JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Capacitance

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

\frac{d}{3}

initially. Suppose the liquid level decreases at a constant speed

V

, the time constant as a function of time

t

Answer: (A)

Solution

At time

t

, the liquid level is at height

h = \frac{d}{3} - V t

.

The capacitor can be viewed as two capacitors in series:

Lower part (with dielectric): height $h$ , capacitance $C_{1} = \frac{K ε _{0}}{h} = \frac{2 ε _{0}}{\frac{d}{3} - V t}$
Upper part (air): height $d - h$ , capacitance $C_{2} = \frac{ε _{0}}{d - h} = \frac{ε _{0}}{\frac{2 d}{3} + V t}$

Equivalent capacitance:

C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{\frac{2 ε _{0}}{\frac{d}{3} - V t} \cdot \frac{ε _{0}}{\frac{2 d}{3} + V t}}{\frac{2 ε _{0}}{\frac{d}{3} - V t} + \frac{ε _{0}}{\frac{2 d}{3} + V t}}

Simplifying gives the time constant

τ = C_{e q} R = \frac{6 ε _{0} R}{5 d + 3 V t}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option