A parallel-plate capacitor is placed in such a way that its plates are horizontal and the lower plate is dipped into a liquid of dielectric constant `K` and density `rho`. Each plate has an area `A`. The plates are now connected to a battery which supplies a positive charge of magnitude `Q` to the upper plate. Find the rise in the level of the liquid in the space between the plates. (figure)

Explanation

The situation is shown in figure. A charge `Q (1- 1/K)` is induced on the upper surface of the liquid and `Q (1- 1/K)` at the surface in contact with the lower plate. The net charge on the lower plate is `-Q + Q (1 - 1/K) = - Q/K`. Consider the equilibrium of the liquid in the volume ABCD. The forcese on this liquid are (a) the force due to the electric field at CD.
(b) the weight of the liquid.
(c) the force due to atmospheric pressure and
(d) the force due to the pressuure of the liquid below `AB`. As AB is in the same horizontal level as the outside surface, the pressure here is the same as the atmospheric pressure. The forces in (c) and (d), therefore, balance each other. Hence, for equilibrium, the forces in (a) and (b) should balance each other.
The electric field at CD due to the charge `Q` is
`E_1 = Q/(2Aepsiolon_(0))`
in the downward direction. The field at CD due to the charge `-Q//K` is
`E_2 = Q/(2Aepsiolon_(0) K)`
also in the downward direction. The net field at CD is `E_1 + E_2 = ((K+1) Q)/(2 Aepsilon_(0) K)` .
The force on the charge `-Q (1 - 1/K)` at CD is
`F = Q (1 - 1/K) ((K+1) Q)/(2 Aepsilon_(0) K)`
`= ((K^2 - 1) Q^(2))/(2 Aepsilon_(0) K^(2)`
in the upward direction. The weight of the liquid considerd is `hArhog`. Thus,
`hArhog = ((K^(2) - 1)Q^(2)/(2 Aepsilon_(0) K^(2))`
or, `h = ((K^2 - 1)Q^(2))/(2 A^(2) K^(2) epsilon_(0) rhog)`