Capacitors

Let $C_1$ = 2 μF and $C_2$ = 8 μF. The charge stored in the capacitor before discharging is Q = $C_1V$ = 2µC E = $\frac{Q^2}{2C_1}$ = 1 µJ After closing switch 2, at study state, we have $\frac{Q-q}{2\mu F}$ = $\frac{q}{8\mu F}$ However, Q = 2 μC. Therefore, $1-\frac{q}{2\mu F}$ = $\frac{q}{8\mu F}$ or q = $\frac{8}{5}\,\mu C$ Hence, the energy stored in capacitors $C_1$ and $C_2$ is E' = $\frac{(Q-q)^2}{2C_1} + \frac{q^2}{2C_2}$ = $\frac{\left[2-\frac{8}{5}\right]^2}{2\times 2} + \frac{\left(\frac{8}{5}\right)^2}{2\times 8}$ = $\left(\frac{1}{5}\right)\,\mu J$ Therefore, the percent of the energy dissipated is $\frac{E'-E}{E} \times 100$ = $\frac{\frac{1}{5} - 1}{1}$ × 100 = 80%