Solution:

Initially, the capacitor stores energy \(U_i = \frac{1}{2}CV^2\) with charge \(Q_i = CV\). When connected with reverse polarity, the capacitor eventually charges to (-V\), and final stored energy is \(U_f = \frac{1}{2}C(-V)^2 = \frac{1}{2}CV^2\). The net change in stored energy is \(0\). The charge that flows from the battery is \(Q_f - Q_i = (-CV) - (CV) = -2CV\), meaning (2CV) charge flows. The work done by the battery is \(W_B = (2CV) \cdot V = 2CV^2\). Since \(W_B = \Delta U + Q_{\text{loss}}\), and ( \Delta U = 0\), the loss of energy is \(Q_{\text{loss}} = W_B = 2CV^2\). Both A and R are true and R explains A in this specific case.