Question

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of:

(i) capacitance

(ii) potential difference between the plates

(iii) electric field between the plates

(iv) the energy stored in the capacitor .

Answer 1

Answer:

Let the original capacitance, potential difference be C and V respectively.

then the new capacitance will be C'=Ck (where k is the dielectric value).

The new Potential Difference will be V'=V/k.

And the energy stored in the capacitor will be=((Ck)$V'^{2}$)/2$k^{2}$