I saw that in potential due to charge distribution at some far points is composed of monopole,dipole,quadrupole etc terms.Does that mean the infinite charges in that distribution distributes as multipoles?
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2$\begingroup$ Hi Arun. I don't understand what Does that mean the infinite charges in that distribution distributes as multipoles? means. Can you clarify what you are asking? Note that any field can be represented as a sum of fields from the various multipoles. This is just a mathematical trick that makes calculations simpler in some circumstances. $\endgroup$– John RennieCommented Sep 29, 2016 at 6:55
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$\begingroup$ And, yes, you might need $n$-poles for $n$ charges. $\endgroup$– Jon CusterCommented Sep 29, 2016 at 12:45
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$\begingroup$ i mean,can i take any charge distribution into infinite number of discrete charged particles?and if i can then will those charged particles arrange themselves as multipoles? $\endgroup$– Arun KancharlaCommented Sep 30, 2016 at 17:54
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A generic charge distribution need not be a monopole or a dipole. Multipole expansion of potential illustrates that the potential due to an arbitrary continuous charge distribution, is in general, a sum of the contributions from an infinite number of terms monopole, dipole, quadrupole, octupole, hexadecapole etc. However, if the field point is far away from the charge distribution, the potential can be approximated by retaining only a finite number of terms in the sum.
I do not understand what did you mean by infinite charges in the distribution. The charge distribution can be confined to a finite region, and if it does not possess spherical symmetry, it will, in general, have all multipole contributions. If you have point charges in mind, then yes. Any finite continuous charge distribution will contain an infinite number of points, mathematically. But this is neither true physically nor is it of any relevance in this case. Even for a set of discrete charges, you can have all multipole contributions in absence of any symmetry.
