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

Answer 1

Answer: (A)

$C_{\text{equivielet }}=\frac{\frac{2{\epsilon }_0}{\frac{d}{3}-vt}\cdot \frac{{\epsilon }_0}{\frac{2d}{3}+vt}}{\frac{2{\epsilon }_0}{\frac{d}{3}-vt}+\frac{{\epsilon }_0}{\frac{2d}{3}+vt}}$
$\therefore \tau =C_{\text{equivalent }}R$