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Electromagnetism at an Interface

Module by: Paul Padley. E-mail the author

Summary: We derive the properties of E and B fields at an interface between two media.

EM at an interface

We want to understand with Electromagnetism what happens at a surface. From Maxwell's equations we can understand what happens to the components of the E E and B B fields: First lets look at the E E field using Gauss' law. Recall ε E d s = ρ V ε E d s = ρ V

Figure 1
Figure 1 (Gauss-E-at-interface_small.png)
Consider the diagram, the field on the incident side is E i + E r E i + E r . On the transmission side, the field is E t E t . We can collapse the cylinder down so that it is a pancake with an infinitely small height. When we do this there are no field lines through the side of the cylinder. Thus there is only a flux through the top and the bottom of the cylinder and we have; ε E d s = [ ε i ( E i + E r ) ε t E t ] s = 0. ε E d s = [ ε i ( E i + E r ) ε t E t ] s = 0. I have set ρ V = 0 ρ V = 0 since we will only consider cases without free charges. So we have ε i E i + ε i E r = ε t E t ε i E i + ε i E r = ε t E t if u ̂ n u ̂ n is a unit vector normal to the surface this can be written ε i u ̂ n E i + ε i u ̂ n E r = ε t u ̂ n E t ε i u ̂ n E i + ε i u ̂ n E r = ε t u ̂ n E t

Similarly Gauss' law of Magnetism B d s = 0 B d s = 0 gives B i + B r = B t B i + B r = B t or u ̂ n B i + u ̂ n B r = u ̂ n B t u ̂ n B i + u ̂ n B r = u ̂ n B t

Amperes law can also be applied to an interface.Then

Figure 2
Figure 2 (Ampere-B-at-interface_small.png)
B μ d l = j d s + t ε E d s B μ d l = j d s + t ε E d s (note that in this case d s d s is perpendicular to the page)

Now we will not consider cases with surface currents. Also we can shrink the vertical ends of the loop so that the area of the box is 0 so that ε E d s = 0 ε E d s = 0 . Thus we get at a surface B i + B r μ i = B t μ t B i + B r μ i = B t μ t or u ̂ n × B i μ i + u ̂ n × B r μ r = u ̂ n × B t μ t u ̂ n × B i μ i + u ̂ n × B r μ r = u ̂ n × B t μ t

Similarly we can use Faraday's law E d l = t B d s E d l = t B d s and play the same game with the edges to get E i + E r = E t E i + E r = E t or u ̂ n × E i + u ̂ n × E r = u ̂ n × E t u ̂ n × E i + u ̂ n × E r = u ̂ n × E t (notice ε ε does not appear)

In summary we have derived what happens to the E E and B B fields at the interface between two media: ε i u ̂ n E i + ε i u ̂ n E r = ε t u ̂ n E t ε i u ̂ n E i + ε i u ̂ n E r = ε t u ̂ n E t

u ̂ n B i + u ̂ n B r = u ̂ n B t u ̂ n B i + u ̂ n B r = u ̂ n B t

u ̂ n × B i μ i + u ̂ n × B r μ i = u ̂ n × B t μ t u ̂ n × B i μ i + u ̂ n × B r μ i = u ̂ n × B t μ t

u ̂ n × E i + u ̂ n × E r = u ̂ n × E t u ̂ n × E i + u ̂ n × E r = u ̂ n × E t

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