Question

A potential difference is set up between the plates of a parallel plate capacitor by a battery and then the battery is removed. If the distance between the plates is decreased, then how the (a) charge (b) potential difference (c) electric field (d) energy and (e) energy density will change?

Answer 1

The capacitance of a parallel plate capacitor is given by C=εAd, where A is the area and d is the separation of the plates. Now since the capacitor is connected with the cell the potential will not change. But capacity will increase if we decrease the separation.

a)Q=CV
as Cincreases, charge Q also increase

b) Energy will increase as the energy is given by E=1/2 CV^2

c) Potential will remain same and will be equal to the emf of cell.

d) Electric field will decrease since the potential difference will remain same and distance decreases

e) Energy density will also increase since the energy is increase