Expansion for the potential of a ring of charge

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SUMMARY

The discussion focuses on deriving the potential expansion for a uniformly charged ring of radius b, specifically in the region where r < b. The charge density is defined as λ = q / (2πb). The previous problem addressed the potential for r > b, yielding a quadrupole expansion. The key insight is that a Taylor expansion in the parameter b/r is valid only for r > b, prompting the need for a new expansion applicable when r < b.

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  • Knowledge of Taylor series and convergence criteria
  • Basic calculus for integration and series expansion
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Homework Statement


A total charge q is distributed uniformly along a ring of radius b. The ring is in the x-y plane centered on the origin. The multipole expansion is not valid for r<b. Find an expansion for the potential valid in this region

Homework Equations


The charge density is just \lambda=\frac{q}{2\pi b}.
The previous problem was about the region r>b, and the book gives the quadrupole-moment tensor components as Q_{xx}=Q_{yy}=\frac{b^2}{2}q, Q_{zz}=-b^2q, Q_{xy}=Q_{yz}=Q_{xz}=0 and the potential as \Phi =\frac{q}{r}+0+\frac{1}{r^5}\frac{b^2q}{4}(x^2+y^2-2z^2).

The Attempt at a Solution


Frankly, I don't understand the problem. Obviously at z=0 we can just integrate \Phi=\int_{0}^{2\pi}\lambda d\theta=\frac{q}{b}, but that's a pretty trivial introductory level problem. Any thoughts on what exactly this problem is asking me to do?
 
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Well, the multipole expansion for large r is basically a Taylor expansion in the parameter b/r. The series only converges if the value of the parameter is less than 1, or equivalently, only if r > b. So I think what this problem wants you to do is come up with a Taylor expansion in a different parameter, one that will be less than 1 when r < b, so that this new expansion will be valid in the region r < b (not just at r = 0).
 

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