 Fixed-voltage capacitor energy balance

For a quasistatic change of capacitance at fixed terminal electric potential difference, the capacitor energy is $U=C{V}^2/2$ and plate electric charge is $Q=CV$. Electrical work supplied to the electric field by the battery is $VdQ$, so mechanical work done on the capacitor is $d{W}_{\mathrm{mech}}=dU-VdQ=-V^{2}dC/2$. Increasing the separation of a parallel-plate capacitor makes $dC<0$: mechanical work is positive even though electrostatic energy decreases, because electrical energy returns to the source. Conversely, when $dC>0$, the battery supplies $V^{2}dC$: half increases capacitor energy and half is available as mechanical work done by the electric field. Differentiating capacitor energy alone at fixed terminal electric potential difference omits this exchange with the battery.