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Multipole Expansion

  1. Nov 30, 2005 #1
    Question:

    Assume the chrages to be on the z axis with the midway between them. Find the potential exactly for a field point on the z axis.

    Okay, so I found the potential which is v = k*p/(z^2-0.25*l^2)

    k is the constant 1/4*pi*epsilon, l stands for the length between the two point charges

    The next part of the question asks

    How large must z be in order that one can approximate the exact potential on the z axis with the dipole term to an accuracy of 1 percent?


    The answer is supposed to be z > 5l, but I don't understand what the question is asking me to do...
     
  2. jcsd
  3. Nov 30, 2005 #2

    Tide

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    Perhaps you could clarify what the problem is asking. You seem to have left out some details. How, exactly, did you arrive at the potential you showed?
     
  4. Nov 30, 2005 #3

    Physics Monkey

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    Presumably you have a positive charge at [tex] x=+ L/2 [/tex] and a negative charge at [tex] x=- L/2 [/tex]. Is this more or less the setup? If so, then you should proceed as follows. You can calculate the exact potential of the charge distribution, right? However, you also know from the multipole expansion that since the system is charge neutral, the leading term in the "far field" should be the dipole term (monopole term is zero). The question is asking you to figure out what constitutes the "far field." All you want to do is find z such that the relative difference between the dipole potential and the true potential is less than one percent. Can you set up that equation?

    Also, take a look at Tide's comment, I don't think you have your potential right. Although, perhaps I have misinterpreted the system.
     
    Last edited: Nov 30, 2005
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