A 900 pF capacitor is charged by 100V battery.
(a) How much electrostatic energy is strored by the capacitor ? The capacitor is disconnected from the battery and connected to another 900 pF capacitor. How much is the electrostatic energy stored in the system ?

Explanation

(a) The charge on the capacitor is
`Q = CV = 900 xx 10^(-12) F xx 100 V = 9 xx 10^(-8)C`
The energy stored by the capacitor is
`= (1//2) CV^(2) = (1//2)QV`
`= (1//2) xx 9 xx 10^(-8) C xx 100 V = 4.5 xx 10^(-6) J`
(b) In the steady situation, the two capacitors have their positive plates at the same potential, and their negative plates at the same potential. Let the common potential difference be V′. The charge on each capacitor is then Q ′ = CV ′ . By charge conservation, Q ′ = Q/2. This implies V ′ = V/2. The total energy
of the system is `= 2 xx (1)/(2)Q'V' =(1)/(4)QV = 2.25 xx 10^(-6)J`
Thus in going from (a) to (b), though no charge is lost, the final energy is only half the initial energy. Where has the remaining energy gone?
There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.