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

• A. $\frac{6{ϵ}_{o}R}{5d+3Vt}$
• B. $\frac{(15d+9Vt){ϵ}_{o}R}{2d^2-3dVt-9V^2{t}^2}$
• C. $\frac{6{ϵ}_{o}R}{5d-3Vt}$
• D. $\frac{(15d-9Vt){ϵ}_{o}R}{2d^2+3dVt-9V^2{t}^2}$

Answer 1

The correct option is A $\frac{6{ϵ}_{o}R}{5d+3Vt}$

At any time $t$ , height of liquid will be $d_1=\frac{d}{3}-Vt$, hence remaining height will be $d_2=d-(\frac{d}{3}-Vt)=\frac{2d}{3}+Vt$

$\frac{1}{C_{eq}}=\frac{(\frac{d}{3}-Vt)}{2{ϵ}_{o}A}+\frac{(\frac{2d}{3}+Vt)}{{ϵ}_{o}A}$
$=\frac{1}{2{ϵ}_{o}A}(\frac{d}{3}-Vt+\frac{4d}{3}+2Vt)$
$\Rightarrow \frac{1}{C_{eq}}=\frac{1}{2{ϵ}_{o}A}(\frac{5d}{3}+Vt)$
$\Rightarrow C_{eq}=\frac{6{ϵ}_{o}A}{(5d+3Vt)}$
Now, time constant in $RC$ circut is given by $\tau =R{C}_{eq}$
$\Rightarrow \tau =\frac{6{ϵ}_{o}AR}{(5d+3Vt)}$
Given that plates have unit area, so
$\Rightarrow \tau =\frac{6{ϵ}_{o}R}{(5d+3Vt)}$