Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 18B / Solution

Using $E=-\nabla \varphi$ and the electrostatic Poisson equation, $\rho =-{ϵ}_0{\nabla }^2\varphi$, integration by parts givesThe boundary potential is zero, so the surface term vanishes. Hence $\boxed{W=({ϵ}_0/2){\int }_D{E}^{2}dV}$.

For the capacitor use the usual parallel-plate approximation, neglecting edge fields. The two gaps have widths $H+a$ and $H-a$, and their field magnitudes are $V/(H+a)$ and $V/(H-a)$. The field-energy expression givesThe two faces of the middle plate carry charges $Q_{-}={ϵ}_{0}AV/(H+a)$ and $Q_{+}={ϵ}_{0}AV/(H-a)$. Its total charge is their sum; the grounded plates have zero potential and contribute zero to $\frac{1}{2}\sum Q_j{\varphi }_j$. The charge-potential expression therefore gives $W=\frac{1}{2}V(Q_{-}+Q_{+})$, exactly the same result. Since $H^2-a^2$ is largest at $a=0$, the energy at fixed $V$ is minimized when the middle plate is centered, with value ${ϵ}_{0}A{V}^2/H$.

This is the electrostatic energy of a three-plate capacitor in the intended uniform-field model. Finite circular plates have fringing corrections; no small-gap-to-radius ratio is explicitly stated in the source, so the simple formula should not be described as an exact finite-disc solution.