the capacitor by the battery is
Q C
V
1
C 100 V
This is also the total charge stored in the parallel com bination w hen this charged
capacitor is connected in parallel w ith an uncharged 10.0- F capacitor. Thus, if V
the resulting voltage across the parallel com bination, Q C p
C 100 V
and
16.60
C
C 10.0 F 30.0 V or 70.0 V C
30.0 V
10.0 F
70.0 V
V
2
2
is
gives
30.0 V 10.0 F
4.29 F
(a) The 1.0- C is located 0.50 m from point P, so its contribution to the potential at P is
V1
ke
q1
r1
(b) The potential at P d ue to the
V2
ke
q2
r2
1.0 10 6 C
0.50 m
8.99 109 N m 2 C 2
1.8 104 V
2.0- C charge located 0.50 m aw ay is
2.0 10 6 C
0.50 m
8.99 109 N m 2 C 2
(c) The total potential at point P is VP
V1 V2
3.6 10 4 V
1.8 3.6
104 V
1.8 104 V
(d ) The w ork required to m ove a charge q 3.0 C to point P from infinity is
W
q V
q VP V
3.0 10
6
C
1.8 104 V 0
5.4 10
2
J
Electrical Energy and Capacitance
16.61
The stages for the red uction of this circuit are show n below .
Thus, Ceq
16.62
80
6.25 F
(a) Due to spherical sym m etry, the charge on each of the concentric spherical shells w ill
be uniform ly d istributed over that shell. Insid e a spherical surface having a uniform
charge d istribution, the electric field d ue to the charge on that surface is zero. Thus ,
in this region, the potential d ue to the charge on that surface is constant and equal
to the potential at the surface. Outsid e a spherical surface having a uniform charge
ke q
d istribution, the potential d ue to the charge on that surface is given by V
r
w here r is the d istance from the center of that surface and q is the charge on that
surface.
In the region betw een a pair of concentric spherical shells, w ith the inner shell
having charge Q and the outer shell having rad ius b and charge Q , the total
electric potential is given by
V
Vdue to
Vdue to
inner shell
outer shell
ke Q
r
ke
Q
b
ke Q
1
r
1
b
The potential d ifference betw een the tw o shells is therefore,
V
V
r a
V
r b
ke Q
1
a
1
b
keQ
1
b
The capacitance of this device is given by
C
Q
V
ab
ke b a
1
b
keQ
b a
ab
Electrical Energy and Capacitance
(b) When b a , then b a b . Thus, in the lim it as b
above becom es
ab
ke b
C
16.63
4
0
From Q C
C0
2 300 J
2W
C
, the capacitance found
a
The energy stored in a charged capacitor is W
V
16.64
a
ke
81
30.0 10-6 F
1
C
2
4.47 103 V
V
2
. H ence,
4.47 kV
V , the capacitance of the capacitor w ith air betw een the plates is
Q0
V
150 C
V
After the d ielectric is inserted , the potential d ifference is held to the original value, but
the charge changes to Q Q0 200 C=350 C . Thus, the capacitance w ith the d ielectric
slab in place is
C
Q
V
350 C
V
The d ielectric constant of the d ielectric slab is th erefore
C
C0
16.65
350 C
V
V
150 C
350
150
2.33
The charges initially stored on the capacitors are
and
Q1 C1
V
Q2
V
C2
6.0 F 250 V
i
i
2.0 F 250 V
1.5 103 C
5.0 102 C
When the capacitors are connected in parallel, w ith the negative plate of one connected
to the positive plate of the other, the net stored charge is
Q Q1 Q2 1.5 103 C 5.0 102 C=1.0 103 C
Electrical Energy and Capacitance
The equivalent capacitance of the parallel com bination is Ceq
C1 C2
82
8.0 F . Thus, the
final potential d ifference across each of the capacitors is
1.0 103 C
125 V
8.0 F
Q
Ceq
V
and the final charge on each capacitor is
and
16.66
Q1
C1
V
6.0 F 125 V
750 C
0.75 mC
Q2
C2
V
2.0 F 125 V
250 C
0.25 mC
The energy required to m elt the lead sam ple is
W
m cPb
T
6.00 10
Lf
6
kg
128 J kg C 327.3 C 20.0 C
24.5 103 J kg
0.383 J
The energy stored in a capacitor is W
V
16.67
2W
C
2 0.383 J
52.0 10-6 F
1
C
2
V
2
, so the required potential d ifference is
121 V
When excess charge resid es on a spherical surface that is far rem oved from any other
charge, this excess charge is uniform ly d istributed over the spherical surface, an d the
electric potential at the surface is the sam e as if all the excess charge w ere concentrated
at the center of the spherical surface.
In the given situation, w e have tw o charged spheres, initially isolated from each other,
w ith charges and potentials of: Q1
6.00 C , V1 keQ1 R1 w here R1 12.0 cm ,
Q2
4.00 C , and V2 keQ2 R2 w ith R2 18.0 cm .
Electrical Energy and Capacitance
83
When these spheres are then connected by a long cond ucting thread , the charges are
red istributed yielding charges of Q1 and Q2 respectively until the tw o surfaces com e to a
com m on potential V1
kQ1 R1 V2
From conservation of charge:
kQ1
R1
From equal potentials:
16.68
kQ2 R2 . When equilibrium is established , w e have:
Q1 Q2
Q1 Q2
kQ2
R2
Q1 Q2
R2
Q1
R1
Q2
2.00 C
or
2.00 C
2.50
Substituting Equation [2] into [1] gives:
Q1
Then, Equation [2] gives:
Q2 1.50 0.800 C
Q2 1.50Q1
[1]
[2]
0.800 C
1.20 C
The electric field betw een the plates is d irected dow nw ard w ith m agnitu d e
V
d
Ey
100 V
2.00 10-3 m
5.00 104 N m
Since the gravitational force experienced by the electron is negligible in com parison to
the electrical force acting on it, the vertical acceleration is
ay
Fy
me
1.60 10
qEy
me
19
C
9.11 10
5.00 104 N m
31
kg
(a) At the closest approach to the bottom plate, vy
from point O is found from v
y
0
v0 sin
2 ay
2
0
2
y
2
0y
v
8.78 1015 m s2
2 ay
0 . Thu s, the vertical displacem ent
y as
5.6 106 m s sin 45
2 8.78 1015 m s 2
2
0.89 mm
The m inim um d istance above the bottom plate is then
d
D
2
y 1.00 mm 0.89 mm
0.11 mm
Electrical Energy and Capacitance
(b) The tim e for the electron to go from point O to the upper plate is found from
1 2
y v0 y t
a y t as
2
1.00 10
3
m
5.6 106
m
sin 45 t
s
1
m
8.78 1015 2 t 2
2
s
Solving for t gives a positive solution of t 1.11 10 9 s . The horizontal
d isplacem ent from point O at this tim e is
x v0 xt
5.6 106 m s cos 45
1.11 10
9
s
4.4 mm
84