Download Chapter 24 Capacitance

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Chapter 24
Capacitance
Conceptual Problems
5
•
A parallel-plate capacitor is connected to a battery. The space between
the two plates is empty. If the separation between the capacitor plates is tripled
while the capacitor remains connected to the battery, what is the ratio of the final
stored energy to the initial stored energy?
Determine the Concept The energy stored in a capacitor is given by
U = 12 QV and the capacitance of a parallel-plate capacitor by C = ∈0 A d . We can
combine these relationships, using the definition of capacitance and the condition
that the potential difference across the capacitor is constant, to express U as a
function of d.
Express the energy stored in the
capacitor:
U = 12 QV
Use the definition of capacitance to
express the charge of the capacitor:
Q = CV
Express the capacitance of a
parallel-plate capacitor in terms of
the separation d of its plates:
C=
∈0 A
U=
∈0 AV 2
(1)
d
where A is the area of one plate.
Substituting for Q and C in equation
(1) yields:
2d
1
, tripling the separation of the plates will reduce the energy stored
d
in the capacitor to one-third its previous value. Hence the ratio of the final stored
energy to the initial stored energy is 1 / 3 .
Because U ∝
9
•
A dielectric is inserted between the plates of a parallel-plate capacitor,
completely filling the region between the plates. Air initially filled the region
between the two plates. The capacitor was connected to a battery during the entire
process. True or false:
(a)
(b)
The capacitance value of the capacitor increases as the dielectric is inserted
between the plates.
The charge on the capacitor plates decreases as the dielectric is inserted
between the plates.
Deleted: 69
71
72
(c)
(d)
Chapter 24
The electric field between the plates does not change as the dielectric is
inserted between the plates.
The energy storage of the capacitor decreases as the dielectric is inserted
between the plates.
Determine the Concept The capacitance of the capacitor is given by
κ ∈0 A
, the charge on the capacitor is given by Q = CV , and the energy
C=
d
stored in the capacitor is given by U = 12 CV 2 .
(a) True. As the dielectric material is inserted, κ increases from 1 (air) to its value
for the given dielectric material.
(b) False. Q does not change. C increases and V decreases, but their product
( Q = CV ) is constant.
(c) False. Because E = V/d and V decreases, E must decrease.
(d) False. The energy stored in the capacitor is given by U = 12 CV 2 . With the
battery connected as the dielectric is inserted, V remains constant while Q
increases. Therefore U increases.
(a) Two identical capacitors are connected in parallel. This
11 ••
combination is then connected across the terminals of a battery. How does the
total energy stored in the parallel combination of the two capacitors compare to
the total energy stored if just one of the capacitors were connected across the
terminals of the same battery? (b) Two identical capacitors that have been
discharged are connected in series. This combination is then connected across the
terminals of a battery. How does the total energy stored in the series combination
of the two capacitors compare to the total energy stored if just one of the
capacitors were connected across the terminals of the same battery?
Picture the Problem The energy stored in a capacitor whose capacitance is C and
across which there is a potential difference V is given by U = 12 CV 2 . Let C0
represent the capacitance of the each of the two identical capacitors.
(a) The energy stored in the parallel
system is given by:
U parallel = 12 CeqV 2
When the capacitors are connected in
parallel, their equivalent capacitance
is:
C parallel = C 0 + C 0 = 2C 0
Deleted: (b) False. Because Q = CV,
and C increases, Q must increase.¶
Deleted: True
Deleted: ,
Deleted: where d is the plate
separation.
Deleted: (d) False. The energy stored in
the capacitor is given by
U = 12 CV 2 . Because V is constant
and C increases, U increases.
Deleted: (d) False. The energy storage
of a capacitor is independent of the
presence of dielectric and is given by
U = 12 QV .
