$(a)$ A $900 \;p\,F$ capacitor is charged by $100 \;V$ battery [Figure $(a)$]. How much electrostatic energy is stored by the capacitor?

$(b)$ The capacitor is disconnected from the battery and connected to another $90\; p\,F$ capacitor [Figure $(b)$]. What is the electrostatic energy stored by the system?

$(a)$ 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

$=(1 / 2) C V^{2}=(1 / 2) Q V$

$=(1 / 2) \times 9 \times 10^{-8} C \times 100 V =4.5 \times 10^{-6} \,J$

$(b)$ 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 $V ^{\prime} .$ The charge on each capacitor is then $g^{\prime}=C V^{\prime} .$ By charge conservation, $Q ^{\prime}= \beta / 2 .$ This implies $V^{\prime}=V / 2 .$ The total energy of the system is

$=2 \times \frac{1}{2} Q^{\prime} V^{\prime}=\frac{1}{4} Q V$$=2.25 \times 10^{-6} \,J$