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Dipole Moment

  1. Nov 20, 2006 #1
    Griffith' E&M problem 3.28 page 151

    Given a spherical surface of radius R which carries a surface charge [itex] \simga = k \cos\theta [/itex]

    Calculate the dipole moment of this charge distribtuion

    well using this equation

    [tex] \vec{p} = \int \vec{r'} \sigma(\theta') dA' = \int Rk \cos\theta R^2 \sin\theta d\theta d\phi [/tex]

    but i was told that this setup is wrong that - tat the first term in the integration which i have as R should be [itex] R \cos\theta[/itex] why is that??

    what about my area element is that correct?

    Please help

    thank you in advance!
     
  2. jcsd
  3. Nov 20, 2006 #2

    OlderDan

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    A simple dipole moment of two charges is the magnitude of the charge times the distance between them pointing from negative to positive. Taken from their center of charge that would be the sum of the products of the charges times the position vector from the center. You are adding a bunch of charges that are positive in the upper half space and negative in the lower half with cylindrical symmetry. You need the z coordinates of each bit of charge times the charge. The z coordinate is R*cosθ. The area element is OK.
     
    Last edited: Nov 20, 2006
  4. Nov 20, 2006 #3
    that makes sense now thanks!

    do you have griffith's textbook
    what do you think are 'good' questions to do in the 'more problem' section for chapter 3?
    15 questions seems rather arduous since i would liek to go onto chapter 4...
     
  5. Nov 20, 2006 #4

    OlderDan

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    No I don't have the book
     
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