I Appropriate coordinates for a given electric field

[kirito]

TL;DR Summary

I was facing a problem where I have an electric field and I have to find the charge density , I noticed that the field outside is that of a dipole and because I decided to use spherical coordinates and reached an appropriate answer yet using divergence in cylindrical lead me to different outputs

this is the field I was provided
and this is the charge density that I have reached

I tried to use this yet the output was different
I also used Cartesian it gave me the same output as the spherical ones

 

[TSny]

The value of the divergence of a field at a point is independent of the coordinate system. So, if you get zero in spherical coordinates you should also get zero in cylindrical coordinates.

 

[kirito]

The value of the divergence of a field at a point is independent of the coordinate system. So, if you get zero in spherical coordinates you should also get zero in cylindrical coordinates.

thats actually why I even made this post , I got weirded out by the calculation using cylindrical coordinates since there is no component in the theta direction whereas the z results in 0 and the component in the r direction results in something other than zero unless I had to do something to the coordinates I changed from cylindrical to spherical so changed the z hat appropriately in the first case I may have missed something but got no clue what it may be

 

[Vanadium 50]

since there is no component in the theta direction whereas the z results in 0 and the component in the r direction results in something other than zero

What is the divergence in cylindrical (or spherical) coordinates? Hint: it is not the same as for Cartesian replacing variable names.

 

[Meir Achuz]

It's easier to do it without coordinates:
$${\bf E}=\frac{3{\bf (r\cdot \mu)r}}{r^5}-\frac{\mu}{r^3}$$
$$\nabla\cdot{\bf E}=\frac{3[\mu\cdot{\bf r}+3{\bf (\mu\cdot r)]}}{r^5}-\frac{15\mu\cdot{\bf r}}{r^5}+\frac{3\mu\cdot{\bf r}}{r^5}=0.$$
Sorry. I don't know what's wrong with my latex.