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