Q.Derive an expression for potential energy of a charged capacitor.

Charging a capacitor to charge Q and voltage V stores energy $U=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV$ in its field -- exactly HALF the total energy $QV$ that the charging battery actually supplies, since the battery delivers every increment of charge at its own full, constant voltage V while the capacitor's own voltage rises only gradually from 0 to V during charging. The other half is unavoidably dissipated as heat in the connecting wires and the battery's own internal resistance. …