Electrical Energy and Capacitance
2.
57
Because electric forces are conservative, the kinetic energy gained is equal to the decrease in
electrical potential energy, or
)
KE = − PE = − q ( ∆V ) = − ( −1 e ) ( +1.0 × 10 4 V = +1.0 × 10 4 eV
so the correct choice is (a).
3.
From conservation of energy, KE f + PE f = KEi + PEi , or
or
v f = vi2 +
=
1
2
mv 2f =
1
2
mvi2 + qVi − qV f
2q( Vi − V f )
m
2  2 (1.660 × 10 −19 C ) (1.50 − 4.00 ) × 10 3 V 
2
5
6
.
20
10
+
= 3.78 × 10 5 m s
m
s
×
(
)
6.63 × 10 −27 kg
Thus, the correct answer is choice (b).
4.
In a uniform electric field, the change in electric potential is ∆V = − Ex ( ∆x ), giving
Ex = −
(
(
)
)
V f − Vi
(190 V − 120 V)
∆V
=−
=−
= −35 V m = −35 N C
∆x
( 5.0 m − 3.0 m )
x f − xi
and it is seen that the correct choice is (d).
5.
With the given specifications, the capacitance of this parallel plate capacitor will be
C=
κ ∈0 A (1.00 × 10
=
d
2
) ( 8.85 × 10
)
)
C2 N ⋅ m 2 (1.0 cm 2  1 m 2 
 10 4 cm 2 
1.0 × 10 −3 m
−12
= 8.85 × 10 −11 F = 88.5 × 10 −12 F = 88.5 pF
and the correct choice is (a).
6.
The total potential at a point due to a set of point charges qi is V = ∑ kqi ri, where ri is the
i
distance from the point of interest to the location of the charge qi . Note that in this case, the point
at the center of the circle is equidistant from the 4 point charges located on the rim of the circle.
Note also that q2 + q3 + q4 = ( +1.5 − 1.0 − 0.5 ) µC = 0 , so we have
Vcenter =
ke q1 ke q2 ke q3 ke q4 ke
k
kq
+
+
+
= ( q1 + q2 + q3 + q4 ) = e ( q1 + 0 ) = e 1 = V1 = 4.5 × 10 4 V
r
r
r
r
r
r
r
or the total potential at the center of the circle is just that due to the first charge alone, and the correct answer is choice (b).
7.
In a series combination of capacitors, the equivalent capacitance is always less than any individual
capacitance in the combination, meaning that choice (a) is false. Also, for a series combination
of capacitors, the magnitude of the charge is the same on all plates of capacitors in the combination, making both choices (d) and (e) false. The potential difference across the capacitance Ci is
∆Vi = Q Ci , where Q is the common charge on each capacitor in the combination. Thus, the largest potential difference (voltage) appears across the capacitor with the least capacitance, making
choice (b) the correct answer.
8.
Keeping the capacitor connected to the battery means that the potential difference between the
plates is kept at a constant value equal to the voltage of the battery. Since the capacitance of a
parallel plate capacitor is C = κ ∈0 A d , doubling the plate separation d, while holding other
characteristics of the capacitor constant, means the capacitance will be decreased by a factor of 2.
56157_16_ch16_p039-085.indd 57
3/18/08 11:28:10 PM
58
Chapter 16
The energy stored in a capacitor may be expressed as U = 1 2 C ( ∆V ) , so when the potential difference ∆V is held constant while the capacitance is decreased by a factor of 2, the stored energy
decreases by a factor of 2, making (c) the correct choice for this question.
2
9.
When the battery is disconnected, there is no longer a path for charges to use in moving onto or
off of the plates of the capacitor. This means that the charge Q is constant. The capacitance of
a parallel plate capacitor is C = κ ∈0 A d and the dielectric constant is κ ≈ 1 when the capacitor
is air filled. When a dielectric with dielectric constant κ = 2 is inserted between the plates, the
capacitance is doubled C f = 2Ci . Thus, with Q constant, the potential difference between the
plates, ∆V = Q C , is decreased by a factor of 2, meaning that choice (a) is a true statement. The
electric field between the plates of a parallel plate capacitor is E = ∆V d and decreases when ∆V
decreases, making choice (e) false and leaving (a) as the only correct choice for this question.
(
)
10.
Once the capacitor is disconnected from the battery, there is no path for charges to move onto or
off of the plates, so the charges on the plates are constant, and choice (e) can be eliminated. The
capacitance of a parallel plate capacitor is C = κ ∈0 A d , so the capacitance decreases when the
plate separation d is increased. With Q constant and C decreasing, the energy stored in the capacitor, U = Q 2 2C increases, making choice (a) false and choice (b) true. The potential difference
between the plates, ∆V = Q C = Q ⋅ d κ ∈0 A, increases and the electric field between the plates,
E = ∆V d = Q κ ∈0 A, is constant. This means that both choices (c) and (d) are false and leaves
choice (b) as the only correct response.
11.
Capacitances connected in parallel all have the same potential difference across them and the
equivalent capacitance, Ceq = C1 + C2 + C3 + L, is larger than the capacitance of any one of the
capacitors in the combination. Thus, choice (c) is a true statement. The charge on a capacitor is
Q = C ( ∆V ), so with ∆V constant, but the capacitances different, the capacitors all store different
charges that are proportional to the capacitances, making choices (a), (b), (d), and (e) all false.
Therefore, (c) is the only correct answer.
12.
For a series combination of capacitors, the magnitude of the charge is the same on all plates of
capacitors in the combination. Also, the equivalent capacitance is always less than any individual
capacitance in the combination. Therefore, choice (a) is true while choices (b) and (c) are both
false. The potential difference across a capacitor is ∆V = Q C , so with Q constant, capacitors
having different capacitances will have different potential differences across them, with the
largest potential difference being across the capacitor with the smallest capacitance. This means
that choice (d) is false and choice (e) is true. Thus, both choices (a) and (e) are true statements.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
2.
Changing the area will change the capacitance and maximum charge but not the maximum voltage. The question does not allow you to increase the plate separation. You can increase the maximum operating voltage by inserting a material with higher dielectric strength between the plates.
4.
Electric potential V is a measure of the potential energy per unit charge. Electrical potential
energy, PE ! QV, gives the energy of the total charge Q.
6.
A sharp point on a charged conductor would produce a large electric field in the region near the
point. An electric discharge could most easily take place at the point.
56157_16_ch16_p039-085.indd 58
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Electrical Energy and Capacitance
8.
59
There are eight different combinations that use all three capacitors in the circuit. These combinations and their equivalent capacitances are:
 1
1
1
All three capacitors in series - Ceq =  +
+ 
 C1 C2 C3 
−1
All three capacitors in parallel - Ceq = C1 + C2 + C3
One capacitor in series with a parallel combination of the other two:
−1
−1
 1
 1
 1
1
1
1
Ceq = 
+  , Ceq = 
+  , Ceq = 
+ 
 C1 + C2 C3 
 C3 + C1 C2 
 C2 + C3 C1 
−1
One capacitor in parallel with a series combination of the other two:
 CC 
 CC 
 CC 
Ceq =  1 2  + C3 , Ceq =  3 1  + C2, Ceq =  2 3  + C1
 C1 + C2 
 C3 + C1 
 C2 + C 3 
10.
Nothing happens to the charge if the wires are disconnected. If the wires are connected to each
other, the charge rapidly recombines, leaving the capacitor uncharged.
12.
All connections of capacitors are not simple combinations of series and parallel circuits. As an
example of such a complex circuit, consider the network of five capacitors C1, C2, C3, C4, and C5
shown below.
This combination cannot be reduced to a simple equivalent by the techniques of combining series
and parallel capacitors.
14.
The material of the dielectric may be able to withstand a larger electric field than air can withstand
before breaking down to pass a spark between the capacitor plates.
PROBLEM SOLUTIONS
16.1
(a)
Because the electron has a negative charge, it experiences a force in the direction opposite
to the field and, when released from rest, will move in the negative x direction. The work
done on the electron by the field is
)
)
W = Fx ( ∆ x ) = ( qE x ) ∆ 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 work done on the particle
by the field. Thus,
∆PE = −W = −1.92 × 10 −18 J
continued on next page
56157_16_ch16_p039-085.indd 59
3/18/08 10:31:59 PM
60
Chapter 16
(c)
Since the Coulomb force is a conservative force, conservation of energy gives
∆KE + ∆PE = 0, or KE f = me v 2f 2 − ∆PE = 0 − ∆PE, and
16.2
(a)
−2 ( −1.92 × 10 −18 J
−2 ( ∆PE )
=
me
vf =
9.11 × 10
−31
kg
) = 2.05 × 10
6
m s in the −x direction
The change in the electric potential energy is the negative of the work done on the particle
by the field. Thus,
∆PE = −W = −  qEx ( ∆x ) + qEy ( ∆y )
)
)
= −  q ( 0 ) ∆x + ( 5.40 × 10 −6 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 =
16.3
∆PE +5.65 × 10 −4 J
=
= +105 V
+5.40 × 10 −6 C
q
The work done by the agent moving the charge out of the cell is
Winput = −Wfield = − ( − ∆PEe ) = + q ( ∆V )
J

