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Which of these laws is more fundamental or forms the basis of electrostatics? I started off with Coulomb's law and then I studied Gauss' law. I was wondering which one is more universal?

My professor derived Gauss' law using Coulomb's law but didn't do it the other way, so is Coulomb's law more fundamental? And can Gauss' law be used to prove the other?

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    $\begingroup$ Just apply Gauss' law to a point charge. $\endgroup$
    – ACuriousMind ♦
    Mar 1, 2015 at 18:38
  • $\begingroup$ @ACuriousMind That was quite silly of me . Thanks . Got the derivation of each of them using the other one . What about the more fundamental one ? $\endgroup$
    – Klosew
    Mar 1, 2015 at 18:44
  • $\begingroup$ With the addition of $\vec{F}_E = q \vec{E}$ you can derive either from the other. They are formally equivalent and choosing one to be "more fundamental" is a philosophical statement. $\endgroup$
    – dmckee --- ex-moderator kitten
    Mar 1, 2015 at 18:44
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    $\begingroup$ Klosew, I assume you didn't read this before posting your question? en.wikipedia.org/wiki/Gauss%27s_law#Relation_to_Coulomb.27s_law $\endgroup$
    – Alfred Centauri
    Mar 1, 2015 at 18:52
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    $\begingroup$ @Klosew, I would say that Gauss' law is more general than Coulomb's law which is what I think you mean by "more fundamental". $\endgroup$
    – Alfred Centauri
    Mar 1, 2015 at 19:22

3 Answers 3

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Because Gauss's law applies for both moving and stationary charges, while Coulomb's law applies only for stationary charges, Gauss's law can be considered more fundamental. This is why Gauss's law is one of the four Maxwell equations. The derivation of Gauss's law from Coulomb's law only works for stationary charges; for moving charges the derivation is invalid yet Gauss's law still holds. However, Gauss's law along with the information from Maxwell's third equation that the $curl E = 0$ for stationary charges (since then $B$ will be constant), can be used to derive Coulomb's equation. In short, Gauss's law can be considered more fundamental because it applies to both stationary and moving charges, while Coulomb's law applies only to stationary charges.

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  • $\begingroup$ To add on to the question, 1) does Gauss's Law guarantee field to be square inversely decaying? My guess is, it shouldn't as it also describes moving charges who don't decay with square power (I am saying this because I read in Feynman that a part of em waves decay with the inverse of distance, not the inverse of the square). 2) So, the only difference is that Coulomb's law is for stationary charge, treated as points? $\endgroup$
    – Prabhat
    Jan 10, 2022 at 12:18
  • $\begingroup$ Why does Gauss law work for moving charges? An experimental fact? $\endgroup$
    – Cheng
    Feb 19 at 13:41
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If we are only considering three spatial dimensions, then Coulomb's and Gauss's laws are completely mathematically equivalent and there is no basis to consider either to be more fundamental than the other. But in several dimensions other than three, they are no longer equivalent, and when theorists consider generalizing electromagnetism to other numbers of dimensions, they almost always keep Gauss's law the same and modify Coulomb's law. So in that very weak sense, one could consider Gauss's law to be more fundamental. I discuss here why they do so. At the end of the day it boils down to a philosophical preference for mathematical elegance; until we find a universe with a different number of dimensions, there is no "right" answer.

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From the Feynman Lectures on Physics (I would have made this a comment but I don't have enough points)

From our derivation you see that Gauss' law follows from the fact that the exponent in Coulomb's law is exactly two. A $1/r^3$ field, or any $1/r^n$ field with $n≠2$, would not give Gauss' law. So Gauss' law is just an expression, in a different form, of the Coulomb law of forces between two charges. In fact, working back from Gauss' law, you can derive Coulomb's law. The two are quite equivalent so long as we keep in mind the rule that the forces between charges are radial.

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  • $\begingroup$ So both can be called equally fundamental ? $\endgroup$
    – Klosew
    Mar 1, 2015 at 18:42
  • $\begingroup$ They are in a sense equivalent. They describe the same phenomenon from different points of view. Similar to the different formulation of classical mechanics. I guess you could call them equally fundamental, but I don't really like the term. $\endgroup$
    – Gonate
    Mar 1, 2015 at 18:45
  • $\begingroup$ Isn't Gauss's law true for moving charges, unlike Coloumb's law, which works solely for stationary charges. $\endgroup$
    – user70720
    Mar 1, 2015 at 18:50
  • $\begingroup$ @Andy, No, Gauss Law is true for all charges. What's more, Gauss's Law is just the Divergence Theorem applied to the electric field. The Divergence Theorem is a the 3D equivalent of the Fundamental Theorem of Calculus and Green's Theorem. The general formulation is called Stoke's Theorem. $\endgroup$
    – Gonate
    Mar 1, 2015 at 19:02
  • $\begingroup$ @Gonate I ment that Gauss's Law works for all charges, moving and stationary, while Coloumb's law works solely for stationary charges. $\endgroup$
    – user70720
    Mar 1, 2015 at 19:07

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