# Maxwell equation boundary conditions on a conducting sheet

I'm having difficulties solving boundary conditions for an infinitely thin conducting layer in a presence of an alternating field.

I use the Maxwell equations:
$\nabla \cdot \mathbf B = 0$
$\nabla \cdot \mathbf D = \rho$
$\nabla \times \mathbf E = i \omega \mathbf B$
$\nabla \times \mathbf H = \mathbf j - i \omega \mathbf D$
The problem arises with the last two ones. For stationary fields the terms $i \omega \mathbf B$ and $i \omega \mu \mathbf D$ vanish, and I can use the "textbook" boundary conditions like: $\hat n \times (\mathbf H_1 - \mathbf H_2)=\mathbf j$
Do those equations get an additional term $-i \omega \mathbf D$ in the case of time-varying fields?
If they do, those terms are different above and below the sheet, and I am not sure what value should be used.
The current on the sheet $\mathbf j = \sigma \mathbf E$ turns out to be discontinued as well.

What field (current) should be used as the field on the very boundary?

I would assume it's the average of the fields directly above and below the boundary, but I'm mostly guessing and don't know an exact proof. I am deeply grateful for any help.

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The treatments I've seen assume the frequencies are low enough that the displacement current is negligible with respect to the voltage: $ωϵ>>σ$. (This simplification can be considered the definition of a "bad" conductor.) That simplification adds the $iωD$ term.

The boundary condition comes from differentiating Maxwell #4 over a circle with edges inside and outside the resistor, and you should get consistent results no matter which circle you use. If you neglect displacement current you get a more simple condition than the standard one you quote. For an infinitely thick battery you can't get an edge "outside"; instead you get an equation relating lower H field, higher H field, and volume current (+ surface displacement current if you don't neglect it).

You probably want to over-determine the problem: for example, if the lower H-field and volume current are not specified, the total H-field is then left as an exercise to the reader.

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The treatments I've seen assume the frequencies are low enough that the displacement current is negligible with respect to the conduction current: $\omega \epsilon << \sigma$. (This simplification can be considered the definition of a "good" conductor.) That simplification eliminates the $i \omega \boldsymbol{D}$ term.

The boundary condition comes from integrating Maxwell #4 over a rectangle with edges inside and outside the conductor, and you should get consistent results no matter which rectangle you use. If you do not neglect displacement current you get a more complicated condition than the standard one you quote. For an infinitely thin conductor you can't get an edge "inside"; instead you get an equation relating upper H field, lower H field, and surface current (+ surface displacement current if you don't neglect it).

You don't want to over-determine the problem: for example, if the upper H-field and surface current are specified, the lower H-field is then determined.

update: Actually, with a finite conductivity $\sigma$, you can't get a non-zero surface current in an infinitely thing conductor: the resistance goes infinite in that limit. I think you either have to consider a perfect conductor or a finite conductor thickness. Jackson, Classical Electrodynamics, Section 8.1 has a treatment.

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