241
C
APACITANCE AND
D
IELECTRICS
24.1.
I
DENTIFY
:
ab
Q
C
V
=
S
ET
U
P
:
6
1
F
10
F
μ
−
=
E
XECUTE
:
6
4
(7.28
10
F)(25.0 V)
1.82
10
C
182 C
ab
Q
CV
μ
−
−
=
=
×
=
×
=
E
VALUATE
:
One plate has charge
Q
+
and the other has charge
Q
−
.
24.2.
I
DENTIFY
and
S
ET
U
P
:
0
A
C
d
=
P
,
Q
C
V
=
and
V
Ed
=
.
(a)
2
0
0
0.00122 m
3.29 pF
0.00328 m
A
C
d
=
=
=
P
P
(b)
8
12
4.35
10
C
13.2 kV
3.29
10
F
Q
V
C
−
−
×
=
=
=
×
(c)
3
6
13.2
10
V
4.02
10
V/m
0.00328 m
V
E
d
×
=
=
=
×
E
VALUATE
:
The electric field is uniform between the plates, at points that aren't close to the edges.
24.3.
I
DENTIFY
and
S
ET
U
P
:
It is a parallelplate air capacitor, so we can apply the equations of Sections 24.1.
E
XECUTE
:
(a)
6
12
0.148
10
C
so
604 V
245
10
F
ab
ab
Q
Q
C
V
V
C
−
−
×
=
=
=
=
×
(b)
0
so
A
C
d
=
P
(
)(
)
12
3
3
2
2
12
2
2
0
245
10
F
0.328
10
m
9.08
10
m
90.8 cm
8.854
10
C / N m
Cd
A
−
−
−
−
×
×
=
=
=
×
=
×
⋅
P
(c)
6
3
604 V
so
1.84
10
V/m
0.328
10
m
ab
ab
V
V
Ed
E
d
−
=
=
=
=
×
×
(d)
0
so
E
σ
=
P
(
)(
)
6
12
2
2
5
2
0
1.84
10
V/m
8.854
10
C
/ N
m
1.63
10
C/m
E
σ
−
−
=
=
×
×
⋅
=
×
P
E
VALUATE
:
We could also calculate
σ
directly as
Q/A
.
6
5
2
3
2
0.148
10
C
1.63
10
C/m ,
9.08
10
m
Q
A
σ
−
−
−
×
=
=
=
×
×
which checks.
24.4.
I
DENTIFY
:
0
A
C
d
=
P
when there is air between the plates.
S
ET
U
P
:
2
2
(3.0
10
m)
A
−
=
×
is the area of each plate.
E
XECUTE
:
12
2
2
12
3
(8.854
10
F/m)(3.0
10
m)
1.59
10
F
1.59 pF
5.0
10
m
C
−
−
−
−
×
×
=
=
×
=
×
E
VALUATE
:
C
increases when
A
increases and
C
increases when
d
decreases.
24.5.
I
DENTIFY
:
ab
Q
C
V
=
.
0
A
C
d
=
P
.
S
ET
U
P
:
When the capacitor is connected to the battery,
12.0 V
ab
V
=
.
E
XECUTE
:
(a)
6
4
(10.0
10
F)(12.0 V)
1.20
10
C
120 C
ab
Q
CV
μ
−
−
=
=
×
=
×
=
(b)
When
d
is doubled
C
is halved, so
Q
is halved.
60 C
Q
μ
=
.
(c)
If
r
is doubled,
A
increases by a factor of 4.
C
increases by a factor of 4 and
Q
increases by a factor of 4.
480 C.
Q
μ
=
E
VALUATE
:
When the plates are moved apart, less charge on the plates is required to produce the same potential
difference. With the separation of the plates constant, the electric field must remain constant to produce the same
potential difference. The electric field depends on the surface charge density,
σ
. To produce the same
σ
, more
charge is required when the area increases.
24
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242
Chapter 24
24.6.
I
DENTIFY
:
ab
Q
C
V
=
.
0
A
C
d
=
P
.
S
ET
U
P
:
When the capacitor is connected to the battery, enough charge flows onto the plates to make
12.0 V.
ab
V
=
E
XECUTE
:
(a)
12.0 V
(b)
(i) When
d
is doubled,
C
is halved.
ab
Q
V
C
=
and
Q
is constant, so
V
doubles.
24.0 V
V
=
.
(ii) When
r
is doubled,
A
increases by a factor of 4.
V
decreases by a factor of 4 and
3.0 V
V
=
.
E
VALUATE
:
The electric field between the plates is
0
/
E
Q
A
=
P
.
ab
V
Ed
=
. When
d
is doubled
E
is unchanged and
V
doubles. When
A
is increased by a factor of 4,
E
decreases by a factor of 4 so
V
decreases by a factor of 4.
24.7.
I
DENTIFY
:
0
A
C
d
=
P
. Solve for
d
.
S
ET
U
P
:
Estimate
1.0 cm
r
=
.
2
A
r
π
=
.
E
XECUTE
:
0
A
C
d
=
P
so
2
2
0
0
12
(0.010 m)
2.8 mm
1.00
10
F
r
d
C
π
π
−
=
=
=
×
P
P
.
E
VALUATE
:
The separation between the pennies is nearly a factor of 10 smaller than the diameter of a penny, so it
is a reasonable approximation to treat them as infinite sheets.
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 Spring '11
 Shaefer
 Energy, Capacitors, Dielectrics, Ceq

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