Question

A parallel-plate capacitor has plate area A. A battery is used to charge the capacitor so that the magnitude of charge on each plate is Q, and then is disconnected. Initially, the capacitor has a plate separation of d. At this separation the capacitor contains energy U. The plates are then moved to a separation of 2d without disturbing the charge. What is the energy of the capacitor at this larger plate separation?

Answer 1

`U_f=2U_o`

When we double the separation between the plates of the capacitor the electric Energy is doubled too.

Step-by-step explanation:

First we analyse the Capacitance.

Initial:
`C_o=\epsilon*S/d`

Final:
`C_f=\epsilon*S/(2d)=C_o/2`

Energy:

Initial:
`U_o=Q^2/(2*C_o)`

Initial:
`U_f=Q^2/(2*C_f)=2*Q^2/(2*C_o)=2U_o`

When we double the separation between the plates of the capacitor the Energy electric is doubled too.