Question

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 $t$ is expressed as $\frac{n{ϵ}_{0}R}{5d+3Vt}$. Then the value of $n$ is

Answer 1

$\frac{1}{C_{e{q}^n}}=\frac{(\frac{d}{3}-vt)}{2{\epsilon }_{0}A}+\frac{(\frac{2d}{3}+vt)}{{\epsilon }_{0}A}$ $=\frac{1}{2{\epsilon }_{0}A}(\frac{d}{3}-Vt+\frac{4d}{3}+2Vt)$
$\frac{1}{C_{{eq}^n}}=\frac{1}{2{\epsilon }_{0}A}(\frac{5d}{3}+vt)\Rightarrow C_{eq}^n=\frac{6{\epsilon }_{0}A}{(5d+3Vt)}\Rightarrow \tau =\frac{6{\epsilon }_{0}AR}{(5d+3Vt)}$
As $A$ = 1, thus $\tau$ = $\frac{6{ϵ}_{0}R}{5d+Vt}$
$⟹$ n = 6