# 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?

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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 $1/r$. – NowIGetToLearnWhatAHeadIs 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