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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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3 Answers

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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Your problem is singular, so you need to use distribution. Vector field operators reads:

div(B)={div(B)}+n12.(B2-B1)*delta(S) curl(E)={curl(E)}+n12x(E2-E1)*delta(S)

{div(B)} is the function derivative, and delta(S) is the 2D delta distribution located on S. Then you equate the regular part with the regular part and the singular part (with the delta(S) )with singular part in each equation.

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