A parallel plate capacitor (air filled) is connected to a 25 volt battery and left alone for a long time. The charge on the capacitor is 221 nanoCoulombs. The battery is then disconnected, and the plates are pulled apart so that their separation distance is 3 times the original. What is the change in energy stored in the capacitor in micro-Joules

Answer: 5.52 * 10^-6J

Explanation: In order to explain this problem we have to take into account the expression for the stored energy in a capacitor which is given by:

U=Q^2/2*C=2.76*10^-6 J (initial stored energy)

We also know that C=Q/V=221*10^-9/25 V=8.84*10^-9 F

When the separation between the plates is increased 3 times, the capacitanve decrease a factor 1/3 then the stored energy is:

U=Q^2 / (2*C/3) = 3*Q^2 / (2*C) = 3 * (221*10^-9) ^2/2*8.84*10^-9=8.28*10^-6J (final stored energy)

The change in the stored energy is : 8.28*10^-6J-2.76*10^-6 J=

=5.52 * 10^-6 J