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Need help on Quadrupole potential

  1. Jan 25, 2005 #1
    I need help on how to do this problem: Carry out the calculation on the simplest quadrupole: Two point dipoles are oppositely oriented along the z-axis, separated by distance a. The potential due to one dipole is V = (p cos(theta) / (4pi epsilon r^2). The result I should get for the quadrupole is: V = a p (3 cos^2(theta) - 1) / (4 pi epsilon r^3).

    I may need to use tensors in here, but I don't know how to use them. Thanks.
  2. jcsd
  3. Jan 25, 2005 #2
    yes and no, quadrupole expansion requires a little bit tensors, but in your problem, you don't

    add up the potential of two dipole, and ignore the r^4 terms, you will get the answer, the calculation is straight forward. nothing tricky, you could able to handle them

    PS.. always draw the graph first.... the graph helps a lot
  4. Jan 25, 2005 #3


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    Do you know the procedure used to derive the potential of a dipole? Use the exact same procedure here... except that instead of using V=q/(4*pi*epsilon*r^2) (potential of a point charge), you'll instead use the V formula for a dipole.

    It involves Taylor series...
  5. Jan 25, 2005 #4
    My book uses Legendre polynomials to do the expansion for the potential of N point charges, and it produces a monopole term, dipole term, and quadrupole term... But I'm still not very clear what's going on. Are you saying if I do this multipole expansion using the V for the dipole, the resulting quadrupole term will be the answer? Thanks.
  6. Jan 25, 2005 #5


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    No, I believe if you do the expansion with V for the dipole, then the dipole term will be your answer. The monopole term should come out to zero. I'm not familiar with Legendre polynomials... but I'm guessing it's like Taylor series... Not that the two dipoles have opposite p. p1=-p2.
  7. Jan 25, 2005 #6
    I just realized since for this particular case there is a r^2 in the denominator as opposed to r in the general case, the expansion won't produce Legendre polynomials. I have to figure out how to do this expansion from scratch -- I will try Taylor series to see if it works.
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