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Field due to charge on surface of dielectric

  1. Sep 18, 2009 #1
    We have an infinite half-space filled with a dielectric. The other half-space is vacuum. We take a sheet of charge of density \sigma and place it on the surface of the dielectric. What are the D & E fields everywhere?

    We clearly have D_1 + D_2 = \sigma. But what other equation do we have? Can we find D_1 & D_2 without knowing the properties of the dielectric?

    It seems like this should be very simple, but it's tripping me up. I don't think we can claim D_1 = D_2, because we don't have symmetry. Or am I wrong about this?
     
  2. jcsd
  3. Sep 18, 2009 #2
    Since no one has responded yet, I will give the reasoning I've come up with for the case of a linear dielectric. We know that the effect of the polarization will be to introduce a bound surface charge density on the surface of the dielectric. Instead of working with the D field and the free charge, we can work with the E field and the total (free plus bound) charge. If we do that, we will find we have a total charge density at the surface of (\sigma - \sigma_b) and the E-field will be the same on either side.

    We can then go back to D_1 + D_2 = \eps_0 E_1 + \eps E_2 = E(\eps_0 + \eps) = \sigma and we get E = \frac{\sigma}{\eps_0 + \eps} on both sides (opposite directions of course).

    This has the right limiting value of E=0 on each side in the case where we take the case of a conductor and let \eps -> \infty.

    This tells me that to find the values of the D field in this case you need to know something about the properties of the dielectric.
     
  4. Sep 18, 2009 #3
    It wasn't actually a homework question, but I agree that it sounds like it could be one :smile:
     
  5. Sep 19, 2009 #4
    you do understand that the only effect of the polarization of the dielectric will be to lessen the strength of the electric field within the dielectric itself? right? just calculate what it would be if there was no polarization then multiply/divide by the appropriate number.
     
  6. Sep 19, 2009 #5
    So I think you are saying that we can assume that the D field is the same on each side. But that would imply that the E field outside the dielectric would be a constant independent of the dielectric constant. And I don't think that's right, because in the limit of a very large dielectric constant (which I think is equivalent to being a conductor in the electrostatic case) the field should be zero everywhere, right?
     
  7. Sep 20, 2009 #6
    its not clear to me what would happen.
     
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