A 4 μ F capacitor is charged by a 200 ~V battery. It is then...

Charge stored at the capacitor $q=C_1 V_1=4 \times 200=800 \mu \mathrm{C}$ When this capacitor is connected with a uncharged capacitor, then common potential on both capacitors $V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{800+0}{4+2}=\frac{800}{6} \mathrm{~V}$ Loss in energy = Initial energy - Final energy $$\begin{aligned} = & \frac{1}{2} C_1 V_1^2-\frac{1}{2}\left(C_1+C_2\right) V^2 \\ = & \frac{1}{2} \times 4 \times 10^{-6} \times(200)^2 \\ & \quad-\frac{1}{2}(4+2) \times 10^{-6} \times\left(\frac{800}{6}\right)^2 \\ = & 2 \times 10^{-6} \times 4 \times 10^4-\frac{3 \times 10^{-6} \times 64 \times 10^4}{36} \\ = & 8 \times 10^{-2}-\frac{64}{12} \times 10^{-2} \\ = & 8 \times 10^{-2}-5.33 \times 10^{-2} \\ = & 2.67 \times 10^{-2} \mathrm{~J} \end{aligned}$$