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Multipole approximation outside conducting sphere

  1. May 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A dipole is placed next to a sphere (see image), at a large distance what is E proportional to?

    3. The attempt at a solution or lack thereof
    I'm having trouble figuring out what's happening in any variations of these. How does the dipole affect the sphere's charge layout? Scavenging through Griffith's but not been able to find this in there and am at quite a loss.
     

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  2. jcsd
  3. May 30, 2014 #2

    SammyS

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    Here is a link to Wikipedia.
     
  4. May 30, 2014 #3
    Isn't that just for a dipole inside the sphere?
     
  5. May 30, 2014 #4

    SammyS

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    No.

    It is a bit strange to have the dipole inside. That simply means that the conducting sphere may be replaced by an image dipole outside the sphere.

    If a dipole identical to the above image dipole is outside the sphere in the same location and orientation as the image dipole above, then there is an image dipole matching the above dipole inside the sphere.
     
  6. May 30, 2014 #5
    Oh right of course, thank you.
     
  7. May 31, 2014 #6
    Having some trouble getting any intuitive understanding from the wiki. What I was thinking (which may be way off) for the first image was that the spheres charges will rearrange so there are more positive charges on the right creating a dipole moment pointing in the same direction as the dipole. And E is proportional to a dipole in that case.

    But the grounded case? No idea what's going on
     
  8. Jun 1, 2014 #7

    SammyS

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    (Sorry for the long delay in responding.)


    It's clear that the answers to this exercise are qualitative rather than quantitative due to the limited information given.

    I think the easiest case to handle is the following, the rightmost of the attached figures in the OP.

    attachment.php?attachmentid=70171&d=1401471955.jpg

    Unfortunately, the choice of variable names in that Wikipedia page aren't too appropriate when discussing a dipole with dipole moment of ##\vec{p}##, due to some conflicts in variable naming.


    Model the dipole as consisting of charges, Q & -Q arranged vertically and separated by some distance, Δ. Both of these are equal distance from the center of the sphere. Let q be the image charge for, Q. It will have a sign opposite that of Q and will be smaller in magnitude. Then the image charge for -Q is -q. The separation, δ, for the image charges is smaller than Δ. The resulting dipole formed by the image chrges is smaller than ##\vec{p}## and has opposite orientation.

    How does that "look" from far far away ?
     
    Last edited: Jun 1, 2014
  9. Jun 1, 2014 #8

    SammyS

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    The other two cases:
    attachment.php?attachmentid=70169&d=1401471955.jpg   . . .     attachment.php?attachmentid=70170&d=14014719.jpg

    The Wikipedia treatment of the image of a dipole can be helpful for these two cases, but I think it's also helpful to look at the dipole as a physical dipole with two charges, a positive charge, Q, and a negative charge, -Q, again separated by distance, Δ, the positive charge being farther from the sphere, the negative charge being closer.

    Consider the locations, the signs, and the magnitudes of the resulting image charges.

    The neutral sphere (non-grounded) case is actually a bit more complicated to analyze than the grounded case.
     
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