Consider a neutral hollow spherical conducting shell. The shape of the shell is described with respect to its center $O$ in spherical coordinates as the locus $a<r<b$ (inner radius $a$, out radius $b$).
Suppose we place a charge $q$ at the center $O$ and we wish to calculate the electric field outside of the conducting sphere at the distance $r=b$. By symmetry, the electric field is spherical, i.e it only has a radial component, and thus taking a Gaussian surface centered at $O$ of radius $b$, we deduce that $E*4 \pi b^2 = q/\epsilon_0$.
Now suppose we displace the charge $q$, so that in Cartesian coordinates it is located at $(2a/3,0,0)$. Let's try to evaluate the electric field strength at the distance $r=b$ on the $x$ axis, i.e at the point $(b,0,0)$.
Apparently, we get the same answer because the field is apparently still spherically symmetric for $r \ge b$, which makes no sense to me.
What am I missing?
What I would really like to do is determine the field for all points along the $x$ axis, but I'm not sure if it's possible to describe nicely.