Capacitance
Substituting for Ceq and simplifying
U parallel =
1
2
(2C 0 )V 2 = C 0V 2
73
(1)
yields:
If just one capacitor is connected to
the same battery the stored energy is:
Dividing equation (1) by equation
(2) and simplifying yields:
U 1 capacitor = 12 C 0V 2
U parallel
=
U 1 capacitor
(2)
C 0V 2
=2
2
1
2 C 0V
or
U parallel = 2U 1 capacitor
(b) The energy stored in the series
system is given by:
U series = 12 C eqV 2
When the capacitors are connected in
series, their equivalent capacitance
is:
C series = 12 C 0
Substituting for Ceq and simplifying
U series =
1
2
( 12 C0 )V 2 = 14 C0V 2
(3)
yields:
Dividing equation (3) by equation
(2) and simplifying yields:
1
U series
CV2
= 14 0 2 =
U1 capacitor 2 C0V
or
U series =
1
2
1
2
U1 capacitor
Estimation and Approximation
13 •• Disconnect the coaxial cable from a television or other device and
estimate the diameter of the inner conductor and the diameter of the shield.
Assume a plausible value (see Table 24–1) for the dielectric constant of the
dielectric separating the two conductors and estimate the capacitance per unit
length of the cable.
Picture the Problem The outer diameter of a "typical" coaxial cable is about
5 mm, while the inner diameter is about 1 mm. From Table 24-1 we see that a
reasonable range of values for κ is 3-5. We can use the expression for the
capacitance of a cylindrical capacitor to estimate the capacitance per unit length
of a coaxial cable.
74
Chapter 24
The capacitance of a cylindrical
dielectric-filled capacitor is given
by:
2πκ ∈0 L
⎛R ⎞
ln⎜⎜ 2 ⎟⎟
⎝ R1 ⎠
where L is the length of the capacitor,
R1 is the radius of the inner conductor,
and R2 is the radius of the second
(outer) conductor.
Divide both sides by L to obtain an
expression for the capacitance per
unit length of the cable:
κ
C 2πκ ∈0
=
=
L
⎛R ⎞
⎛R ⎞
ln⎜⎜ 2 ⎟⎟ 2k ln⎜⎜ 2 ⎟⎟
⎝ R1 ⎠
⎝ R1 ⎠
C=
If κ = 3:
C
=
L
3
⎛ 2.5 mm ⎞
⎟⎟
2 8.988 × 10 N ⋅ m / C ln⎜⎜
⎝ 0.5 mm ⎠
(
9
2
2
)
≈ 0.1 nF/m
If κ = 5:
C
=
L
5
⎛ 2.5 mm ⎞
⎟⎟
2 8.988 ×10 N ⋅ m / C ln⎜⎜
⎝ 0.5 mm ⎠
(
9
A reasonable range of values for C/L,
corresponding to 3 ≤ κ ≤ 5, is:
2
2
)
0.1nF/m ≤
≈ 0.2 nF/m
C
≤ 0.2 nF/m
L
15 ••
Estimate the capacitance of the Leyden jar shown in the Figure 24-34.
The figure of a man is one-tenth the height of an average man.
Picture the Problem Modeling the Leyden jar as a parallel-plate capacitor, we
can use the equation for the capacitance of a dielectric-filled parallel-plate
capacitor that relates its capacitance to the area A of its plates and their separation
(the thickness of the glass) d to estimate the capacitance of the jar. See Table 24-1
for the dielectric constants of various materials.
The capacitance of a dielectric-filled
parallel-plate capacitor is given by:
C=
κ ∈0 A
d
where κ is the dielectric constant.
Comment [EPM1]: DAVID: The
solution neglects the fact that the
conductor covers the bottom as well
as the sides of the jar.
Capacitance
Let the plate area be the sum of the
area of the lateral surface of the jar
and its base:
Substitute for A and simplify to
obtain:
75
A = Alateral + Abase = 2πRh + πR 2
area
where h is the height of the jar and R is
its inside radius.
