Argument for electric field inside spherical shell If I have a conducting spherical shell of outer radius $R$ and inner radius $r$, the electric field inside is $0$. The argument commonly given is:
Pick a point $P$ inside the shell (i.e. radius < $r$) and draw a sphere such that $P$ lies on its boundary. Then the charge contained by this sphere is $0$, so the electric through it is $0$ by Gauss' Law, and so the electric field at this point must be $0$.
Isn't this argument faulty? By this logic, I can pick any point anywhere in space (even outside the charged spherical shell), and draw SOME sphere which passes through it but doesn't intersect with the charged shell, and argue that since the flux through this other sphere is $0$, the electric field at this point (which is outside our charged shell) is $0$, which isn't the case? What am I missing?
 A: For Gauss' law to be applied straightforwardly, your point of interest must lie on an equipotential surface with some symmetry. For the case of the point inside the sphere, it is easy to visualize a Gaussian sphere with radius $r<R$ with said point on its boundary. The charge contained inside the Gaussian sphere would be zero -- the flux through the Gaussian sphere would be zero. You can then argue that, since the field lines should be normal to this Gaussian sphere, the flux $\phi_E = 4 \pi r^2 \textbf{E} \cdot \hat{r} = 0$ implies $\textbf{E} = 0$ inside.
It's quite different outside the charged sphere, however. If you try to hypothesize a similar Gaussian surface, you'll fail, because the electric field due to the charged sphere varies with position, and no Gaussian sphere would be normal to the field lines, unless you envelop the charged sphere completely (which would amount to computing the field at $r>R$.) The summary is that you cannot equate the flux to zero and infer the field is zero as well, like in the previous case. The net flux would be zero, sure, but that would involve a complicated mess of integrals and dot products with the radial vector and the normal vectors of our Gaussian surface.
Perhaps it is more illuminating to look at what is actually happening to the spherical shell, which we'll assume is perfectly conducting. In the absence of an external electric field, the free charges in the shell will align themselves such that the net electric field in the shell is zero; otherwise, they will simply move around until that occurs. On the other hand, if the shell is placed in an external field $\textbf{E}_\text{ext},$ this will prompt the free charges in the shell to reorient themselves such that the induced field $\textbf{E}_\text{ind}$ cancels out the external field perfectly everywhere inside the shell: $$\textbf{E}_\text{ext} + \textbf{E}_\text{ind} = 0$$
I hope this clears your doubt.
A: If the shell is charged and your gausian surface enclosed that shell hence electric field at this point is not 0.
A: Gauss law can help you find the field only when the surface is symmetrical about the point charge.
Taking a sphere around that point in the cavity with the centre of the sphere being the centre of the shell, you get a sphere that is symmetric wrt every charge in the spherical shell.
Hence you can use gauss law here to find field.
Taking any random outer sphere,flux through it is 0, yes.
However writing it directly as E×surface area of sphere is faulty, as E is variable with every point. So it involves more complex math and obviously would not come out to be 0.
