The Wet Condenser?!

Energy before water started to flows in is:

${E}_{1} = \frac{{q}^{2}}{2{C}_{1}}$ Where ${C}_{1}=\frac{{e}_{0}S}{d}$

When the water in steady state we have two capacitator

${C}_{2}= \frac {{e}_{0}{e}_{r}S}{h}$ and ${C}_{3}=\frac{{e}_{0}S}{d-h}$

each with charge $q$

They have energies ${E}_{2}=\frac{{q}^{2}}{2{C}_{2}}$ and ${E}_{3}=\frac{{q}^{2}}{2{C}_{3}}$

Also the water has potentional energy ${E}_{p}=\frac {mgh}{2}$

By conversation law we now ${E}_{1}={E}_{2}+{E}_{3}+{E}_{p}$ and finally we get :

$h=(\frac{q}{S})^2*\frac{(1-\frac{1}{\epsilon_r})}{(\ro g \epsilon_o)}$

The electric field due to the induced charges(these charges are induced on the top and bottom surfaces of the dielectric) superimposes with the already present electric field and reduces it by a factor $\epsilon_r$ . The magnitude of induced surface charge density will be:

$\sigma_{induced} = \large \frac{\sigma (\epsilon_r - 1)}{\epsilon_r}$

Hence, putting $\epsilon_r = 80$ we get $\large \boxed{\sigma_{induced} = \frac{79\sigma}{80}}$

The top surface of water(dielectric) will rise till the net force due to the following is zero:

1. The weight of the water (which is pulling down the top surface)

2. Electrostatic force of attraction due to the charge induced on the lowermost surface of the dielectric (which is again pulling down the top most surface down)

3. And the electrostatic force acting in the upward direction on the top layer of water due to the electric field of the condenser.

Therefore, $(\frac{F_{1}}{A})_{\text{due to condenser}} = (\frac{V\rho_{water}g}{A})_{\text{due to weight of water}} + (\frac{F_{2}}{A})_{\text{due attraction between induced layer of charges on the dielectric}}$

$(\frac{\sigma}{\epsilon_{0}})*\sigma_{induced} = h\rho g + (\frac{\sigma_{induced}}{2\epsilon_{0}})*\sigma_{induced}$

Now after putting the values for

$\sigma = 2*10^{-6}C$ ;

$\sigma_{induced} = \frac{79\sigma}{80}$ ;

$\rho=1000kg/m^{3}$ ;

$g=10m/s^{-2}$

we get the answer as:

$h=2.27 * 10^{-3}cm$