A parallel plate capacitor is located horizontally such that one of the plates is submerged in a liquid while the other is above the liquid surface.  When plates are charged the level of liquid.

When the capacitor gets charged electric field will be induced between the plates. Since water is dielectric, charge will gets distributed in it.
As we know the total charge induced in the dielectric is  ${\sigma }^{\text{'}}=\sigma (1-\frac{1}{ϵ})$
Assuming the top plate is positively charged, there will be electric field from top to bottom. Simillarly electric field in water will be from bottom to top.
The electric field due to water will pull the water surface upward due to which the water level rises to some height h Now calculating net electric field,   ${\text{E}}_1=\frac{\sigma }{2{ϵ}_0}\text{ (downward)}$
$$\begin{gathered} {\text{E}}_3=\frac{{\sigma }^{\text{'}}}{2{ϵ}_0}=\frac{\sigma (1-\frac{1}{ϵ})}{2{ϵ}_0}\text{ (upward) } \\ {\text{E}}_2=\frac{\sigma }{2{ϵ}_0}(\text{ downward) } \\ {\text{E}}_{\text{Net}}=({\text{E}}_1+{\text{E}}_2)-{\text{E}}_3=\frac{\sigma }{2{ϵ}_0}+\frac{\sigma }{2{ϵ}_0}=\frac{\sigma (1-\frac{1}{ϵ})}{2{ϵ}_0} \\ =\frac{\sigma }{{ϵ}_0}[1-\frac{1}{2}+\frac{1}{2ϵ}]=\frac{\sigma }{{ϵ}_0}[\frac{1}{2}+\frac{1}{2ϵ}] \\ {\text{E}}_{\text{Net}}=\frac{\sigma }{2{ϵ}_0}[1+\frac{1}{ϵ}]\text{ (downward)} \end{gathered}$$

For calculating force,   $F=qEAs,q={\sigma }^{\text{'}}A$

$$\begin{gathered} \frac{F}{A}={\sigma }^{\text{'}}E=\sigma (1-\frac{1}{ϵ})\frac{\sigma }{2ϵ}[1+\frac{1}{{ϵ}_0}] \\ \frac{F}{A}=\frac{{\sigma }^2}{2{ϵ}_0}(1-\frac{1}{{ϵ}^2})=P\text{ (upward)} \end{gathered}$$

As this in equilibrium then,  $\rho gh=\frac{{\sigma }^2}{2{ϵ}_0}(1-\frac{1}{{ϵ}^2})$

$h=\frac{{\sigma }^2}{2\rho g{ϵ}_0^0}(1-\frac{1}{{ϵ}^2})=\frac{{\sigma }^2}{2\rho g{ϵ}_0}\left(\frac{{ϵ}^2-1}{{ϵ}^2}\right)$