= (1.60 × 10 −19 C  + 90 × 10 −3  = 1.4 × 10 −20 J

C
)
)
(
∆PEe
−1.92 × 10 −17 J
=
= − 3.20 × 10 −19 C
+60.0 J C
V f − Vi
16.4
∆PEe = q ( ∆V ) = q V f − Vi , so q =
16.5
E=
16.6
Since potential difference is work per unit charge ∆V =
∆V
25 000 J C
=
= 1.7 × 10 6 N C
d
1.5 × 10 −2 m
W
, the work done is
q
)
W = q ( ∆V ) = ( 3.6 × 10 5 C ( +12 J C ) = 4.3 × 10 6 J
16.7
∆V
600 J C
=
= 1.13 × 10 5 N C
d
5.33 × 10 −3 m
(a)
E=
(b)
F = q E = (1.60 × 10 −19 C (1.13 × 10 5 N C = 1.80 × 10 −14 N
(c)
W = F ⋅ s cos θ
)
)
)
= (1.80 × 10 −14 N ( 5.33 − 2.90 ) × 10 −3 m  cos 0° = 4.38 × 10 −17 J
56157_16_ch16_p039-085.indd 60
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Electrical Energy and Capacitance
16.8
(a)
61
Using conservation of energy, ∆KE + ∆PE = 0, with KE f = 0 since the particle is
“stopped,” we have
1
1


∆PE = − ∆KE = −  0 − me vi2  = + ( 9.11 × 10 −31 kg ( 2.85 × 10 7 m s


2
2
)
)
2
= + 3.70 × 10 −16 J
The required stopping potential is then
∆V =
(b)
Being more massive than electrons, protons traveling at the same initial speed will have
more initial kinetic energy and require a greater magnitude stopping potentiial .
(c)
Since ∆Vstopping =
∆PE − ∆KE − mv 2 2
=
=
, the ratio of the stopping potential for a proton to
q
q
q
that for an electron having the same initial speed is
∆Vp
∆Ve
16.9
∆PE +3.70 × 10 −16 J
=
= −2.31 × 10 3 V = −2.31 kV
q
−1.60 × 10 −19 C
(a)
=
− mp vi2 2(+ e)
− me vi2 2(− e)
= − mp me
Use conservation of energy
( KE + PEs + PEe ) f = ( KE + PEs + PEe )i
or
∆ ( KE ) + ∆ ( PEs ) + ∆ ( PEe ) = 0
∆( KE ) = 0 since the block is at rest at both beginning and end.
∆ ( PEs ) =
1 2
kxmax − 0, where xmax is the maximum stretch of the spring.
2
∆ ( PEe ) = −W = − (QE ) xmax
Thus, 0 +
1 2
kxmax − (QE ) xmax = 0 , giving
2
x max =
(b)
−6
4
2 QE 2 ( 35.0 × 10 C ) ( 4.86 × 10 V m )
= 4.36 × 10 −2 m = 4.36 cm
=
k
78.0 N m
At equilibrium, ΣF = Fs + Fe = 0, or − kxeq + QE = 0
Therefore,
xeq =
QE 1
= xmax = 2.18 cm
k
2
The amplitude is the distance 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 .
continued on next page
56157_16_ch16_p039-085.indd 61
3/19/08 1:54:16 AM
62
Chapter 16
(c)
1 2
From conservation of energy, ∆ ( KE ) + ∆ ( PEs ) + ∆ ( PEe ) = 0 + kxmax
+ Q ( ∆V ) = 0. Since
2
xmax = 2 A, this gives
∆V = −
16.10
Using ∆y = v0 y t +
0 = v0 y t +
2
k (2 A)
kxmax
=−
2Q
2Q
2
∆V = −
or
2 kA 2
Q
1 2
ay t for the full flight gives
2
1 2
ay t ,
2
ay =
or
−2v0 y
t
Then, using v y2 = v02 y + 2 ay ( ∆y ) for the upward part of the flight gives
( ∆y )max =
0 − v02 y
2 ay
=
(
− v02 y
2 −2 v0 y t
From Newton’s second law, ay =
)
ΣFy
m
=
=
v0 y t
4
=
( 20.1 m s ) ( 4.10 s ) = 20.6 m
4
− mg − qE
qE 

= −g +
 . Equating this to the earlier

m
m
qE  −2 v0 y

, so the electric field strength is
result gives ay = −  g +
=

m
t

 m   2 v0 y
  2.00 kg   2 ( 20.1 m s )
− g = 
E= 
− 9.80 m s 2  = 1.95 × 10 3 N C
 
−6

 q t
5.00 × 10 C  4.10 s


)
Thus, ( ∆V )max = ( ∆ymax ) E = ( 20.6 m ) (1.95 × 10 3 N C = 4.02 × 10 4 V = 40.2 kV
16.11
(a)
9
2
2
−19
ke q (8.99 × 10 N ⋅ m C ) ( −1.60 × 10 C )
−7
VA =
=
= −5.75 × 10 V
rA
0.250 × 10 −2 m
(b)
VB =
9
2
2
−19
ke q (8.99 × 10 N ⋅ m C ) ( −1.60 × 10 C )
=
= −1.92 × 10 −7 V
rB
0.750 × 10 −2 m
∆V = VB − VA = −1.92 × 10 −7 V − ( −5.75 × 10 −7 V ) = +3.83 × 10 −7 V
16.12
(c)
The original electron will be repelled by the negatively charged particle which suddenly
appears at point A. Unless the electron is fixed in place, it will move in the opposite direction, away from points A and B, thereby lowering the potential difference between these
points.
(a)
At the origin, the total potential is
Vorigin =
kq1 kq2
+
r1
r2
 4.50 × 10 −6 C ( −2.24 × 10 −6 C ) 
6
= (8.99 × 10 9 N ⋅ m 2 C2 ) 
+
 = 2.12 × 10 V
−2
1.80 × 10 −2 m 
 1.25 × 10 m
continued on next page
56157_16_ch16_p039-085.indd 62
3/19/08 2:15:55 AM
63
Electrical Energy and Capacitance
(b)
At point B located at (1.50 cm, 0 ), the needed distances are
2
r1 =
2
x B − x1 + yB − y1 =
(1.50 cm )2 + (1.25 cm )2
= 1.95 cm
(1.50 cm )2 + (1.80 cm )2
= 2.34 cm
and
2
r2 =
x B − x2 + yB − y2
2
=
giving
VB =
16.13
(a)
 4.50 × 10 −6 C ( −2.24 × 10 −6 C ) 
ke q1 ke q2
6
+
= (8.99 × 10 9 N ⋅ m C2 ) 
+
 = 1.21 × 10 V
−2
r1
r2
2.34 × 10 −2 m 
 1.95 × 10 m
Calling the 2.00 µC charge q3,
V =∑
i
q q
ke qi
q3 
= ke  1 + 2 +

ri
r12 + r22 
 r1 r2


N ⋅ m 2  8.00 × 10 −6 C 4.00 × 10 −6 C

+
=  8.99 × 10 9
+
2

C   0.060 0 m
0.030 0 m


2 .00 × 10 −6 C
( 0.060 0 )2 + ( 0.030 0 )2


m 
V = 2 .67 × 10 6 V
(b)
Replacing 2.00 × 10 −6 C by − 2.00 × 10 −6 C in part (a) yields
6
V = 2.13 × 10 V
16.14
(
)
W = q ( ∆V ) = q V f − Vi , and
V f = 0 since the 8.00 µC is infinite distance from other charges.

