A capacitor of $2\mu F$ is charged as shown in the diagram. When the switch $S$ is turned to position 2 , the percentage of its stored energy dissipated is

 

$\text{ Initially, the energy stored in }2\mu \mathrm{F} \text{ capacitor is }$

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

initially, the charge stored in  $2\mu \mathrm{F}$ capacitor is

$Q_i=CV=\left(2\times {10}^{-6}\right)V=2V\times {10}^{-6}$ coulomb. When switch S is turned to position 2, the charge fl ows and both the capacitors share charges till a common potential VC is reached.

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

$\text{ Finally, the energy stored in both the capacitors }$

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

$\text{\% loss of energy, }\Delta U=\frac{U_i-U_f}{U_i}\times 10\%$

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