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enter image description here

I came up with a conclusion where I find half of the result on the picture. Here is how can the half of the result:

I thought the same gaussian surface. But since we know that, charges outside the gaussian surface contributes nothing to the electric flux. thus we disregard their electric field too. Thus, our surface charge has E field toward both outside and inside and taking it from here , we find $2*E*A=Q/e0$ where $Q=$surface density of charges * A

EDIT: CONDUCTOR IS CLOSED SHAPE

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  • $\begingroup$ There is no electric field inside a conductor. $\endgroup$
    – Farcher
    Jul 5, 2022 at 7:19
  • $\begingroup$ yeah, I know that. what I am asking is, the way I find the half result seems legit. Electric field inside the conductor is zero if you regard all of the charges around the conductor. I am saying that, lets group those charges in 2. 1st group is charges inside the gausssian surface and the 2nd group is charges outside the gaussian surface. total electric flux is the summation of electric flux of the 1st and 2nd group. 2nd group has zero electric flux since they are outside. thus 1st group gives the half result $\endgroup$
    – ozgun can
    Jul 5, 2022 at 7:32
  • $\begingroup$ The charges are either on the conductor's surface or off it. Where else can they be? You need to produce a diagram to show the distribution of charges relative to the Gaussian surface. $\endgroup$
    – Farcher
    Jul 5, 2022 at 7:39
  • $\begingroup$ irrelevant response to my point $\endgroup$
    – ozgun can
    Jul 5, 2022 at 7:42
  • $\begingroup$ So do you mean that the charges are distributed all over a conductor with the diagram only showing part of the conductor? $\endgroup$
    – Farcher
    Jul 5, 2022 at 8:05

2 Answers 2

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I think I understand it. Reason my calculation is wrong is not the train of thought but the the last calculation step which is considering the E field of the charges inside the gauss surface. for a charged 2-D surface, if you were to calculate the surface integral no matter how close to the charged surface, since E field on the surface that is to be integrated are not perpendicular at everypoint to the normal vector of the surface, you cant simply say it is EA for upper surface(for instance) therefore with my method(superposition theorem), you can only set a boundary which shows the max flux and that is 2E*A in this case. whatever the final flux is, it must be less than that, which is parallel to the final conclusion.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jul 5, 2022 at 9:13
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For simplicity let us consider a hollow spherical charged conducting shell. When you take the conductor as whole then surface charge density is whole charge divided by whole area ie. Q/A =sigma (say) which is same as if calculated for a small portion inside pill box ie deltaQ/deltaA . As charge is residing on one surface only so we do not take sigma as Q/2A or deltaQ/2deltaA.

But once you want to view the problem as two separate portions of the charged conductor and you want to neglect the presence of the non contributing portion of the charged conductor lying outside the Gaussian surface then you need to make one more change that the charge deltaQ will now reside on both sides of the portion so that new surface charge density of this small separate portion case will be deltaQ/2deltaA.

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