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), but in the case of the conductor the electric field inside of it would be $\vec{0}$ and so $\vec{E_1}$ should be $\frac{\sigma}{\epsilon_0}$.
I cannot see where I've gone wrong.

  • $\begingroup$ Why would E = 0 in your calc? $\endgroup$ – ggcg Dec 22 '18 at 11:42
  • $\begingroup$ conductor in electrostatic equilibrium $\endgroup$ – user184836 Dec 22 '18 at 12:42
  • $\begingroup$ That does not answer my question. $\endgroup$ – ggcg Dec 22 '18 at 13:35
  • $\begingroup$ Well in a conductor which is in electrostatic equilibrium the electric field inside is 0 $\endgroup$ – user184836 Dec 22 '18 at 17:08
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    $\begingroup$ I'm glad it helped. $\endgroup$ – ggcg Dec 22 '18 at 20:29

You are not alone about being confused about this topic and in part is is because of the use the same symbol $\sigma$ being used to mean two different things; sheet charged density and surface charge density.

On the HyperPhysics website there is a derivation Electric Field: Sheet of Charge as shown below

enter image description here

The sheet charge density $\sigma$ is related to the total charge residing on both surfaces of a piece of conducting sheet not the charge residing on one surface of a piece of conducting sheet.

Note that $\sigma$ has not been called the surface charge density in the HyperPhyics derivation.

Let me change the definition of a symbol.
In the diagram below the sheet charge density is $\Sigma$ per unit area.

enter image description here

So the total charge on the sheet (with charges residing above and below the sheet) is $\Sigma A$.

In this case the surface charge density is $\sigma = \dfrac{\Sigma A}{2A} = \dfrac {\Sigma}{2}$

In your example you are dealing with only one surface which has a surface charge density which is double the surface charge density that was used in the HyperPhysics derivation and so you should expect the electric field to be twice as large.

  • $\begingroup$ Although I accepted the answer I still would like to restate my problem : When you take a cylinder as the gaussian surface to find the $\vec{E}$ near the conductor well it (the cylinder) has an end inside of it (the conductor), and the $\vec{E}.\vec{dA}$ inside would be $0$ since $\vec{E}=0$ inside right? $\endgroup$ – user184836 Dec 22 '18 at 18:24
  • $\begingroup$ @archaic Yes so the total flux through the two ends of the Gaussian cylinder is $\sigma A+0$ and the total charge within the Gaussian cylinder is $\sigma A$. $\endgroup$ – Farcher Dec 22 '18 at 21:01
  • $\begingroup$ Yes well it's not supposed to be that, the whole field created by the conductor should be equal to that but the contribution from the element of surface close to our point (the one we're assimilating to be an infinite sheet) should be $\frac{\sigma}{2\epsilon_0}$ and this is why I'm confused $\endgroup$ – user184836 Dec 23 '18 at 6:39
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    $\begingroup$ @archaic Does not the “plane” that you are considering have the same thickness as the rest of the conductor? $\endgroup$ – Farcher Dec 23 '18 at 7:26
  • $\begingroup$ yes of course . $\endgroup$ – user184836 Dec 23 '18 at 11:27

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