Solution:

Let \(C\) be the capacitance of each capacitor. For capacitor A, initial energy \(U_A = \frac{Q_A^2}{2C} = 4\text{J}\), so \(Q_A = \sqrt{8C}\). Capacitor B is uncharged, so \(Q_B = 0\). When connected in parallel, total charge is conserved: \(Q_{total} = Q_A + Q_B = \sqrt{8C}\). The equivalent capacitance is \(C_{eq} = C + C = 2C\). The final total energy is \(U_{total} = \frac{Q_{total}^2}{2C_{eq}} = \frac{(\sqrt{8C})^2}{2(2C)} = \frac{8C}{4C} = 2\text{J}\).