A 900 pF capacitor is charged by 100 V battery. The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic energy stored by the system?
 
 

 $Q=CV=(900\times {10}^{-12}F)\left(100V\right)=9\times {10}^{-8}C$
Energy stored by the capacitor,
$U=\frac{1}{2}QV=\frac{1}{2}(9\times {10}^{-8}C)\left(100V\right)$   $=4.5\times {10}^{-6}J$
In steady state, pd across each capacitor, $V\text{'}=\frac{v}{2}$ 
Charge on each capacitor,  $Q\text{'}=CV\text{'}=CV/2=Q/2$
Total energy of the system,  $U\text{'}=2(\frac{1}{2}Q\text{'}V\text{'})$  $=\left[\frac{1}{2}\left(\frac{Q}{2}\right)\left(\frac{V}{2}\right)\right]$  $=\frac{1}{4}QV=\frac{1}{2}U$

$=2.25\times {10}^{-6}J$