homework and exercises - How come divergence of $vec E$ is zero in this case where $vec E=xifrac{left[x,yright]}{sqrt{(x^2+y^2)^3}},?$

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.

Answer

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: $$\nabla\cdot\vec{E}\propto\frac{1}{x^2+y^2}-\frac{2 x^2}{\left(x^2+y^2\right)^2}+\frac{1}{x^2+y^2}-\frac{2 y^2}{\left(x^2+y^2\right)^2}=0$$