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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
    λ
    . 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)?

    dErad=−14πϵ0rλdθ(r2+z2)rˆ

    Erad=−14πϵ0rλ(r2+z2)∫2π0rˆdθ
     
  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]

    So
    [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

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    Here's a picture of cylindrical coordinates:
    CylindricalCoordinates_1001.gif

    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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