A capacitor is charged by connecting a battery across its plates. It stores energy $$U.$$ Now the battery is disconnected and another identical capacitor is connected across it, then the energy stored by both capacitors of the system will be

Solution :
When a capacitor is charged by connecting a battery across its plates, the initial energy stored,
$$U = \frac{{{q^2}}}{{2C}}$$
When the battery is disconnected, then the charge remains constant i.e. $$q =$$  constant. Now, another identical capacitor is connected across it i.e. the capacitors are connected in parallel, so the equivalent capacitance
$$\eqalign{ & {C_{{\text{eq}}}} = {C_1} + {C_2} \cr & = C + C = 2C \cr} $$
Thus, final energy stored by the system of capacitors,
$$\eqalign{ & U' = \frac{{{q^2}}}{{2{C_{{\text{eq}}}}}} \cr & U' = \frac{{{q^2}}}{{2\left( {2C} \right)}} \cr & U' = \frac{1}{2}U \cr & \therefore U' = \frac{U}{2} \cr} $$