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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?

Kyle Kanos
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Klosew
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    Just apply Gauss' law to a point charge. – ACuriousMind Mar 01 '15 at 18:38
  • @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 ? – Klosew Mar 01 '15 at 18:44
  • 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. – dmckee --- ex-moderator kitten Mar 01 '15 at 18:44
  • But doesn't Gauss's law work for moving charges while coloumb's law applies solely for stationary charges? –  Mar 01 '15 at 18:51
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    Klosew, I assume you didn't read this before posting your question? http://en.wikipedia.org/wiki/Gauss%27s_law#Relation_to_Coulomb.27s_law – Alfred Centauri Mar 01 '15 at 18:52
  • @AlfredCentauri Thank you . But I got the derivation after ACuriousMind's comment . So Gauss Law is more fundamental as it is applicable for moving charges ? – Klosew Mar 01 '15 at 19:03
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    @Klosew, I would say that Gauss' law is more general than Coulomb's law which is what I think you mean by "more fundamental". – Alfred Centauri Mar 01 '15 at 19:22
  • @AlfredCentauri Yes I meant that . I was wondering that in Bohr Model we used Coulombs Law to get centripetal force between the nuclues and electron . So here the law was applied on a moving charge and it was valid too . – Klosew Mar 01 '15 at 19:38
  • I'd think Gauss' Law. It's valid for time varying charge distributions as mentioned elsewhere, but a small deviation of it is valid in a curved space-time – R. Romero Feb 09 '22 at 17:50

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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.

Devraj bahl
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  • 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? – Prabhat Jan 10 '22 at 12:18
  • Why does Gauss law work for moving charges? An experimental fact? – Cheng Feb 19 '23 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.

Devraj bahl
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tparker
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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.

Gonate
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  • So both can be called equally fundamental ? – Klosew Mar 01 '15 at 18:42
  • 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. – Gonate Mar 01 '15 at 18:45
  • Isn't Gauss's law true for moving charges, unlike Coloumb's law, which works solely for stationary charges. –  Mar 01 '15 at 18:50
  • @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. – Gonate Mar 01 '15 at 19:02
  • @Gonate I ment that Gauss's Law works for all charges, moving and stationary, while Coloumb's law works solely for stationary charges. –  Mar 01 '15 at 19:07
  • @Andy My fault, you are perfectly correct. – Gonate Mar 01 '15 at 19:12
  • @Klosew Coulomb's law applies to charges at rest i.e., time-independent charge densities. Gauss' law derived from Coulomb's law applies to time-independent charge densities. But Gauss' law in its generalized form $\nabla\cdot\textbf{E}(\textbf{r},t)=\rho(\textbf{r},t)/\epsilon_0$ constitute one of the Maxwell's equation, and applies to time-dependent charge densities as well. – SRS Aug 09 '17 at 08:44