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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 $\mathbf{D}={ϵ}_0\mathbf{E}$. Integrating the equation gives Ampère's circuital law with displacement current; in a steady state the displacement term vanishes.

Take $z$ upward and define $V={\varphi }_{\mathrm{upper}}-{\varphi }_{\mathrm{lower}}=V_0\sin \omega t$. Neglect fringing and use the low-frequency thin-capacitor approximation. ThenInside the gap, a circular Ampère loop encloses only displacement current:Because $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$, the Poynting vector isIt points inward while the electrical energy is increasing and outward while it decreases. The net power entering the curved side isThis is the leading quasistatic energy balance. The time-averaged field energies areFor 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 giveThe solution regular at the axis and normalized to the central voltage is approximatelyAt $\omega =\pi c/a$, the radial profile is $J_0(\pi r/a)$: its first zero is at $r\simeq 0.765a$, 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.

Fresnel diffraction retains quadratic aperture-dependent propagation phases and generally depends on both aperture and screen distance. Fraunhofer diffraction neglects those phases and gives the Fourier transform of the aperture in angle space, either sufficiently far away or in a suitable lens's focal plane.

The calculations below use the Fraunhofer, paraxial approximation. For an overall aperture scale $D\sim 1000\lambda$, free propagation requires $L\gg D^2/\lambda \sim 10^6\lambda$. The printed $L\gg 1000\lambda$ alone only makes the separation small relative to distance; it is not by itself the full far-field condition.