Electric field due to a charged conductor and sheet of charge Electric field due to sheet of charge = σ/2ε
Electric field due to charged conductor = σ/ε

*

*For the conductor, when using Gauss's law, why the other side of the conductor is not considered.


*For conductor, charge will be on the outside and hence should be on both the sides. I am confused why while using Gauss law for sheet of charge, Gaussian surface goes through the sheet while in conductor it just stops in the sheet.
 A: The electric field is zero inside of a conductor. Therefore, there will be no electric flux through the side of the Gaussian surface that faces in towards the "meat" of the conductor. The case for the sheet of charge is different because there is an electric field on both sides of the sheet of charge, while there is only electric field in the exterior of the conductor.
Let's compute the electric field of the conductor:
$$
\iint_S E\cdot dA = \underbrace{E\cdot A  + 0}_{\text{contribution from outside/inside of conductor}}= \frac{\sigma A}{\epsilon_0} \\
\therefore E = \sigma/\epsilon_0.
$$
Meanwhile, that for the sheet of charge is
$$
\iint_S E\cdot dA = \underbrace{E\cdot A  + E\cdot A}_{\text{contribution from top/bottom of sheet}}= \frac{\sigma A}{\epsilon_0} \\
\therefore E = \sigma/2\epsilon_0.
$$
A: The Gaussian surface you use is up to you, it's just an imaginary surface chosen for purposes of convenience.  For the conductor, you choose to put it inside because you know the field is zero there.  For the sheet of charge, there is no middle, but you can use symmetry.  I think the basic answer to your question is that the meaning of sigma for the conductor is the charge per unit area on one face of the conductor--- it's not counting the charge on the other side, a charge that is there to insure the field is zero in the center.
A: The sheet u r talking about is probably being used in a capacitor, but the basic problem lies in the distribution of the charges on the sheet.
Let's assume that we have a cylindrical Gaussian surface(axis perpendicular to the sheet). The length of this cylinder is less than the thickness of the sheet, the electric flux will pass through the 2 bases(considering that half of the cylinder is inside and other half is outside the sheet)
Now flux= closed integral of E.ds
Flux total=2* electric field* area of base
q/£=2E.A
Further E=q/2A£(note distance from the sheet doesn't matter)
Even if combinations of these sheets are taken the NET electric field should always be 0(The distance between sheets doesn't matter)
If only one sheet is taken into account then both the faces have same charge density and field can easily be seen as0 
If more than1 sheet is considered the net field inside results in 0