C=
=
κ ∈ 0 (2πRh + πR 2 )
d
πκ ∈ 0 R(2h + R )
d
If the glass of the Leyden jar is Bakelite of thickness 2.0 mm and the radius and
height of the jar are 4.0 cm and 40 cm, respectively, then:
⎛
C=
C2 ⎞
⎟ [2(40 cm ) + 4.0 cm]
N ⋅ m 2 ⎟⎠
=
2.0 mm
π (4.9)(4.0 cm )⎜⎜ 8.854 ×10−12
⎝
2.3 nF
The Storage of Electrical Energy
(a) The potential difference between the plates of a 3.00-μF capacitor
19 •
is 100 V. How much energy is stored in the capacitor? (b) How much additional
energy is required to increase the potential difference between the plates from
100 V to 200 V?
Picture the Problem Of the three equivalent expressions for the energy stored in
a charged capacitor, the one that relates U to C and V is U = 12 CV 2 .
(a) Express the energy stored in the
capacitor as a function of C and V:
U = 12 CV 2
Substitute numerical values and
evaluate U:
U=
(b) Express the additional energy
required as the difference between
the energy stored in the capacitor at
200 V and the energy stored at
100 V:
1
2
(3.00 μF)(100 V )2 =
15.0 mJ
ΔU = U (200 V ) − U (100 V )
=
1
2
(3.00 μF)(200 V )2 − 15.0 mJ
= 45.0 mJ
76
Chapter 24
23 ••
An air-gap parallel-plate capacitor that has a plate area of 2.00 m2 and
a separation of 1.00 mm is charged to 100 V. (a) What is the electric field
between the plates? (b) What is the electric energy density between the plates?
(c) Find the total energy by multiplying your answer from Part (b) by the volume
between the plates. (d) Determine the capacitance of this arrangement.
(e) Calculate the total energy from U = 12 CV 2 , and compare your answer with
your result from Part (c).
Picture the Problem Knowing the potential difference between the plates, we
can use E = V/d to find the electric field between them. The energy per unit
volume is given by u = 12 ∈0 E 2 and we can find the capacitance of the parallelplate capacitor using C =∈0 A d .
V
100 V
=
= 100 kV/m
d 1.00 mm
(a) Express the electric field
between the plates in terms of their
separation and the potential
difference between them:
E=
(b) Express the energy per unit
volume in an electric field:
u = 12 ∈0 E 2
Substitute numerical values and
evaluate u:
⎛
C2 ⎞
⎟ (100 kV/m )2
u = 12 ⎜⎜ 8.854 × 10 −12
2 ⎟
N
m
⋅
⎝
⎠
= 44.27 mJ/m 3 = 44.3 mJ/m 3
(c) The total energy is given by:
U = uV = uAd
(
)(
)
= 44.27 mJ/m 3 2.00 m 2 (1.00 mm )
= 88.5 μJ
∈0 A
(d) The capacitance of a parallelplate capacitor is given by:
C=
Substitute numerical values and
evaluate C:
⎛
C2
⎜⎜ 8.854 × 10 −12
N ⋅ m2
⎝
C=
1.00 mm
d
= 17.71 nF = 17.7 nF
(e) The total energy is given by:
U = 12 CV 2
⎞
⎟⎟ 2.00 m 2
⎠
(
)
Capacitance
Substitute numerical values and
evaluate U:
U=
1
2
77
(17.71nF)(100 V )2
= 88.5 μJ, in agreement with (c).
Spherical Capacitors
49 ••
A spherical capacitor consists of a thin spherical shell that has a
radius R1 and a thin, concentric spherical shell that has a radius R2, where R2 > R1.
(a) Show that the capacitance is given by C = 4π∈0R1R2/(R2 – R1). (b) Show that
when the radii of the shells are nearly equal, the capacitance approximately is
given by the expression for the capacitance of a parallel-plate capacitor,
C = ∈0A/d, where A is the area of the sphere and d = R2 – R1.
Picture the Problem We can use the definition of capacitance and the expression
for the potential difference between charged concentric spherical shells to show
that C = 4π ∈ 0 R1 R2 (R2 − R1 ).
