$C_1$ = 600 pF, $V_1$ = 200 V , $C_2$ = 600 pF , $V_2$ = 0 On connecting charged capacitor to uncharged capacitor, the common potential V across the capacitors is V = $\frac{C_1V_1+C_2V_2}{C_1+C_2}$ = $\frac{600 \times 10^{-12} \times 200 + 0}{(600+600) \times 10^{-12}}$ or V = 100 V Energy stored in capacitors before connection is $U_i$ = $\frac{1}{2} C_1 V_1^2$ + 0 = $\frac{1}{2}$ × 600 × $10^{-12} \times 200^2$ or $U_i$ = 12 µJ and energy stored in capacitors after connection is $U_f$ = $\frac{1}{2} (C_1+C_2) V^2$ = $\frac{1}{2}$ (600 + 600) × $10^{-12} \times 100^2$ or $U_f$ = 6 µJ Hence the energy lost in the process is Δ = $U_f-U_i$ = (6 - 12) µJ or ΔU = - 6 µJ