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

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?

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.

$\rho(\vec{s})$ 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 $r=s$ .

• Thanks Prof. Shonku for the explanation. I am bit confused due to the following. Jan 12, 2016 at 4:30
• Which part of the answer confuses you. $\rho(r)$ is for any $r$ in the space. If you put $\rho(r)=0$ then it will mean it is zero at $r=s$ also which is not true. $r$ is written is the sense of a variable it does not stand for a particular position like $r=r_0$. Jan 12, 2016 at 4:37
• Thanks Prof. Shonku for the explanation. I am bit confused due to the following. In the above problem we are asked to find out the divergence of the electric field $E(r)$. $r$ is the position vector of the point at which $E$ is calculated. The charge distribution which generates the field $E(r)$ is at a distance $s$ from the origin. Hence the charge distribution is written as $\rho(s)$. (cont.) Jan 12, 2016 at 4:49
• (cont.) Now when we finally find the expression of $\nabla\cdot E(r)$ it comes out to be $\rho(r)/\epsilon_0$, which is a function of $r$ not $s$. this says that the $\nabla\cdot E(r)$ is independent of the source coordinates. Suppose $\rho(s)$ is a finite quantity but $\rho(r)=0$. then $\nabla\cdot E(r)$ will be 0. There is no effect of $\rho(s)$ on $\nabla.E(r)$. But $\rho(s)$ is the cause of $E(r)$. How is it possible that the cause of $E(r)$ has no effect on the $\nabla\cdot E(r)$? Jan 12, 2016 at 4:49
• Suppose there is a point inside an extended source having charge density rho. We want to calculate div.E at that point. Will it be same as that for a point outside the extended source? Jan 12, 2016 at 5:11

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

• If $E_{out}=\frac{\rho R^3}{3\epsilon_0 r^2}$ and $\rho=0$ then $E_{out}=0$ but is it? Jan 12, 2016 at 8:46
• If there is no charge distribution on the sphere, then its a neutral sphere. So E(out) will be 0 and E(in) will also be 0. But here the Dirac delta function plays the role by assigning rho(r)=rho for r<R and rho(r)=0 for r>R. I think we are talking of the same thing but in different language. Thanks Jan 12, 2016 at 9:31
• Okay.Basically you showed that $\rho(r)\neq 0$ for $r<R$ and $\rho(r)=0$ for $r>R$ which is equivalent to saying $\rho(r)=\frac{Q}{4\pi r^2}\delta(r-s)$ for a single charge situated at $r=s$ which I said earlier. Jan 12, 2016 at 9:34
• yes.That's what I said.That was my point that the information is not lost because the explicit expression for $\rho(r)$ contains the information where the charge is situated. Jan 12, 2016 at 9:36
• It still is not sinking into me that how a simple notation "rho(r)" without an integration or summation short of sign can convey about the charge density for the whole range of r. But I very well understand now that here the Dirac delta function plays the role by assigning relevance, only to the charge density, at that particular point where we are calculating E(r). Jan 12, 2016 at 9:41