A capacitor of 2\ is charged as shown in the diagram. When the...

Initial energy stored in the capacitor $C_1$ is:

$U_i = \dfrac{1}{2}C_1 V^2$

When the switch is turned to position 2, the charge is shared between $C_1$ and $C_2$. The common potential $V_c$ is:

$V_c = \dfrac{C_1 V}{C_1 + C_2}$

The final energy of the system is:

$U_f = \dfrac{1}{2}(C_1 + C_2)V_c^2 = \dfrac{1}{2}(C_1 + C_2)\left(\dfrac{C_1 V}{C_1 + C_2}\right)^2 = \dfrac{1}{2}\dfrac{C_1^2 V^2}{C_1 + C_2}$

The energy dissipated is:

$\Delta U = U_i - U_f = \dfrac{1}{2}C_1 V^2 - \dfrac{1}{2}\dfrac{C_1^2 V^2}{C_1 + C_2} = \dfrac{1}{2}\dfrac{C_1 C_2}{C_1 + C_2}V^2$

The percentage of stored energy dissipated is:

$\text{Percentage} = \dfrac{\Delta U}{U_i} \times 100 = \dfrac{\dfrac{1}{2}\dfrac{C_1 C_2}{C_1 + C_2}V^2}{\dfrac{1}{2}C_1 V^2} \times 100 = \dfrac{C_2}{C_1 + C_2} \times 100$

Given $C_1 = 2\ \mu\text{F}$ and $C_2 = 8\ \mu\text{F}$:

$\text{Percentage} = \dfrac{8}{2 + 8} \times 100 = \dfrac{8}{10} \times 100 = 80\%$

Answer: $80\ \%$