A capacitor of capacitance 50pF is charged by 100V source. It is...

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 $\dfrac{V}{2}$.

Now, loss of energy, $\DeltaH = {U}_{i} - {U}_{f}$

$⇒ \DeltaH = \dfrac{1}{2}C{V}^{2} - \dfrac{1}{2}\left(2C\right) \times {\left(\dfrac{V}{2}\right)}^{2}$

$⇒ \DeltaH = \dfrac{1}{2}C{V}^{2} - \dfrac{1}{4}C{V}^{2}$ $⇒ \DeltaH = \dfrac{1}{4}C{V}^{2} = \dfrac{1}{4} \times 50 \times {10}^{- 12} \times {100}^{2} = 125 \times {10}^{- 9}J = 125nJ$