electrostaticsgauss-law
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.
The basic concept is to see the charge distribution among them. From these example I hope your doubt will be quite clear
$$E\cdot \mathrm ds + E\cdot \mathrm ds + 0 = \frac{1}{\varepsilon_0}(\sigma\cdot \mathrm ds) $$
For curved surface it will be zero and let us take sigma distribution on sheet. See Figure 1.
Here Charges will distribute automatically. So now Let us take one gaussian cylinder passing through that sheet but not completely like till half of it sheet as shown in diagram. NOTE: It is not completely passed from the other end.
$$E\cdot \mathrm ds = \frac{1}{\varepsilon_0}(\sigma\cdot \mathrm ds)$$
Reason : Because at the middle of the sheet there is flat surface of our gaussian cylinder and at that surface there will be 2 forces acting on it. First one acting left hand side and other on right hand side due to charge distribution at both of the surfaces equally. See figure 2.
Now again let us take gaussian cylinder and this time it will pass from other end completely. Now just before going to question let us take one example. Imagine that if we put one cylinder inside sheet then we will see that 2 layers sheet will come out. First from front of the sheet and another from the back side. Why I said this because since charge is equally distributed on that conducting sheet so charges will be in front and back side so that's why if we take one gaussian cylinder then 2 charge layer will act at that time means there will be change in the above equation which we are writing.
So the equation would be like
$$2E\cdot \mathrm ds = \frac{1}{\varepsilon_0} (2\sigma \cdot \mathrm ds)$$
Which will make our equation
$$E=\frac{\sigma}{\varepsilon_0}$$
Do the field lines move inside the conductor due to the charges present on the surface
No, electric field lines never move inside the conductor. That's why electric field inside the conductor is zero as conductors are equipotential.
See, here the reasoning is such that you go from the fact that there is no charge inside the conductor to using Gauss's law to state that the electric field inside the conductor is 0 everywhere. However, this is faulty. The very premise of your reasoning should be that there is no electric field inside the conductor. Think about this, if there is an electric field inside the field then the free electrons of the conductor will start moving and a current will be created although there is no voltage applied. This is impossible and hence E=0 everywhere inside the conductor.
Now, use Gauss's law to get the fact that there can be no charge inside the conductor as any closed surface inside the conductor will have zero flux coming out of it( No electric field linked with the surface area). Hence, any charge provided to the conductor must reside on the surface. This is the simplest possible explanation.
I can not see how you used Coulomb's law to get a field inside the wire( Remember, Gauss's law is a far more fundamental law than Coulomb's law).This should hold for a really long wire as well as it is a conductor as well as long as you do not apply potential difference across its ends. Apply the same logic as above. It should be easy to see the truth.
Best Answer
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 $$ \iint_S E\cdot dA = \underbrace{E\cdot A + E\cdot A}_{\text{contribution from top/bottom of sheet}}= \frac{\sigma A}{\epsilon_0} \\ \therefore E = \sigma/2\epsilon_0. $$