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mcconnell (kam2342) – homework12 – Turner – (60230) 1
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001 10.0 points
A capacitor network with air-filled 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 right-hand 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 right-hand 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

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
.