Help with FLP argument of non-uniformly distributed surface charges

  • #1
Hello,
I'm reading FLP vol II, and I would appreciate some help to understand the argument supporting Figure 6-6.
Basically they claim if a sphere has non-uniform charge distribution whose surface density is proportional to the cosine of polar angle, then this surface charge distribution is equivalent to two solid spheres with equal-but-opposite uniform volume density, separated by a small gap.

Intuitively this is reasonable, but I can't prove this rigorously.

Similarly, in section 14-4, similar argument was applied to cylindrical non-uniform surface charge distributions (whose surface density is proportional to the cosine of azimuth angle, the claim is it's equivalent to two equal-but-opposite solid uniform volume-charged cylinders separated by a small gap).

Any help would be greatly appreciated
Thanks
 

Answers and Replies

  • #2
vanhees71
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Just calculate the surface charge of the two spheres and expand wrt. the distance of their centers. That's equivalent to the multipole expansion. The linear order of the expansion is equivalent to the dipole approximation. It's of course also very illuminating to directly treat the boundary-value problem using spherical harmonics.
 
  • #3
Thanks for the explanation. So it is a non trivial mathematical consequence which was not explained in the textbook. Usually FLP will either say just trust the claim because the proof is non trivial and beside the point, or give a semi proof. I guess it's not the case here
 

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