A $2\mathrm{\mu F}$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is

When switch S is connected to terminal 1, the potential difference across the $2\mathrm{\mu F}$ capacitor is V volt. Therefore, energy stored in the system is

$\begin{array}{l}{\mathrm{U}}_1=\frac{1}{2}{\mathrm{C}}_1{\mathrm{V}}^2=\frac{1}{2}\times 2\times {\mathrm{V}}^2\\ ={\mathrm{V}}^2\mathrm{\mu J}\end{array}$

When switch S is turned to terminal 2, the charge will flow from $2\mathrm{\mu F}$ capacitor to $8\mathrm{\mu F}$ capacitor until their potentials are equalized. The common potential is

$\begin{array}{l}{\mathrm{V}}^2=\frac{\mathrm{q}}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{{\mathrm{C}}_1\mathrm{V}}{{\mathrm{C}}_1+{\mathrm{C}}_2}\\ =\frac{2\mathrm{V}}{(2+8)}=\frac{\mathrm{V}}{5}\text{ volt }\end{array}$

$\therefore$ Energy stored in the system now will be

$\begin{array}{l}{\mathrm{U}}_2=\frac{1}{2}\left({\mathrm{C}}_1+{\mathrm{C}}_2\right){\mathrm{V}}_2^2\\ =\frac{1}{2}(2+8)\times {\left(\frac{\mathrm{V}}{5}\right)}^2=\frac{{\mathrm{V}}^2}{5}\mathrm{\mu J}\end{array}$

$\therefore$Percentage loss of energy is

$\frac{{\mathrm{U}}_1-{\mathrm{U}}_2}{{\mathrm{U}}_1}\times 100=\frac{\left({\mathrm{V}}^2-\frac{{\mathrm{V}}^2}{5}\right)}{{\mathrm{V}}^2}\times 100=80\mathrm{\%}$