A parallel plate capacitor is located horizontally, so that one of its plates is just submerged into liquid while the other is over the surface. The dielectric constant of the liquid is equal to k. Its density is equal to $\mathrm{\rho }$ . To what height will the level of the liquid in the capacitor rise after its plates get a charge of surface density$\sigma$?

Induced charge on the upper face of the liquid =$$\begin{gathered} \sigma "=+\sigma (1-\frac{1}{k}) and that on the lower face of the liquid is -\sigma ". Here charge density of the upper plate of the capacitor is -\sigma \\ and that on the lower plkate of the capacitor is +\sigma . \end{gathered}$$Therefore net upward force on the liquid  F = [$A[\frac{\sigma }{2{\in }_0}\sigma "+\frac{\sigma }{2{\in }_0}\sigma "-\frac{\sigma "}{2{\in }_0}\sigma "]=\frac{{\sigma }^2(k^2-1)A}{2k{\in }_0}$

For equilibrium of the liquid column inside the capacitor , F =($\rho Ah)g$

Therefore h = $\frac{{\sigma }^2(k^2-1)}{2{\in }_{0}k\rho g}$