A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0U0U_0. The capacitor remains connected to the battery while the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is UUU. Find the ratio U/U0U/U0.

Explanation:

Energy Stored in Capacitors

The capacitance of a parallel-plate capacitor is given by

Where is the permitivity of the dielectric, A is the area of the plates and d is their separation. If the separation was doubled, the new capacitance would be

The energy stored in the initial condition is

And the energy stored when the separation doubles is

Thus the ratio

The energy is half the initial energy

more than, less

less than, more