A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates and (iii) the energy stored in the capacitor, be affected ? Justify your answer.

Explanation

When battery is removed, charge Q remains constant.
(i) Capacity C increases.
(ii) Electric field `E=V/d` decreases as `V=Q/C` decreases.
(iii) Energy stored `=1/2 Q^(2)/C` decreases.