\(U_{i}=\frac{1}{2} C V^{2}=\frac{1}{2}\left(2 \times 10^{-6}\right)\, V^{2}=V^{2} \times 10^{-6} \,\mathrm{J}\)

Initially, the charge stored in \(2 \,\mu \mathrm{F}\) capacitor is

\(Q_{i}=C V=\left(2 \times 10^{-6}\right) V=2 V \times 10^{-6}\) coulomb. When switch \(S\) is turned to position \(2,\) the charge flows and both the capacitors share charges till a common potential \(V_{c}\) is reached.

\(V_{C}=\frac{\text { total charge }}{\text { total capacitance }}=\frac{2 V \times 10^{-6}}{(2+8) \times 10^{-6}}=\frac{V}{5} \text { volt }\)

Finally, the energy stored in both the capacitors

\(U_{f}=\frac{1}{2}\left[(2+8) \times 10^{-6}\right]\left(\frac{V}{5}\right)^{2}=\frac{V^{2}}{5} \times 10^{-6}\, \mathrm{J}\)

\(\%\) loss of energy, \(\Delta U=\frac{U_{i}-U_{f}}{U_{i}} \times 100 \%\)

\(=\frac{\left(V^{2}-V^{2} / 5\right) \times 10^{-6}}{V^{2} \times 10^{-6}} \times 100 \%=80 \%\)