A parallel plate capacitor is located horizontally, so that one of its plates is 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 $\rho$ . To what height will the level of the liquid in the capacitor rise after its plates get a charges of surface density $\sigma$ ?

Because of charge of the plates of capacitor , a charge will induce on the upper and lower layers ( which is in contact with negative plate of capacitor) of the liquid. The force on upper layer of liquid id $F=EQ$  in upward direction. Because of this 
Force, liquid rises till force becomes equal to weight of liquid risen. 
Thus  $F=E{Q}^1=mg$
Where  $Q^1=Q(1-\frac{1}{k})andmg=Ah\rho g$
Charge on lower plate 

$=Q+Q^1=-Q+Q(1+\frac{1}{k})=-\frac{Q}{k}$

The electric field due to both the plates of capacitor 

$E=\frac{\sigma }{2{\epsilon }_0}+\frac{{\sigma }^{\text{'}}}{2{\epsilon }_0}$

$=\frac{Q}{2{\epsilon }_{0}A}+\frac{Q}{2k{\epsilon }_{0}A}=\frac{Q}{2{\epsilon }_{0}A}(1+\frac{1}{k})$
We have  $EQ=mg$
$\because \frac{Q}{2{\epsilon }_{0}A}(1+\frac{1}{k})Q(1-\frac{1}{k})=Ah\rho g$
Or  $h=\frac{(k^2-1Q^2)}{2A^{2}k{\epsilon }_0\rho g}$