A 600 pF capacitor is charged by a 200 V supply.

Solution:

Capacitance of the capacitor, C = 600 pF

Potential difference, V = 200 V

Electrostatic energy stored in the capacitor is given by,

$E=\frac{1}{2} C V^{2}$

$=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2}$

$=1.2 \times 10^{-5} \mathrm{~J}$

If supply is disconnected from the capacitor and another capacitor of capacitance $C=600 \mathrm{pF}$ is connected to it, then equivalent capacitance (C) of the combination is given by,$\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{C}$

$\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{C}$

$=\frac{1}{600}+\frac{1}{600}=\frac{2}{600}=\frac{1}{300}$

$\therefore C^{\prime}=300 \mathrm{pF}$

New electrostatic energy can be calculated as

$E^{\prime}=\frac{1}{2} \times C^{\prime} \times V^{2}$

$=\frac{1}{2} \times 300 \times(200)^{2}$

$=0.6 \times 10^{-5} \mathrm{~J}$

Loss in electrostatic energy $=E-E^{\prime}$

$=1.2 \times 10^{-5}-0.6 \times 10^{-5}$

$=0.6 \times 10^{-5}$

$=6 \times 10^{-6} \mathrm{~J}$

Therefore, the electrostatic energy lost in the process is $6 \times 10^{-6} \mathrm{~J}$.