Electric field inside a conductor non zero

Question

I have a spherical conductor with a charge $+q$ place inside the cavity, now the charges redistribute as shown, If I apply gauss law where my guassian surface is such the $q$ inside is non zero now , $\oint \vec{E}.\vec{da}= \dfrac{q}{\epsilon} $ we can say since $q≠0$ , $\vec{E}≠0$. Now this contradicts the fact we already know that electric field inside a conductor is zero , please tell where I went,is it something wrong with my guassian surface (why?)any help would be appreciated, Thanks.

$\textbf{EDIT}$: To avoid any confusion my guassian surface is only about the boundary of conductor it neither goes inside the cavity nor outside the conductor but it includes the charges present at the periphery of conductor that is some positive charge and a lesser negative charge making overall net charge non zero.

Answer 1

The charge density on the conductor surface is singular, so gauss' law is not well defined if the surface you draw goes through the conductor surface. You can put the surface slightly within the conductor, and the surface charge will not be picked up, giving no field inside the conductor, or you can put the surface just outside the conductor and the charge will be picked up, reflecting that the surface charge creates a field outside the conductor.

Answer 2

To include the surface charge densities, the Gaussian surface must be just outside of the surfaces. The flux in will be proportional to the included negative charge on that surface, and the flux out is proportional to the included positive charge on the other surface (with no field or flux in the conductor).

Answer 3

Gauss' law tells the total charge inside a surface. In special cases you can also draw conclusions on the value of $\bf E$ at the surface. In your example it is not possible to draw the conclusion that $\bf E\neq 0$ inside the conductor.

Note that the statement that $\bf E = 0$ inside a conductor is only true at a scale where the underlying ionic system can be treated as a continuum. At this scale the surface charge has zero thickness. At atomic scale $\bf E = 0$ only on average over large distances.

Answer 4

Gauss law only can only be used to evaluate electric field of charges contained inside gaussian surface. The contribution due to charges outside always dies.

You can prove that any external field outside the gaussian surfaces dies on taking the integral. Therefore, you can't speak of net field using gauss law.

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OP in comments:

You are working the maths wrong here in the counter example , firstly you are taking the E out of the integral considering it constant which is not true , the fact is neither is the magnitude of E nor the direction is same at all points odlf the guassian surface so you cant perform the maths so simply.

Consider Electric field defined on the gaussian surface decomposed the following way: Enet=E+E′ E is due to charge inside the gaussian surface and E′ is due to the charge outside. Consider the case of removing the charge inside and evaluating the flux across the boundary due to the external charge, in this case we find that the flux due to external charge is zero by gauss law. Now, reinsert the old charge, we find that ∫Enet⋅dS=∫EdS and we can evaluate it's field as needed by pulling out the E.

Ultimately the point is that gauss law is useless for making any claims for fields caused by charges outside the gaussian surface because they will always cause zero flux independent of the distribution