Gauss' Law in 2D? At first I thought it was $$∮E.dl=Q/ϵ.$$
So i've read through some sources here and on the internet and most of them said that 
$$∮E.dl=2πq.$$
But I'm confused. Can anyone explain where does the $2πq$ come from?
Also someone answer this question with a result: $$∮E.dl=λ/ϵ.$$ So does this means that $λ/ϵ=2πq $? Can anyone show me a proof for this ?
Edit: I'm trying to solve this problem https://www.chegg.com/homework-help/questions-and-answers/5-problem-apply-gauss-law-2-dimension-calculate-electric-field-charge-distribution-2d-syst-q45861111 hence the question about Gauss' Law in 2-D
 A: In 2D one can define an entity analogous to flux by the integral of $E\cdot dl$ where the $dl$ is a segment of curve in the vicinity of a charge and is represented by a vector perpendicular to the curve and directed away from the charge.  (No area is involved.) In 2D the $E$ field is spreading in two directions rather than three. For a point charge, it would be $E = s\frac{q}{r}$, where the $s$ is a constant determined experimentally by someone living in 2D.  If you take a circular Gaussian loop around a point charge the “flux” becomes:   $$\frac{sq}{r}2\pi r = (2\pi s)q$$  (One might choose units for $q$ that make $s = 1$, analogous to so-called "Gaussian" units in 3D.)  By working with a small angle measured at $q$, one can show that this result is valid for any size or shape of the Gaussian loop.  
If you go to 3D, a cylindrical Gaussian surface around a line of length $L$ of charge of density $\lambda$ (coulombs/m) gives 
$$k\lambda L = 2\pi r L  E$$
Cancel the $L$ and this resembles the 2D situation.
Notice that in your problem they give the charge density as a function of $r'$.
