# Gauss law at the surface of a conductor

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

• There is no electric field inside a conductor. Jul 5, 2022 at 7:19
• 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 Jul 5, 2022 at 7:32
• 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. Jul 5, 2022 at 7:39
• irrelevant response to my point Jul 5, 2022 at 7:42
• So do you mean that the charges are distributed all over a conductor with the diagram only showing part of the conductor? Jul 5, 2022 at 8:05