(a) Using its definition, relate the
capacitance of the concentric
spherical shells to their charge Q and
the potential difference V between
their surfaces:
C=
Express the potential difference
between the conductors:
⎛1
R − R1
1 ⎞
V = kQ⎜⎜ − ⎟⎟ = kQ 2
R1R2
⎝ R1 R2 ⎠
Substitute for V and simplify to
obtain:
C=
=
(b) Because R2 = R1 + d:
Q
V
R1R2
Q
=
R2 − R1 k (R2 − R1 )
kQ
R1 R2
4π ∈0 R1R2
R2 − R1
R1 R2 = R1 (R1 + d ) = R12 + R1d
≈ R12 = R 2
because d is small.
Substitute for R1R2 and R2 − R1 to
obtain:
C≈
4π ∈0 R 2
∈ A
= 0
d
d
78
Chapter 24
Disconnected and Reconnected Capacitors
A 100-pF capacitor and a 400-pF capacitor are both charged to
53 ••
2.00 kV. They are then disconnected from the voltage source and are connected
together, positive plate to negative plate and negative plate to positive plate.
(a) Find the resulting potential difference across each capacitor. (b) Find the
energy dissipated when the connections are made.
Picture the Problem (a) Just after the two capacitors are disconnected from the
voltage source, the 100-pF capacitor carries a charge of 200 nC and the 400-pF
capacitor carries a charge of 800 nC. After switches S1 and S2 in the circuit are
closed, the capacitors are in parallel between points a and b, and the equivalent
capacitance of the system is C eq = C100 + C 400 . The plates to the right in the
diagram below form a single conductor with a charge of 600 nC, and the plates to
the left form a conductor with charge −Q = −600 nC. The potential difference
across each capacitor is V = Q C eq . In Part (b) we can find the energy dissipated
when the connections are made by subtracting the energy stored in the system
after they are connected from the energy stored in the system before they are
connected.
C1 = 100 pF
+200 nC
C2 = 400 pF
−200 nC
+800 nC
− 800 nC
C1 = 100 pF
+200 nC
−200 nC
a
b
S1
S2
− 800 nC
+800 nC
C2 = 400 pF
(a) When the switches are closed and
the capacitors are connected together,
their initial charges redistribute and
the final charge on the two-capacitor
system is 600 nC and the equivalent
capacitance is 500 pF:
The potential difference across each
capacitor is the potential difference
across the equivalent capacitor:
Ceq = 500 pF
− 600 nC
V=
+600 nC
Q
600 nC
=
= 1.20 kV
Ceq 500 pF
Capacitance
(b) The energy dissipated when the
capacitors are connected is the
difference between the energy stored
after they are connected and the
energy stored before they were
connected:
U dissipated = U before − U after
U before is given by:
U before = U 1 + U 2
79
(1)
= 12 Q1V1 + 12 Q2V2
=
1
2
(Q1 + Q2 )Vi
where Vi is the charging voltage.
U after is given by:
U after = 12 QV
where Q is the total charge stored after
the capacitors have been connected and
V is the voltage found in Part (a).
Substitute for U before and U after in
equation (1) and simplify to obtain:
U dissipated =
1
2
(Q1 + Q2 )Vi − 12 QV
Substitute numerical values and evaluate U dissipated :
U dissipated =
1
2
(200 nC + 800 nC)(2.00 kV ) − 12 (600 nC)(1.20 kV ) =
640 μJ
59 •• Capacitors 1, 2 and 3, have capacitances equal to 2.00 μF, 4.00 μF, and
6.00 μF, respectively. The capacitors are connected in parallel, and the parallel
combination is connected across the terminals of a 200-V source. The capacitors
are then disconnected from both the voltage source and each other, and are
connected to three switches as shown in Figure 24-42. (a) What is the potential
difference across each capacitor when switches S1 and S2 are closed but switch S3
remains open? (b) After switch S3 is closed, what is the final charge on the
leftmost plate of each capacitor? (c) Give the final potential difference across each
capacitor after switch S3 is closed.
Picture the Problem Let lower case qs refer to the charges before S3 is closed
and upper case Qs refer to the charges after this switch is closed. We can use
conservation of charge to relate the charges on the capacitors before S3 is closed
to their charges when this switch is closed. We also know that the sum of the
potential differences around the circuit when S3 is closed must be zero and can
use this to obtain a fourth equation relating the charges on the capacitors after the
switch is closed to their capacitances. Solving these equations simultaneously will
yield the charges q1, q2, and q3. Knowing these charges, we can use the definition
Deleted: Q1
Deleted: Q2
Deleted: Q3
80
Chapter 24
of capacitance to find the potential difference across each of the capacitors.
