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Consider the integral form of gauss's law
[itex] \oint \vec{E}.\vec{d \sigma}=\frac{q}{\epsilon_0} [/itex]
Then let's write q in terms of the volume and surface charge densities [itex] \rho [/itex] and [itex] \delta [/itex] and let's 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 let's 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
[itex] \oint \vec{E}.\vec{d \sigma}=\frac{q}{\epsilon_0} [/itex]
Then let's write q in terms of the volume and surface charge densities [itex] \rho [/itex] and [itex] \delta [/itex] and let's 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 let's 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