A parallel-plate capacitor was lowered into water in a horizontal position, with water filling up the gap between the plates $d=1.0mm$ wide. Then a constant voltage $V=500V$ was applied to the capacitor. Find the water pressure increment in the gap. Dielectric constant of water is 81 .

$F=\frac{dU}{dx}$ (when battery remains connected)

 $F=-\frac{dU}{dx}$ (when battery is disconnected)

Let at an instant a thickness $x$ of gap between plates is filled with water shown in fig.

$$\begin{gathered} C=\frac{{\epsilon }_{0}S}{{\displaystyle \frac{x}{\epsilon }}+(d-x)}=\frac{{\epsilon }_{0}S}{x\left({\displaystyle \frac{1}{\epsilon }}-1\right)+d} \\ \therefore U=\frac{1}{2}C{V}^2 \\ or U=\frac{1}{2}\frac{{\epsilon }_{0}S\epsilon V^2}{x(1-\epsilon )+\epsilon d} \end{gathered}$$

$\therefore F=\frac{dU}{dx}. (\sin ce, battery remains connected)$

$\therefore F=-\frac{1}{2}\frac{{\epsilon }_{0}S\epsilon V^2(1-\epsilon )}{{\left\{x\right(1-\epsilon )+\epsilon d\}}^2}$

When water is filled in the gap between plates of capacitor $x=d$

$$\begin{gathered} \therefore F=\frac{{\displaystyle \frac{1}{2}}{\epsilon }_{0}S\epsilon V^2(\epsilon -1)}{{\left\{d\right(1-\epsilon )+\epsilon d\}}^2}=\frac{{\displaystyle \frac{1}{2}{\epsilon }_{0}S\epsilon V^2(\epsilon -1)}}{{\{d-d\epsilon +\epsilon d\}}^2}=\frac{{\displaystyle \frac{1}{2}{\epsilon }_{0}S\epsilon V^2(\epsilon -1)}}{d^2} \\ \therefore Excess pressure is ∆P=\frac{F}{S}=\frac{{\epsilon }_0\epsilon V^2(\epsilon -1)}{2d^2} \end{gathered}$$       For water                        $\epsilon =81$                              

       n putting the values, we get $∆p=7.17kPa=0.07atm.$