A capacitor holding 4.0 J of stored energy is connected to an...

Let the capacitance of each capacitor be $C$ and the initial charge on the first capacitor be $Q$.

Initial energy of the first capacitor is $U_{i} = \dfrac{Q^2}{2C} = 4.0 \text{ J}$.

When it is connected to an identical uncharged capacitor, the total charge $Q$ is conserved.

The equivalent capacitance of the parallel combination becomes $C_{eq} = C + C = 2C$.

The final total energy stored in the system is $U_{f} = \dfrac{Q^2}{2C_{eq}} = \dfrac{Q^2}{2(2C)}$.

$U_{f} = \dfrac{1}{2} \left( \dfrac{Q^2}{2C} \right) = \dfrac{1}{2} U_{i}$.

Substituting the given value, $U_{f} = \dfrac{1}{2} \times 4.0 \text{ J} = 2.0 \text{ J}$.