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Homework Help: Electric Fields in circular wire

  1. Oct 26, 2003 #1
    I'm stumped by the following problem:

    A circular ring of fine wire carries a uniformly distributed positive charge, q. Find the magnitude and direction of the electric field at the center of the ring caused by just the charge on a portion of the ring subtending an angle [the] at the center, in terms of q, [the]and radius r.

    The uniform line charge [lamb] = dq/dl
    l = r[the] so dl = rd[the]

    E = kq/r^2
    dE = kdq/r^2 the dE's in the y direction cancel each other out because of symmetry.
    E = k[lamb]r/r^2[inte]cos[the]d[the] from -[the]/2 to [the]/2

    k = 1/(4[pi][ee])

    so E= [q/(2r[pi][ee]]sin([the]/2)

    The book show the answer to be: [q/(4[pi]^2[ee]r^2]sin([the]/2)

    Can someone please point out where I'm going wrong?

  2. jcsd
  3. Oct 26, 2003 #2
    You replaced λ with q.

    Your result
    E= [q/(2rΠε]sin(Θ/2)

    should be
    E= [λ/(2rΠε]sin(Θ/2)

    Now, λ = q/(2Πr) so make that substitution & you'll get the book's answer
  4. Oct 27, 2003 #3
    Thanks Gnome.

    Boy that was simple. I guess I should read more carefully. I completely forgot that for a uniform line charge [lamb] = q/l.

    Thanks again.
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