(a) With S1 and S2 closed, but S3
open, the charges on and the potential
differences across the capacitors do
not change. Hence:
V1 = V2 = V3 = 200 V
(b) When S3 is closed, the charges
can redistribute; express the
conditions on the charges that must
be satisfied as a result of this
redistribution:
q2 − q1 = Q2 − Q1 ,
q3 − q2 = Q3 − Q2 ,
and
q1 − q3 = Q1 − Q3 .
V1 + V2 + V3 = 0
Express the condition on the
potential differences that must be
satisfied when S3 is closed:
where the subscripts refer to the three
capacitors.
Use the definition of capacitance to
eliminate the potential differences:
Q1 Q2 Q3
+
+
=0
C1 C 2 C3
Use the definition of capacitance to
find the initial charge on each
capacitor:
(1)
q1 = C1V = (2.00 μF)(200 V ) = 400 μC ,
q 2 = C 2V = (4.00 μF)(200 V ) = 800 μC ,
and
q3 = C3V = (6.00 μF)(200 V ) = 1200 μC
Let q = q1. Then:
q2 = 2q and q3 = 3q
Express q2 and q3 in terms of q1 and
q:
q 2 = q + q1
(2)
and
q3 = q1 + 2q
(3)
Substitute in equation (1) to obtain:
q1 q + q1 q1 + 2q
+
+
=0
C1
C2
C3
or
q1
q + q1
q + 2q
+
+ 1
=0
2.00 μF 4.00 μF 6.00 μF
Solving for and evaluating q1 gives:
q1 = − 117 q = − 117 (400 μC ) = − 255 μC
Formatted Table
Capacitance
Substitute in equation (2) to obtain:
q 2 = 400 μC − 255 μC = 145 μC
Substitute in equation (3) to obtain:
q3 = −255 μC + 2(400 μC ) = 545 μC
(c) Use the definition of capacitance
to find the potential difference across
each capacitor with S3 closed:
V1 =
q1 − 255 μC
=
= − 127 V ,
2.00 μF
C1
V2 =
q 2 145 μC
=
= 36.3 V ,
C 2 4.00 μF
and
V3 =
81
q3 545 μC
=
= 90.8 V
C3 6.00 μF
Deleted: (a) With S1 and S2 closed, but
S3 open, the charges on and the potential
differences across the capacitors do not
change. Hence:¶
... [1]
General Problems
A parallel combination of two identical 2.00-μF parallel-plate
79 ••
capacitors (no dielectric is in the space between the plates) is connected to a
100-V battery. The battery is then removed and the separation between the plates
of one of the capacitors is doubled. Find the charge on the positively charged plate
of each of the capacitors.
Picture the Problem When the battery is removed, after having initially charged
both capacitors, and the separation of one of the capacitors is doubled, the charge
is redistributed subject to the condition that the total charge remains constant; that
is, Q = Q1 + Q2 where Q is the initial charge on both capacitors and Q2 is the
charge on the capacitor whose plate separation has been doubled. We can use the
conservation of charge during the plate separation process and the fact that,
because the capacitors are in parallel, they share a common potential difference.