q q  
N ⋅ m 2  2 .00 × 10 −6 C

Vi = ke  1 + 2  =  8.99 × 10 9
+
C2   0.030 0 m
 r1 r2  

4.00 × 10 −6 C
( 0.030 0 )2 + ( 0.0060 0 )2


m 
= 1.135 × 10 6 V
Thus,
56157_16_ch16_p039-085.indd 63
)
)
W = ( 8.00 × 10 −6 C ( 0 − 1.135 × 10 6 V = − 9.08 J
3/19/08 2:16:33 AM
64
16.15
Chapter 16
(a)
V =∑
i
ke qi
ri

N ⋅ m 2   5.00 × 10 −9 C 3.00 × 10 −9 C 
=  8.99 × 10 9
−
= 103 V
C2   0.175 m
0.175 m 

(b)
PE =
ke qi q2
r12
)
)
−9
−9

N ⋅ m 2  ( 5.00 × 10 C ( − 3.00 × 10 C
=  8.99 × 10 9
= − 3.85 × 10 − 7 J
C2 
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).
16.16
The potential at distance r = 0.300 m from a charge Q = + 9.00 × 10 −9 C is
V=
)
)
9
2
2
−9
keQ (8.99 × 10 N ⋅ m C ( 9.00 × 10 C
=
= +270 V
r
0.300 m
Thus, the work required to carry a charge q = 3.00 × 10 −9 C from infinity to this location is
)
W = qV = ( 3.00 × 10 −9 C ( +270 V) = 8.09 × 10 −7 J
16.17
The Pythagorean theorem gives the distance from the midpoint 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 −2 m
Then, the potential at the midpoint of the base is V = ∑ ke qi ri , or
i
)
)
)
−9
−9
−9

N ⋅ m 2   ( −7.00 × 10 C ( −7.00 × 10 C ( +7.00 × 10 C 
V =  8.99 × 10 9
+
+


0.010 0 m
C2   0.010 0 m

15 × 10 −2 m 
= −1.10 × 10 4 V = − 11.0 kV
56157_16_ch16_p039-085.indd 64
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Electrical Energy and Capacitance
16.18
65
Outside the spherical charge distribution, the potential is the same as for a point charge at the
center of the sphere,
V = keQ r , where Q = 1.00 × 10 −9 C
 1 1
Thus, ∆ ( PEe ) = q ( ∆V ) = − ekeQ  − 
 rf ri 
and from conservation of energy, ∆ ( KE ) = − ∆ ( PEe ) ,
or

 1 1 
1
me v 2 − 0 = −  − ekeQ  −  . This gives v =
2
 rf ri  

v=
2 keQe  1 1 
−
, or
me  rf ri 

N ⋅ m2 
2  8.99 × 10 9
(1.00 × 10−9 C (1.60 × 10−19 C  1
C2 


1
−
−31

9.11 × 10 kg
 0.020 0 m 0.030 0 m 
)
)
v = 7.25 × 10 6 m s
16.19
(a)
When the charge configuration consists of only the
two protons ( q1 and q2 in the sketch ) , the potential
energy of the configuration is
)
( 8.99 × 109 N ⋅ m 2 C2 (1.60 × 10−19 C
kqq
PEa = e 1 2 =
6.00 × 10 −15 m
r12
or
(b)
)
2
PEa = 3.84 × 10 −14 J
When the alpha particle ( q3 in the sketch ) is added to the configuration, there are three
distinct pairs of particles, each of which possesses potential energy. The total potential
energy of the configuration is now
)
 ke ( 2 e 2 
ke q1q2 ke q1q3 ke q2 q3
+
+
= PEa + 2 

r12
r13
r23
 r13 
PEb =
where use has been made of the facts that q1q3 = q2 q3 = e ( 2 e ) = 2 e2 and
r13 = r23 =
( 3.00 fm )2 + ( 3.00 fm )2
= 4.24 fm = 4.24 × 10 −15 m. Also, note that the first
term in this computation is just the potential energy computed in part (a). Thus,
PEb = PEa +
4 ke e 2
r13
= 3.84 × 10
−14
J+
)
4 ( 8.99 × 10 9 N ⋅ m 2 C2 (1.60 × 10 −19 C
4.24 × 10 −15 m
)
2
= 2.55 × 10 −13 J
continued on next page
56157_16_ch16_p039-085.indd 65
3/18/08 1:23:37 PM
66
Chapter 16
(c)
If we start with the three-particle system of part (b) and allow the alpha particle to escape to
infinity [thereby returning us to the two-particle system of part (a)], the change in electric
potential energy will 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 mα vα2 2 − 0 = −∆PE, or
vα =
(e)
−2 ( ∆PE )
=
mα
6.64 × 10
−27
kg
)=
8.08 × 10 6 m s
When, starting with the three-particle system, the two protons are both allowed to escape
to infinity, there will be no remaining pairs of particles and hence no remaining 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 identical particles, this
increase in kinetic energy is split equally between them giving
KEproton =
or
16.20
−2 ( −2.17 × 10 −13 J
vp =
1
1
m p v 2p = ( PEb )
2
2
PEb
2.55 × 10 −13 J
=
= 1.24 × 10 7 m s
mp
1.67 × 10 −27 kg
(a)
If a proton and an alpha particle, initially at rest 4.00 fm apart, are released and allowed to
recede to infinity, the final speeds of the two particles will differ because of the difference in
the masses of the particles. Thus, attempting to solve for the final speeds by use of conservation of energy alone leads to a situation of having one equation with two unknowns , and
does not permit a solution.
(b)
In the situation described in part (a) above, one can obtain a second equation with the two
unknown final speeds by using conservation of linear momentum . Then, one would have
two equations which could be solved simultaneously both unknowns.
continued on next page
56157_16_ch16_p039-085.indd 66
3/18/08 1:23:39 PM
67
Electrical Energy and Capacitance
(c)
kq q 
1
 1
 

From conservation of energy:  mα vα2 + m p v 2p  − 0  +  0 − e α p  = 0


ri 
2
 2
 
mα vα2 + m p v 2p =
or
2 ke qα q p
=
ri
)
)
2 ( 8.99 × 10 9 N ⋅ m 2 C2 ( 3.20 × 10 −19 C (1.60 × 10 −19 C
4.00 × 10
−15
m
mα vα2 + m p v 2p = 2.30 × 10 −13 J
yielding
)
[1]
From conservation of linear momentum,
mα vα + m p v p = 0
or
 mp 
vα = 
vp
 mα 
[2]
Substituting Equation [2] into Equation [1] gives
2
 mp  2
mα 
v p + m p v 2p = 2.30 × 10 −13 J
 m 
or
α
 mp

2
−13
 m + 1 m p v p = 2.30 × 10 J
α
and
vp =
(
2.30 × 10 −13 J
=
m p mα + 1 m p
)
(1.67 × 10
−27
2.30 × 10 −13 J
= 1.05 × 10 7 m s
6.64 × 10 −27 +1 (1.67 × 10 −27 kg
)
)
Then, Equation [2] gives the final speed of the alpha particle as
 mp 
 1.67 × 10 −27 kg 
7
6
=
vα = 
v
p
 6.64 × 10 −27 kg  (1.05 × 10 m s = 2.64 × 10 m s
 mα 
)
16.21
V=
keQ
r
so
r=
)
)
9
2
2
−9
71.9 V ⋅ m
keQ ( 8.99 × 10 N ⋅ m C ( 8.00 × 10 C
=
=
V
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 radii are inversely proportional to the potential.
16.22
56157_16_ch16_p039-085.indd 67
By definition, the work required to move a charge from one point to any other point on an equipotential surface is zero. From the definition of work, W = ( F cos θ ) ⋅ s , the work is zero only if
s = 0 or F cos θ = 0. The displacement s cannot be assumed to be zero in all cases. Thus, one
must require that F cos θ = 0. The force F is given by F = qE and neither the charge q nor the
field strength E can be assumed to be zero in all cases. Therefore, the only way the work can be
zero in all cases is if cos θ = 0. But if cos θ = 0, then θ = 90° or the force (and hence the electric
field) must be perpendicular to the displacement s (which is tangent to the surface). That is, the
field must be perpendicular to the equipotential surface at all points on that surface.
3/18/08 1:23:39 PM
68
16.23
Chapter 16
From conservation of energy, ( KE + PEe ) f = ( KE + PEe )i , which gives
0+
keQq 1
= mα vi2 + 0
rf
2
rf =
16.24
(a)
or
rf =
2 keQq 2 ke ( 79 e ) ( 2 e )
=
mα vi2
mα vi2

