Answer :

Let the battery connected to the capacitor be of potential *V*.

Let the length of the part of the slab inside the capacitor be *x*.

b – Width of plates

The capacitor can be considered to be two capacitors which are connected in parallel.

The capacitances of the two capacitors in parallel is given by –

*C*_{1} is the part of the capacitor having the dielectric inserted in it and *C*_{2} is the capacitance of the part of the capacitor without dielectric.

As, *C*_{1} and *C*_{2} are in parallel therefore, the net capacitance is given by

This dielectric slab is attracted by the electric field of the capacitor and applies a force

The direction of force is in left direction.

Let assume that electric force of magnitude *F* pulls the slab toward left direction.

Let there be an differential displacement *dx* towards the left direction by the force *F*.

The work done by the force

Let *V*_{1} and *V*_{2} be the potential of the battery connected to the left capacitor and that of the battery connected with the right capacitor

With increase in the displacement of slab, the capacitance will increase, hence the energy stored in the capacitor will also increase.

In order to maintain constant voltage, the battery will supply extra charge, and gets damage .

Therefore the battery will do work.

Now,

Work done by the battery

= Energy change of capacitor + work done by the force *F* on the capacitor

1)

Let’s take the differential charge d*q* is supplied by the battery, and the change in the capacitor be *dC*

We know that energy in capacitor dW_{B}

we know q = cv

⇒ 2)

And force F is given by,

From 1) and 2)

In order to keep the dielectric slab in equilibrium, the electrostatic force acting on it must be balanced by the weight of the block attached.

Therefore,

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