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If the electric charge is constant the law of Coulomb says that:

$E(r) = kQ/r^2$.

The question is, if $Q = Q(t)$, can I consider that:

$E(r,t) = kQ(t)/r^2$?

Update:

It appears that the first equation in Maxwell's system is the law of Coulomb put by Gauss in an equivalent mathematical form. However, in Maxwell's equations $E = E(r,t)$ not $E = E(r)$ like in the law of Coulomb. This is the reason I asked if $E(r,t) = kQ(t)/r^2$.

enter image description here Source: Wikipedia

hazdrubal
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    This question seems somewhat ill-posed because it doesn't indicate why the charge is changing as a function of time. Physically, charge cannot just 'appear' out of nowhere. The local continuity equation must be satisfied. So, there should be some current associated with the fact that the charge is changing and we have to account for that. – Kevin Driscoll Nov 23 '15 at 08:44
  • I did not say that the charge appears out of nothing. You can account for the currents associated with a varying charge. – hazdrubal Nov 23 '15 at 09:26

2 Answers2

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First, the Wikipedia article already says on the derivation of Gauss' law from Coulomb's law:

Note that since Coulomb's law only applies to stationary charges, there is no reason to expect Gauss's law to hold for moving charges based on this derivation alone. In fact, Gauss's law does hold for moving charges, and in this respect Gauss's law is more general than Coulomb's law.

And indeed, Gauss' law does hold for general charges since Gauss' law together with Ampere's circuital law are the equations of motion for the electromagnetic field, e.g. derived from the Euler-Lagrange equations for the Lagrangian of electrodynamics $$ L[A,J] = -\frac{1}{4\mu_0}F^{\mu\nu}F_{\mu\nu} - A_\mu J^\mu$$ for the four-current $J(\vec x,t)$ (with $J^0 \propto \rho$ and the spatial part the usual current), $A(\vec x,t)$ the four-potential and $F(\vec x,t)$ the field strength tensor.

Everything there is time-dependent and nevertheless, we get Gauss' law. However, to derive Coulomb's law, you must assume a static charge distribution and that the field of a static charge distribution is curl-free (or get that information from the Maxwell-Faraday equation).

Thus, Coulomb's law is not valid for moving charges, because deriving it from Gauss' law requires the assumption of electrostatics, and Gauss' law and Coulomb's law are not equivalent in full electrodynamics. However, Coulomb's law together with special relativity is equivalent to the full Maxwell equations, see this question.

Community
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ACuriousMind
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  • Gauss' law for the electric field is exactly Coulomb's law. "The first equation is just Coulomb's law of electrostatics, manipulated very elegantly ... by Gauss." (see the demonstration: http://www.conservapedia.com/Maxwell's_Equations ). – hazdrubal Nov 23 '15 at 21:42
  • @hazdrubal Conservapedia is just mistaken here, not about the derivation but about the underlying logic. Gauss' Law is not just Coulomb's law. What the demonstration there shows is that Coulomb's law is CONSISTENT with Gauss' law, namely that the electric field predicted by Coulomb's law satisfies also Gauss' law.

    So, we have proven that "If the electric field satisfies Coulomb's Law, then it satisfies Gauss' law." More abstractly, we proved $ P \rightarrow Q$ for $P$ = Satisfies Coulomb's Law and $Q$ = Satisfies Gauss' law.

     – Kevin Driscoll Nov 24 '15 at 00:17
  • However, $P \rightarrow Q$ does NOT imply $Q \rightarrow P$ and it also does not imply the claimed statement "Gauss' Law is exactly Coulomb's Law," which in this model is symbolized by $P \iff Q$. – Kevin Driscoll Nov 24 '15 at 00:27
  • @hazdrubal And so, for example, any situation where there are time-varying charges and currents will NOT generally be described by Coulomb's law. Instead Coulomb's law is replaced by the series of much more complicated Jefimenko equations , provided that there are no external fields being applied to these charges and currents. – Kevin Driscoll Nov 24 '15 at 00:29
  • There are numerous sources like "Atomic Spectroscopy and Radiative Processes" which state that: "The first equation is just Coulomb's law for electrical charges in its differential form". There is no doubt that Maxwell's first eq. is in fact Coulomb's law. – hazdrubal Nov 24 '15 at 04:29
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    @hazdrubal: Gauss' law is valid for moving charges, Coulomb's law is not. KevinDriscoll correctly linked the Jefimenko equations which are the proper general formulation for the electric field in terms of moving sources and currents. I don't know why you insist "There is no doubt that Maxwell's first eq. is in fact Coulomb's law.", but although some people might say it is, it is just not true. If they were the same, then you could derive Coulomb's law from Gauss' law without the assumptions of electrostatics. You have not presented such a derivation. – ACuriousMind Nov 24 '15 at 11:58
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Simply, no!

Light takes some time to travel, so the effect of chancing a charge at one point should be retarded further away.

If you assume such instantaneous force, it is called quasistatic approximation of Maxwell equations and can for example be used in Mie scattering theory, where a particle is much smaller than the wavelength of light.

Mikael Kuisma
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