= Solution
Using $E=-\nabla\phi$ and the electrostatic <Poisson equation>, $\rho=-\epsilon_0\nabla^2\phi$, <integration by parts> gives
$$
W=-\frac{\epsilon_0}{2}\int_D\phi\nabla^2\phi\,dV
=\frac{\epsilon_0}{2}\int_D|\nabla\phi|^2dV
-\frac{\epsilon_0}{2}\int_{\partial D}\phi\,\partial_n\phi\,dS.
$$
The boundary potential is zero, so the surface term vanishes. Hence \b[$\boxed{W=(\epsilon_0/2)\int_DE^2dV}$].
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 gives
$$
\boxed{W=\frac{\epsilon_0AV^2}{2}\left(\frac1{H+a}+\frac1{H-a}\right)
=\frac{\epsilon_0AV^2H}{H^2-a^2}.}
$$
The two faces of the middle plate carry charges $Q_-=\epsilon_0AV/(H+a)$ and $Q_+=\epsilon_0AV/(H-a)$. Its total charge is their sum; the grounded plates have zero potential and contribute zero to $\frac12\sum Q_j\phi_j$. The charge-potential expression therefore gives $W=\frac12V(Q_-+Q_+)$, exactly the same result. Since $H^2-a^2$ is largest at $a=0$, \b[the <energy> at fixed $V$ is minimized when the middle plate is centered], with value $\epsilon_0AV^2/H$.
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.
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