A capacitor of \( 2\text{ }\mu\text{F} \) is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its storage energy dissipated is

Initial energy \( U_i = \frac{1}{2} C_1 V^2 \). When connected to an uncharged capacitor \( C_2 = 8\text{ }\mu\text{F} \), common potential is \( V' = \frac{C_1 V}{C_1 + C_2} \). Final energy \( U_f = \frac{1}{2}(C_1 + C_2)V'^2 = \frac{1}{2}\frac{C_1^2 V^2}{C_1 + C_2} = U_i (\frac{2}{2+8}) = 0.2 U_i \). Dissipated energy = \( 80\% \).