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

Explanation

The charge on the capacitor is
`Q=CV=900 xx 10^(-12) xx 100=9 xx 10^(-8)C`
The energy stored by the capcitor is
`=(1/2)CV^(2)=(1/2)QV=(1/2) xx 9 xx 10^(-8) xx 100=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 potentia. Let the common potential difference be V.. The charge on each capacitor is then Q.=CV.. By charge conservation 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`. 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.