A body of capacity $4\mathit{\mu }F$ is charged to 80V and another body of capacity $6\mathit{\mu }F$ is charged to 30V. When they are connected, the energy lost by $4\mathit{\mu }F$ is

$V_{\text{com }}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

$=\frac{4\times {10}^{-6}\times 80+6\times {10}^{-6}\times 30}{4\times {10}^{-6}+6\times {10}^{-6}}=30+18=50$  V

Loss in energy of $4\mathit{\mu }F$

Capacitor is $\mathrm{\Delta }U=\frac{1}{2}C\left[V^2-V_{com}^2\right]$

$\frac{1}{2}\times 4\times {10}^{-6}\left[{80}^2-{50}^2\right]=7.8\times {10}^{-3}\mathrm{J}$

= 7.8 mJ