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capacitors and dielectric
1. A battery with emf
ε
= 20 V is used to charge two identical capacitors
C
1
=
C
2
= 6
μF as shown in the diagram. The switch is initially closed
.
Calculate the resulting charge
Q
2
on capacitor
C
2.
The voltage across the capacitor
C
2
is
V
2
= 20 V . (since it is in parallel with the battery).
Hence the resulting charge is,
22
2
2
2
(6
) (20 )
120
QC
V
QF
V
µ
=×
=
2. The switch is now open. What is the total energy stored in the system?
It is the same energy store before opening the switch.
The capacitors are in parallel. Hence the equivalent capacitance of the system is:
12
66
12
eq
eq
eq
CC
C
CF
F
=+
=
The energy stored in the system is then:
11
(12
F)(20 V)
2400
2.4 m
µµ
==
=
=
UC
V
J
J
3. While the switch is open, a dielectric with
κ
= 1.5 is inserted between the plates of
C
2
, filling all the available space between the plates. Due to this, the charge on
C
1
:
a. Increases.
*b. Decreases.
c. Stays the same.
When the dielectric is inserted between the plates of
C
2
, the capacitance of the capacitor
2 increases. Due to this, capacitor 2 draws more charge (more capacitance means it is
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View Full Document“easier” to store charge there now), and since the system is isolated and disconnected
from the battery, the additional charge in capacitor 2 must come from capacitor 1. Thus,
the charge in
C
1
must decrease.
4. Let
V
1
and
V
2
be the magnitudes of the potential difference across the plates of
C
1
and
C
2
, respectively,
after the slab is inserted. Which of the following is true?
a.
V
1
<
V
2
*b.
V
1
=
V
2
c.
V
1
>
V
2
The capacitors are in parallel. Hence the correct answer is b)
V
1
=
V
2
Also, the potential in a conductor in equilibrium is constant. Thus all the points in the red
section below have the same potential, and all the points in the blue section have the
same potential, and therefore the potential difference between the plates of both
capacitors is 
V
blue

V
red

5. And how does
V
2
compare to
ε
, the emf of the battery?
*a.
V
2
<
b.
V
2
=
c.
V
2
>
It is hard to say if we look at
C
2
, because both its charge and its capacitance increased, so
it is not clear what happened to
V
2
=
Q
2
/
C
2
.
However,
V
2
=
V
1.
Capacitor 1 has the same capacitance as before but less charge (see
question 3), so the potential difference across its plates must have decreased from what it
was before (
).
6. Determine the final charge on capacitor
C
1
after the dielectric is inserted if
κ
=1.5.
Before the dielectric is inserted, the charges on each capacitor are
C
1
C
2
12
(20 V)(6 F)
120 C
QQ
µ
==
=
Let
Q’
1
and
Q’
2
be the charges on the capacitors after the dielectric is inserted. The
system is isolated, so the total charge must be the same as before the dielectric is inserted,
which means
12 12
(1)
QQ QQ
′′
+=+
Also, from question 4, we know that the potential difference across both capacitors must
be the same. In terms of the charges and capacitances, this is:
(2)
CC
κ
=
(1) and (2) are a system of two equations and two unknowns. The solution is:
1
2
1
2
1
2
240 C
96 C
11
.
5
1
240 C
144 C
1
1
1
1.5
Q
C
C
Q
C
C
+
′
===
+
+
+
′
+
+
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View Full Documentparallel capacitors
1. Two capacitors
C
1
and
C
2
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This homework help was uploaded on 03/26/2008 for the course PHYS 221 taught by Professor Herrerasiklody during the Spring '08 term at Iowa State.
 Spring '08
 HerreraSiklody
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