I am trying to figure out how the equation for the field if a Gaussian surface was applied to a 2D plane. I did see another question already asked on the subject, but I didn't get a particular explanation and the question was a couple of years ago.
So - if I draw the 2D version of a Gaussian sphere (a circle) around a charge, and use the analog of Gauss' law as Flux = (Field)(Circumference), then I know what I have on the left hand side of the equation, namely, $E(2\pi r)$, but I'm not sure what I must equate this to in order to get the relation $E = q/r$ which is apparently what I'm looking for.
EDIT: Here is the link to the other Phys.SE question: Electric field and electric potential of a point charge in 2D and 1D
The answer by the user 'user1504' describes how $E(2\pi r)$ = $2\pi q$. I don't understand how the expression $2\pi q$ on the right hand side of the equation was arrived at.