
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 2V.
Now, loss of energy, ΔH=Ui−Uf
⇒ΔH=21CV2−21(2C)×(2V)2
⇒ΔH=21CV2−41CV2 ⇒ΔH=41CV2=41×50×10−12×1002=125×10−9J=125nJ