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) After time t, thickness of liquid will remain $\left( \frac{d}{3} - vt \right).$
Now, time constant as function of time
$\tau_{c} = CR$
$\ \ \ \ \ = \frac{\varepsilon_{0}(1).R}{\left( d - \frac{d}{3} + vt \right) + \frac{d/3 - vt}{2}}\ \left( Applying\ C = \frac{\varepsilon_{0}A}{d - t + \frac{t}{k}} \right)$
$\ \ \ \ \ = \frac{6\varepsilon_{0}R}{5d + 3vt}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.