Question

A parallel plate capacitor is charged to a potential difference V by d.c.
source and then disconnected. The distance between the plates is then
halved. Then write the change in
(1) Electric potential difference between the plates
(2) Electric field intensity
(3) Capacitance of parallel plate capacitor.​

Answer 1

Answer:

Since the charge as well as the area of the plates does not change, the electric field between the plates also does not change. Let the initial capacitance be C and the final capacitance be C'. Hence, the capacitance of the capacitor gets halved when the distance between the plates is doubled.

Explanation:

Hope this helps

Answer 2

Answer:

(i) Electric field between the plates decrease as E=Vd

(ii) Capacitance becomes half i.e., decreases as C=ε0Ad

(iii) Energy stored increases as E=1Q22C.