A cylindrical capacitor of inner $a,$ outer radius $b$ and length $L$ is kept with its axis vertical. Lower cross section of the capacitor is sealed with very thin dielectric material and the annular space is filled completely with the oil of dielectric constant $K$. Plates are connected to battery of emf $V.$ Suddenly an orifice of cross section Area A (of negligible resistance) is made at the bottom. Find the current i in the circuit as a function of time.

Let after time $t$ the liquid column inside the cylinder occupies a height $x.$Therefore the air column of height $(L-x)$.

The arrangement is now parallel combination of two capacitors of capacity $c_1$ and $c_2$ where:

   $$\begin{gathered} C_1=\frac{2\pi {\epsilon }_{0}xk}{\ln \left[{\displaystyle \frac{b}{a}}\right]} ......\left(1\right) \\ {C}_2=\frac{2\pi {\epsilon }_0\left[L-x\right]}{\ln \left[{\displaystyle \frac{b}{a}}\right]} ......\left(2\right) \end{gathered}$$

Therefore, equivalent capacitance,

  $C=C_1+C_2=\frac{2\pi {\epsilon }_0}{\ln \left[{\displaystyle \frac{b}{a}}\right]} \left[L+x(K-1)\right] .....\left(3\right)$

The cross-section area of cylinder in which liquid is present,

    $A\text{'}=\pi (b^2-a^2)$ m²

Applying equation of continuity,

    $A\sqrt{2gx}=-A\text{'}\frac{dx}{dt}\Rightarrow dt=\frac{-A\text{'}}{A}\frac{dx}{\sqrt{2gx}}$

Integrating,  ${\int }_0^{t}dt=\frac{-A\text{'}}{A\sqrt{2g}}{\int }_L^x\frac{dx}{\sqrt{x}}=\frac{-A\text{'}}{A\sqrt{2g}}2{\left[\sqrt{x}\right]}_L^x=\frac{A\text{'}}{A}\sqrt{\frac{2}{g}} \left(\sqrt{L}-\sqrt{x}\right)$

$$\begin{gathered} \Rightarrow t=\frac{A\text{'}}{A}\sqrt{\frac{2}{g}}\left(\sqrt{L}-\sqrt{x}\right) \Rightarrow x={\left[\sqrt{L}-\frac{At}{A\text{'}}\sqrt{\frac{g}{2}}\right]}^2 \\ x=L-\frac{2A}{A\text{'}}\sqrt{\frac{gL}{2}}t+\frac{A^{2}g{t}^2}{2A{\text{'}}^2} ......\left(4\right) \end{gathered}$$

Putting $x$ from equation(4) in equation(3), we get

$$\begin{gathered} C=\frac{2\pi {\epsilon }_0}{\ln \left[{\displaystyle \frac{b}{a}}\right]}\left[L+\left(L-\frac{2A}{A\text{'}}\sqrt{\frac{gL}{2}}t+\frac{A^{2}g{t}^2}{2A{\text{'}}^2}\right)(K-1)\right] \\ \Rightarrow C=\frac{2\pi {\epsilon }_0}{\ln \left[{\displaystyle \frac{b}{a}}\right]}\left[KL-\frac{A}{A\text{'}}\sqrt{2gL}(K-1)t+{\left(\frac{A}{A\text{'}}\right)}^2\frac{g}{2}(K-1)t^2\right] \end{gathered}$$

Charge on capacitor $Q=CV$

Current in the connecting wires:

     $I=\frac{dQ}{dt}=\frac{dC}{dt}V$ (as $V$ is constant)

$$\begin{gathered} \Rightarrow \frac{dC}{dt}=\frac{2\pi {\epsilon }_0}{\ln {\displaystyle \frac{b}{a}}}(K-1)\left\{{\left(\frac{A}{A\text{'}}\right)}^{2}gt-\frac{A}{A\text{'}}\sqrt{2gL}\right\} \\ \Rightarrow i=\frac{2\pi {\epsilon }_0}{\ln {\displaystyle \frac{b}{a}}}(K-1)\left[\frac{A^2}{A{\text{'}}^2}gt-\frac{A}{A\text{'}}\sqrt{2gL}\right]V \end{gathered}$$