Capacitance of the capacitor, \(\mathrm{C}=600 \mathrm{pF}\)
Potential difference, \(\mathrm{V}=200 \mathrm{~V}\)
Electrostatic energy stored in the capacitor is given by,
$$
E_{1}=\frac{1}{2} C V^{2}=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2} \mathrm{~J}=1.2 \times 10^{-5} \mathrm{~J}
$$
If supply is disconnected from the capacitor and another capacitor of capacitance \(=600 \mathrm{pF}\) is connected to it, then equivalent capacitance (Ceq) of the combination i given by,
$$
\frac{1}{C_{e q}}=\frac{1}{C}+\frac{1}{C}
$$
$$
\begin{aligned}
&\Rightarrow \frac{1}{C_{e q}}=\frac{1}{600}+\frac{1}{600}=\frac{2}{600}=\frac{1}{300} \\
&\Rightarrow C_{e q}=300 \mathrm{pF}
\end{aligned}
$$
New electrostatic energy can be calculated as
$$
E_{2}=\frac{1}{2} C_{e q} V^{2}=\frac{1}{2} \times 300 \times(200)^{2} \mathrm{~J}=0.6 \times 10^{-5} \mathrm{~J}
$$
Loss in electrostatic energy \(=\mathrm{E}_{1}-\mathrm{E}_{2}\)
$$
=1.2 \times 10^{-5}-0.6 \times 10^{-5} \mathrm{~J}=0.6 \times 10^{-5} \mathrm{~J}=6 \times 10^{-6} \mathrm{~J}
$$
Therefore, the electrostatic energy lost in the process is \(6 \times 10^{-6} \mathrm{~J}\).
