Question

A capacitor is connected to a battery. If dielectric slab of strength k is now inserted between the plates to fill the space completely, then ( U is initial energy stored)​

Answer 1

Answer:

See below.

Explanation:

Since the battery is connected after the insertion of the capacitor, the potential difference across the capacitor remains unchanged.

We know that the energy stored in a capacitor is given by,

$U=\dfrac{1}{2}CV^{2}$

After the insertion of the dielectric, the capacitance is KC where K is the dielectric constant of the slab. Hence, the energy stored becomes,

$U'=\dfrac{1}{2}KCV^{2}$

Note, that the value of V does not change.

The change in energy stored is,

$\begin{aligned}U'-U=\dfrac{1}{2}CV^{2}\left( K-1\right) \\ =U\left( K-1\right) \end{aligned}$

Since K > 1, there will be an increase in the energy stored.

Hence, option 1 is correct.

Tip:

On inserting a dielectric,

If the battery/voltage supply is connected,  V ( P.D across the capacitor ) remains constant

If the battery/voltage supply is not connected,  Q ( charge on the capacitor ) remains constant.