N ⋅ m2 
2  8.99 × 10 9
(158 ) (1.60 × 10 −19 C
C2 

)
( 6.64 × 10−27 kg ( 2 .00 × 10 7 m s
)
)
2
2
= 2 .74 × 10 −14 m
The distance from any one of the corners of the square to the point at the center is one half
the length of the diagonal of the square, or
r=
a2 + a2 a 2
a
=
=
2
2
2
diagonal
=
2
Since the charges have equal magnitudes and are all the same distance from the center of
the square, they make equal contributions to the total potential. Thus,
Vtotal = 4Vsingle = 4
charge
(b)
keQ
kQ
Q
= 4 e = 4 2 ke
r
a
a 2
The work 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
qQ
Q

W = PEcenter − PE∞ = qVtotal − 0 = q  4 2 ke  = 4 2 ke

a
a
16.25
)
2
6
A 
C2  (1.0 × 10 m
= 1.1 × 10 −8 F
=  8.85 × 10 −12
d 
N ⋅ m 2 
( 800 m )
(a)
C = ∈0
(b)
Qmax = C ( ∆V )max = C ( Emax d )
)
)
= (1.11 × 10 −8 F ( 3.0 × 10 6 N C ( 800 m ) = 27 C
16.26
16.27
Q
27.0 µC
=
= 3.00 µ F
∆V
9.00 V
(a)
C=
(b)
Q = C ( ∆V ) = ( 3.00 µ F ) (12.0 V) = 36.0 µC
(a)
The capacitance of this air filled ( dielectric constant, κ = 1.00 ) parallel plate capacitor is
C=
)
)
−12
2
2
−4
2
k ∈0 A (1.00 ) ( 8.85 × 10 C N ⋅ m ( 2.30 × 10 m
=
= 1.36 × 10 −12 F = 1.36 pF
d
1.50 × 10 −3 m
continued on next page
56157_16_ch16_p039-085.indd 68
3/18/08 1:23:40 PM
Electrical Energy and Capacitance
16.28
16.29
69
)
(b)
Q = C ( ∆V ) = (1.36 × 10 −12 F (12.0 V) = 1.63 × 10 −11 C = 16.3 × 10 −12 C = 16.3 pC
(c)
E=
(a)
Q = C ( ∆V ) = ( 4.00 × 10 −6 F (12.0 V) = 48.0 × 10 −6 C = 48.0 µC
(b)
Q = C ( ∆V ) = ( 4.00 × 10 −6 F (1.50 V) = 6.00 × 10 −6 C = 6.00 µC
(a)
E=
∆V
20.0 V
=
= 1.11 × 10 4 V m = 11.1 kV m directed toward the negative plate
d
1.80 × 10 −3 m
(b)
C=
−12
2
2
−4
2
∈0 A ( 8.85 × 10 C N ⋅ m ( 7.60 × 10 m
=
−3
d
1.80 × 10 m
∆V
12.0 V
=
= 8.00 × 10 3 V m = 8.00 × 10 3 N C
d
1.50 × 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.
16.30
C=
∈0 A
, so
d
)
)
−12
2
2
−12
2
∈0 A ( 8.85 × 10 C N ⋅ m ( 21.0 × 10 m
d=
=
= 3.10 × 10 −9 m
C
60.0 × 10 −15 F
 1Å 
d = ( 3.10 × 10 −9 m )  −10  = 31.0 Å
 10 m 
16.31
(a)
Assuming the capacitor is air-filled (κ = 1), the capacitance is
C=
56157_16_ch16_p039-085.indd 69
)
)
−12
2
2
2
∈0 A ( 8.85 × 10 C N ⋅ m ( 0.200 m
=
= 5.90 × 10 −10 F
d
3.00 × 10 −3 m
)
(b)
Q = C ( ∆V ) = ( 5.90 × 10 −10 F ( 6.00 V) = 3.54 × 10 −9 C
(c)
E=
∆V
6.00 V
=
= 2.00 × 10 3 V m = 2.00 × 10 3 N C
d
3.00 × 10 −3 m
(d)
σ=
Q 3.54 × 10 −9 C
=
= 1.77 × 10 −8 C m 2
A
0.200 m 2
(e)
Increasing the distance separating the plates decreases the capacitance, the charge
stored, and the electric field strength between the plates. This means that all of the
previous answers will be decreased .
3/19/08 1:56:18 AM
70
16.32
Chapter 16
ΣFy = 0 ⇒ T cos 15.0° = mg or T =
mg
cos 15.0°
ΣFx = 0 ⇒ qE = T sin 15.0° = mg tan 15.0°
or
E=
mg tan 15.0°
q
mgd tan 15.0°
q
∆V = Ed =
∆V =
16.33
(a)
( 350 × 10
−6
)
)
kg ( 9.80 m s 2 ( 0.040 0 m ) tan 15.0°
30.0 × 10 −9 C
= 1.23 × 10 3 V = 1.23 kV
Capacitors in a series combination store the same charge, Q = Ceq ( ∆V ), where Ceq is the
equivalent capacitance and ∆V is the potential difference maintained across the series
combination. The equivalent capacitance for the given series combination is
CC
1
1
1
, or Ceq = 1 2 , giving
=
+
C1 + C2
Ceq C1 C2
Ceq =
( 2.50 µF ) ( 6.25 µF ) = 1.79 µF
2.50 µ F + 6.25 µ F
so the charge stored on each capacitor in the series combination is
Q = Ceq ( ∆V ) = (1.79 µ F ) ( 6.00 V) = 10.7 µC
(b)
16.34
(a)
When connected in parallel, each capacitor has the same potential difference, ∆V = 6.00 V,
maintained across it. The charge stored on each capacitor is then
For C1 = 2.50 µ F :
Q1 = C1 ( ∆V ) = ( 2.50 µ F ) ( 6.00 V) = 15.0 µC
For C2 = 6.25 µ F :
Q2 = C2 ( ∆V ) = ( 6.25 µ F ) ( 6.00 V) = 37.5 µC
When connected in series, the equivalent capacitance is 1 = 1 + 1 , or
Ceq C1 C2
Ceq =
(b)
C1C2
( 4.20 µF ) ( 8.50 µF ) = 2.81 µF
=
C1 + C2
4.20 µ F + 8.50 µ F
When connected in parallel, the equivalent capacitance is
Ceq = C1 + C2 = 4.20 µ F + 8.50 µ F = 12.7 µ F
56157_16_ch16_p039-085.indd 70
3/18/08 10:41:19 PM
71
Electrical Energy and Capacitance
16.35
(a)
First, we replace the parallel combination
between points b and c by its equivalent capacitance, Cbc = 2.00 µ F + 6.00 µ F = 8.00 µ F.
Then, we have three capacitors in series
between points a and d. The equivalent
capacitance for this circuit is therefore
1
1
1
1
3
=
+
+
=
Ceq Cab Cbc Ccd 8.00 µ F
giving
(b)
Ceq =
8.00 µ F
= 2.67 µ F
3
The charge stored on each capacitor in the series combination is
Qab = Qbc = Qcd = Ceq ( ∆Vad ) = ( 2.67 µ F ) ( 9.00 V) = 24.0 µC
Q
24.0 µC
Then, note that ∆Vbc = bc =
= 3.00 V . The charge on each capacitor in the original
Cbc 8.00 µ F
circuit is:
(c)
On the 8.00 µ F between a and b:
Q8 = Qab = 24.0 µC
On the 8.00 µ F between c and d:
Q8 = Qcd = 24.0 µC
On the 2.00 µ F between b and c:
Q2 = C2 ( ∆Vbc ) = ( 2.00 µ F ) ( 3.00 V) = 6.00 µC
On the 6.00 µ F between b and c:
Q6 = C6 ( ∆Vbc ) = ( 6.00 µ F ) ( 3.00 V) = 18.0 µC
Note that ∆Vab = Qab Cab = 24.0 µC 8.00 µ F = 3.00 V , and that
