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Non conducting hollow ball

  1. Oct 21, 2014 #1

    NYK

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    1. The problem statement, all variables and given/known data
    https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-xaf1/v/t1.0-9/10155063_1485929725016579_7557468199677155885_n.jpg?oh=d575aa48176de7ecda27201b7ce35a5b&oe=54F7F368&__gda__=1425195231_44abd00cc231109df2fbe8d44cef9869

    2. Relevant equations
    ∫E⋅da⋅n = 4πkQenc
    ΔV(voltage) = -∫Edr
    3. The attempt at a solution
    To be honest I am having trouble starting the problem and defining the Qenc for the three areas of the sphere.

    I do believe that for r < R ⇒ E(r) = 2QK/r2

    and for r > R ⇒ E(r) = KQ/r2

    I know that the answer is listed for R<r<2R but i cant seem to come up with that answer.

    I think im just having problems with defining the Qenc

    for R<r<2R does Qenc = 2Qpt charge - Qinduced = Q sound right?
     
  2. jcsd
  3. Oct 24, 2014 #2

    vela

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    That's right.

    No. The ball is non-conducting, so there won't be any induced charges. You have a charge -Q spread out uniformly within the ball. When R < r < 2R, you have to figure out what fraction of the ball is enclosed inside of the sphere of radius r. It's a geometry problem.
     
  4. Oct 24, 2014 #3

    NYK

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    I understand that, so if it was a conducting sphere the charge would be Qenc = Qpt charge +λ(π(r2-R2))?
     
  5. Oct 24, 2014 #4

    vela

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    No. How did you come up with that expression?
     
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