electrostatics - Laplacian of $1/r^2$ (context: electromagnetism and Poisson equation)

We know that a point charge $q$ located at the origin $r=0$ produces a potential $\sim \frac{q}{r}$, and this is consistent with the fact that the Laplacian of $\frac{q}{r}$ is

$$\nabla^2\frac{q}{r}~=~-4\pi q~ \delta^3(\vec{r}).$$

My question is, what is the Laplacian of $\frac{1}{r^2}$ (at the origin!)? Is there a charge distribution that would cause this potential?

Answer

The electric field from your potential is:

$$E(r) = {2\over r^3}$$

Using Gauss's law, the total charge in a sphere of radius R is:

$$Q(r) = \oint E \cdot dS = 4\pi r^2 {2\over r^3} = {8\pi\over r}$$

The total charge is decreasing with r, so there is a negative charge cloud of density

$$ \rho(r) = {1\over 4\pi r^2} {dQ\over dr} = - {4\over r^4}$$

But the total charge at infinity is zero, so there is a positive charge at the origin, cancelling the negative charge cloud, of a divergent magnitude. If you assume this charge is a sphere of infinitesimal radius $\epsilon$, the positive charge at the origin is

$$Q_0 = \int_\epsilon^\infty 4\pi r^2 {4\over r^4} = {16\pi \over \epsilon}$$

This is not a distribution in the mathematical sense, but it is certainly ok to work with, so long as you keep the $\epsilon$ around and take the limit $\epsilon$ goes to zero at the end of the day. Mathematicians have not had the last word on the class of appropriate generalized solutions yet.