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omnyx
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Let's say there is a conducting plane of conductivity [itex]\sigma[/itex]. Boundary conditions for above and below such plane are:
Now, let's look at the same problem only this time we externally drive constant surface current [itex]K[/itex] through the plane. Tangential component of the electric field should still be continuous, but wouldn't the in plane field now be [itex]K/\sigma[/itex]? What would be the surface charge density [itex]\rho[/itex] in such case? I think the reflection coefficient should depend on K.
Your thoughts are appreciated
[itex]E_1^t = E_2^t [/itex]
[itex](D_2 - D_1 ) \cdot n = \rho [/itex]
i.e. tangential components of E field are continuous and perpendicular components of D are discontinuous because of surface charge density [itex]\rho[/itex]. But, what exactly does [itex]\rho[/itex] refer to? If there is no free surface current (i.e. only the one induced by incoming E field) then one is tempted to write, assuming plane wave vector k and frequency w[itex](D_2 - D_1 ) \cdot n = \rho [/itex]
[itex]\rho = j k/\omega [/itex]
[itex]j = \sigma E^t [/itex]
where j is the "induced" surface current, and [itex]E^t[/itex] is the tangential component of the electric field (which we said is continuous above/below the surface). I.e. the in plane electric field is equal to the tangential electric field below (or above) the plane. From these two equations one can get the reflection coefficient off a conducting plane.[itex]j = \sigma E^t [/itex]
Now, let's look at the same problem only this time we externally drive constant surface current [itex]K[/itex] through the plane. Tangential component of the electric field should still be continuous, but wouldn't the in plane field now be [itex]K/\sigma[/itex]? What would be the surface charge density [itex]\rho[/itex] in such case? I think the reflection coefficient should depend on K.
Your thoughts are appreciated