Okay, let me point out something. Inside a spherical shell of uniform charge density, there is no electric field, assuming there are no other charges anywhere (Griffiths, Problem 2.7). This is what is usually shown with Gauss’s Law, and it’s a classic result of electrostatics. I, MonkeyMesch, John Mace, MC Master of Ceremonies, MikeS, and Desmostylus were addressing this issue.
Also, a conductor is a kind of object which will redistribute its surface charge to cancel out any E-field (Griffiths, Chapter 2.5). As a result of this, there will be no E-field inside a solid conductor, and a hollow conductor will “mask” any outside charges so that their cavities will not feel an external E-field. This cannot be shown with Gauss’s Law in general, which requires a high degree of symmetry. CurtC, Crafter_Man (in the other thread), panamajack, and matt were addressing this issue.
I think these two problems are being confused, so I just want to point out that they don’t necessarily have anything to do with each other. But, they are linked, because if you want a shell of uniform charge density, what better way than to just use a conductor? This is what Qwertyasdfg’s first post got into.