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Find the electrostatic potential due to a hoop of charge

  1. Jan 22, 2009 #1
    1. The problem statement, all variables and given/known data

    (Part 1) A hoop of charge of radius k lies in the y, z plane, centred on the x axis, so that
    it occupies the points (0, y, z) with y^2 + z^2 = k^2. If the (linear) charge density
    in the hoop is j, calculate the electrostatic potential Fi at all points on the
    x axis, and show that, far from the hoop on the x axis, Fi (x, 0, 0) = (2*pi*j)/|x|.
    Explain briefly why this result was only to be expected.

    (Part 2) More generally, what do you expect the leading behaviour of Fi(r) to be, far
    from the hoop? (You do not need to give any detailed calculations.) Use your
    answer to deduce the limiting forms of the equipotential surfaces and field lines,
    again far from the hoop.

    2. Relevant equations

    3. The attempt at a solution

    (Part 1)

    Total Charge Q = Integrating j * dl with interval [0 , 2*pi]
    = 2*pi*k*j

    By Symmetry, Fi (x) = 2*pi*k*j / |x| (Is this wrong?)

    And now I'm stuck for the rest of the question.

    Please help! Thank you!
  2. jcsd
  3. Jan 22, 2009 #2
    Integrating the charge elements and using symmetry is involved for solving this problem.

    [tex]E = E_x =\int\mbox{dEcos}\theta\mbox{d}\theta[/tex]


    [tex]dE =\frac{Kd\rho}{r^2}[/tex]

    The sine components of E cancel due to symmetry, and

    [tex]cos\theta = \frac{x}{\sqrt{k^2+x^2}}[/tex]

    The charge element is

    [tex]d\rho = jkd\phi \ \mbox{and}\ d\phi \ \mbox{ is integrated from 0 to} \ 2\pi[/tex]

    Once E is found, the potential is found from integrating E from infinity to x along the x axis.
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