Find the equivalent capacitance of
the two 2.00-μF parallel-plate
capacitors connected in parallel:
Use the definition of capacitance to
find the charge on the equivalent
capacitor:
Relate this total charge to charges
distributed on capacitors 1 and 2
when the battery is removed and the
Ceq = 2.00 μF + 2.00 μF = 4.00 μF
Q = CeqV = (4.00 μF)(100 V ) = 400 μC
Q = Q1 + Q2
(1)
82
Chapter 24
separation of the plates of capacitor
2 is doubled:
Because the capacitors are in
parallel:
V1 = V2 and
Q1 Q2
Q
2Q2
=
= 1 2 =
C1 C2 ' 2 C2
C2
Solve for Q1 to obtain:
⎛C ⎞
Q1 = 2⎜⎜ 1 ⎟⎟Q2
⎝ C2 ⎠
Substitute equation (2) in equation
(1) and solve for Q2 to obtain:
Q2 =
Q
2(C1 C2 ) + 1
Substitute numerical values and
evaluate Q2:
Q2 =
400 μC
= 133 μC
2(2.00 μF 2.00 μF) + 1
Substitute numerical values in
equation (1) or equation (2) and
evaluate Q1:
Q1 = 267 μC
(2)
85 ••• An electrically isolated capacitor that has charge Q on its positively
charged plate is partly filled with a dielectric substance as shown in Figure 24-51.
The capacitor consists of two rectangular plates that have edge lengths a and b
and are separated by distance d. The dielectric is inserted into the gap a distance
x. (a) What is the energy stored in the capacitor? Hint: the capacitor can be
modeled as two capacitors connected in parallel. (b) Because the energy of the
capacitor decreases as x increases, the electric field must be doing work on the
dielectric, meaning that there must be an electric force pulling it in. Calculate this
force by examining how the stored energy varies with x. (c) Express the force in
terms of the capacitance and potential difference V between the plates. (d) From
where does this force originate?
Picture the Problem We can model this capacitor as the equivalent of two
capacitors connected in parallel, one with an air gap and other filled with a
dielectric of constantκ. Let the numeral 1 denote the capacitor with the dielectric
material whose constant is κ and the numeral 2 the air-filled capacitor.
(a) Using the hint, express the energy
stored in the capacitor as a function
of the equivalent capacitance Ceq :
The capacitances of the two
capacitors are:
U=
C1 =
Deleted: (a) Using the hint, express the
energy stored in the capacitor as a
function of the equivalent capacitance
2
1Q
2 Ceq
κ ∈ 0 bx
d
Ceq :¶
and C 2 =
∈ 0 b(a − x )
d
... [2]
Capacitance
Because the capacitors are in
parallel, the equivalent
capacitance, C ( x ) , is the sum of C1
and C2:
Substitute for Ceq in the expression
for U and simplify to obtain:
(b) The force exerted by the electric
field is given by:
κ ∈ 0 bx ∈ 0 b(a − x )
C ( x ) = C1 + C 2 =
=
∈0 b
=
∈0 b
U=
d
d
83
d
+
d
(κx + a − x )
[(κ − 1)x + a]
Q2d
2 ∈ 0 b[(κ − 1)x + a ]
dU
dx
⎤
d ⎡1
Q2d
=− ⎢
⎥
dx ⎣ 2 ∈ 0 b[(κ − 1)x + a ]⎦
F =−
{
=−
Q2d d
[(κ − 1)x + a ]−1
2 ∈ 0 b dx
=
(κ − 1)Q 2 d
2
2b ∈ 0 [(κ − 1)x + a ]
}
(κ − 1)Q 2 ⎛⎜ b ∈ 0 ⎞⎟
(c) Rewrite the result in (b) to obtain:
F=
⎝ d ⎠
2
⎛ b∈ ⎞
2
2⎜ 0 ⎟ [(κ − 1)x + a ]
d
⎠
⎝
(κ − 1)Q 2 ⎛⎜ b ∈ 0 ⎞⎟
⎝ d ⎠
=
2
2C eq
=
(κ − 1)b ∈ 0 V 2
2d
Note that this expression is independent
of x.
(d) The force originates from the fringing fields around the edges of the capacitor.
The effect of the force is to pull the polarized dielectric into the space between the
capacitor plates.
84
Chapter 24
Page 81: [1] Deleted
David Mills
(a) With S1 and S2 closed, but S3
open, the charges on and the potential
differences across the capacitors do
not change. Hence:
V1 = V2 = V3 = 200 V
(b) When S3 is closed, the charges
can redistribute; express the
conditions on the charges that must
be satisfied as a result of this
redistribution:
q2 − q1 = Q2 − Q1 ,
q3 − q2 = Q3 − Q2 ,
11/6/2007 1:45:00 PM
and
q1 − q3 = Q1 − Q3 .
