As uncharged capacitor is connected parallel

So, \(\quad \mathrm{C}^{\prime}=2 \mathrm{C}\)

and \(\quad \mathrm{V}_{\mathrm{c}}=\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\)

\(\mathrm{V}_{\mathrm{c}}=\frac{\mathrm{q}+0}{\mathrm{C}+\mathrm{C}} \Rightarrow \mathrm{V}_{\mathrm{c}}=\frac{\mathrm{V}}{2}\)

Initial Energy of system, \(\mathrm{U}_{\mathrm{i}}=\frac{1}{2} \mathrm{CV}^{2} \ldots(\mathrm{i})\)

Final energy of system, \(\mathrm{U}_{\mathrm{f}}=\frac{1}{2}(2 \mathrm{C})\left(\frac{\mathrm{V}}{2}\right)^{2}\)

\(=\frac{1}{2} \mathrm{CV}^{2}\left(\frac{1}{2}\right).....(ii)\)

From equation \((i)\) and \((ii)\)

\(\mathrm{U}_{\mathrm{f}}=1 / 2 \mathrm{U}_{\mathrm{i}}\)

i.e., Total electrostatic energy of resulting system decreases by a factor of \(2\)