Answer :

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

Considering the left capacitor -

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

The left capacitor can be considered to be two capacitors in parallel.

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 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

Therefore, the potential energy stored in the left capacitor will be

1)

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

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.

Let us consider a small displacement *dx* of the slab towards the inward direction.

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)

Solving for voltages V_{1} and V_{2} -

Similarly, for the right side the voltage of the battery is given by-

Now, the ratio of the voltages is given by-

Thus, the ratio of the emfs of the left battery to the right battery is given by -

Rate this question :

A capacitor is maPhysics - Exemplar

A parallel plate Physics - Exemplar

The battery remaiPhysics - Exemplar

A parallel-plate HC Verma - Concepts of Physics Part 2

How many time conHC Verma - Concepts of Physics Part 2

The plates of a cHC Verma - Concepts of Physics Part 2

A capacitor of caHC Verma - Concepts of Physics Part 2