∆Vcd = Qcd Ccd = 24.0 µC 8.00 µ F = 3.00 V. We earlier found
that ∆Vbc = 3.00 V, so we conclude that the potential difference
across each capacitor in the circuit is
∆V8 = ∆V2 = ∆V6 = ∆V8 = 3.00 V
Cparallel = C1 + C2 = 9.00 pF ⇒
16.36
1
Cseries
=
1
1
+
C1 C2
⇒
Thus, using Equation [1],
Cseries =
C1 = 9.00 pF − C2
[1]
C1C2
= 2.000 pF
C1 + C2
Cseries =
( 9.00 pF − C2 ) C2
( 9.00 pF − C2 ) + C2
C22 − ( 9.00 pF ) C2 + 18.0 ( pF ) = 0 ,
2
or
= 2 .00 pF, which reduces to
(C2 − 6.00 pF ) (C2 − 3.00 pF ) = 0
Therefore, either C2 = 6.00 pF and, from Equation [1], C1 = 3.00 pF
or
C2 = 3.00 pF and C1 = 6.00 pF.
We conclude that the two capacitances are 3.00 pF and 6.00 pF .
56157_16_ch16_p039-085.indd 71
3/18/08 10:45:04 PM
72
16.37
Chapter 16
(a)
The equivalent capacitance of the series combination in the
upper branch is
1
Cupper
or
=
3.00 mF 6.00 mF
1
1
2 +1
+
=
3.00 µ F 6.00 µ F 6.00 µ F
2.00 mF 4.00 mF
Cupper = 2.00 µ F
Likewise, the equivalent capacitance of the series combination in the lower branch is
1
Clower
=
90.0 V
1
1
2 +1
or Clower = 1.33 µ F
+
=
2.00 µ F 4.00 µ F 4.00 µ F
These two equivalent capacitances are connected in parallel with each other, so the equivalent capacitance for the entire circuit is
Ceq = Cupper + Clower = 2.00 µ F + 1.33 µ F = 3.33 µ F
(b)
Note that the same potential difference, equal to the potential difference of the battery,
exists across both the upper and lower branches. The charge stored on each capacitor in the
series combination in the upper branch is
Q3 = Q6 = Qupper = Cupper ( ∆V ) = ( 2.00 µ F ) ( 90.0 V) = 180 µC
and the charge stored on each capacitor in the series combination in the lower branch is
Q3 = Q6 = Qlower = Clower ( ∆V ) = (1.33 µ F ) ( 90.0 V) = 120 µC
(c)
56157_16_ch16_p039-085.indd 72
The potential difference across each of the capacitors in the circuit is:
∆V2 =
Q2 120 µC
=
= 60.0 V
C2 2.00 µ F
∆V4 =
Q4 120 µC
=
= 30.0 V
C4 4.00 µ F
∆V3 =
Q3 180 µC
=
= 60.0 V
C3 3.00 µ F
∆V6 =
Q6 180 µC
=
= 30.0 V
C6 6.00 µ F
3/18/08 1:23:44 PM
Electrical Energy and Capacitance
16.38
(a)
73
The equivalent capacitance of the series
combination in the rightmost branch of the
circuit is
1
Cright
=
1
1
1+ 3
+
=
24.0 µ F 8.00 µ F 24.0 µ F
Figure P16.38
Cright = 6.00 µ F
or
(b)
The equivalent capacitance of the three
capacitors now connected in parallel with
each other and with the battery is
C eq = 4.00 µ F + 2.00 µ F + 6.00 µ F
= 12.0 µ F
(c)
Diagram 1
The total charge stored in this circuit is
Qtotal = Ceq ( ∆V ) = (12.0 µ F ) ( 36.0 V)
or
Qtotal = 432 µC
Diagram 2
(d)
The charges on the three capacitors shown in Diagram 1 are:
Q4 = C4 ( ∆V ) = ( 4.00 µ F ) ( 36.0 V) = 144 µC
Q2 = C2 ( ∆V ) = ( 2.00 µ F ) ( 36.0 V) = 72 µC
Qright = Cright ( ∆V ) = ( 6.00 µ F ) ( 36.0 V) = 216 µC
Yes. Q4 + Q2 + Qright = Qtotal as it should.
(e)
The charge on each capacitor in the series combination in the rightmost branch of the original
circuit (Figure P16.38) is
Q24 = Q8 = Qright = 216 µC
56157_16_ch16_p039-085.indd 73
(f)
∆V24 =
Q24 216 µC
=
= 9.00 V
C24 24.0 µ F
(g)
∆V8 =
Q8 216 µC
=
= 27.0 V
C8 8.00 µ F
Note that ∆V8 + ∆V24 = ∆V = 36.0 V as it should.
3/18/08 11:43:10 PM
74
Chapter 16
16.39
Figure 1
Figure 2
Figure 3
The circuit may be reduced in steps as shown above.
Using Figure 3,
Then, in Figure 2,
and
Qac = ( 4.00 µ F ) ( 24.0 V) = 9 6.0 µC
( ∆V )ab =
Qac 9 6.0 µC
=
= 16.0 V
Cab 6.00 µ F
( ∆V )bc = ( ∆V )ac − ( ∆V )ab = 24.0 V − 16.0 V = 8.00 V
Finally, using Figure 1,
Q1 = C1 ( ∆V )ab = (1.00 µ F ) (16.0 V) = 16.0 µC
Q5 = ( 5.00 µ F ) ( ∆V )ab = 80.0 µC ,
and
16.40
Q8 = ( 8.00 µ F ) ( ∆V )bc = 64.0 µC
Q4 = ( 4.00 µ F ) ( ∆V )bc = 32.0 µC
From Q = C ( ∆V ), the initial charge of each capacitor is
Q10 = (10.0 µ F ) (12 .0 V) = 120 µC and Qx = C x ( 0 ) = 0
After the capacitors are connected in parallel, the potential difference across each is
∆V ′ = 3.00 V, and the total charge of Q = Q10 + Qx = 120 µC is divided between the two
capacitors as
Q10′ = (10.0 µ F ) ( 3 .00 V) = 30.0 µC and
Qx′ = Q − Q10′ = 120 µC − 30.0 µC = 90.0 µC
Thus, C x =
56157_16_ch16_p039-085.indd 74
Qx′
90.0 µC
=
= 30.0 µ F
∆V ′
3.00 V
3/19/08 2:00:04 AM
Electrical Energy and Capacitance
16.41
(a)
From Q = C ( ∆V ) , Q25 = ( 25.0 µ F ) ( 50.0 V) = 1.25 × 10 3 µC = 1.25 mC
and
(b)
75
Q40 = ( 40.0 µ F ) ( 50.0 V) = 2 .00 × 10 3 µC = 2.00 mC
When the two capacitors are connected in parallel, the equivalent capacitance is
Ceq = C1 + C2 = 25.0 µ F + 40.0 µ F = 65.0 µ F.
Since the negative plate of one was connected to the positive plate of the other, the total
charge stored in the parallel combination is
Q = Q40 − Q25 = 2 .00 × 10 3 µC − 1.25 × 10 3 µC = 750 µC
The potential difference across each capacitor of the parallel combination is
∆V =
Q
750 µC
=
= 11.5 V
Ceq 65.0 µ F
and the final charge stored in each capacitor is
Q25′ = C1 ( ∆V ) = ( 25.0 µ F ) (11.5 V) = 288 µC
and
16.42
(a)
Q40
′ = Q − Q25′ = 750 µC − 288 µC = 462 µC
The original circuit reduces to a single equivalent capacitor in the steps shown below.
 1
1
Cs =  + 
 C1 C2 
−1

