Why does Gauss law give E = σ/ε for a parallel plate capacitor, not 2σ/ε?

[hokhani]

we know that the electric field between the planes of planar capacitor is σ/ε (according to gauss law)
however we have two conducting plate that each plate produe this electric field that are in the same direction therefore we must have E=2σ/ε
what is the reason for this contradiction.

 

[Born2bwire]

It seems to be my recollection that a sheet of charge has an electric field of

$$E = \frac{\sigma}{2\epsilon}$$

It's trivial to derive this by hand.

$$\oint \mathbf{E}\cdot d\mathbf{a} = \frac{1}{\epsilon} Q = \frac{\sigma A}{\epsilon}$$
$$\oint \mathbf{E}\cdot d\mathbf{a} = 2A \left| \mathbf{E} \right| = \frac{\sigma A}{\epsilon}$$
$$\mathbf{E} = \frac{\sigma}{2\epsilon} \hat{n}$$
where \hat{n} is the normal direction of the sheet.

 

[hokhani]

Ok
If we have non conducting sheet we can take a cylindrical gauss surface around two sides of the sheet and you are right but if we have conducting sheet and take a cylindrical gauss surface in such that on of the circular surface of gauss surface take between two sides of the sheet(as shown in the picture), the electric field will be:
E=σ/ε
Because the electric field will be zero in the conductor

 

[Born2bwire]

Ok
If we have non conducting sheet we can take a cylindrical gauss surface around two sides of the sheet and you are right but if we have conducting sheet and take a cylindrical gauss surface in such that on of the circular surface of gauss surface take between two sides of the sheet(as shown in the picture), the electric field will be:
E=σ/ε
Because the electric field will be zero in the conductor

Ahh, I got it. However, if we have a charged conducting sheet then the charge is spread evenly across its surface. Neglecting the charge on the edges (assuming a thin sheet) then the top side will have half the charge and the bottom side would have the other half. Thus, when you take your Guassian surface that has one surface terminating inside the volume then the enclosed charge is actually

$$Q = \sigma A$$
$$Q_{upper} = \frac{1}{2}Q = \frac{\sigma A}{2}$$

since you are only enclosing one surface of the conductor. So it still works out ok. If it didn't work out the same, then we come to a problem where we now have a problem. If we have an infinitesimally thin sheet of charge we get the factor of 1/2. If a conducting sheet gave us twice the E-field then what happens when we take the limit of the sheet to have zero thickness? We arrive at a contradiction.

 

[hokhani]

thank you very very much.