# Why does Griffiths's book say that there can be no surface current since this would require an infinite electric field for an incident wave?

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.

• hint: what is the 3d density of a surface current? Jan 3 '18 at 13:46
• We need Dirac's Delta, don't we? Jan 3 '18 at 23:43
• In AC circuits, the skin depth can be arbitrarily small.
– user4552
Jan 28 '19 at 15:13

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.