Fixed-voltage capacitor energy balance
= Fixed-voltage capacitor energy balance
{title2=$dU_{\rm field}=dW_{\rm mech}+V\,dQ$}
For a quasistatic change of <capacitance> at fixed voltage, the <capacitor energy> is $U=CV^2/2$ and charge is $Q=CV$. Electrical work supplied to the field is $V\,dQ$, so mechanical work is $dU-V\,dQ=-V^2dC/2$. 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.