This preview shows pages 1–3. Sign up to view the full content.
241
C
APACITANCE AND
D
IELECTRICS
24.1.
IDENTIFY:
ab
Q
C
V
=
SET UP:
6
1 F
10 F
μ
−
=
EXECUTE:
64
(7.28 10 F)(25.0 V)
1.82 10 C
182
C
ab
QC
V
−−
==×
=
×
=
EVALUATE:
One plate has charge
Q
+
and the other has charge
Q
−
.
24.2.
IDENTIFY
and
SET UP:
0
A
C
d
=
P
,
Q
C
V
=
and
VE
d
=
.
(a)
2
00
0.00122 m
3.29 pF
0.00328 m
A
C
d
==
=
PP
(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
×
=
×
EVALUATE:
The electric field is uniform between the plates, at points that aren't close to the edges.
24.3.
IDENTIFY
and
SET UP:
It is a parallelplate air capacitor, so we can apply the equations of Sections 24.1.
EXECUTE:
(a)
6
12
0.148 10 C
so
604 V
245 10
F
ab
ab
QQ
CV
VC
−
−
×
=
=
×
(b)
0
so
A
C
d
=
P
()
(
)
12
3
32
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
dE
d
−
=
=
×
×
(d)
0
so
E
σ
=
P
(
)
61
2
2
2
5
2
0
1.84 10 V/m 8.854 10
C / N m
1.63 10 C/m
E
×
×
⋅ = ×
P
EVALUATE:
We could also calculate
directly as
Q/A
.
6
52
1.63 10 C/m ,
Q
A
−
−
−
×
=
×
×
which checks.
24.4.
IDENTIFY:
0
A
C
d
=
P
when there is air between the plates.
SET UP:
22
(3.0 10 m)
A
−
=×
is the area of each plate.
EXECUTE:
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
−
−
×
=
×
EVALUATE:
C
increases when
A
increases and
C
increases when
d
decreases.
24.5.
IDENTIFY:
ab
Q
C
V
=
.
0
A
C
d
=
P
.
SET UP:
When the capacitor is connected to the battery,
12.0 V
ab
V
=
.
EXECUTE:
(a)
(10.0 10 F)(12.0 V)
1.20 10 C
120
C
ab
V
=
×
=
(b)
When
d
is doubled
C
is halved, so
Q
is halved.
60 C
Q
=
.
(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
=
EVALUATE:
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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document 242
Chapter 24
24.6.
IDENTIFY:
ab
Q
C
V
=
.
0
A
C
d
=
P
.
SET UP:
When the capacitor is connected to the battery, enough charge flows onto the plates to make
12.0 V.
ab
V
=
EXECUTE:
(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
=
.
EVALUATE:
The electric field between the plates is
0
/
EQ A
=
P
.
ab
VE
d
=
. 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.
IDENTIFY:
0
A
C
d
=
P
. Solve for
d
.
SET UP:
Estimate
1.0 cm
r
=
.
2
Ar
π
=
.
EXECUTE:
0
A
C
d
=
P
so
22
00
12
(0.010 m)
2.8 mm
1.00 10
F
r
d
C
ππ
−
==
=
×
PP
.
EVALUATE:
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.
24.8.
INCREASE:
ab
Q
C
V
=
.
ab
d
=
.
0
A
C
d
=
P
.
SET UP:
We want
4
1.00 10 N/C
E
=×
when
100 V
V
=
.
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.
 Spring '11
 Shaefer

Click to edit the document details