$(a)$ $A$ $900 \; pF$ capacitor is charged by a $100 \; V$ battery [Figure $(a)$]. How much electrostatic energy is stored by the capacitor?
$(b)$ The capacitor is disconnected from the battery and connected to another $900 \; pF$ capacitor [Figure $(b)$]. What is the electrostatic energy stored by the system?

(N/A) The charge on the capacitor is $Q = C V = 900 \times 10^{-12} \; F \times 100 \; V = 9 \times 10^{-8} \; C$.
The energy stored by the capacitor is given by:
$U = \frac{1}{2} C V^2 = \frac{1}{2} Q V$
$U = \frac{1}{2} \times 9 \times 10^{-8} \; C \times 100 \; V = 4.5 \times 10^{-6} \; J$.
$(b)$ In the steady state,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 $Q' = C V'$.
By charge conservation,the total charge $Q$ is shared equally between the two identical capacitors,so $Q' = Q / 2$.
This implies $V' = V / 2 = 50 \; V$.
The total energy of the system is:
$U_{total} = 2 \times \left( \frac{1}{2} C (V')^2 \right) = 2 \times \frac{1}{2} \times (900 \times 10^{-12} \; F) \times (50 \; V)^2$
$U_{total} = 900 \times 10^{-12} \times 2500 = 2.25 \times 10^{-6} \; J$.