= Solution
Here $\mathbf H$ is <magnetic field intensity>, $\mathbf J$ free <electric current> <density>, and $\mathbf D$ electric displacement. In vacuum, $\mathbf B=\mu_0\mathbf H$ and $\mathbf D=\epsilon_0\mathbf E$. Integrating the equation gives <Ampère's circuital law> with <displacement current>; in a <steady state> the displacement term vanishes.
Take $z$ upward and define $V=\phi_{\rm upper}-\phi_{\rm lower}=V_0\sin\omega t$. Neglect fringing and use the low-frequency thin-capacitor approximation. Then
$$
\boxed{\mathbf E=E_z(t)\hat{\mathbf z},\qquad
E_z=-\frac{V_0}{L}\sin\omega t,\qquad E_r=E_\phi=0.}
$$
Inside the gap, a circular Ampère loop encloses only <displacement current>:
$$
2\pi rH_\phi=\epsilon_0\pi r^2\dot E_z,
\qquad
\boxed{\mathbf H=\frac{\epsilon_0r}{2}\dot E_z\,\hat{\boldsymbol\phi}
=-\frac{\epsilon_0r\omega V_0}{2L}\cos\omega t\,\hat{\boldsymbol\phi}.}
$$
Because $\hat{\mathbf z}\times\hat{\boldsymbol\phi}=-\hat{\mathbf r}$, the <Poynting vector> is
$$
\boxed{\mathbf S=-\frac{\epsilon_0r}{2}E_z\dot E_z\,\hat{\mathbf r}.}
$$
It points inward while the electrical <energy> is increasing and outward while it decreases. The net <power> entering the curved side is
$$
-2\pi aL S_r(a)=\epsilon_0\pi a^2L E_z\dot E_z
=\frac d{dt}\left(\frac{\epsilon_0\pi a^2L}{2}E_z^2\right)
=\dot U_E.
$$
This is the leading quasistatic <energy> balance. The time-averaged field energies are
$$
\overline U_E=\frac{\epsilon_0\pi a^2V_0^2}{4L},\qquad
\overline U_B=\frac{\mu_0\epsilon_0^2\omega^2\pi a^4V_0^2}{32L},
\qquad
\boxed{\frac{\overline U_B}{\overline U_E}=\frac18\left(\frac{\omega a}{c}\right)^2.}
$$
For the <axisymmetric high-frequency capacitor field> when $\omega a/c$ is not small, induction and radial propagation prevent a spatially uniform $E_z$. In an ideal axially uniform axisymmetric interior model, Faraday's and Ampère's equations give
$$
\frac1r\frac d{dr}\left(r\frac{dE_z}{dr}\right)+\frac{\omega^2}{c^2}E_z=0.
$$
The solution regular at the axis and normalized to the central <voltage> is approximately
$$
\boxed{E_z(r,t)=-\frac{V_0}{L}J_0(\omega r/c)\sin\omega t.}
$$
At $\omega=\pi c/a$, the radial profile is $J_0(\pi r/a)$: its first zero is at $r\simeq0.765a$, and the edge has opposite <phase>, $J_0(\pi)\simeq-0.304$. This is an illustrative thin-disc interior profile; the exact driven high-frequency field also depends on the feed and edge/radiation <boundary conditions>. Continuing the uniform quasistatic field would be inconsistent.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-gpp2014-capacitor-profile.png]
{title=Original ideal axisymmetric capacitor field profiles, comparing the uniform quasistatic limit with the Bessel radial profile at angular frequency pi times c over a}
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