Appropriate coordinates for a given electric field

  • #1
kirito
77
9
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
Screenshot 2024-10-13 at 16.19.16.png

this is the field I was provided
and this is the charge density that I have reached
Screenshot 2024-10-13 at 16.21.52.png

Screenshot 2024-10-13 at 18.40.38.png

I tried to use this yet the output was different
I also used Cartesian it gave me the same output as the spherical ones
 
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  • #2
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.
 
  • #3
TSny said:
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
 
  • #4
kirito said:
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.
 
  • #5
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.
 
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