Parallel-Plate Vacuum Capacitor

LostinPhysics

If a parallel-plate vacuum capacitor, with its plates separated by a distance of x1 has an amount of energy equal to U stored in it. The separation is then decreased to x2 .

What is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed? I know that I am suppose to use the formula C= Q/V but how do I put U in terms of x1 and x2?

mukundpa

As the battery is disconnected, the charge on the capacitor plates becomes constant. Now by changing capacitance the potential difference will change and hence it is batter to use the formula Q^2 /2C of the energy.

Capacitance in the two situations are to be calculated and Q is calculated by first situation only.