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Potential of a Quadrupole (Far Away)

  1. Oct 24, 2012 #1
    So we're currently covering multipole expansion techniques in my EDM class. The multipole expansion is a summation of integrals each treating a different configuration of discrete charges (monopole, dipole... and so on). The term in the summation that corresponds to the physical configuration that your dealing with, whether it be a monopole, dipole, or whatever, is the dominant term. My first question is: Does that mean it is safe to ignore the other terms in the summation and treat, say, a quadrupole with only the quadrupole term? Secondly, there is a simple formula to express the potential of a discrete charge configuration at a large distance where the configuration appears as a single point charge. The problem I'm looking at has a charge configuration like this: a +3q charge at point z=a, a +q charge at z=-a, a -2q charge at y=a and y=-a, this forms a quadrupole and the problem wants an expression for potential that is valid at large distances from the quadrupole. It uses the word 'simple' to describe the expression for potential so I imagine it's not the multipole expansion formula. Would I use the formula for large distances to treat this problem? The only problem I see with that would be the fact that the net charge would=0 and then the potential would= 0 as well. How could that be the case for this quadrupole?
     
  2. jcsd
  3. Oct 24, 2012 #2
    A multipole expansion is just a expansion of the potential in terms of powers of ##1/|r-r'|##. If you can verify that a certain collection of charges has neither a net charge nor a dipole moment and that the quadrupole moment is nonzero, it is guaranteed that the quadrupole contribution is dominant over all others far away from the charge distribution.
     
  4. Oct 25, 2012 #3

    mfb

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    It depends on the required precision. If you want to know the orbital period of moon to calculate the days of full/new moon, it is fine to treat earth as a monopole. If you want to know its position with an accurary of centimeters, you have to add more terms.

    What about the dipole moment?

    I think "simple" means the highest order is sufficient.
     
  5. Oct 27, 2012 #4
    OK, thank you for the replies they give some new ways to think about the problem.
     
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