1
1 
=
+
 5.00 µ F 10.0 µ F 
−1
= 3.33 µ F
C p1 = Cs + C3 + Cs = 2 ( 3.33 µ F ) + 2 .00 µ F = 8.66 µ F
C p2 = C2 + C2 = 2 (10.0 µ F ) = 20.0 µ F
 1
1 
+
Ceq = 

 C p1 C p 2 
−1

1
1 
=
+
 8.66 µ F 20.0 µ F 
−1
= 6.04 µ F
continued on next page
56157_16_ch16_p039-085.indd 75
3/18/08 1:23:48 PM
76
Chapter 16
(b)
The total charge stored between points a and b is
Qtotal = Ceq ( ∆V )ab = ( 6.04 µ F ) ( 60.0 V) = 362 µC
Then, looking at the third figure, observe that the charges of the series capacitors of that figure are Q p1 = Q p 2 = Qtotal = 362 µC. Thus, the potential difference across the upper parallel
combination shown in the second figure is
( ∆V ) p1 =
Q p1
C p1
=
362 µC
= 41.8 V
8.66 µ F
Finally, the charge on C3 is
Q3 = C3 ( ∆V ) p1 = ( 2 .00 µ F ) ( 41.8 V) = 83.6 µC
16.43
From Q = C ( ∆V ), the initial charge of each capacitor is
Q1 = (1.00 µ F ) (10.0 V) = 10.0 µC
and
Q2 = ( 2 .00 µ F ) ( 0 ) = 0
After the capacitors are connected in parallel, the potential difference across one is the same as
that across the other. This gives
∆V =
Q1′
Q2′
=
1.00 µ F 2 .00 µ F
or
Q2′ = 2 Q1′
[1]
From conservation of charge, Q1′ + Q2′ = Q1 + Q2 = 10.0 µC . Then, substituting from Equation [1],
this becomes
Q1′ + 2 Q1′ = 10.0 µC ,
giving
Finally, from Equation [1],
16.44
Q1′ =
10
µC
3
Q2′ =
20
µC
3
Recognize that the 7.00 µ F and the 5.00 µ F of the
center branch are connected in series. The total capacitance of that branch is
1 
 1
Cs = 
+
 5.00 7.00 
−1
= 2 .92 µ F
Then recognize that this capacitor, the 4.00 µ F
capacitor, and the 6.00 µ F capacitor are all connected
in parallel between points a and b. Thus, the equivalent
capacitance between points a and b is
Ceq = 4.00 µ F + 2 .92 µ F + 6.00 µ F = 12 .9 µ F
56157_16_ch16_p039-085.indd 76
3/18/08 1:23:49 PM
Electrical Energy and Capacitance
16.45
Energy stored =
16.46
(a)
77
Q2 1
1
2
2
= C ( ∆V ) = ( 4.50 × 10 −6 F (12.0 V) = 3.24 × 10 −4 J
2C 2
2
)
The equivalent capacitance of a series combination of C1 and C2 is
1
1
1
2 +1
=
+
=
Ceq 18.0 µ F 36.0 µ F 36.0 µ F
Ceq = 12.0 µ F
or
When this series combination is connected to a 12.0-V battery, the total stored energy is
Total energy stored =
(b)
1
1
2
2
Ceq ( ∆V ) = (12.0 × 10 −6 F (12.0 V) = 8.64 × 10 −4 J
2
2
)
The charge stored on each of the two capacitors in the series combination is
Q1 = Q2 = Qtotal = Ceq ( ∆V ) = (12.0 µ F ) (12.0 V) = 144 µC = 1.44 × 10 −4 C
and the energy stored in each of the individual capacitors is
)
(1.44 × 10−4 C = 5.76 × 10−4 J
Q2
Energy stored in C1 = 1 =
2C1 2 (18.0 × 10 −6 F
and
Energy stored in C2 =
2
)
)
(1.44 × 10−4 C = 2.88 × 10−4 J
Q22
=
2C2 2 ( 36.0 × 10 −6 F
2
)
Energy stored in C1 + Energy stored in C2 = 5.76 × 10 −4 J + 2.88 × 10 −4 J = 8.64 × 10 −4 J ,
which is the same as the total stored energy found in part (a). This must be true
if the computed equivalent capacitance is trruly equivalent to the original combinationn .
(c)
If C1 and C2 had been connected in parallel rather than in series, the equivalent capacitance
would have been Ceq = C1 + C2 = 18.0 µ F + 36.0 µ F = 54.0 µ F . If the total energy stored
2
1
 2 Ceq ( ∆V )  in this parallel combination is to be the same as was stored in the original
series combination, it is necessary that
∆V =
2 ( Total energy stored )
=
Ceq
2 ( 8.64 × 10 −4 J
54.0 × 10 −6 F
)=
5.66 V
Since the two capacitors in parallel have the same potential difference across them, the
2
1
energy stored in the individual capacitors  C ( ∆V )  is directly proportional to their
2

capacitances. The larger capacitor, C2 , stores the most energy in this case.
56157_16_ch16_p039-085.indd 77
3/18/08 1:23:50 PM
78
16.47
Chapter 16
(a)
The energy initially stored in the capacitor is
( Energy stored )1 =
(b)
Qi2
1
1
2
2
= Ci ( ∆V )i = ( 3.00 µ F ) ( 6.00 V) = 54.0 µ J
2 Ci 2
2
When the capacitor is disconnected from the battery, the stored charge becomes isolated
with no way off the plates. Thus, the charge remains constant at the value Qi as long as
the capacitor remains disconnected. Since the capacitance of a parallel plate capacitor
is C = κ ∈0 A d, when the distance d separating the plates is doubled, the capacitance
is decreased by a factor of 2 i.e., C f = Ci 2 = 1.50 µ F . The stored energy (with Q
unchanged) becomes
)
(
 Q2 
Qi2
Qi2
=
= 2  i  = 2 ( Energy stored )1 = 108 µ J
2 C f 2 ( Ci 2 )
 2C f 
( Energy stored )2 =
(c)
When the capacitor is reconnected to the battery, the potential difference between the
plates is reestablished at the original value of ∆V = ( ∆V )i = 6.00 V, while the capacitance
remains at C f = Ci 2 = 1.50 µ F . The energy stored under these conditions is
( Energy stored )3 =
16.48
1
1
2
2
C f ( ∆V )i = (1.50 µ F ) ( 6.00 V) = 27.0 µ J
2
2
The energy transferred to the water is
W =
)
8
1 1
 ( 50.0 C ) (1.00 × 10 V
Q
∆
V
= 2 .50 × 10 7 J
=
( )
200
100  2

Thus, if m is the mass of water boiled away,
W = m  c ( ∆T ) + Lv  becomes


J 
2 .50 × 10 7 J = m  4186
(100°C − 30.0°C ) + 2 .26 × 10 6 J kg 

kg ⋅ °C 


giving
16.49
(a)
m=
2 .50 × 10 7 J
= 9.79 kg
2 .55 J kg
Note that the charge on the plates remains constant at the original value, Q0, as the dielectric is inserted. Thus, the change in the potential difference, ∆V = Q C , is due to a change
in capacitance alone. The ratio of the final and initial capacitances is
Cf
Ci
=
κ ∈0 A d
=κ
∈0 A d
and
Cf
Ci
=
Q0 ( ∆V ) f
Q0 ( ∆V )i
=
( ∆V )i
( ∆V ) f
=
85.0 V
= 3.40
25.0 V
Thus, the dielectric constant of the inserted material is κ = 3.40 , and the material is probably nylon (see Table 16.1).
(b)
56157_16_ch16_p039-085.indd 78
If the dielectric only partially filled the space between the plates, leaving the remaining
space air-filled, the equivalent dielectric constant would be somewhere between κ = 1.00
(air) and κ = 3.40 . The resulting potential difference would then lie somewhere between
( ∆V )i = 85.0 V and ( ∆V ) f = 25.0 V.
3/18/08 1:23:51 PM
Electrical Energy and Capacitance
16.50
(a)
79
The capacitance of the capacitor while air-filled is
C0 =
)
)
−12
2
2
−4
2
∈0 A ( 8.85 × 10 C N ⋅ m ( 25.0 × 10 m
= 1.48 × 10 −12 F = 1.48 pF
=
d
1.50 × 10 −2 m
The original charge stored on the plates is
)
)
Q0 = C0 ( ∆V )0 = (1.48 × 10 −12 F ( 2.50 × 10 2 V = 370 × 10 −12 C = 370 pC
Since distilled water is an insulator, introducing it between the isolated capacitor plates
does not allow the charge to change. Thus, the final charge is Q f = 370 pC .
(b)
After immersion distilled water (κ = 80
see Table 16.1), the new capacitance is
C f = κ C0 = ( 80 ) (1.48 pF ) = 118 pF
and the new potential difference is ( ∆V ) f =
(c)
Qf
Cf
=
370 pC
= 3.14 V .
118 pF
The energy stored in a capacitor is: Energy stored = Q 2 2C. Thus, the change in the stored
energy due to immersion in the distilled water is
Q2  Q2   1
1  ( 3700 × 10
− =
∆E =
− 0 = 0 
2 C f 2 Ci  2   C f Ci 
2
Q 2f
−12
)
2
C 
1
1

