A capacitor of 2 μF is charged as shown in the diagram. When the switch S is turned to position 2 , the…

First position,

$C_1=2\text{μF}$
$C_2=8\text{μF}$
$Q=C_{1}V$
$U_i=\frac{Q^2}{2C_1}$

Second position,

$\frac{Q-q}{C_1}=\frac{q}{C_2}$
$C_{2}Q-C_{2}q=C_{1}q$
$q=\frac{C_{2}Q}{C_1+C_2}$
$\&Q-q=\frac{C_1}{C_1+C_2}Q$
Energy on $2 \mathrm{\mu F}=\frac{1}{2}\frac{{(Q-q)}^2}{C_1}$
${\left(U_{\mathrm{f}}\right)}_1=\frac{1}{2}\frac{C_1}{{(C_1+C_2)}^2}{Q}^2$
Energy on $8 \mathrm{\mu F}=\frac{1}{2}\frac{q^2}{C}$
${\left(U_f\right)}_2=\frac{1}{2}\frac{C_2}{{\left(C_1+C_2\right)}^2}{Q}^2$
Total energy $$\begin{gathered} =\frac{1}{2}\frac{C_1}{{(C_1+C_2)}^2}{Q}^2+\frac{1}{2}\frac{C_2}{{(C_1+C_2)}^2}{Q}^2 \\ =\frac{1}{2}\frac{Q^2}{(C_1+C_2)} \end{gathered}$$
Energy dissipated $=U_{\mathrm{f}}-U_{\mathrm{i}}=\frac{1}{2}\frac{C_2}{C_1(C_1+C_2)}{Q}^2$ $=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}{V}^2$

Percentage of energy dissipated, $=\frac{C_2}{C_1+C_2}\times 100=\frac{8}{2+8}\times 100=80\%$