$C_1$ = 4 µF , $V_1$ = 200 V , $C_2$ = 2 µF , $V_2$ = 0 So, common potential difference across the two capacitors after connection is V = $\frac{C_1V_1+C_2V_2}{C_1+C_2}$ = $\frac{4\times10^{-6}\times200+0}{(4+2)\times10^{-6}}$ = 133.33 V Initially, total energy stored in capacitors before connection is $U_i$ = $\frac{1}{2} C_1V_1^2$ = $\frac{1}{2}$ × 4 × $10^{-6} \times 200^2$ = 0.08 J and total energy stored in capacitors after connection is $U_f$ = $\frac{1}{2} (C_1+C_2)V^2$ = $\frac{1}{2}$ (4 + 2) × $10^{-6} \times 133.33^2$ or $U_f$ = 0.053 J So, energy lost due to connection is ΔU = $U_f-U_i$ = 0.053 - 0.08 or ΔU = - 0.027 J