Question

Which of the following would increase the capacitance of a parallel-plate capacitor? I. Insert a dielectric between the plates. II. Increase the surface area of each plate. III. Increase the separation distance between the plates. O I and II only OII and III only All of the above. A capacitor is charged with a battery to a voltage V and then disconnected from the battery. A dielectric is inserted between the plates. When the dielectric is inserted, what happens to the electrostatic potential energy stored in the capacitor? O The stored energy increases. O The stored energy decreases. The stored energy remains constant.

Answer 1

1) correct option is I and II only

we know the relation

C=k$\varepsilon$_oA/d

so if we increase the surface area it increases the capacitance.

when we introduce a dielectric between the plates of the capacitor the charge in the capacitor increases because the dielectric creates its own electric field due to polarization

and since C=q/v

capacitance increases.

2) the correct option is -the stored energy decreases.

the energy in the capacitor is given U=Q²/2C

as we have discussed in previous question, when dielectric is inserted then the capacitance increases which decreses the potential energy as capacitance is inversely proportional to potential energy.