Question

A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while the capacitor remains connected to the battery?

Answer 1

new energy of the capacitor is half that of initial energy

Step-by-step explanation:

As we know that the energy stored in the capacitor is given as

`U = (1)/(2)CV^2`

here we know that capacitance of parallel plate capacitor is given as

`C = (\epsilon_0 A)/(d)`

V = voltage of battery

so now we have capacitor remains connected to the same battery and the separation between the plates is doubled

so we have

`C' = (\epsilon_0 A)/(2d)`

so now the energy stored between the plates is

`U = (1)/(2)((C)/(2))V^2`

so new energy of the capacitor is half