Consider the integral form of gauss's law(adsbygoogle = window.adsbygoogle || []).push({});

[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

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# Differential form of gauss's law and surface charge distributions

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