Question

A charged parallel plate capacitor of distance (d)
has U, energy. A slab of dielectric constant (K) and
thickness (d) is then introduced between the plates
of the capacitor. The new energy of the system is
given by :
(1) KU. (2) KĽU. (3) (4)​

Answer 1

Answer:

Explanation

C becomes K times of C

Thus C'= KC

And V' = V/K

New energy would be Unot / K

Answer 2

The new energy of the system is  given by :

(1) KU.

Let for the capacitor having distance 'd' have its capacitance as C.

After inserting a slab with dielectric constant K, let us assume that the new capacitance becomes C'.

For this situation, we know that the new capacitance C' is:

C' = KC

We know, energy of the capacitor with capacitance C is equal to (1/2)CV².

U = (1/2)CV².

Thus, energy of the capacitor with capacitance C' will be (1/2)C'V².

U' = (1/2)C'V².

As, C' = KC, substituting it in U' gives,

U' = (1/2)KCV².

U' = K(1/2)CV² = KU.            [As, U = (1/2)CV²]

Thus, U' = KU.