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For a circular air-filled parallel-plate capacitor with voltage phasor $V_0{e}^{-i\omega t}$, uniform gap $d$ and radius $a$, the Ampère-Maxwell equation on a circular loop of radius $s<a$ gives the displayed magnetic field. The enclosed displacement current is proportional to $s^2$, whereas the circulation is $2\pi s{B}_{\varphi }$. Consequently its amplitude grows linearly from zero on the axis. This assumes negligible fringing fields and a quasistatic approximation $\omega a/c\ll 1$.

Neglect fringing field near the edges of the parallel-plate capacitor. Put the plates at $z=0,d$, with prescribed electric potentials $V_A,V_B$. Translational symmetry parallel to a plane makes the electrostatic potential independent of the coordinates parallel to the plates. Thus Laplace's equation reduces to $d^{2}V/d{z}^2=0$. Its general solution is $V=A+Bz$, and the two boundary conditions determine both constants:The electric field is uniform and points toward decreasing electric potential. No plate voltages are specified, so a numerical electric potential cannot be fixed; choosing one plate as the reference only fixes the arbitrary additive constant.

The continuity equation for electric charge is $\nabla \cdot J=-{\partial }_t\rho$. Taking the divergence of the magnetostatic Ampère's law would instead require $\nabla \cdot J=0$. The displacement current repairs this conflict: using Gauss's law, $\nabla \cdot ({ϵ}_0{\partial }_{t}E)={\partial }_t\rho$, so the Ampère-Maxwell equation isIn the parallel-plate capacitor, neglect fringing fields and take $E_z=V_0{e}^{-i\omega t}/d$. A circular Ampère loop of radius $s<a$ encloses displacement current $-i\omega {ϵ}_0\pi s^2{V}_0{e}^{-i\omega t}/d$. Therefore the magnetic field between circular capacitor plates has phasor and amplitudeIt is zero on the axis, grows linearly with radius, and is azimuthal by the right-hand rule. This is a quasistatic approximation: in addition to $d\ll a$, the uniform-field model requires $\omega a/c\ll 1$ and excludes the edge region. The answer gives the oscillation amplitude, rather than a time-independent instantaneous magnitude.

A scalar field unchanged by every translation parallel to a specified plane depends only on the coordinate normal to that plane. For electrostatics between ideal infinite parallel plates, this reduces Laplace's equation to $d^{2}V/d{z}^2=0$. Finite plate edges break the exact symmetry and produce fringing field.