Percentage of Stored Energy Dissipated in Connected Capacitors

When the switch $S$ is initially in position 1, the capacitor $C_1 = 2\ \mu\text{F}$ is connected across a source of potential difference $V$.

The initial energy stored in the capacitor $C_1$ is:
$U_i = \frac{1}{2} C_1 V^2$

The initial charge acquired by the capacitor $C_1$ is:
$Q = C_1 V$

When the switch $S$ is turned to position 2, the source of potential difference is disconnected, and the charged capacitor $C_1$ is connected in parallel with an uncharged capacitor $C_2 = 8\ \mu\text{F}$.

By the principle of conservation of charge, the total charge remains constant:
$Q = (C_1 + C_2) V_f$

where $V_f$ is the common final potential across the two capacitors. Solving for $V_f$:
$V_f = \frac{C_1 V}{C_1 + C_2}$

Substituting the given values $C_1 = 2\ \mu\text{F}$ and $C_2 = 8\ \mu\text{F}$:
$V_f = \frac{2}{2 + 8} V = \frac{2}{10} V = \frac{V}{5}$

The total final energy stored in the system of two capacitors is:
$U_f = \frac{1}{2} (C_1 + C_2) V_f^2$

Substituting $V_f = \frac{C_1 V}{C_1 + C_2}$ into the equation:
$U_f = \frac{1}{2} (C_1 + C_2) \left(\frac{C_1 V}{C_1 + C_2}\right)^2 = \frac{1}{2} \frac{C_1^2 V^2}{C_1 + C_2} = \left(\frac{C_1}{C_1 + C_2}\right) U_i$

Calculating $U_f$ in terms of $U_i$:
$U_f = \frac{2}{2 + 8} U_i = \frac{2}{10} U_i = 0.2 U_i$

The energy dissipated during this process ($\Delta U$) is the difference between the initial and final stored energy:
$\Delta U = U_i - U_f = U_i - 0.2 U_i = 0.8 U_i$

Therefore, the percentage of stored energy dissipated is:
$\text{Percentage dissipated} = \frac{\Delta U}{U_i} \times 100\% = \frac{0.8 U_i}{U_i} \times 100\% = 80\%$

Thus, the correct option is D.