Gauss' theorem and inverse square law

[physics user1]

So, I know that the gauss law states that the Flux of the electric field through a closed surface is Q/ε , but does the gauss theorem works also for non inverse square law Fields?

I think not because in order to not have a Flux depending on distance but a constant one we need that r^2 of the surface has to cancel with something that is r^2 too, am I right?

And why then is correct to calculate the electric field caused by a infinite long charged rod or an infinite plane charged surface, the electric field there doesn't obey to the inveresults square law but we assume that the gauss theorem works...

I thought about this and I came to the conclusion that individually charges on the rod or the surface obeys to the inverse square law, it's just the sum of all these electric fields of all the charged enclosed in the gauss surface that are not inverse square so it's OK because that, is my "theory" Right?

 

[vanhees71]

Gauss's Law reads (in Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{E}=\rho.$$
For electrostatics you also have
$$\vec{\nabla} \times \vec{E}=0$$
and this implies that there exists (in any simply connected region of space) an electric potential such that
$$\vec{E}=-\vec{\nabla} \phi.$$
This implies
$$\Delta \phi=-\rho,$$
and the Treen's function for the (negative) Laplace operator in 3D Euclidean space is
$$G(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|}.$$
From a long distance you can characterize any charge distribution just by its total charge, and the corresponding monopole contribution to the potential reads
$$\phi(\vec{x})=\frac{Q}{4 \pi |\vec{x}|} \; \Rightarrow \; \vec{E}=-\vec{\nabla} \phi=\frac{Q}{4 \pi |\vec{x}|^2} \frac{\vec{x}}{|\vec{x}|}.$$
So the leading order of the multipole expansion indeed always decays with $1/r^2$ with distance (in three dimensions).

Of course, this does not hold for the full Maxwell equations, i.e., for time-dependnet fields. There the leading order of the em. waves goes like $1/r$.

 

[USeptim]

The radiated electric field depends on 1/r but it's perpendicular to the direction of propagation so the total flux is zero. If you take an sphere centered on an accelerating particle the electric field will be perpendicular to the surface everywhere so the flux will be zero.