# A 12 pF capacitor is connected to a 50V battery.

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, = 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 $1.5 \times 10^{-8} \mathrm{~J}$.

Leave a comment