1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

EM reflection coefficient off a conducting plane

  1. Sep 21, 2011 #1
    Let's say there is a conducting plane of conductivity [itex]\sigma[/itex]. Boundary conditions for above and below such plane are:
    [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]\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.

    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
  2. jcsd
  3. Sep 22, 2011 #2
    You may already realize this, but unless the conductor is very thin or very lossy, to a very good approximation conductors are perfect reflectors.
  4. Sep 23, 2011 #3


    User Avatar
    Science Advisor
    Gold Member

    The charge density \rho is the free charge that has collected on the surface. The boundary conditions implicitly take into account the induced bound charges. That is, by requiring that the normal component of the electric flux density is continuous across a boundary (when in a source free region), we imply that the normal component of the electric field is discontinuous across the boundary (assuming that the permittivity has a step discontinuity). This means that there is a surface charge that is built up on the boundary to cause this discontinuous electric field. So any discontinuity in the flux density has to be brought about by a previously impressed surface charge.

    So, in a source free region (which is probably the case you are actually considering), we assume that the tangential electric field is continuous and the normal electric flux density is continuous. This is fully sufficient (along with the dual condition for the magnetic field) to define the reflection and transmission coefficients of the fields for any material. This is because the conductivity of the material is encapsulated into the imaginary part of the permittivity.

    As chrisbaird already mentioned, a very conductive medium can usually be approximated as being a perfect electrical conductor under many circumstances. One can also make the approximation of it being a good conductor (where the permittivity is purely imaginary). That is, in general,

    [tex]\epsilon = \epsilon'+i\frac{\sigma}{\omega} [/tex]

    For perfect electrical conductors we assume that the conductivity, \sigma, is infinite. For good conductors we assume that \epsilon is purely imaginary due to the dominance of \sigma.

    Now impressed sources change things. A current affects the magnetic bondary conditions while charge affects the electric. However, the reflection coefficients of the fields depend on their polarization with respect to the surface of the interface. As such, one can expect in general that the resulting reflection of the electric fields can depend upon both the source surface charge and currents.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook