Question

A parallel plane 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. $\frac{6{\epsilon }_{0}R}{5d+3vt}$
• B. $\frac{(15d+9vt){\epsilon }_{0}R}{2d^2-3dvt-9v^2{t}^2}$
• C. $\frac{6\pi {\epsilon }_{0}R}{5d-3vt}$
• D. $\frac{(15d-9vt){\epsilon }_{0}R}{2d^2+3dvt-9v^2{t}^2}$

Answer 1

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

At any time $t$ , height of liquid will be $d_1=d/3-vt$ and hence remaining height will be $d_2=d-(\frac{d}{3}-vt)=\frac{2d}{3}+vt$

since capacitance $C=\frac{{ϵ}_{0}A}{d}$

$C_1=\frac{{ϵ}_0}{2d/3+vt}$

$C_2=\frac{2{ϵ}_0}{d/3-vt}$, since $A=1$ and $k=1$

since both capacitor will act as connected in series , $\frac{1}{C}=\frac{1}{c_1}+\frac{1}{C_2}$

hence we get , $C=\frac{6{ϵ}_0}{5d+3vt}$

and time constant is given by $\tau =CR=\frac{6{ϵ}_{0}R}{5d+3vt}$