−


−12
118 × 10 F 1.48 × 10 −12 F 
= − 4.57 × 10 −8 J = − 45.7 × 10 −9 J = − 45.7 nJ
16.51
(a)
The dielectric constant for Teflon® is κ = 2 .1, so the capacitance is
C=
−12
2
2
−4
2
κ ∈0 A ( 2 .1) ( 8.85 × 10 C N ⋅ m ) (175 × 10 m )
=
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 × 10 6 V m ( 0.040 0 × 10 −3 m
)
Vmax = 2 .40 × 10 3 V = 2 .40 kV
16.52
Before the capacitor is rolled, the capacitance of this parallel plate capacitor is
C=
κ ∈0 A κ ∈0 ( w × L )
=
d
d
where A is the surface area of one side of a foil strip. Thus, the required length is
L=
56157_16_ch16_p039-085.indd 79
)
)
( 9.50 × 10−8 F ( 0.025 0 × 10−3 m
C ⋅d
=
= 1.04 m
κ ∈0 w ( 3.70 ) ( 8.85 × 10 −12 C2 N ⋅ m 2 ( 7.00 × 10 −2 m
)
)
3/18/08 1:23:52 PM
80
16.53
Chapter 16
(a)
V=
m 1.00 × 10 −12 kg
=
= 9.09 × 10 −16 m 3
ρ
1100 kg m 3
13
4π r 3
 3V 
Since V =
, the radius is r =   , and the surface area is
3
 4π 
 3V 
A = 4π r = 4π  
 4π 
2
(b)
C=
)
 3 ( 9.09 × 10 −16 m 3 
= 4π 

4π


2 3
= 4.54 × 10 −10 m 2
κ ∈0 A
d
=
(c)
2 3
( 5.00 ) ( 8.85 × 10 −12
)
C2 N ⋅ m 2 ( 4.54 × 10 −10 m 2
100 × 10
−9
m
)
)=
2 .01 × 10 −13 F
)
Q = C ( ∆V ) = ( 2 .01 × 10 −13 F (100 × 10 −3 V = 2 .01 × 10 −14 C
and the number of electronic charges is
n=
16.54
Q 2 .01 × 10 −14 C
=
= 1.26 × 10 5
e 1.60 × 10 −19 C
Since the capacitors are in parallel, the equivalent capacitance is
Ceq = C1 + C2 + C3 =
or
16.55
Ceq =
∈0 A1 ∈0 A2 ∈0 A3 ∈0 ( A1 + A2 + A3 )
+
+
=
d
d
d
d
∈0 A
where A = A1 + A2 + A3
d
Since the capacitors are in series, the equivalent capacitance is given by
1
1
1
1
d
d
d
d + d2 + d 3
=
+
+
= 1 + 2 + 3 = 1
Ceq C1 C2 C3 ∈0 A ∈0 A ∈0 A
∈0 A
or
16.56
(a)
Ceq =
∈0 A
where d = d1 + d2 + d3
d
Please refer to the solution of Problem 16.37, where the
following results were obtained:
Ceq = 3.33 µ F
Q3 = Q6 = 180 µC
Q2 = Q4 = 120 µC
The total energy stored in the full circuit is then
( Energy stored )total
3.00 mF 6.00 mF
1
1
2
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
2.00 mF 4.00 mF
)
90.0 V
continued on next page
56157_16_ch16_p039-085.indd 80
3/18/08 1:23:52 PM
Electrical Energy and Capacitance
(b)
The energy stored in each individual capacitor is
For 2.00 µ F:
( Energy stored )2 =
−6
−6
( Energy stored )3
For 4.00 µ F:
( Energy stored )4 =
( Energy stored )6
)
(120 × 10 C = 3.60 × 10−3 J = 3.60 mJ
Q22
=
2C2 2 ( 2.00 × 10 −6 F
)
2
)
(180 × 10 C = 5.40 × 10−3 J = 5.40 mJ
Q2
= 3 =
2C3 2 ( 3.00 × 10 −6 F
For 3.00 µ F:
For 6.00 µ F:
(c)
81
2
)
)
(120 × 10 C = 1.80 × 10−3 J = 1.80 mJ
Q42
=
2C4 2 ( 4.00 × 10 −6 F
−6
)
2
)
(180 × 10 C = 2.70 × 10−3 J = 2.70 mJ
Q2
= 6 =
2C6 2 ( 6.00 × 10 −6 F
−6
2
)
The total energy stored in the individual capacitors is
Energy stored = ( 3.60 + 5.40 + 1.80 + 2.70 ) mJ = 13.5 mJ = ( Energy stored )total
Thus, the sums of the energies stored in the individual capacitors equals the total energy
stored by the system.
16.57
In the absence of a dielectric, the capacitance of the
parallel plate capacitor is
∈ A
C0 = 0
d
1
d
3
1
d
3
k
2
k
c1
d
d
With the dielectric inserted,
3
it fills one-third of the gap
2
between the plates as shown
d
c2
3
in sketch (a) at the right.
We model this situation
as consisting of a pair of
(a)
capacitors, C1 and C2 , connected in series as shown
(b)
in sketch (b) at the right. In reality, the lower plate of C1 and
the upper plate of C2 are one and the same, consisting of the
lower surface of the dielectric shown in sketch (a). The capacitances in the model of sketch (b)
are given by:
C1 =
κ ∈0 A 3κ ∈0 A
=
d 3
d
and
C2 =
∈0 A 3 ∈0 A
=
2d 3
2d
and the equivalent capacitance of the series combination is
1
d
2d
 2κ + 1 1
 2κ + 1 d
1
  d   2κ + 1 d
=
=
=
+
=  + 2 
=



  3 ∈0 A   κ  3 ∈0 A  3κ  ∈0 A  3κ  C0
Ceq 3κ ∈0 A 3 ∈0 A  κ
and
56157_16_ch16_p039-085.indd 81
 3κ 
Ceq = 
C
 2κ + 1 0
3/18/08 1:23:53 PM
82
16.58
Chapter 16
For the parallel combination: C p = C1 + C2 which gives
For the series combination:
Thus, we have C2 =
C p − C1 =
1
1
1
=
+
Cs C1 C2
[1]
1
1
1 C1 − Cs
=
−
=
C2 Cs C1
Cs C1
or
Cs C1
and equating this to Equation [1] above gives
C1 − Cs
Cs C1
C1 − Cs
We write this result as :
or
C p C1 − C p Cs − C12 + Cs C1 = Cs C1
C12 − C p C1 + C p Cs = 0
and use the quadratic formula to obtain
Then, Equation [1] gives
16.59
C2 = C p − C1
C2 =
C1 =
1
1 2
Cp ±
C p − C p Cs
2
4
1
1 2
Cp ∓
C p − C p Cs
2
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 combination when this charged capacitor is
connected in parallel with an uncharged 10.0-µ F capacitor. Thus, if ( ∆V )2 is the resulting voltage
across the parallel combination, Q = C p ( ∆V )2 gives
C (100 V) = ( C + 10.0 µ F ) ( 30.0 V)
16.60
( 70.0 V) C = ( 30.0 V) (10.0 µF )
or
and
 30.0 V 
C=
(10.0 µF ) = 4.29 µF
 70.0 V 
(a)
The 1.0-µC is located 0.50 m from point P, so its contribution to the potential at P is
V1 = ke
(b)
 1.0 × 10 −6 C 
q1
= 1.8 × 10 4 V
= ( 8.99 × 10 9 N ⋅ m 2 C2 
r1
 0.50 m 
)
The potential at P due to the −2.0-µC charge located 0.50 m away is
V2 = ke
 − 2.0 × 10 −6 C 
q2
= ( 8.99 × 10 9 N ⋅ m 2 C2 
= − 3.6 × 10 4 V
r2
0.50 m 

