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Chapter 24
Electrostatic Energy and Capacitance
*1 •
Determine the Concept The capacitance of a parallel-plate capacitor is a function of the
surface area of its plates, the separation of these plates, and the electrical properties of the
matter between them. The capacitance is, therefore, independent of the voltage across the
capacitor. (c) is correct.
2
•
Determine the Concept The capacitance of a parallel-plate capacitor is a function of the
surface area of its plates, the separation of these plates, and the electrical properties of the
matter between them. The capacitance is, therefore, independent of the charge of the
capacitor. (c) is correct.
4
•
Picture the Problem The energy stored in the electric field of a parallel-plate capacitor is
related to the potential difference across the capacitor by U  12 QV .
Relate the potential energy stored in
the electric field of the capacitor to the
potential difference across the
capacitor:
U  12 QV
With Q constant, U is directly proportion al to V . Hence, doubling V doubles U .
*5 ••
Picture the Problem 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:
Use the definition of capacitance to
express the charge of the capacitor:
U  12 QV
Q  CV
Substitute to obtain:
U  12 CV 2
Express the capacitance of a
parallel-plate capacitor in terms of
the separation d of its plates:
C
Substitute to obtain:
Because U 
0 A
d
where A is the area of one plate.
U
0 AV 2
2d
1
, doubling the
d
separation of the plates will reduce
the energy stored in the capacitor to
1/2 its previous value:
(d ) is correct.
6
••
Picture the Problem Let V represent the initial potential difference between the plates, U
the energy stored in the capacitor initially, d the initial separation of the plates, and V , U
, and d  these physical quantities when the plate separation has been doubled. We can
use U  12 QV to relate the energy stored in the capacitor to the potential difference
across it and V = Ed to relate the potential difference to the separation of the plates.
Express the energy stored in the
capacitor before the doubling of the
separation of the plates:
U  12 QV
Express the energy stored in the
capacitor after the doubling of the
separation of the plates:
U'  12 QV'
because the charge on the plates does not
change.
Express the ratio of U to U:
U' V'

U V
Express the potential differences
across the capacitor plates before
and after the plate separation in
terms of the electric field E between
the plates:
V  Ed
and
V'  Ed'
because E depends solely on the charge on
the plates and, as observed above, the
charge does not change during the
separation process.
Substitute to obtain:
U' Ed' d'


U
Ed d
For d  = 2d:
U ' 2d

 2 and (b) is correct
U
d
11 •
(a) False. The capacitance of a parallel-plate capacitor is defined to be the ratio of the
charge on the capacitor to the potential difference across it.
(b) False. The capacitance of a parallel-plate capacitor depends on the area of its plates A,
their separation d, and the dielectric constant  of the material between the plates
according to C   0 A d .
(c) False. As in part (b), the capacitance of a parallel-plate capacitor depends on the area
of its plates A, their separation d, and the dielectric constant  of the material between the
plates according to C   0 A d .
*31 ••
Picture the Problem We can relate the charge Q on the positive plate of the capacitor to
the charge density of the plate  using its definition. The charge density, in turn, is
related to the electric field between the plates according to  0 E and the electric field
can be found from E = V/d. We can use U  12 QV in part (b) to find the increase
in the energy stored due to the movement of the plates.
(a) Express the charge Q on the
positive plate of the capacitor in
terms of the plate’s charge density 
and surface area A:
Relate  to the electric field E
between the plates of the capacitor:
Q  A
 0 E
V
d
Express E in terms of the change in
V as the plates are separated a
distance d:
E
Substitute for  and E to obtain:
Q 0 EA 0 A
Substitute numerical values and evaluate Q:
V
d


Q  8.85  1012 C2 /N  m 2 500 cm 2
100 V
 0.4

cm
(b) Express the change in the
electrostatic energy in terms of the
change in the potential difference:
U  12 QV
Substitute numerical values and
evaluate U:
U 
1
2
11.1nC
11.1nC 100 V 
0.553 J
*55 ••
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  4 0 R1R2 R2  R1 .
Q
V
(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 1 
R  R1
V  kQ    kQ 2
R1R2
 R1 R2 
Substitute to obtain:
C

(b) Because R2 = R1 + d:
Q
R1R2

R  R1 k R2  R1 
kQ 2
R1R2
4 0 R1R2
R2  R1
R1R2  R1 R1  d 
 R12  R1d
 R12  R 2
because d is small.
Substitute to obtain:
C
4 0 R 2
 A
 0
d
d
61 ••
Picture the Problem Let C1 represent the capacitance of the 1.2-F capacitor and C2 the
capacitance of the 2nd capacitor. Note that when they are connected as described in the
problem statement they are in parallel and, hence, share a common potential difference.
We can use the equation for the equivalent capacitance of two capacitors in parallel and
the definition of capacitance to relate C2 to C1 and to the charge stored in and the
potential difference across the equivalent capacitor. In part (b) we can use U  12 CV 2 to
find the energy before and after the connection was made and, hence, the energy lost
when the connection was made.
Q1  C1V  1.2 F30 V  36 C
(a) Using the definition of
capacitance, find the charge on
capacitor C1:
Ceq  C1  C2
Express the equivalent capacitance
of the two-capacitor system and
solve for C2:
and
Using the definition of capacitance,
express Ceq in terms of Q2 and V2:
Ceq 
C2  Ceq  C1
Q2 Q1

V2 V2
where V2 is the common potential
difference (they are in parallel) across the
two capacitors.
Substitute to obtain:
C2 
Q1
 C1
V2
Substitute numerical values and
evaluate C2:
C2 
36 C
 1.2 F  2.40 F
10 V
U  U before  U after
(b) Express the energy lost when the
connections are made in terms of the
energy stored in the capacitors
before and after their connection:
 12 C1V12  12 CeqVf2

1
2
C V
2
1 1
 CeqVf2

Substitute numerical values and evaluate U:


U  12 1.2 F30 V  3.6 F10 V  360 J
107 •••
2
2
Picture the Problem Note that, with switch S closed, C1 and C2 are in parallel and we
can use U closed  12 CeqV 2 and Ceq  C1  C2 to obtain an equation we can solve for C2.
We can use the definition of capacitance to express Q2 in terms of V2 and C2 and
Uopen  12 C1V12  12 C2V22 to obtain an equation from which we can determine V2.
Express the energy stored in the
capacitors after the switch is closed:
U closed  12 CeqV 2
Express the equivalent capacitance
of C1 and C2 in parallel:
Ceq  C1  C2
Substitute to obtain:
U closed 
Solve for C2:
C2 
2U closed
 C1
V2
Substitute numerical values and
evaluate C2:
C2 
2960 J 
 0.2 F  0.100 F
80 V 2
Express the charge on C2 when the
switch is open:
Q2  C2V2
Express the energy stored in the
capacitors with the switch open:
Uopen  12 C1V12  12 C2V22
Solve for V2 to obtain:
Substitute in equation (1) to obtain:
V2 
1
2
C1  C2 V 2
(1)
2U open  C1V12
Q2  C2
C2
2U open  C1V12

C2
 C2 2U open  C1V12

Substitute numerical values and evaluate Q2:
Q2 
0.1 F21440 J   0.2 F40 V 2  
16.0 C