Solution:

Given,

Capacitance, $C=600 \mathrm{pF}$

Potential difference, $V=200 v$

Electrostatic energy stored in the capacitor is given by:

$E_{1}=\frac{1}{2} C V^{2}=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2} \mathrm{~J}=1.2 \times 10^{-5} \mathrm{~J}$

According to the question, if the source is unplugged from the $600 pF$ capacitor and connected to another capacitor of the same value, the equivalent capacitance $C_eq$ of the combination is provided by the equation,

$\frac{1}{C_{e q}}=\frac{1}{C}+\frac{1}{C}$

$\frac{1}{C_{e q}}=\frac{1}{600}+\frac{1}{600}$

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

$\mathrm{C}_{\mathrm{eq}}=300 \mathrm{pF}$

The following formula can be used to compute new electrostatic energy:

$E_{2}=\frac{1}{2} C V^{2}=\frac{1}{2} \times 300 \times 10^{-12} \times(200)^{2} \mathrm{~J}=0.6 \times 10^{-5} \mathrm{~J}$

Electrostatic energy is lost as a result of this.

$E=E_{1}-E_{2}$

$E=1.2 \times 10^{-5}-0.6 \times 10^{-5} \mathrm{~J}=0.6 \times 10^{-5} \mathrm{~J}=6 \times 10^{-6} \mathrm{~J}$

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