Why is the divergence of $v=\frac{1}{r^2} \hat{r}$ zero everywhere (except the origin)? Manipulating the equations, I understand how to get $0$, but is there a more intuitive/physical explanation? Moreover, why isn't the behavior of the divergence of $\frac{1}{r^2}\hat{r}$ the same as say, $\frac{1}{r}\hat{r}$ or $r^2$?
 A: *

*The flux of $v$ across the sphere $|x|=r$ is equal to $(4\pi r^2 )( r^{-2}) = 4\pi$, i.e., is independent of $R$. It follows that the integral of divergence over every spherical shell $R_1<|x|<R_2$ is zero. 

*Since the field is rotationally invariant, so it its divergence: that is, the divergence  depends on $|x|$ only. 

*Putting 1 and 2 together, we can conclude that the divergence is identically zero, except  possibly at the origin. At the origin we have the point source of intensity $4\pi$. 

*The above only works because $r^{-2}$ cancels $r^2$ in the formula for the area  of a sphere. Any other exponent would not cancel out, leading the the flux being dependent on $r$ and consequently the divergence being nonzero. 

*Generalization: in $n$ dimensions,  $r^{1-n}\hat r$ has the above properties. 
A: In Electrostatic, $\hat{r}/r^{2}$ is the electric field produced by a charge $+1$ at the origin of coordinates. The charge density associated to that charge is represented by a 
Dirac Delta Function. Namely, $\delta\left(\vec{r}\right) = \nabla\cdot\left[\left(\hat{r}/r^{2}\right)/\left(4\pi\right)\right]$.
