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

## Best Answer

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

And indeed, Gauss' law

doeshold for general charges since Gauss' law together with Ampere's circuital law are theequations of motionfor 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

notvalid for moving charges, because deriving it from Gauss' law requires the assumption of electrostatics, and Gauss' law and Coulomb's law arenotequivalent in full electrodynamics. However, Coulomb's lawtogether with special relativityis equivalent to the full Maxwell equations, see this question.