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In sec. 9.4.2 Griffiths shows the well known boundary conditions for E and B fields, one of them is this:

$$\frac{1}{\mu_{1}}\textbf{B}_{1}^{\parallel}-\frac{1}{\mu_{2}}\textbf{B}_{2}^{\parallel}=\textbf{K}_{f}\times\hat{\textbf{n}}$$

Where $\textbf{K}_{f}$ is the free surface current. Griffiths says in this section:

"... For Ohmic conductors ($\textbf{J}_{f}=\sigma\textbf{E}$) there can be no surface current, since this would require an infinite electric field at the boundary."

I just can't understand it yet. Why is this true?

I have another question: The boundary for the normal component of E is $$\epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp}=\sigma_{f}$$ Where $\sigma_{f}$ is the free surface charge.

When we treat with incident EM waves on a conductor, is it necessary to consider $\textbf{K}_{f}$ and $\sigma_{f}$ different to zero? I am asking this because in this section of the book Griffiths made $\textbf{K}_{f}=0$ and $\sigma_{f}$ vanishes naturally because he only studies normal incidence, but my question goes to the most general case in which the normal component of E is nonzero.

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    $\begingroup$ hint: what is the 3d density of a surface current? $\endgroup$
    – hyportnex
    Jan 3 '18 at 13:46
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    $\begingroup$ We need Dirac's Delta, don't we? $\endgroup$
    – Amadeus
    Jan 3 '18 at 23:43
  • $\begingroup$ In AC circuits, the skin depth can be arbitrarily small. $\endgroup$
    – user4552
    Jan 28 '19 at 15:13
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I think this is so because for finite conductivity and for ohmic conductors, J=$\sigma$E would require that the current density be parallel electric field. Since for conductors, electric field is perpendicular to the surface, so J (current) would also be normal to surface. But the boundary condition n $\times$ H = K requires K not to be normal to surface (as it should be perpendicular to the normal), thus there would be no surface current.

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