About applicability of Gauss’ law in electrostatics

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?

• 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
– user65081
Nov 4, 2021 at 15:45
• 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. Nov 4, 2021 at 15:45
• @NickD: That looks like a pretty good answer to me. Nov 5, 2021 at 20:52

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 $$r^2$$.