Divergence of electrostatic field?

[terryds]

Homework Statement

By Gauss' law, how is it able to obtain $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$ ?

By Coulomb's law, $\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}$

I calculate the divergence of $\frac{1}{r^2} \hat{r}$ and get the result is zero

That means the divergence of $\frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}$ is also zero too.

How come it is $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$ ?

Homework Equations

Gauss' Law
Coulomb's Law

The Attempt at a Solution

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[Orodruin]

Your computation is only valid for $r > 0$ where the charge density is zero. The divergence is a delta function at the origin, just as the charge density.

 

[terryds]

Your computation is only valid for $r > 0$ where the charge density is zero. The divergence is a delta function at the origin, just as the charge density.

So, what happens at point (0,0,0) ?The divergence is infinity or $\frac{\rho}{\epsilon_0}$?

 

[Orodruin]

The divergence is $q \delta^{(3)}(\vec x)/\epsilon_0$ and so is the charge density/epsilon0. Both are "infinite" at the origin. (The proper treatment is to consider them distributions rather than functions)