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Today I encountered a question that had the solution: electric field outside of a charged nonconductor is

$$ E=\frac{\sigma}{2\epsilon_0} $$

which had the diagram below:

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Fig. 1

and for the electric field outside of a conductor is:

$$ E=\frac{\sigma}{2\epsilon_0} $$

and had the diagram like below:

enter image description here

Fig. 2 (Cylindrical gaussian surface, shown from side)

The point that I couldn't understand is what is the role of conductivity in determining the gaussian surface's position? And what determines our cylindrical gaussian surface to whether go through the surface (like in Fig. 1), or to place its one end in the middle of the charged object(like in Fig. 2)? I couldn't catch the nuance what exactly distinguishes the both cases so that we do our calculation through different approaches.

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  • $\begingroup$ I think in the second case $E=\sigma /\epsilon_0$, no? $\endgroup$
    – garyp
    Commented Feb 17, 2016 at 19:25
  • $\begingroup$ @garyp no, sorry, it says $E=\frac{\sigma}{2\epsilon_0}$. However that's not the actual point. It sometimes in some (similar) questions turn out to be $E=\frac{\sigma}{\epsilon_0}$, but how do we actually determine the position of the gaussian surface? why in the second case we embedded it? $\endgroup$
    – Burak.
    Commented Feb 17, 2016 at 19:28
  • $\begingroup$ One face of the surface is taken to be inside the conductor because the field is zero there. There is flux only on one face of the cylindrical surface. That's why there's no factor of 2. You could take the surface so that both faces are outside, as in the first case. But you still get the same answer: no factor of 2. The field from a charged sheet does have the factor of 2. I'll bet these two cases did not come from the same source, and someone somewhere was talking about a charged sheet. $\endgroup$
    – garyp
    Commented Feb 17, 2016 at 19:37

1 Answer 1

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enter image description here

only thing to remember --- find symmetry so that E comes out of integral I.E. must be constant at every point of surface , and remember that density of nonconducting is not surface density but charge contained in the plane per unit area

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  • $\begingroup$ First of all thank you. However, it still does not provide the answer of main question. Your work shows how conducting and nonconducting surfaces differ in calculation. But my point was why and where do we use the embedded gaussian cylinder surface in our calculations (Fig. 2 in my post). $\endgroup$
    – Burak.
    Commented Feb 18, 2016 at 0:59
  • $\begingroup$ i just used the same figure in both the parts , so, basically, you can use any figure in any situation as long as electric field is constant at a particular area so, we can take E out of integral, it is your job to figure out symmetry by thinking about how line of forces would vary around it by putting an imaginary charge nearby it. concept is that electric field is zero inside the conductor due to obvious reasons and it is not zero inside a solid non conducting surface, $\endgroup$
    – Mrigank
    Commented Feb 18, 2016 at 7:29
  • $\begingroup$ but, it is zero right at its centre, so we could have taken a surface with one circle outside it and one right at it's centre parallel to plane, and flux passing through the latter would be zero $\endgroup$
    – Mrigank
    Commented Feb 18, 2016 at 7:31
  • $\begingroup$ Sorry but I have to dumb the situation down a bit, because I want to get convinced. You're saying that for nonconducting charged sheet, whole material is charged. Ok everything is fine so far. So when we take the cylindrical gaussian surface shouldn't it affect both of the sides instead of one? I mean If the whole thing is charged , its back must be charged as well, we need to take into account of the electric field on the back I believe, however, in your answer we're just taking only one side of the gaussian surface. Why? $\endgroup$
    – Burak.
    Commented Feb 18, 2016 at 22:50
  • $\begingroup$ I meant taking the surface for the first one that you have taken for 2nd one, the one you are talking about is the one I have used in my answer with one circle on each side symmetrically outside the material. $\endgroup$
    – Mrigank
    Commented Feb 19, 2016 at 0:37

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