PHYS1200
PHYSICS II
SPRING 2006
HOMEWORK SOLUTIONS
CLASS 11.
32.14, 32.19, 32.53; Q15.01, Q15.02
32.14.
Since the radius of the circular loop is greater than the radius of the capacitor plates, the
displacement current through the loop is the entire displacement current between the
plates. Other than that, the size of the loop has nothing to do with the problem.
dt
dE
R
dt
dE
A
dt
EA
d
dt
d
i
E
d
2
0
0
0
0
)
(
π
ε
ε
ε
ε
=
=
=
Φ
=
, where
R
is the radius of the capacitor
plates, since the electric field is confined to the space between the plates. Then,
2
2
2
12

2
0
m)
10
.
0
(
)
m
/N
C
10
(8.85
A
0
.
2
π
π
ε
⋅
×
=
=
R
i
dt
dE
d
s
V/m
10
2
.
7
12
⋅
×
=
dt
dE
32.19.
)
m
6
.
1
m)(
/
F
10
85
.
8
(
)
(
2
12
0
0
0
dt
dE
dt
dE
A
dt
EA
d
dt
d
i
E
d

×
=
=
=
Φ
=
ε
ε
ε
A
8
.
2
s)
m
/
V
10
0
.
2
V)(
/
s
m
A
10
42
.
1
(
Then
s.
m
/
V
10
0
.
2
s
10
0
.
2
m)
/
V
10
)
0
.
4
0
.
0
(
c)
0
Then
.
0
b)
A
71
.
0
s)
m
/
V
10
0
.
5
V)(
/
s
m
A
10
42
.
1
(
Then
s.
m
/
V
10
0
.
5
s
10
0
.
4
m)
/
V
10
)
0
.
6
0
.
4
(
a)
V)
/
s
m
A
10
42
.
1
(
11
11
11
6
5
10
11
10
6
5
11
=
⋅
×

⋅
⋅
×
=
⋅
×

=
×
×

=
=
=
=
⋅
×

⋅
⋅
×
=
⋅
×

=
×
×

=
⋅
⋅
×
=





d
d
d
d
d
d
i
i
dt
dE
i
dt
dE
i
i
dt
dE
dt
dE
i
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 Spring '06
 Stoler
 Physics, Current, Work, DT DT DT

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