mcconnell (kam2342) – homework12 – Turner – (60230)
1
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11
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001
10.0 points
A capacitor network with airfilled capacitors
as shown below.
85 V
27
.
2
μ
F
27
.
2
μ
F
27
.
2
μ
F
27
.
2
μ
F
b
a
c
d
When the top righthand capacitor is filled
with a material of dielectric constant
κ
, the
charge on this capacitor is increases by a fac
tor of 1
.
48.
Find the dielectric constant
κ
of the mate
rial inserted into the top righthand capaci
tor.
Correct answer: 2
.
84615.
Explanation:
Let :
C
1
=
C
= 27
.
2
μ
F
,
C
2
=
C
= 27
.
2
μ
F
,
C
3
=
C
= 27
.
2
μ
F
,
C
4
=
C
= 27
.
2
μ
F
,
E
B
= 85 V
,
and
Q
′
= 1
.
5
Q .
E
B
C
1
C
3
C
2
C
4
b
a
c
d
The capacitors
C
3
and
C
4
have nothing to
do with this problems. In addition, the capac
itances are all equal and their specific values
are immaterial. Furthermore, the electric po
tential of the battery is not required.
C
1
=
C
2
=
C
3
=
C
4
, where
Q
and
Q
′
are the initial and final charges on
C
2
and
Q
′
Q
≡
α
=ratio of final to initial charge on
C
2
.
We know the charges on
C
1
and
C
2
are the
same. Initially,
V
ab
=
V
1
+
V
2
=
Q
C
1
+
Q
C
2
=
Q
C
+
Q
C
= 2
Q
C
.
(1)
Therefore
Q
=
1
2
V
ab
C .
After the dielectric material is inserted in
C
2
,
the capacitance becomes
C
′
2
=
κ C
.
There
fore,
V
ab
=
V
′
1
+
V
′
2
=
Q
′
C
1
+
Q
′
C
′
2
=
Q
′
C
+
Q
′
κ C
=
κ
+ 1
κ
Q
′
C
,
and using Eq. (1) and solving for
Q
′
, we have
2
Q
C
=
κ
+ 1
κ
Q
′
C
Q
′
=
κ
κ
+ 1
V
ab
C
=
κ
κ
+ 1
2
Q
Q
′
Q
≡
α
=
2
κ
κ
+ 1
= 1
.
48
.
Solving for
κ
, we have
κ
=
α
2

α
=
1
.
48
2

1
.
48
=
2
.
84615
.
002
10.0 points
Consider the two cases shown below. In Case
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mcconnell (kam2342) – homework12 – Turner – (60230)
2
One two identical capacitors are connected to
a battery with emf
V
.
In Case Two, a di
electric slab with dielectric constant
κ
fills the
gap of capacitor
C
2
. Let
C
12
be the resultant
capacitance for Case One and
C
′
12
the resul
tant capacitance for Case Two.
Case One
V
C
1
C
2
Case Two
V
C
1
C
′
2
κ
The ratio
C
′
12
C
12
of the resultant capacitances is
1.
None of these
2.
C
′
12
C
12
=
κ
2
.
3.
C
′
12
C
12
= 1 +
κ .
4.
C
′
12
C
12
= 2
κ .
5.
C
′
12
C
12
=
1 +
κ
2
κ
.
6.
C
′
12
C
12
=
2
1 +
κ
.
7.
C
′
12
C
12
=
2
κ
1 +
κ
.
correct
8.
C
′
12
C
12
=
κ .
9.
C
′
12
C
12
=
1 +
κ
2
.
10.
C
′
12
C
12
=
1
κ
.
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 Spring '10
 Turner
 Electrostatics, Electric charge, McConnell, μF, c12

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