)
(c)
The total potential at point P is VP = V1 + V2 = ( +1.8 − 3.6 ) × 10 4 V = − 1.8 × 10 4 V
(d)
The work required to move a charge q = 3.0 µC to point P from infinity is
)
)
W = q∆V = q (VP − V∞) = ( 3.0 × 10 −6 C ( −1.8 × 10 4 V − 0 = − 5.4 × 10 −2 J
56157_16_ch16_p039-085.indd 82
3/18/08 1:23:54 PM
Electrical Energy and Capacitance
16.61
The stages for the reduction of this circuit are shown below.
Thus,
16.62
83
(a)
Ceq = 6.25 µ F
Due to spherical symmetry, the charge on each of the concentric spherical shells will be
uniformly distributed over that shell. Inside a spherical surface having a uniform charge
distribution, the electric field due to the charge on that surface is zero. Thus, in this region,
the potential due to the charge on that surface is constant and equal to the potential at the
surface. Outside a spherical surface having a uniform charge distribution, the potential due
kq
to the charge on that surface is given by V = e , where r is the distance from the center of
r
that surface and q is the charge on that surface.
In the region between a pair of concentric spherical shells, with the inner shell having
charge + Q and the outer shell having radius b and charge − Q, the total electric potential is
given by
V = Vdue to
inner shell
+ Vdue to
outer shell
=
keQ ke ( −Q )
 1 1
+
= keQ  − 
 r b
r
b
The potential difference between the two shells is therefore
 1 1
 b − a
 1 1
∆V = V r = a − V r =b = keQ  −  − keQ  −  = keQ 
 b b
 ab 
 a b
The capacitance of this device is given by
C=
(b)
Q
ab
=
∆V
ke ( b − a )
When b >> a , then b − a ≈ b . Thus, in the limit as b → ∞, the capacitance found above
becomes
C→
16.63
The energy stored in a charged capacitor is W =
∆V =
56157_16_ch16_p039-085.indd 83
ab
a
=
= 4π ∈0 a
ke ( b ) ke
2W
=
C
1
2
C ( ∆V ) . Hence,
2
2 ( 300 J )
= 4.47 × 10 3 V = 4.47 kV
30.0 × 10 −6 F
3/18/08 1:23:56 PM
84
16.64
Chapter 16
From Q = C ( ∆V ), the capacitance of the capacitor with air between the plates is
C0 =
Q0 150 µC
=
∆V
∆V
After the dielectric is inserted, the potential difference is held to the original value, but the charge
changes to Q = Q0 + 200 µC = 350 µC. Thus, the capacitance with the dielectric slab in place is
C=
Q
350 µC
=
∆V
∆V
The dielectric constant of the dielectric slab is therefore
κ=
16.65
C  350 µC   ∆V  350
=
= 2 .33
=

C0  ∆V   150 µC  150
The charges initially stored on the capacitors are
Q1 = C1 ( ∆V )i = ( 6.0 µ F ) ( 250 V) = 1.5 × 10 3 µC
and Q2 = C2 ( ∆V )i = ( 2.0 µ F ) ( 250 V) = 5.0 × 10 2 µC
When the capacitors are connected in parallel, with the negative plate of one connected to the
positive plate of the other, the net stored charge is
Q = Q1 − Q2 = 1.5 × 10 3 µC − 5.0 × 10 2 µC = 1.0 × 10 3 µC
The equivalent capacitance of the parallel combination is Ceq = C1 + C2 = 8.0 µ F. Thus, the final
potential difference across each of the capacitors is
( ∆V )′ =
Q 1.0 × 10 3 µC
=
= 125 V
8.0 µ F
Ceq
and the final charge on each capacitor is
Q1′ = C1 ( ∆V )′ = ( 6.0 µ F ) (125 V) = 750 µC = 0.75 mC
and Q2′ = C2 ( ∆V )′ = ( 2.0 µ F ) (125 V) = 250 µC = 0.25 mC
16.66
The energy required to melt the lead sample is
W = m  cPb ( ∆T ) + L f 
)
= ( 6.00 × 10 −6 kg (128 J kg ⋅ °C ) ( 327.3°C − 20.0°C ) + 24.5 × 10 3 J kg 
= 0.383 J
The energy stored in a capacitor is W =
∆V =
56157_16_ch16_p039-085.indd 84
2W
=
C
1
2
C ( ∆V ) , so the required potential difference is
2
2 ( 0.383 J )
= 121 V
52.0 × 10 −6 F
3/18/08 1:23:56 PM
85
Electrical Energy and Capacitance
16.67
When excess charge resides on a spherical surface that is far removed from any other charge, this
excess charge is uniformly distributed over the spherical surface, and the electric potential at the
surface is the same as if all the excess charge were concentrated at the center of the spherical surface.
In the given situation, we have two charged spheres, initially isolated from each other, with
charges and potentials of: Q1 = + 6.00 µC, V1 = keQ1 R1 where R1 = 12.0 cm, Q2 = − 4.00 µC ,
and V2 = keQ2 R2 with R2 = 18.0 cm.
When these spheres are then connected by a long conducting thread, the charges are redistributed
( yielding charges of Q1′ and Q2′, respectively ) until the two surfaces come to a common potential
(V1′ = kQ1′ R1 = V2′ = kQ2′ R2 ) . When equilibrium is established, we have:
From conservation of charge:
From equal potentials:
Q1′ + Q2′ = Q1 + Q2
kQ1′ kQ2′
=
R1
R2
R 
Q2′ =  2  Q1′
 R1 
⇒
Q1′ =
Substituting Equation [2] into [1] gives:
Then, Equation [2] gives:
16.68
⇒
Q1′ + Q2′ = +2.00 µC
[1]
Q2′ = 1.50 Q1′
[2]
or
+2.00 µC
= 0.800 µC
2.50
Q2′ = 1.50 ( 0.800 µC ) = 1.20 µC
The electric field between the plates is directed downward with magnitude
Ey =
∆V
100 V
=
= 5.00 × 10 4 N m
2.00 × 10 −3 m
d
Since the gravitational force experienced by the electron is negligible in comparison to the electrical force acting on it, the vertical acceleration is
ay =
(a)
Fy
me
=
qEy
me
=
( −1.60 × 10
−19
)
C ( −5.00 × 10 4 N m
9.11 × 10
−31
kg
) = + 8.78 × 10
15
m s2
At the closest approach to the bottom plate, v y = 0. Thus, the vertical displacement from
point O is found from v y2 = v02 y + 2 ay ( ∆y ) as
)
2
−  − ( 5.6 × 10 6 m s sin 45° 
0 − ( v0 sin θ 0 )
∆y =
=
= − 0.89 mm
2 ay
2 ( 8.78 × 1015 m s 2
2
)
The minimum distance above the bottom plate is then
d=
(b)
D
+ ∆y = 1.00 mm − 0.89 mm = 0.11 mm
2
The time for the electron to go from point O to the upper plate is found from
1
∆y = v0 y t + ay t 2 as
2
1
m

 
15 m  2
+1.00 × 10 −3 m =  −  5.6 × 10 6
 sin 45°  t +  8.78 × 10
t

2
s
s2 


Solving for t gives a positive solution of t = 1.11 × 10 −9 s. The horizontal displacement from
point O at this time is
)
)
∆ x = v0 x t = ( 5.6 × 10 6 m s cos 45°  (1.11 × 10 −9 s = 4.4 mm
56157_16_ch16_p039-085.indd 85
3/18/08 1:23:57 PM