A capacitor of capacitance 50 pF is charged by 100   V source. It is then connected to another…

Initial charge on first capacitor will be, $Q=CV$

As the second capacitor is identical to the first one, hence potential drop across both capacitor will be equal to $\frac{V}{2}$.

Now, loss of energy, $\Delta H=U_i-U_f$

$\Rightarrow \Delta H=\frac{1}{2}C{V}^2-\frac{1}{2}\left(2C\right)\times {\left(\frac{V}{2}\right)}^2$

$\Rightarrow \Delta H=\frac{1}{2}C{V}^2-\frac{1}{4}C{V}^2$  $\Rightarrow \Delta H=\frac{1}{4}C{V}^2=\frac{1}{4}\times 50\times {10}^{-12}\times {100}^2=125\times {10}^{-9} \mathrm{J}=125 \mathrm{nJ}$