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Electric surface current on a PEC

  1. Oct 1, 2014 #1
    Hello!
    When considering the boundary conditions for the electromagnetic field [itex]\mathbf{E}, \mathbf{H}[/itex] on the surface of a Perfect Eletric Conductor we have:
    • [itex]\mathbf{E} \times \mathbf{\hat{n}} = 0[/itex]
    • [itex]\mathbf{J}_S = \mathbf{\hat{n}} \times \mathbf{H}[/itex]
    the tangential electric field should vahish, while the tangential magnetic field is discontinuous and the discontinuity generates the surface current density.

    The current density flows on the surface a PEC, so we can consider as an equivalent situation the superposition of [itex]\mathbf{J}_S[/itex] and its image current, which is exactly opposite of [itex]\mathbf{J}_S[/itex]: the net current is 0 (this argument is used in order to prove that an electric current flowing on a PEC does not radiate).

    But doesn't this affect the boundary conditions on the magnetic field? If the current is zero, why the relative boundary condition is not written as [itex]0 = \mathbf{\hat{n}} \times \mathbf{H}[/itex]?!

    Emily
     
  2. jcsd
  3. Oct 6, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Oct 6, 2014 #3

    jasonRF

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    Not quite. I think you are confusing impressed currents and induced currents. impressed currents are not carried by the charges in the PEC; they are carried by external agents. When you impress an electric arbitrarily close to a PEC (like you might do when invoking equivalence theorems), it induces a current in the PEC that acts to effectively short out the current (reciprocity is a nice way to prove this).

    For the rest of your question, consider the simple case of a plane wave normally incident on a plane PEC. The total H field on the surface is twice the incident field, which should allow you to compute the electric surface current on the PEC. This surface current on the PEC is radiating the reflected electromagnetic wave.

    jason
     
  5. Oct 7, 2014 #4

    Meir Achuz

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    What does this mean? There is no "image current".
     
  6. Jul 15, 2015 #5
    Thank you for your answers and sorry for the great delay. It was misleading to me the fact that for several days there were no replies.
    Thank you jasonRF for your complete and useful answers. For Meir Achuz: as said by jasonRF, I made a little confusion between impressed currents and induced currents.
     
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