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

Answer:

new energy of the capacitor is half that of initial energy

Explanation:

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

$U = \frac{1}{2}CV^2$

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

$C = \frac{\epsilon_0 A}{d}$

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

$C' = \frac{\epsilon_0 A}{2d}$

so now the energy stored between the plates is

$U = \frac{1}{2}(\frac{C}{2})V^2$

so new energy of the capacitor is half