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

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.

At fixed electric potential difference $V$, a change of capacitance has electrical work input $VdQ=V^{2}dC$ and field-energy increase $dU=V^{2}dC/2$. The remaining half is mechanical work done by the field, so the force in the direction of increasing overlap is $F_x=V^2{C}^{\prime }(x)/2$. Equivalently the effective mechanical potential is $-C{V}^2/2$. For a partially filled coaxial capacitor, this gives $F_x=\pi {ϵ}_0({ϵ}_r-1)V^2/\ln (b/a)$, directed inward when ${ϵ}_r>1$. The fringing field of a capacitor supplies the local field gradient and transmits this force even when its small correction to total capacitance is neglected.