# Electric flux passing through a plate charge

My textbook describes the electric flux that passes through a flat charged sheet when using a perpendicular cylinder as the Gaussian surface to be double (i.e. 2EA). At first, it made sense to me because the electric field will leave in antiparallel directions. but, in other examples, like the uniform line of charge, we didn't consider the electric field leaving (positively charged line) in both directions when using a cylinder (whose axis is along the line) as the Gaussian surface.

my question is can someone explain why they did 2EA in the plate example but only EA in the line example. why isn't symmetry considered? thank you.

edit: thinking about it, the Gaussian surface encloses the charge so if I'm correct if a large cylinder was used to enclose the plate it would be EA. doesn't that mean we are using 2 cylinders when we "extend" one cylinder if we are only using a portion of the plate charge density to get 2EA.

• -1 A diagram would be useful to explain what you mean. And/or the explanations given by your textbook. – sammy gerbil Feb 6 '18 at 21:57

I think that the key is that the flux is E times the total area. For the cylindrical Gaussian surface and a plane, A is the area of one end of the cylinder only, e.g., $A = \pi*r^2$, so the total area is $2A$.
For a cylindrical Gaussian surface about a line, I suspect that the area you calculated is the full area already,e.g., $A=2\pi*r*l$, so there is no extra factor of 2.