A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the electrostatic energy stored in the combined system to that stored initially in the single capacitor.

Explanation

Let parallel plate capacitor of capacitacne C be charged to a potential V. Then the energy stored in this capacitor `u_i = 1/2 CV^2`
Now, let the charged capacitor is connected to another uncharged capacitor of same capacitance C. Obviously two capacitors shere the charge of first capacitor and as a result the common potential of capacitors `= V. = Q/(2C) = (CV)/(2C) = V/2`
`:.` Total energy stored by the capacitors `u_f = 2[1/2 CV._2] = 2 xx 1/2 C xx (V/2)^2 =1/4 CV^2`
`rArr u_f/u_i = (1/4CV^2)/(1/2CV^2) =1/2`