I have this grave confusion that I have been having since a while. When we calculate the electric field due to an infinite plane sheet of charge then the answer comes out to be $σ/2ε$. In this case we take a cylindrical Gaussian surface which extends to both the sides and that is why we have halved the electric field to get the answer. Now when I try to get the same for a charged conducting surface then why does the value double itself? The diagram in my book shows a rectangular slab kind of a conductor in which one flat surface of the cylinder (the Gaussian surface) is embedded inside the conductor and the other is outside. According to the diagram,then , there should be a back side of the conductor. So if I construct a Gaussian surface such that it extends out from both the sides ,then I should get the same answer as that for a charged sheet! Why do we not do this in this case?
All I know is that for a charged conductor the field inside is 0 and the charge resides on the surface.
Also considering any one surface of the conductor ,the field lines move outward. Do the field lines move inside the conductor due to the charges present on the surface?That is -if I have a slab like conductor with a point S inside it, then is there any field due to the surface charge at that point.