How come divergence of $\vec E$ is zero in this case where $\vec E=\xi\frac{\left[x,y\right]}{\sqrt{(x^2+y^2)^3}}\,?$

I hope you could help me clearing some doubts about Gauss' law of the electric field that states $\epsilon_0\nabla\cdot\vec E=\rho$. Take for instance the case of a point charge in the origin in empty space. The law states $\nabla\cdot\vec E = 0$ anywhere but the origin, but it's enough to compute $\nabla\cdot\vec E$ to get to a contradiction: $$\vec E=\xi\frac{\left[x,y\right]}{\sqrt{(x^2+y^2)}^3}$$ where I dropped every constant to $\xi$ because that's my favourite greek letter, and considered the 2D case since in my opinion it's a good compromise between complexity and meaningfulness $$\nabla\cdot\vec E=-\frac{x^2+y^2}{\sqrt{{(x^2+y^2)}^5}}=(x^2+y^2)^{1-\frac{5}{2}}$$ which is clearly not zero except at infinity.

• your divergence is wrong - you did not calculate the derivative correctly – Sanya Sep 21 '16 at 16:45
• You misunderstand Gauss's Law - you have given the differential form, but have ignored that it comes from the divergence of the E field, meaning an integral over a surface enclosing a volume. – Jon Custer Sep 21 '16 at 16:45
• You can't just use any number of dimensions. Coulomb's law changes in 2D. – Javier Sep 21 '16 at 17:20
• Jon Custer, may you explain further this point? I know this law comes from the "global" one which states that the electric flux through a closed surface is the inner charge (which I have understood well, actually), I also know the mathematical derivation of the "local" law, but it appears i haven't understood what does it actually mean – J. Joestar Sep 21 '16 at 18:18
• Sanya, Javier: even without doing calculations this doesen't sound right to me: the electric field due to a charge is made of radial vectors pointing toward or outward the charge, with amplitude decreasing with distance. Since the amplitude is decreasing, mathematically speaking every point in space will have both $\vec E$ and its derivatives different from 0, so there's no way $\nabla\cdot\vec E$ is also zero. From the intuitive point of view, every point has a polarized direction through which more field enters than it leaves, thus having a nonzero divergence. – J. Joestar Sep 21 '16 at 18:28

As pointed out in the comment, Gauss' law has a precise geometric meaning and so is strongly dependent on the dimension of the space. Let's work in 2D as in your example and apply Gauss' law: the flux of the electric field through a circle centered in the point charge is $2\pi rE$. This must be equal to the total charge which is $e/\epsilon_0$. This gives for the modulus of the electric field: $$E=\frac{e}{2\pi\epsilon_0 r}$$ and for the cartesian components: $$\vec{E}=\frac{e}{2\pi\epsilon_0}(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})$$ As you can verify the divergence of this is zero: $$\nabla\cdot\vec{E}\propto\frac{1}{x^2+y^2}-\frac{2 x^2}{\left(x^2+y^2\right)^2}+\frac{1}{x^2+y^2}-\frac{2 y^2}{\left(x^2+y^2\right)^2}=0$$