Why is the divergence of $\widehat{r}/r^2$ equal to $0$? I have read that $\nabla\cdot\dfrac{\widehat{r}}{r^2}$ is equal to $0$. 
But I cannot understand why. I tried but I cannot solve it. Can anyone explain it please?
 A: If you put it into spherical coordinates (link to Wikipedia article "Del in cylindrical and spherical coordinates"), 
$$
\nabla\cdot{\hat{r}\over r^2}=1/r^2{\partial\over\partial r} {r^2\hat{r}\over r^2}
$$
$$
=1/r^2{\partial\over\partial r} \hat{r}
$$
Assuming $r\neq 0$, this is zero because $\hat{r}$ doesn't change with $r$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\color{#66f}{\large\nabla\cdot\pars{\hat{r} \over r^{2}}}&
=\nabla\cdot\pars{{1 \over r^{3}}\,\vec{r}}
=\nabla\pars{1 \over r^{3}}\cdot\vec{r} + {1 \over r^{3}}\,\nabla\cdot\vec{r}
=\bracks{\hat{r}\,\totald{}{r}\pars{1 \over r^{3}}}\cdot\vec{r}
+ {1 \over r^{3}}\times 3
\\[5mm]&=\bracks{{\vec{r} \over r}\pars{-\,{3 \over r^{4}}}}\cdot\vec{r}
+{3 \over r^{3}}
=-\,{3 \over r^{3}}+ {3 \over r^{3}}=\color{#66f}{\Large 0}\,,\qquad \vec{r} \not= 0
\end{align}
A: This is not zero everywhere.  Use the integral definition of divergence.  Let $M$ be a volume centered on a point $p$.  Let $V$ be the volume of this region.
$$\left. \nabla \cdot \frac{\hat r}{r^2} \right|_p = \lim_{V \to 0} \frac{1}{V} \oint_{p \in M} \frac{\hat r}{r^2} \cdot \hat n \, dS$$
In particular, when $p$ is the origin, use spherical coordinates and a spherical integration volume.  Then $\hat n = \hat r$ and $dS = r^2 \sin \theta \, d\theta \, d\phi$, and we get
$$\left . \nabla \cdot \frac{\hat r}{r^2}  \right|_\vec{0} = \lim_{R \to 0} \frac{4\pi R}{4\pi R^3/3} $$
which diverges.  This only happens at the origin. At all other points, you can use the differential form of the divergence and get the right answer--which is zero.
