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Gauss' law for dielectrics

  1. Jun 16, 2012 #1
    Inside a dielectric we have:

    ∇[itex]\cdot[/itex]ε0E = ρbound + ρfree , where ρbound refers to the fact that these charges come from polarization.

    We can write this as:

    ∇[itex]\cdot[/itex]ε0E = -∇[itex]\cdot[/itex]P + ρfree

    where P is the polarization of the material. And combing the two divergence terms we get:

    ∇[itex]\cdot[/itex]D = ρfree

    which is Gauss' law for dielectrics which is quite useful sometimes. However wouldn't it only hold for solid, spherically symmetric, dielectrics, where you consider r<R. My speculation comes from the fact that this derivation does NOT consider the bound surface charges than a polarization can result in. This is of course of no problem if you are inside a solid sphere, but I don't see how it wouldn't be a problem in every other case.
     
  2. jcsd
  3. Jun 16, 2012 #2

    Born2bwire

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    I don't quite understand your reasoning here, the divergence of the polarization is the bound charge density. For example, if you look at the boundary conditions between say vacuum and dielectric. The displacement field is continuous (assuming no free charges) and the electric field is discontinous. The discontinuity in the electric field arises from the boundary charges that are induced and of course we see that this must be facilitated by the polariation field for the displacement field to be continuous.
     
  4. Jun 16, 2012 #3
    well consider a hollow sphere with inner radius a and outer radius b. According to theory of polarization bound charges will accumulate both on the inner and outer surface of the sphere. So could you use Gauss' law in the region where a<r<b? Certainly not right? Since there are bound surface charges inside any sphere with radius bigger than a, which are not accomodated for in the above derivation of Gauss' law for D.
     
  5. Jun 16, 2012 #4
    Yes you can. Gauss's law is linear so you can take the total (enclosing all) and subtract a smaller interior sphere (enclosing the inner surface).

    I can't vouch for the above derivation but the shapes of the surfaces don't matter as long as they are closed. The extension to multiply connected domains can be made in the way I laid out above.
     
  6. Jun 16, 2012 #5
    hmm so you mean that if I am inside some weird-shape closed surface with a uniform surface charge then I wont register the field from the surface charges? hmm.. Wouldn't that only hold for a sphere or an infinite cylinder?
     
  7. Jun 18, 2012 #6
    No, it holds for all closed surfaces. The Gauss integral registers only the enclosed charge. The surface could be shaped like an Alien facehugger. If the charge is inside its right hanging sack it will be counted in the integral. If the charge is under one of its knuckles (say on your forehead) then it will contribute zero to the integral.
     
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