Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 4 / 7D (Electromagnetism)

The grounded conducting plane supplies the induced charge needed for a uniform field between the plates and zero field above the upper plate. Gauss's law gives $E_z=-\sigma /{ϵ}_0$ in the gap. With electric potential zero at the grounded plate,Integrating the electrostatic energy density ${ϵ}_0|\mathbf{E}{|}^2/2$ across the gap gives the capacitor energy per unit areaFor an isolated upper plate, charge is fixed, so increasing $d$ to $(1+\alpha )d$ gives exactly $\Delta u=\frac{1}{2}\sigma V\alpha$, with $V$ its initial value. The positive increase equals the mechanical work done against attraction.

At fixed voltage, the new charge density is $\sigma /(1+\alpha )$, and the energy becomes $u/(1+\alpha )$. Consequently $\Delta u=-\frac{1}{2}\sigma V\alpha +O({\alpha }^2)$. Charge $\sigma \alpha +O({\alpha }^2)$ per unit area flows back to the voltage source, which receives energy $\sigma V\alpha +O({\alpha }^2)$. The externally supplied mechanical work is still $\frac{1}{2}\sigma V\alpha +O({\alpha }^2)$: the field loses half this returned electrical energy and the mechanical work supplies the other half. Thus the fixed-voltage capacitor energy balance includes both field and source energy.