24-1
C
APACITANCE AND
D
IELECTRICS
Note:
These solutions use P
0
for the permittivity of free space. P
0
=
ε
0
= 8.85E-12 F/m.
24.1.
I
DENTIFY
:
The capacitance depends on the geometry (area and plate separation) of the plates.
S
ET
U
P
:
For a parallel-plate capacitor,
V
ab
=
Ed,
E
=
Q
!
0
A
, and
C
=
Q
V
ab
.
E
XECUTE
:
(a)
V
ab
=
Ed
=
4
.
00
!
10
6
V/m
(
)
2
.
50
!
10
"
3
m
(
)
=
1
.
00
!
10
4
V.
(b)
Solving for the area gives
A
=
Q
E
!
0
=
80
.
0
"
10
#
9
C
(4
.
00
"
10
6
V/m
)[
8
.
854
"
10
#
12
C
2
/
(N
$
m
2
)]
=
2
.
26
"
10
#
3
m
2
=
22.6 cm
2
.
(c)
C
=
Q
V
ab
=
80
.
0
!
10
"
9
C
1
.
00
!
10
4
V
=
8
.
00
!
10
"
12
F
=
8
.
00 pF.
E
VALUATE
:
The capacitance is reasonable for laboratory capacitors, but the area is rather large.
24.2.
I
DENTIFY
and
S
ET
U
P
:
0
A
C
d
=
P
,
Q
C
V
=
and
V
Ed
=
.
(a)
C = P
0
A/d = P
0
(0.00122m
2
) / (0.00328m) = 3.29 pF
(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.
24.5.
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,
!
. To produce the same
!
, more
charge is required when the area increases.
24

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