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Expressing electric field in cylindrical coordinates.

  1. Mar 6, 2015 #1
    Hi everyone, I am new to the physics forums and I need your help :)

    I understand that depending on the symmetry of the problem, it may be easier to change the coordinate system you are using. My question is, how would I convert the electric field due to a point charge at the origin, from Cartesian coordinates to cylindrical ones?

    I am using the equation [tex] E = \frac{q}{4\pi\epsilon_0}\frac{(r-r')}{(r-r')^3}[/tex]
  2. jcsd
  3. Mar 6, 2015 #2
    If you want to exploit symmetry use spherical coordinates, not cylindrical. If you are in love with cylindrical coordinates :P , then look up on http://mathworld.wolfram.com/CylindricalCoordinates.html the relationships between the unit vectors in cartesian coordinates to cylindrical and the relationship each variable has. Once this is done, you just simply have to use grandiose algebra.

    So far the only time i've used cylindrical coordinates when working with electric fields is when I'm considering an infinite rod with charges inside it and want to calculate the electric field generated by it in the exterior space.
    Last edited: Mar 7, 2015
  4. Mar 6, 2015 #3
    I am in love with cylindrical coordinates :P No I joke, I just want to practice using them.
    I have seen those relationships before, in fact I went through the derivation, so I think I understand them. Where I'm confused is how to apply them to the E field formula. Since we have only r showing up in the formula, do I need to transform only this variable?
  5. Mar 7, 2015 #4
    ##\mathbf{E}=\frac{kq}{x^2+y^2+z^2} * (\frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{\sqrt{x^2+y^2+z^2}})##
    ##=\frac{kq}{(x^2+y^2+z^2)^{3/2}} * (x \mathbf{i} + y \mathbf{j} + z \mathbf{k})##

    From there just use the relationships you've said you know, and you'll get the vector in cylindrical coordinates.
  6. Mar 7, 2015 #5
    In response to your question:

    You have to transform (x,y,z) and the unit base vectors (i,j,k).
  7. Mar 7, 2015 #6
    So to transform the x,y and z I have to use
    And to transform the unit vectors I have to use

    Is this right?
    Last edited: Mar 7, 2015
  8. Mar 7, 2015 #7
    Yup. Thats right.
  9. Mar 7, 2015 #8
    How do I do this? I'm not sure how to use those relationships since I've done very little with matrices.
  10. Mar 7, 2015 #9
    The first matrix is straight forward. x=rcos(th) etc. The next matrix is more complicated. What you have to do is something like this:

    ##\begin{bmatrix} cos\theta \\ -sin\theta \\ 0 \end{bmatrix} \mathbf{x} + \begin{bmatrix} sin\theta \\ cos\theta \\ 0 \end{bmatrix} \mathbf{y} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \mathbf{z} = \begin{bmatrix} \mathbf{r} \\ \theta \\ \mathbf{z} \end{bmatrix} ##

    For some reason the last ##\theta## is not appearing in bold letters.

    But the idea is that these columns behave as vectors. You add them up and multiply them according to their coefficient. ##\mathbf{r}## reads for example:

    ##\mathbf{r} = cos\theta \mathbf{x} + sin\theta \mathbf{y} ##

    Which for example, is a natural expression for the radial unit vector, if you think about it.

    Once you obtain those expressions, like the one for ##\mathbf{r}##, what you have to do is substitution into your electric field equation (for example the one I wrote above in cartesian coordinates).
  11. Mar 7, 2015 #10
    I see, so I will end up with a whole lot of new vectors for x, y, z, xhat, yhat and zhat.
    The I have to do is some algebra and make it look pretty?

    I think I got it now! I really appreciate your help, thank you so much :)
  12. Mar 7, 2015 #11
    I have serious doubts that it will look pretty at all. The coordinate system is not really fit for the field of a point charge.

    And youre welcome.
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