1

this page

proves the gauss theorem. First they prove that there has to be charge allocated in the origo of a function and that this is where you get the gauss formula and therefore you can integrate over any shape and obtain the same value. It is expressed as

$\nabla \cdot E= \frac{q_i}{\varepsilon_o}\delta(r)$ (208)

where

$\delta(r)$

Is a funcion that is 0 everywhere except at the origin where it is 1 to make the fact that it is 0 everywhere than in the origo as proved by equation (192)

$\nabla \cdot E=\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y }+\frac{\partial E_z}{\partial z}=\frac{q}{4\pi\varepsilon_o}\frac{3r^2-3x^2-3y^2-3z^2}{r^5}=0$ (192)

and (192) is only undefined at the origo In (210) they extend this to many charges by the formula

$\nabla \cdot E= \sum_{i=1}^{N}\frac{q_i}{\varepsilon_o}\delta(r-r_i)$ (210)

My problem is that I don't get why the spatial arrangment of the additional charges other places then at the origo don't matter for the electrical dot product. They just extend it to (210) from (208) without a mathematically justification. I wanted a mathematically justification for this generalization. I wanted to ask this to an expert but I am not sure how to pose the question clever enough. Can you help me formulate a mathematical problem for this extention?

torgny
  • 141

1 Answers1

0

That page does not prove Gauss's theorem. It uses Gauss's theorem and pointwise differentiation to show that there is a singularity in $E$ at the origin and then calls that singularity "$\delta(r)$."

To answer your question, though, first recall that the electric field at any location is the sum of the electric fields generated by all the charges. By superposition, its divergence is zero since the divergence of the field generated by each charge is zero. This is why $E$ has the property that away from charges $\nabla\cdot E = 0$.

One way of thinking of $\delta(r-r_0)$ is by drawing a thin spike at $r_0$. Intuitively, $\nabla\cdot E = q\delta(r-r_0)/\epsilon_0$ means that the entire field is emanating from the single point $r_0$ and is not being created anywhere else.

(In fact, mathematicians have actually invented math to justify this physical intuition. Here is Wikipedia's page on the Dirac delta distribution to get you started.)

Neal
  • 32,659
  • Thanks! But assuming that I get the dirac delta function for origo is there a mathematical way to justify it for many additional charges located outside the origo? – torgny Oct 17 '14 at 04:53
  • For every single charge, translate it to the origin and examine there the electric field it generates. It will be identical to your original calculation. – Neal Oct 17 '14 at 04:55
  • Thanks again! Would taking the antidivergence of (210) give the original E? Can one prove this? – torgny Oct 17 '14 at 05:12