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Electric Field of a ring , mathematically

  1. Nov 5, 2011 #1
    This is about the electric field of a ring with radius r, at a distance z from center, along the axis of the ring. The ring carries a uniform line charge
    . We always say that the radial component of the field cancels out due to symmetry. Can somebody tell how to prove it mathematically (using cylindrical coordinate system only)?


  2. jcsd
  3. Nov 6, 2011 #2

    I like Serena

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    Welcome to PF, MinaKaiser! :smile:

    First, let's clean up your formula.

    I get:

    [tex]d\vec E_{rad} = - {1 \over 4\pi \epsilon_o} \cdot {\lambda r d\theta \over r^2 + z^2} \cdot {r \over \sqrt{r^2 + z^2}} \hat r[/tex]

    [tex]\vec E_{rad} = - {1 \over 4\pi \epsilon_o} \cdot {\lambda r d\theta \over r^2 + z^2}\cdot {r \over \sqrt{r^2 + z^2}} \int_0^{2\pi} \hat r d\theta[/tex]

    So what we need is:
    [tex]\int_0^{2\pi} \hat r d\theta = 0[/tex]

    Since [itex]\hat r[/itex] is not a constant vector, we rewrite it in a basis that is constant:

    [tex]\hat r = \hat x \cos \theta + \hat y \sin \theta[/tex]

    [tex]\int_0^{2\pi} \hat r d\theta = \int_0^{2\pi} (\hat x \cos \theta + \hat y \sin \theta) d\theta[/tex]

    Can you see that this is zero?
  4. Nov 6, 2011 #3
    I dont' get it , Why r is not a constant vector ?
  5. Nov 6, 2011 #4

    I like Serena

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    Here's a picture of cylindrical coordinates:

    At some point (r, theta, z) we have a local basis which contains [itex]\hat r[/itex].
    But if theta is increased, the direction of [itex]\hat r[/itex] changes.
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