# Divergence of conservative electric field

I have a little doubt about the following: according Gauss law in the form of Maxwell's equation, we know that:

$${\rm div} (D)~=~ \rho(v)$$

This just tells us that the electric field has nonzero divergence if there is some nonzero charge distribution $\rho(v)$.

But, for a point charge $$D=(qk\hat{r}/r^2)$$ while, in polar coordinates,
$$div(\hat{r}/r^2)=(1/r^2){(d/dr)(r^2/r^2)}= 0$$ why does this apparent contradiction arise?

• The integral form of Maxwell's equations is better than the differential form in the sense that such singularity won't occur. – Pu Zhang Jun 18 '14 at 16:59
• @PuZhang that doesn't make the integral form unconditionally better. In fact one could argue that hiding a singularity is a bad thing, depending on the circumstances. – David Z Jun 18 '14 at 18:11

$$\tag{1} \vec{D}(\vec{r})~=~\frac{1}{4\pi} \frac{\vec{r}}{r^3}, \qquad \vec{r}\neq \vec{0},$$
of a point particle $q$ is not a differentiable function $\vec{D}: \mathbb{R}^3\to \mathbb{R}^3$ at the origin $\vec{r}=\vec{0}$.
$$\tag{2} \vec{\nabla}\cdot\vec{D}(\vec{r})~=~q ~\delta^3(\vec{r})~=~\rho(\vec{r})$$
via the theory of distributions, aka. generalized functions. The charge distribution $\rho(\vec{r})$ for a point charge is (proportional to) the 3D Dirac delta distribution $\delta^3(\vec{r})$. One of the characteristic features of the 3D Dirac delta distribution is that it is only supported in the origin $\vec{r}=\vec{0}$, as we would expect of a point charge.