Question:
A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Solution:
Capacitor of the capacitance, $C=12 \mathrm{pF}=12 \times 10^{-12} \mathrm{~F}$
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
$E=\frac{1}{2} C V^{2}$
$=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^{2}$
$=1.5 \times 10^{-8} \mathrm{~J}$
Therefore, the electrostatic energy stored in the capacitor is $1.5 \times 10^{-8} \mathrm{~J}$.