Point-Source derivative form Gauss's law valid?

In summary, It was discovered during a review of Maxwell's equations that the volume integral of the derivative form of Gauss's equation appears to be 0 for point sources, when it should actually equal Q/(electro constant). This is due to the gradient being equal to 0 and anything else done to 0 resulting in it staying as 0. However, when both versions of Gauss's law were checked on valid field equations without singularities, everything worked out correctly. It was also noted that the equation for the field of a point source is -kq(x/r^3, y/r^3, z/r^3), where k = 1/4/pi/(electro constant) and r^2 = x^2
  • #1
ShamelessGit
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I reviewed Maxwell's equations today and I was upset to find that the volume integral of the derivative form of Gauss's equation seems to be 0 for point sources when it should come out to be Q/(electro constant). This comes from the fact that the gradient is equal to 0 and anything else you do to 0 appears to make it stay zero.

By the way I checked both versions of Gauss's law on valid field equations that did not have singularities and everything seemed to work out fine. (For instance, I used the integral form of gauss's law to create a field equation for gravity inside a sphere of uniform density to check the derivative form of Gauss's law)

What I did (and sorry I don't know how to make the math look pretty like you other guys do):

It looks like the equation for the field of a point source is -kq(x/r^3, y/r^3, z/r^3) where r^2 = x^2 + y^2 + z^2 and k = 1/4/pi/(electro constant). You can use Pythagoras's theorem to check that the magnitude of this vector is kq/r^2. When I took the divergence for x I got -1/(x^2 + y^2 + z^2)^1.5 + 3x^2/(x^2 + y^2 + z^2)^2.5 ... (pattern continues for y and z). Which equals -3/r^3 + 3(x^2 + y^2 + z^2)/r^5 = 3(1/r^3 - 1/r^3) = 0. Maybe you should check to make sure I did this right. This is what you would expect from Gauss's theorem I think because the charge density is zero everywhere except at the point source. However, taking the volume integral of this should give you kq4pi = q/(electro constant).

1. Have I made a mistake?

2. If not, is there a way to resolve this issue with the singularity to make Gauss's law valid in this case?
 
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  • #2
The problem is there is a singularity at the origin. The diivergence is not definable there.

Using spherical coordinates to make life easier, with vectors in bold,

D = Dr r where r = unit vector in r direction,
so
div D = (1/r2)∂/∂r(r2Dr)
which at r = 0 gives ∞ for the 1/r2 part and zero for the partial derivative, and of course Er= Dr0.

So what is zero times infinity? Trouble, that's what ...

Possibly one can fool around with the δ function to get around this. Any takers?

Somehow you did not take div D correctly. I suspect you "used" the ∞ times zero thing and got beached. As you probably know, you can "prove" 1 = 2 by falling in that trap.

P.S. the divergence theorem is OK because it multiplies div D by volume element dv = (4/3)πdr3 surrounding the point charge. dv is infinitesmally small but not zero, whereas a point charge has identically zero volume.
By the same token, the surface element dA surrounding dv is non-zero.
 
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Related to Point-Source derivative form Gauss's law valid?

1. What is the Point-Source derivative form of Gauss's law?

The Point-Source derivative form of Gauss's law is a mathematical expression that relates the electric flux through a closed surface to the charge enclosed within that surface. It is represented by the equation ∇·E = ρ/ε0, where ∇·E is the divergence of the electric field and ρ is the charge density.

2. How is the Point-Source derivative form of Gauss's law different from the integral form?

The Point-Source derivative form is a differential equation, while the integral form is an integral equation. This means that the Point-Source derivative form is expressed in terms of derivatives, while the integral form is expressed in terms of integrals. The Point-Source derivative form is also valid for any charge distribution, while the integral form is only valid for symmetric charge distributions.

3. What conditions must be met for the Point-Source derivative form of Gauss's law to be valid?

The Point-Source derivative form of Gauss's law is valid for any charge distribution, as long as the electric field is continuous and differentiable everywhere. It is also only valid in regions where there are no charges outside the surface of interest that could affect the electric field.

4. What are some applications of the Point-Source derivative form of Gauss's law?

The Point-Source derivative form of Gauss's law is used in a variety of applications, including in the design of electric circuits, in the analysis of electric fields in materials, and in the study of electromagnetic waves. It is also used in the calculation of capacitance and in the derivation of other important equations in electromagnetism.

5. How is the Point-Source derivative form of Gauss's law derived?

The Point-Source derivative form of Gauss's law is derived from the integral form by using the divergence theorem. This theorem states that the volume integral of the divergence of a vector field is equal to the surface integral of the normal component of that vector field over the closed surface. By applying this theorem to the integral form of Gauss's law, we can obtain the Point-Source derivative form.

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