Differential form of gauss's law and surface charge distributions

In summary, the integral form of Gauss's law can be written as \oint \vec{E}.\vec{d \sigma}=\frac{1}{\epsilon_0} \left[ \int \rho d \tau + \oint \delta d \sigma \right], which can be generalized to the differential form by using the divergence theorem. However, when accounting for surface charges, we need to represent them as charge densities in order to maintain the form of Gauss's law.
  • #1
ShayanJ
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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
 
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  • #2
Shyan said:
[itex]\oint \vec{E}.\vec{d \sigma}=\frac{1}{\epsilon_0} \left[ \int \rho d \tau + \oint \delta d \sigma \right] [/itex]

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.
 
  • #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.
 

1. What is the differential form of Gauss's Law?

The differential form of Gauss's Law is a mathematical expression that relates the electric field at a point to the distribution of electric charge in the surrounding space. It is given by the equation ∇⋅E = ρ/ε₀, where ∇⋅E represents the divergence of the electric field, ρ is the charge density, and ε₀ is the permittivity of free space.

2. How is the differential form of Gauss's Law different from the integral form?

The differential form of Gauss's Law is a local equation, meaning it describes the electric field at a specific point. It is derived from the integral form of Gauss's Law, which is a global equation that describes the total electric flux through a closed surface. The differential form is often used for calculations involving continuous charge distributions, while the integral form is more useful for discrete charge distributions.

3. How are surface charge distributions related to Gauss's Law?

Gauss's Law can be used to determine the electric field created by a surface charge distribution. By applying the differential form of Gauss's Law, we can find the electric field at a point near the surface by considering the charge density at that point and integrating over the entire surface. This allows us to calculate the electric field at any point near the surface without having to explicitly consider the individual charges on the surface.

4. Can the differential form of Gauss's Law be used for non-uniform charge distributions?

Yes, the differential form of Gauss's Law can be used for both uniform and non-uniform charge distributions. However, the calculation may be more complex for non-uniform distributions as it involves integrating over the entire charge distribution to determine the total electric field at a particular point. It is often easier to use the integral form of Gauss's Law for non-uniform charge distributions.

5. Can the differential form of Gauss's Law be applied to both conductors and insulators?

Yes, the differential form of Gauss's Law can be applied to both conductors and insulators. However, in conductors, the electric field is always zero inside the material, so the equation simplifies to ∇⋅E = 0. In insulators, the electric field may vary inside the material, but it must still satisfy the differential form of Gauss's Law at all points.

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