Platt, David – Homework 8 – Due: Sep 26 2005, 4:00 am – Inst: Ken Shih
1
This
printout
should
have
12
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
For 003 you need give your answer in unit
of pF (1pF=e12*F).
001
(part 1 of 2) 10 points
Consider a airfilled parallelplate capacitor
with plates of length 8
.
8 cm, width 6
.
78 cm,
spaced a distance 1
.
86 cm apart.
Now imagine that a dielectric slab with di
electric constant 2
.
7 is inserted a length 4
.
8 cm
into the capacitor.
The slab has the same
width as the plates.
The capacitor is com
pletely filled with dielectric material down to
a length of 4
.
8 cm.
A battery is connected to the plates so that
they are at a constant potential 0
.
83 V while
the dielectric is inserted.
8
.
8 cm
4
.
8 cm
1
.
86 cm
dielectric
none  air
bottom
1
.
0
dielectric
constant
top
2
.
7
What is the ratio of the new potential en
ergy to the potential energy before the inser
tion of the dielectric?
Edge effects can be
neglected.
Correct answer: 1
.
92727 .
Explanation:
Let :
w
= 6
.
78 cm
,
d
= 1
.
86 cm
,
L
= 8
.
8 cm
,
‘
= 4
.
8 cm
,
κ
= 2
.
7
,
and
V
= 0
.
83 V
.
The given capacitor can be considered as
two capacitors in parallel, as shown below.
C
1
C
2
The potential energy in a capacitor is
U
=
1
2
C V
2
,
so the ratio is
U
0
U
=
C
0
V
0
2
C V
2
.
Since the electric potential difference (volt
age) across the capacitor is constant,
U
0
U
=
C
0
C
.
(1)
The capacitance of a parallel plate capaci
tor is
C
=
²
0
·
area
separation
=
²
0
w L
d
.
(2)
After the insertion of the dielectric, the ca
pacitor can be considered in two parts:
the
part filled with dielectric, and the part with
no dielectric.
Since the dielectric is inserted with a length
‘ ,
C
0
=
C
0
1
+
C
2
,
where
(3)
C
0
1
=
²
0
w κ ‘
d
C
2
=
²
0
w
(
L

‘
)
d
.
The total capacitance of the new system is
C
0
=
C
0
1
+
C
2
= (
L

‘
+
κ ‘
)
²
0
w
d
.
(4)
Therefore, we have
U
0
U
=
C
0
C
=
n
L
+
‘
[
κ

1]
o
²
0
w
d
L
²
0
w
d
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Platt, David – Homework 8 – Due: Sep 26 2005, 4:00 am – Inst: Ken Shih
2
= 1 +
‘
L
h
κ

1
i
(5)
= 1 +
(4
.
8 cm)
(8
.
8 cm)
h
(2
.
7)

1
i
=
1
.
92727
.
When the dielectric is fully inserted
‘
=
L
,
so
U
0
U
= 1 +
L
L
h
κ

1
i
=
κ .
(6)
002
(part 2 of 2) 10 points
From this, we can deduce that if a person
inserts the slab part way into the conductor,
and then releases it, the capacitor will exert a
force on the dielectric in what direction?
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 Fall '08
 Turner
 Physics, Work, Electric charge, Platt, Ken Shih, David – Homework

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