V1 + V2 + V3 = 0
Express the condition on the
potential differences that must be
satisfied when S3 is closed:
where the subscripts refer to the three
capacitors.
Use the definition of capacitance to
eliminate the potential differences:
Q1 Q2 Q3
+
+
=0
C1 C 2 C3
Use the definition of capacitance to
find the initial charge on each
capacitor:
q1 = C1V = (2.00 μF)(200 V ) = 400 μC ,
q 2 = C 2V = (4.00 μF)(200 V ) = 800 μC ,
and
q3 = C3V = (6.00 μF)(200 V ) = 1200 μC
Let q = q1. Then:
q2 = 2Q and q3 = 3Q
Express Q2 and Q3 in terms of Q1
and Q:
Q2 = Q + Q1
and
Q3 = Q1 + 2Q
Substitute in equation (1) to obtain:
(1)
(2)
(3)
Q1 Q + Q1 Q1 + 2Q
+
+
=0
C1
C2
C3
or
Q1
Q + Q1 Q1 + 2Q
+
+
=0
2.00 μF 4.00 μF 6.00 μF
Solve for and evaluate Q1 to obtain:
Q1 = − 117 Q = − 117 (400 μC ) = − 255 μC
Substitute in equation (2) to obtain:
Q2 = 400 μC − 255 μC = 145 μC
Substitute in equation (3) to obtain:
Q3 = −255 μC + 2(400 μC ) = 545 μC
(c) Use the definition of capacitance
to find the potential difference across
each capacitor with S3 closed:
Q1 − 255 μC
=
= − 127 V ,
2.00 μF
C1
145 μC
Q
= 36.3 V ,
V2 = 2 =
C2 4.00 μF
V1 =
and
Q3 545 μC
=
= 90.8 V
C3 6.00 μF
Q − 255 μC
= − 127 V ,
V1 = 1 =
2.00 μF
C1
Q
145 μC
= 36.4 V ,
V2 = 2 =
C 2 4.00 μF
V3 =
and
V3 =
Page 82: [2] Deleted
(a) Using the hint, express the energy
stored in the capacitor as a function
of the equivalent capacitance Ceq :
Q3 545 μC
=
= 90.9 V
C3 6.00 μF
David Mills
11/6/2007 2:38:00 PM
2
1Q
2 Ceq
U=
κ ∈0 ax
The capacitances of the two
capacitors are:
C1 =
Because the capacitors are in
parallel, Ceq is the sum of C1 and
Ceq = C1 + C2 =
C2:
Substitute for Ceq in the
expression for U and simplify to
obtain:
d
=
∈0 a
=
∈0 a
U=
d
d
and C2 =
∈0 a(a − x )
d
κ ∈0 ax ∈0 a(a − x )
d
+
(κx + a − x )
[(κ − 1)x + a]
Q 2d
2 ∈0 a[(κ − 1)x + a ]
d
(b) The force exerted by the electric
field is given by:
dU
dx
⎤
d ⎡1
Q 2d
=− ⎢
⎥
dx ⎣ 2 ∈0 a[(κ − 1)x + a ]⎦
F =−
{
=−
Q 2d d
[(κ − 1)x + a ]−1
2 ∈0 a dx
=
(κ − 1)Q 2 d
2
2a ∈0 [(κ − 1)x + a ]
}
(κ − 1)Q 2 ⎛⎜ a ∈0 ⎞⎟
(c) Rewrite the result in (b) to obtain:
F=
⎝ d ⎠
2
⎛ a ∈0 ⎞
2
2⎜
⎟ [(κ − 1)x + a ]
d
⎝
⎠
(κ − 1)Q 2 ⎛⎜ a ∈0 ⎞⎟
⎝ d ⎠
=
2Ceq2
=
(κ − 1)a ∈0 V 2
2d
Note that this expression is independent
of x.
(d) The force originates from the fringing fields around the edges of the capacitor.
The effect of the force is to pull the polarized dielectric into the space between the
capacitor plates.