Electric fields and conductors

[archaic]

Hello guys!
The electric field created by a conductor at a point $M$ extremely close to it is $\vec{E}=\vec{E_1}+\vec{E_2}$ where $\vec{E_1}$ is the electric field created by such a tiny bit of the conductor that we can suppose it to be a plane, and since $M$ is extremely close to the conductor such that the distance is really small compared to the size of the plane we further ahead assimilate it to an infinite plane and hence $\vec{E_1}=\frac{\sigma}{2\epsilon_0}$ and this is where I block, when we use Gauss' law on an infinite plane we also account for the electric fields on the other side of the cylinder (here our gaussian surface) i.e the part inside the conductor, but in the case of the conductor the electric field inside of it would be $\vec{0}$ and so $\vec{E_1}$ should be $\frac{\sigma}{\epsilon_0}$ (##E \pi r^2 + 0 + 0= \frac{\sigma \pi r^2}{\epsilon_0}##)
(I have no idea why latex isn't processing this ^)
I cannot see where I've gone wrong.

 

[Charles Link]

I think you are trying to put a Gaussian surface around a conductor that has a surface charge density $\sigma$. Applying Gauss' law, you get $\\$ $E (2 \pi r)L=\frac{\sigma (2 \pi r L)}{\epsilon_o}$ so that $E=\frac{\sigma}{\epsilon_o}$, (pointing radially outward). $\\$ This should come as no surprise, even though for an infinite sheet of charge $E=\frac{\sigma}{2 \epsilon_o}$.

 

[archaic]

I think you are trying to put a Gaussian surface around a conductor that has a surface charge density $\sigma$. Applying Gauss' law, you get $\\$ $E (2 \pi r)L=\frac{\sigma (2 \pi r L)}{\epsilon_o}$ so that $E=\frac{\sigma}{\epsilon_o}$, (pointing radially outward). $\\$ This should come as no surprise, even though for an infinite sheet of charge $E=\frac{\sigma}{2 \epsilon_o}$.

Not exactly that, Coulomb's theorem states that the electric field created by a conductor on a point M extremely close can represented as the sum of the field created by an element of surface (which the point is so close to and thus can be "seen" as an infinite sheet of charge density $\sigma$ resulting in $\vec{E_1}=\frac{\sigma}{2\epsilon_0}$ ) and the rest of the conductor. My problem is with the field created by that one tiny element we've assimilated to be an infinite sheet because when applying Gauss' law with a cylinder as the Gaussian surface, the part of the cylinder that is *inside* the conductor will see no electric field and so, as I see it, $E_1 \pi r^2 + 0 + 0 = \frac{\sigma \pi r^2}{\epsilon_0}$ whereas this should be $\vec{E_1}=\frac{\sigma}{2\epsilon_0}$ apparently

 

[Charles Link]

Not exactly that, Coulomb's theorem states that the electric field created by a conductor on a point M extremely close can represented as the sum of the field created by an element of surface (which the point is so close to and thus can be "seen" as an infinite sheet of charge density $\sigma$ resulting in $\vec{E_1}=\frac{\sigma}{2\epsilon_0}$ ) and the rest of the conductor. My problem is with the field created by that one tiny element we've assimilated to be an infinite sheet because when applying Gauss' law with a cylinder as the Gaussian surface, the part of the cylinder that is *inside* the conductor will see no electric field and so, as I see it, $E_1 \pi r^2 + 0 + 0 = \frac{\sigma \pi r^2}{\epsilon_0}$ whereas this should be $\vec{E_1}=\frac{\sigma}{2\epsilon_0}$ apparently

$E_2$ is more difficult to solve for, but in the case of a cylinder, ir must be $E_2=\frac{\sigma}{2 \epsilon_o}$.