4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ hint: what is the 3d density of a surface current? $\endgroup$ – hyportnex Jan 3 '18 at 13:46
  • 2
    $\begingroup$ 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. $\endgroup$ – Ben Crowell Jan 28 at 15:13
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

If there was surface current at the boundary separating conductor and nonconducting linear medium, then the parallel component of electric field at the boundary (on both the sides) would be responsible to drive the current. Now, in the conducting side $\sigma$ is non-zero, hence finite parallel component of electric field exist that is driving the current, but in the nonconducting linear medium side, $\sigma$ is zero hence for finite current, infinite electric field should be present (since its an insulator). Therefore, on one side we have finite parallel component of E and on the other side it is infinite, but boundary condition require parallel components to be equal, a contradiction. Therefore, surface current at the boundary should be zero. The inference of infinite E as a consequence of $\sigma=0$ can also be made using equation $ J=\sigma E $ as limit of area through which current is flowing is zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.