Physics 21
Fall, 2005
Solution to HW7
24
P59
Two identical capacitors are connected in parallel,
and each acquires a charge
Q
0
when connected to a source
of voltage
V
0
. The voltage source is disconnected and then
a dielectric (
K
=2
.
5) is inserted to Fll the space between
the plates of one of the capacitors. Determine (a) the charge
now on each capacitor, and (b) the voltage now across each
capacitor.
+Q
0
+Q
0
2Q
0

QQ
C
0
C
0
C
0
KC
0
(a)
(b)
±or panel (a), the two capacitors are the same and
Q
0
=
C
0
V
0
.
±or panel (b), the charge on the two upper plates
will rearrange so that the voltages across the two capacitors
(
C
0
on the left and
KC
0
on the right) are equal.
Since
the capacitors are in parallel, the total charge on the two
positive plates does not change. We can Fnd the charge
Q
that ends up on the modiFed capacitor by using the equality
of voltages:
2
Q
0
−
Q
C
0
=
Q
0
Solving this equation for
Q
gives
Q
Q
0
K
K
+1
and
2
Q
0
−
Q
Q
0
1
K
.
The voltage across both capacitors is
V
=
Q
0
=
Q
0
C
0
2
K
=
V
0
2
K
.
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 Fall '07
 Hickman
 Physics, Charge, Volt, High Voltage, Electric charge, Wire

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