To solve this problem, we need to calculate the energy dissipated when the switch is moved from position 1 to position 2.$\\$Initially, the 2 μF capacitor is charged to a voltage $V. \\$The initial energy stored in the 2 μF capacitor is given by:$\\ E_{initial} = \frac{1}{2} C_1 V^2 \\ E_{initial} = \frac{1}{2} \times 2 \times 10^{-6} \times V^2 \\ E_{initial} = 10^{-6} V^2 \\$When the switch is moved to position 2, the 2 μF capacitor is connected in parallel with the 8 μF capacitor.$\\$The total capacitance $C_{total}$ is:$\\ C_{total} = C_1 + C_2 \\ C_{total} = 2 \, \mu F + 8 \, \mu F \\ C_{total} = 10 \, \mu F \\$The charge $Q$ on the 2 μF capacitor is:$\\ Q = C_1 V \\ Q = 2 \times 10^{-6} \times V \\$This charge is now distributed over the total capacitance, so the new voltage $V'$ is:$\\ V' = \frac{Q}{C_{total}} \\ V' = \frac{2 \times 10^{-6} \times V}{10 \times 10^{-6}} \\ V' = \frac{V}{5} \\$The final energy stored in the system is:$\\ E_{final} = \frac{1}{2} C_{total} (V')^2 \\ E_{final} = \frac{1}{2} \times 10 \times 10^{-6} \times \left(\frac{V}{5}\right)^2 \\ E_{final} = \frac{1}{2} \times 10 \times 10^{-6} \times \frac{V^2}{25} \\ E_{final} = \frac{10^{-6} V^2}{5} \\ E_{final} = 2 \times 10^{-7} V^2 \\$The energy dissipated $E_{dissipated}$ is:$\\ E_{dissipated} = E_{initial} - E_{final} \\ E_{dissipated} = 10^{-6} V^2 - 2 \times 10^{-7} V^2 \\ E_{dissipated} = 8 \times 10^{-7} V^2 \\$The percentage of energy dissipated is:$\\$Percentage$= \left(\frac{E_{dissipated}}{E_{initial}}\right) \times 100 \\$Percentage$= \left(\frac{8 \times 10^{-7} V^2}{10^{-6} V^2}\right) \times 100 \\$Percentage$= 80\% \\$Therefore, the percentage of stored energy dissipated is $80\%$, which corresponds to Option 2.