Maxwells' equations and Coulomb's law

Coulomb's law and Maxwell's equations should be consistant as one can be derived from the other.

Say we have a point charge with such a charge that $-kq=1$, meaning that at any point the electric field will have a magnitude of

$$|E|=\frac{1}{r^2}$$

where $r$ is the distance from the origin (were we place our charge), and the vectors point towards the origin at all point. This would be equivalent to the following in cartesian co-ordinates:

$$E=-\hat{x}\frac{x}{(x^2+y^2+z^2)^{3/2}}-\hat{y}\frac{y}{(x^2+y^2+z^2)^{3/2}}-\hat{z}\frac{z}{(x^2+y^2+z^2)^{3/2}}$$ We can verify that

$$|E|=\frac{1}{\sqrt{x^2+y^2+z^2}}$$

Gauss's law in its differential form allows us to calculate a charge distribution that would give rise to such a electric field using the divergence operator:

$$\nabla \cdot E=-\frac{1}{x^2+y^2+z^2}$$

from wolfram alpha: one, two

Which absolutely doesn't make sense to me! Intuitively I would think it would be zero everywhere except (0,0,0). Or at least not to go of to infinity at any point.

Could somebody please explain what's going on?

Actually if you calculate $\boldsymbol{\nabla} \cdot \mathbf{E}$, you get zero except at the origin, where you get infinity. So you can do it more precisely and obtain a delta function. I suspect an error in your calculation.
• The error is he says $|\mathbf{E}|$ goes like $1/r$. – Brian Moths Jan 31 '14 at 0:36
• In other words, he should have done this calculation instead. Adding up gives a denominator of $2x^2 + 2y^2 + 2z^2 - (y^2 + z^2) - (x^2 + z^2) - (x^2 + y^2) = 0$ – lionelbrits Jan 31 '14 at 17:17
What is the problem? Any charged object has a volume, so the denominator never can be zero. So E will never be infinite. The law is $$E = 1 / r^2$$ is correct. But always $$\vec \nabla \cdot \vec E = q/\epsilon_0$$ because kq=1 yuo can immediately calculate: $$k= 1/ (4 \pi \epsilon_0 )$$ $$q = 4 \pi \epsilon_0$$ and verify with $$\vec \nabla \cdot \vec E = \int^0_r\int^0_{2\pi}\int^0_\pi E r \sin(\theta) d r d \theta d \phi= E 4\pi r^2$$ $$E 4\pi r^2 = q /\epsilon_0$$ $$4\pi = q /\epsilon_0$$ $$q = 4 \pi \epsilon_0$$