Why does the the electric field intensity is 0 inside a charged conducting spherical shell? [duplicate]

Question

Given a conducting spherical shell with infinitesimal thickness and the radius of $R$. The shell is uniformly charged with surface density $σ$. Find the electric field intensity vector $\bf E$ at the radius of $r$ from the centre $C$ of the shell ($r<R$).

So most of the answers I read state that: No enclosed electric charge -> Net flux through the gaussian surface (C; r) is 0 -> E is 0. However, I have my doubts.

1. How I understand gauss's law is that if there is no enclosed charge in a gaussian surface then there can't be any electric field line emerging from the inside of the surface and as for the external electric field, they cancel out each other since all external electric field line come in through the surface would come out with different Area ( a video about this would explain further https://www.youtube.com/watch?v=jCuWaC0Oibw). Therefore, for every value of (E.dA) of the field generated by a charge outside the gaussian face and an Area, there would be a certain amount of flux that is equal to the negative of that value (due to the opposite side the surface that the flux come through). So what are cancelling each other are the flux, not the field intensity vector( accounting only a gaussian sphere and a not-enclosed charge). It would make sense that if the net flux through a gaussian surface generated by a single charge is 0 then the entire shell would generate a field with the same net flux through the surface (C;r). But, that does not prove that E is 0.
2. I solved a problem that asked me to calculate the electric field intensity vector at a point on a diameter of a uniformly charged ring (infinitesimal thickness), which is a vector aligned with the diameter and the vector point to the centre, it's length is not 0. I think that I place multiple identical rings with the same diameter axis, they would make a spherical shell and all of the electric field intensity vectors would be identical as long as it is a point on the diameter of all rings. So with a shell, we could split them into rings with the common diameter being the line through the centre and the point we are calculating. Thus the field intensity at that point is not 0.

So can anyone please explain to me whether the field intensity is 0 or not, if not, how can it be calculated, use integral and derivatives if you must.

Answer 1

By symmetry the field, E, at any radius, r, is in the radial direction and has the same value everywhere on the surface of the sphere with radius r. So when calculating the flux, E can be taken outside the integral. You are left with E4$\pi$$r^2$ = 0 implying that E = 0.