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E field of a hemisperical shell

  1. Feb 4, 2012 #1
    My teacher explained a problem of a hemispherical shell in class but i dont understand what he is doing.

    http://img116.imageshack.us/img116/7656/naamloos27mf.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 5, 2012 #2

    tiny-tim

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    welcome to pf!

    hi nosmas! welcome to pf! :wink:

    (he only worked out the z component, because from symmetry the x and y components must be zero)

    he sliced the shell into rings because for any one ring, the z component of the field must be the same …

    so if the total charge of that ring is q, then it has the same effect (on the z component) as a charge q all at one point (instead of spread out around the ring)

    then he multiplied charge x 1/distance2 x cosθ
    (he seems to have unnecessarily put in a lot of r's that then cancelled …

    i expect that's because they were in Eq 23-10)
     
  4. Feb 6, 2012 #3
    Equation 23-10 is dE = k*(dQ/r^2)

    What I am struggeling with is how the distance to the point of interest z = rcos(theta) and how they came up with the charge on the ring?
     
  5. Feb 6, 2012 #4

    tiny-tim

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    hi nosmas! :smile:
    rcosθ is the distance from the centre to the plane of the ring

    the charge is the charge density times the area,

    and the area is the arc-length (rdθ) times the circumference of the ring (2πrsinθ)
     
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