Q.
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) $\frac{6\varepsilon_{0}R}{5d + 3Vt}$
B) $\frac{(15d + 9Vt)\varepsilon_{0}R}{2d^{2} - 3dVt - 9V^{2}t^{2}}$
C) $\frac{6\varepsilon_{0}R}{5d - 3Vt}$
D) $\frac{(15d - 9Vt)\varepsilon_{0}R}{2d^{2} + 3dVt - 9V^{2}t^{2}}$
Explanation
(a) Time constant $= \tau = RC$
$C = \frac{C_{1}C_{2}}{C_{1} + C_{2}} = \frac{\left( \frac{A\varepsilon_{0}}{d - x} \right)\left( \frac{KA\varepsilon_{0}}{x} \right)}{\frac{A\varepsilon_{0}}{d - x} + \frac{KA\varepsilon_{0}}{x}}$
$C = \frac{KA\varepsilon_{0}}{x + K(d - x)} \Rightarrow x = \frac{d}{3} - Vt$
$\tau = \frac{RKA\varepsilon_{0}}{\frac{d}{3} - Vt + K\left( d - \frac{d}{3} + Vt \right)}$
$A = 1,\ K = 2,\ \tau = \frac{3 \times 2R\varepsilon_{0}}{d - 3Vt + 6d - 2d + 6Vt} = \frac{6R\varepsilon_{0}}{5d + 3Vt}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.