[Physics] Maxwell equation boundary conditions on a conducting sheet

boundary conditionselectromagnetism

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.

Best Answer

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.