What does the Dirac delta function physically do while deriving Gauss Law form Coulomb's law?

Question

While doing this derivation, the the source coordinates are mentioned as "$s$" and the coordinate of the point at which field is to be calculated is mentioned as "$r$". Kindly follow this Wikipedia link and click on the "Outline proof" under "Derivation of Gauss Law from Coulomb's law".

Finally it comes out that $$\nabla\cdot E(r)= \frac{\rho(r)}{\epsilon_0}. $$But $\rho$ is actually defined for the "$s$" coordinates and $\rho(r)$, where $r$ is the point at which electric field is calculated is 0. Here I can not understand how the $\nabla\cdot E(r)$ is equal to $\frac{\rho(r)}{\epsilon_0}$.The information about $\rho(s)$ is totally lost in the final equation. What does the Dirac delta function actually do?

Answer 1

Dirac's delta is a function that describes a distribution (of charge, in this instance) which is concentrated at one point: precisely what you need. So essentially, the equations on the given proof outline read in plain english as follows:

(1) Coulomb's law of a point charge (2) Coulomb's law integrated for a smoothly distributed charge with density $\rho$ (putting $\rho=e_0 \delta$ gives you back (1)). Each point contributes $\rho$. (3) Field ${\bf r}/r^3$ desribes a field that originates at a point at the origin, with no other sources. We recognize this term under the integral (2). (4) The sources of E are an integral over contributions of sources with magnitude $\rho$ at each point - which is basically just saying "a smooth blob of charge is just like having a continuous distribution of little point charges". (5) Just restatement of (4) (mathematically, using the delta function definition).

So really, this outline does virtually nothing. It says "we generalized Coulomb's law for a point charge to a continuous charge distribution by adding them together and, oh the surprise, that the source of the resulting electric field is the charge distribution we put in in the first place". If you ask me, this "proof" is kind of circular.

Answer 2

$\rho(\vec{s})$ is function of $\vec{s}$ where $\vec{s}$ is position vector of the location of charge density.Whereas $\vec{r}$ is position at which you want to calculate $E(\vec{r})$. So what you are essentially doing is calculating electric field due to some elementary volume $d^3s$ located at the position $\vec{s}$ and finally integrating over all such contribution for all such $\vec{s}$.

$\textbf{Important:}$ The Law in the differential form and valid for all r and $\rho{(r)}\neq 0$ for all $r$ as $r$ can be any point in space which means even $r=s$.In other words you have to write say for a point charge (spherical symmetry) $$\rho(r)=\frac{Q}{4\pi r^2}\delta(r-s)$$.Writing in this way you are not losing your information that charge is only at $r=s$ .

Answer 3

I think I have found the answer. Let us consider the case of a uniformly charged sphere of radius $R$ and charge density $\rho$. The field inside this sphere is $E_{in}=\frac{\rho\times r}{3\epsilon_0}$. Here $r$ is the distance from the centre and $r < R$. If we calculate the divergence of $E_{in}$ then

$$\nabla.E(r)=\frac{\rho}{\epsilon_0} $$ Kindly note that $\rho$ is actually $\rho(r)$ which is constant for r < R.

The electric field $E_{out}=\frac{\rho\times R^3}{3\times\epsilon_0\times r^2}$ for $r>R$.

Calculating the divergence of this field we get

$\nabla.E(r)=0$, Kindly note that for this point $(r > R)$ $\rho(r)=0$.

This is the reason why I disagree with the interpretation $\rho(r)$ by Prof Shonku It proves that $\nabla.E(r)=\frac{\rho(r)}{\epsilon_0}$ where $\rho(r)$ is the charge density exactly at the point where the field $E(r)$ is measured. If $r$ is such that the point is inside an extended charge distribution then $\nabla.E(r)$ is non zero. If $r$ is such that the point is outside an extended charge distribution, then $\nabla.E(r)$ is zero. Thanks