Problem in understanding Differential form of Gauss's Law I am well aware of the integral form of Gauss's Law and the mathematical deduction through which it is reduced to the differential form.
But I think I have a flaw in my understanding of divergence.

Here I am getting a non-zero divergence even though there is no net charge inside the region.
Where am I making the mistake?
 A: The easiest way to solve this problem is to change from cartesian coordinates $(x,y,z)$ to polar coordinates in the 2-dim. case  $(\rho,\phi)$ or to spherical coordinates $(r,\theta,\phi)$  in the 3-dim. case. 
For simplicity we will first compute the divergence in 3-dim case, because in this case the formulas are as we are used to. So in this case the electrical field strength of a charge $Q$ at a distance $r$ is (using SI-units):
$$E_r = \frac{Q}{4\pi \epsilon r^2}$$
whereas the other components $E_\theta$ and $E_\phi$ are zero.  Remember that in 3-dim. spherical coordinates a vector decomposes $\vec{F} = F_r e_r + F_\theta e_\theta + F_\phi e_\phi$.
Then the divergence in spherical coordinates is : 
$$\vec{\nabla} \cdot \vec{F} = \frac{1}{r^2}\frac{\partial (r^2 F_r)}{\partial r} + \frac{1}{r\sin\theta}\frac{\partial (\sin \theta F_\theta)}{\partial \theta} + \frac{1}{r\sin\theta}\frac{\partial  F_\phi}{\partial \phi}$$
If we plug the only non-zero component of the electrical field in this formula we get 
$$\vec{\nabla} \cdot \vec{E}= \frac{1}{r^2}\frac{\partial (r^2 E_r)}{\partial r} = 0$$
So this is as we wanted. In the 2-dim. case the math is not as we are "used" to. The Coulomb law in 2 dimensions is actually :
$$E_\rho = \frac{Q}{2\pi \epsilon \rho}$$
where as the other component $E_\phi$ is again zero. Also the formula for the divergence changes ($\vec{F} = F_\rho e_\rho + F_\phi e_\phi$).
$${\vec{\nabla} \cdot \vec{F}}= \frac{1}{\rho}\frac{\partial (\rho F_\rho)}{\partial \rho} + \frac{1}{\rho}\frac{\partial  F_\phi}{\partial \phi}$$
But again, if we plug the expression $E_\rho = \frac{Q}{2\pi \epsilon \rho}$ into the precedent formula, the divergence turns out to be zero. 
Just the side remark: Why does Coulomb's law change ? It has to be adapted to the dimension of the space. In general it says: 
$$\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon}$$
In 3 dimensions the surface is  $4\pi r^2$ whereas in 2 dimensions it is $2\pi \rho$. From this you get the corresponding formula for the electrical field strength in 3- and 2 dimensions. 
