Fixed-voltage capacitor energy balance ($d{U}_{\mathrm{field}}=d{W}_{\mathrm{mech}}+VdQ$)

For a quasistatic change of capacitance at fixed voltage, the capacitor energy is $U=C{V}^2/2$ and charge is $Q=CV$. Electrical work supplied to the field is $VdQ$, so mechanical work is $dU-VdQ=-V^{2}dC/2$. Increasing the separation of parallel plates makes $dC<0$: mechanical work is positive even though field energy decreases, because electrical energy returns to the source.