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IIT JEE Advanced 2008 Paper 2
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Section:
Physics
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© examsnet.com
Question : 3
Total: 66
A parallel-plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K = 2. The level of liquid is d/3 initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time t is
[JEE Adv 2008 P2]
6
ε
0
R
5
d
+
3
V
t
(
15
d
+
9
V
t
)
ε
0
R
2
d
2
−
3
d
V
t
−
9
V
2
t
2
6
ε
0
R
5
d
−
3
V
t
(
15
d
−
9
V
t
)
ε
0
R
2
d
2
+
3
d
V
t
−
9
V
2
t
2
Validate
Solution:
The effective capacitance
C
eff
=
A
ε
0
d
−
y
+
(
y
ε
r
)
where A is the area of the capacitor plate, d is the plate separation, y is the thickness of dielectric and
ε
r
is the dielectric constant. It is given that rate of change of dielectric liquid is
d
y
d
t
= - V ⇒ y = -- Vt +
C
1
At t = 0, we have
y =
d
3
⇒
C
1
=
d
3
Therefore,
C
eff
=
ε
0
d
+
V
t
−
d
3
+
[
(
−
V
t
+
(
d
3
)
)
ε
r
]
=
ε
0
ε
r
[
d
(
2
ε
r
+
1
)
+
3
V
(
ε
r
−
1
)
]
3
=
6
ε
0
5
d
+
3
V
t
The time constant is
τ =
R
C
eff
=
6
ε
0
R
5
d
+
3
V
t
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