Question

A parallel plate capacitor is charged by a battery which is then disconnected. A dielectric slab is then inserted to fill the space between the plates. Explain the change in the charge on the plates.​

Answer 1

Answer:

(i) The capacitance increases as the dielectric constant K>1.

ii) Potential difference V= CQ

. As C increases and Q remains the same since the battery is disconnected, the p.d. between the plates decreases.

(iii) Electric field E= v/d

where V is the p.d. and d the separation between the plates. As V decreases and d remains the same, electric field also decreases.