Initially, the energy stored in 2 $\mu$F capacitor is

${U_i} = {1 \over 2}C{V^2} = {1 \over 2}\left( {2 \times {{10}^{ - 6}}} \right){V^2} = {V^2} \times {10^{ - 6}}J$

Initially, the charge stored in 2 $\mu$F capacitor is
Q_i = CV = (2 × 10^–6)V = 2V × 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.

${V_C} = {{{\rm{total }}\,{\mathop{\rm charge}\nolimits} } \over {total\,capacitance}}$

$= {{2V \times {{10}^{ - 6}}} \over {\left( {2 + 8} \right) \times {{10}^{ - 6}}}} = {V \over 5}volt$

Finally, the energy stored in both the capacitors

${U_f} = {1 \over 2}\left[ {\left( {2 + 8} \right) \times {{10}^{ - 6}}} \right]\left( {{V \over 5}} \right)$

$= {{{V^2}} \over 5} \times {10^{ - 6}}J$

% loss of energy, $\Delta U = {{{U_i} - {U_f}} \over {{U_i}}} \times 100\%$

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