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While reading about Gauss’ law in electrostatics, I saw that we do not apply this law to evaluate the fields generated by dipoles, quadrupoles etc. It is only applicable to cases where the fields fall of like $1/r^2$ to preserve the integral where the surface term increases like $r^2$. Now, my confusion is if it is like that, then we do get some fields with the application of Gauss’ law which increases like $r^2$ (Reference: Introduction to Electrodynamics, Fourth Edition, David J. Griffiths, Page 76). I have read several answers on SE as well to understand this thing and many have argued that when we get fields which increase like $r^2$ or anything else other than $1/r^2$ variation, it is due to the fact that we can consider the charge distribution to be a collection of tiny charges and for each individual charge, the field falls off like $1/r^2$ but when we consider the field due to the whole charge distribution, all the fields add up to give the $r^2$ variation or anything other than $1/r^2$.

So, why don’t we apply Gauss’ law to evaluate the field generated by a dipole or quadrupole by applying the Gauss’ law for each individual charge and then adding up? I have tried this, but I didn’t get the $1/r^3$ variation for dipole and $1/r^4$ variation for quadrupole. So, what is the problem here?

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    $\begingroup$ Gauss law still applies. The total charge inside is zero, but it is not useful to calculate E because you don't have a surface in which E is a constant $\endgroup$
    – user65081
    Nov 4, 2021 at 15:45
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    $\begingroup$ It is always applicable: what it says for a dipole is that if you do the surface integral over a surface that surrounds the dipole, you get 0. That is not enough information to deduce the field at any point. The reason that you can get away with it for a point charge is the symmetry: because of spherical symmetry, you can say that the field has exactly the same value at every point of a sphere centered on the point charge (and it is normal to the surface of the sphere), so you can pull it out of the integral. $\endgroup$
    – NickD
    Nov 4, 2021 at 15:45
  • $\begingroup$ @NickD: That looks like a pretty good answer to me. $\endgroup$
    – Michael Seifert
    Nov 5, 2021 at 20:52

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It is only applicable to cases where the fields fall of like $1/r^2$ to preserve the integral where the surface term increases like $r^2$.

Gauss' law applies to any configuration of enclosed charge because the law only says that the net electric flux across the surface enclosing the charge equals the net charge enclosed divided by the electrical permittivity. It does not tell you what the electric field is at any point on the surface, unless there is sufficient symmetry in the arrangement of the enclosed charge.

Examples of the application of the law where there is such symmetry can be found on the fourth slide here:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

Hope this helps.

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