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. $\frac{6{ϵ}_{0}R}{5d+3Vt}$
• B. $\frac{(15d+9Vt){ϵ}_{0}R}{2d^2-3dVt-9V^2{t}^2}$
• C. $\frac{6{ϵ}_{0}R}{5d-3Vt}$
• D. $\frac{(15d-9Vt){ϵ}_{0}R}{2d^2+3dVt-9V^2{t}^2}$

Answer 1

The correct option is A $\frac{6{ϵ}_{0}R}{5d+3Vt}$

At time t the level of liquid is,

$\frac{d}{3}$$-Vt{C}_{eq}=\frac{\frac{{ϵ}_{0}KA}{\frac{d}{3}-Vt}\cdot \frac{{ϵ}_{0}A}{\frac{2d}{3}+Vt}}{\frac{{ϵ}_{0}KA}{\frac{d}{3}-Vt}+\frac{{ϵ}_{0}A}{\frac{2d}{3}+Vt}}=\frac{{ϵ}_{0}KA}{K(\frac{2d}{3}+Vt)+\frac{d}{3}-Vt}=\frac{2{ϵ}_0}{(\frac{4d}{3}+2Vt)+\frac{d}{3}-Vt}=\frac{2{ϵ}_0}{(\frac{5d}{3}+Vt)}=\frac{6{ϵ}_0}{5d+3Vt}\tau =R{C}_{eq}=\frac{6{ϵ}_{0}R}{5d+3Vt}$