Question

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.

• A. $\frac{(K^2-1)Q^2}{2A^2{K}^2{E}_0\rho g}$
  
• B. $\frac{(K^2+1)Q^2}{2A^2{K}^2{E}_0\rho g}$
  
• C. $\frac{(K^2-1)Q^2}{A^2{K}^2{E}_0\rho g}$
  
• D. $\frac{(K^2+1)Q^2}{A^2{K}^2{E}_0\rho g}$
  

Answer 1

The correct option is A $\frac{(K^2-1)Q^2}{2A^2{K}^2{E}_0\rho g}$


The situation is as shown in the figure.


A charge $-Q(1-\frac{1}{K})$ is induced on the upper surface of the liquid and $Q(1-\frac{1}{K})$ at the surface in contact with the lower plate.

So, the net charge on the lower plate is $-Q+Q(1-\frac{1}{K})=\frac{-Q}{K}$.

Consider the equilibrium of the liquid in the volume $ABCD$. The forces on this liquid are :
(a) The force due to the electric field at $CD$
(d) The weight of the liquid
(c) The force due to atmospheric pressure
(d) The force due to the pressure of the liquid below $AB$

As $AB$ is in the same horizontal level as the outer surface, the pressure here is the same as the atmospheric pressure.

The forces at $C$ 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 $E_1=\frac{Q}{2A{\epsilon }_0}$ (downward).

The field at $CD$ due to the charge $\frac{-Q}{K}$ is $E_2=\frac{Q}{2A{\epsilon }_{0}K}$ (downward).

$\therefore$ Net field at $CD$ is

$E_1+E_2=\frac{(K+1)Q}{2A{\epsilon }_{0}K}$

The force on the charge $-Q(1-\frac{1}{K})$ at $CD$ is

$F=Q(1-\frac{1}{K})\frac{(K+1)Q}{2A{\epsilon }_{0}K}$

$\Rightarrow F=\frac{(K^2-1)Q^2}{2A{\epsilon }_0{K}^2}$ (upward)

The weight of the liquid considered is $hA\rho g$.

Thus, at equilibrium

$hA\rho g=\frac{(K^2-1)Q^2}{2A{\epsilon }_0{K}^2}$

$\Rightarrow h=\frac{(K^2-1)Q^2}{2A^2{K}^2{\epsilon }_0\rho g}$

Hence, option (a) is the correct answer.