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Differential form of gauss's law and surface charge distributions

  1. Oct 9, 2012 #1

    ShayanJ

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    Gold Member

    Consider the integral form of gauss's law
    [itex] \oint \vec{E}.\vec{d \sigma}=\frac{q}{\epsilon_0} [/itex]
    Then lets write q in terms of the volume and surface charge densities [itex] \rho [/itex] and [itex] \delta [/itex] and lets assume that the surface charge is distributed over the gaussian surface of the above integral
    [itex]\oint \vec{E}.\vec{d \sigma}=\frac{1}{\epsilon_0} \left[ \int \rho d \tau + \oint \delta d \sigma \right] [/itex]
    Now lets apply gauss's theorem(and one of its corollaries)to the two surface integrals to get:

    [itex] \int \vec{\nabla}.\vec{E} d \tau=\frac{1}{\epsilon_0} \left[ \int \rho d \tau + \int \mid \vec{\nabla} \delta \mid d \tau \right] [/itex]

    Now we can write:
    [itex]
    \vec{\nabla}.\vec{E}=\frac{1}{\epsilon_0} \left[ \rho + \mid \vec{\nabla} \delta \mid \right]
    [/itex]

    The calculations seemed correct to me
    So I think it can be used as a generalization to the differential form of gauss's law for surface charge distributions but of course with its own limitations:
    1-The surface charge should vary with space somehow.
    2-The guassian surface should be the same as the surface on which the surface charge resides.
    I worked it out for [itex] \delta = q \cos{\theta} [/itex](in spherical coordinates)
    The differential equation had a degree of arbitrariness in it,so looks like a third problem arises.
    Any way,with my own choices,I found [itex] \vec{E}=-q \cos{\theta} \hat{\theta} [/itex]
    Which seems to be right
    Every thing seems correct to me but I can find nothing about it to confirm my results
    Any Ideas?
    Thanks
     
  2. jcsd
  3. Oct 9, 2012 #2
    This is in direct contradiction with the divergence theorem. The surface integral on the left is exactly equal to the volume integral on the right, and that is all. [itex]q[/itex] is taken to represent only the volume integral as a result.
     
  4. Oct 9, 2012 #3
    If there's a surface charge density, E is discontinuous on the surface. So the flux integral over the surface doesn't seem particularly well defined in this case, and the conclusions aren't justified.

    If you want to account for surface charges in the differential form of Gauss's law, first think about how we handle point charges. We keep the usual form of Gauss's law, ##\vec{\nabla} \cdot \vec{E} = \rho/\epsilon_0## and say that a point charge Q at position x_0 has a charge density given by a Dirac delta function: ##\rho(\vec{x}) = Q\delta^3(\vec{x}-\vec{x_0})##. We can do a similar thing to represent a uniform surface charge sigma on the z=z_0 plane as a charge density: ##\rho(\vec{x}) = \sigma \delta(z - z_0)##. The case of a line charge is left as an exercise to the reader.
     
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