The slowly charging capacitor is the standard example used to illustrate that the displacement current density is needed in Ampere's law if we want to correctly determine the magnetic field between capacitor plates. However, in any quasi-static situation the magnetic field can also be determined using the Biot-Savart law including only the real current densities. In this work, we will numerically calculate the magnetic field due to the surface currents on the capacitor plates and add it to the magnetic field due to the charging wire and show how they combine to create the correct magnetic field thoughout all space. For regions to the left or right of the capacitor, we find the surprising result that the surface currents replicate the magnetic field that would have been created by the missing section of the charging wire between the plates. For points between the capacitor plates, the magnetic field due to the surface currents mostly cancels the magnetic field from the near-infinite length charging wire, resulting in the well-known reduced field in that interior region. We will also illustrate the impact of finite capacitor plates on these results and briefly comment on how textbook and/or classroom discussions could be improved by carefully discussing these details.

## References

Imagine a third surface with the same shape as $S2$ but with a smaller radius so it cuts across capacitor plates III and IV. That surface would be pierced by the surface currents on the plates and by the displacement currents.

Since introductory textbooks always discuss charging or discharging *RC* circuits, it is likely that students think that the standard solutions in Eqs. (2) and (3) apply to those circuits. However, since those currents are exponentially decreasing with time, those solutions wouldn't satisfy Faraday's Law.

Consider an isolated capacitor whose infinitely large plates have equal and opposite charge densities. Now connect a wire between the plates along their symmetry axis. If we apply Eq. (1) to a point analogous to point P in Fig. 1 and use a circular Amperian loop with surface $S1$, we would conclude that there is no magnetic field to the left or right of the plates. This means that the magnetic field due to the surface currents on the plates perfectly cancels the magnetic field due to the discharging wire across them. Since these surface currents are in exactly the opposite direction from those in the charging capacitor problem we are focusing on here, this also means that the surface currents for the charging plates must create exactly the *same* magnetic field as the missing current element across the plates would have created.

In fact, in the region $z\u2009\u2272\u20092$ cm, the ratio is within $0.002%\u22120.008%$ of 1.0. This difference is so small that we are reaching the limits of the numerical algorithm. If one looks *very* closely at Fig. 3, one can see that curve A lies slightly above 1.0 at $\u223cz=0.8$ cm (by $\u223c0.008%$). This certainly isn't physically possible. We can definitely say that in that small *z* region the magnetic field due to the surface currents is essentially identical to the magnetic field that would have been produced by the missing current element across the capacitor but the exact difference between them isn't accurately calculated here. At larger *z*, the numerical limitations are not relevant and we have an accurate picture of the ratio.

Note that the maximum contribution of the surface currents on the plates to the net magnetic field is 50%. Imagine a geometry where the capacitor plates are infinitely large so that Eq. (2) applies and also assume that $d\u226b\rho \u226bz$. In this geometry, the charging wire to the right of point P is a semi-infinite current so it must contribute exactly half the infinite straight current field. That means that the surface currents on the plates and the charging wire far to the left must contribute the other half. In this limit, the influence of the charging wire on the left will be negligible so the surface currents must create the other 50% of the field. Since we know that the magnetic field due to the surface currents replicates the field that would have been created by the missing current element, we can see this mathematically using Eq. (6). Notice that the prefactor in front is exactly half the infinite straight current field and that both terms in parenthesis are guaranteed to be <1. In this limit, the first term approaches 1.0 and the second approaches zero so the resulting expression is at most 50% of Eq. (2).

Note that the curving of the fringing electric field lines does not affect our use of Eq. (1) since our Amperian loop is equidistant from each capacitor plate. Thus, the electric field lines (and hence the displacement current) are completely perpendicular to the loop.

$E\u2192(r\u2192\u2009)=(1/4\pi \u03f50)\u222bV((\rho (r\u2032\u2192)(r\u2192\u2212r\u2032\u2192)d\tau \u2032)/|r\u2192\u2212r\u2032\u2192|3)$ where *ρ* in this context is the volume charge density.

Of course, Eq. (3) has an exactly linear increase for $\rho \u2264a$ and a $\rho \u22121$ decrease for $\rho \u2265a$ but that is the infinite parallel plate solution and not the correct one for the finite plates which we calculate here.

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