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Griffiths Problem 3.28

  1. Feb 24, 2008 #1
    [SOLVED] Griffiths Problem 3.28

    1. The problem statement, all variables and given/known data
    Please stop reading unless you have Griffiths E and M book.

    In this problem, I found that the approximation agrees with the exact potential. I am not sure what to conclude about higher multipoles. Are they all identically zero or do they all cancel? Is there something about this charge distribution that makes that happen? Could I have predicted that without calculating it?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 24, 2008 #2

    pam

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    It is a pure dipole.
    The cos\theta tells you it is pure P_1(cos\theta).
     
  4. Feb 24, 2008 #3
  5. Feb 25, 2008 #4

    pam

    User Avatar

    That post gives the definition of a dipole used in elementary texts.
    It describes one simple model of a dipole.
    What I meant is that the field outside the sphere has only the dipole term as in the expansion. If you calculate the multipole moments of the sphere, you will find only a diple moment because of the P_1 dilstribution of charge.
    Use Griffith's or some other text for definitions, not the web.
     
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