• # question_answer A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant $\kappa =2$. 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) $\frac{6{{\in }_{0}}R}{5d+3vt}$ B) $\frac{\left( 15d+9vt \right){{\in }_{0}}R}{2{{d}^{2}}-3dvt-9{{v}^{2}}{{t}^{2}}}$ C) $\frac{6{{\in }_{0}}R}{5d-3vt}$ D) $\frac{\left( 15d-9vt \right){{\in }_{0}}R}{2{{d}^{2}}-3dvt-9{{v}^{2}}{{t}^{2}}}$

 at any time, the level of liquid ${{x}_{1}}=\left( \frac{d}{3}-vt \right),$and the thickness of the air ${{x}_{2}}=d-\left( \frac{d}{3}-vt \right)=\left( \frac{2d}{3}+vt \right).$ Now $C=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}};$where ${{C}_{1}}=\frac{{{\in }_{0}}kA}{{{x}_{1}}},$and ${{C}_{2}}=\frac{{{\in }_{0}}A}{{{x}_{2}}}.$ Time constant $\tau =CR=\left[ \frac{6{{\in }_{0}}R}{5d+3vt} \right]$