Charge inside a charged spherical shell 
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*If I were to put a negative charge inside a negatively charged spherical shell, will it move to the center? 

*Electric field inside the shell due to the shell is zero (Gauss's Law), would that mean the charge inside the sphere faces no force?
But, that doesn't make intuitive sense to me. If the negative charge was near the walls of the sphere, wouldn't the charges on the near wall push the negative charge to the centre as the force due to the charges on the wall closest to it is higher than that form the walls further away from it.


*What about in the case of a ring? Will the charge move towards the center? 

 A: 
First and foremost, I don't think you could ever prove Gauss's Law wrong. If it says the field (and hence force) is zero, it is indeed zero, and accordingly its consequences. Now if you want the physical situation that causes the force experienced inside to be zero, it is crudely represented in the diagram. Consider the major bowl and the minor bowl (I like calling them that!). While the major bowl has more charge (due to larger surface area), it is further away. In the case of minor bowl, the surface area (and hence charge) is lesser, but so is its distance. For both the bowls, the increase/decrease in charge cancels out the decrease/increase in the distance from the point charge, such that the resultant fields due to each bowl are equal, and opposite, which then cancel each other, and hence field(force) is zero. Similar argument for center or any other point inside the sphere may be given.
P.S. the lines represent the direction of force due to each elemental charge (repulsive, as both are negatively charged). I forgot to draw similar lines for the minor bowl (in the opposite direction).
A: It would experience no net force inside shell. It is free to move.
A: Since there are no charges inside a charged spherical shell . This means the net charge is equal to zero. So magnitude of electric field $E$=0. So there is no net force. So magnitude of net force =0. No movement towards centre.
A: So, this is an interesting property of the mathematics of a force that diminishes like $1/r^2$ in 3D-space: if you have a uniform charge distributed over a sphere, that charge exerts no forces inside the sphere; they all balance out. Furthermore the field outside the sphere behaves as if all of the charge on the sphere was concentrated at a point at the center of the sphere. It is also, for example, true of the gravitational force: a "spherical shell" of matter exerts no gravitational forces within, but acts like a force field from a point outside.
The basic mechanism can be understood like this: a "sheet" of charge has a force which does not diminish with distance because the $1/r^2$ effectively competes with the $r^2$ of surface area that you can "see" (e.g. if we assume that the entire contribution comes from a fixed solid angle); so the space between two equally charged plates works out to have exactly 0 force due to those plates. (There may be other forces due to stuff outside the plates, of course, if the plates are not conducting and are uniformly charged.)
In the case of a ring, we could probably expect that yes, the ring would push the particle towards the center; however the equilibrium is unstable as if it deviates in any way off the plane of the ring, it will receive a push further in that direction.
A: How about this?
1) There is a charged spherical shell. The origin of the sphere must not have any electric field due to symmetry. 
$${\bf E}({\bf 0}) = {\bf 0}$$.
2) Now take a point from the to the origin at $\bf r$. Due to symmetry of the problem the electric field has to be radial (points away from the origin), but can (still) have a magnitude $A_r$. 
$${\bf E}({\bf r}) = A_r(|{\bf r}|) \hat {\bf r}$$
3) Apply the differential form of Gauss Law (with knowledge, that there is no charges, thus the charge density needs to vanish)
$$ \nabla \cdot {\bf E} = \rho({\bf r}) = 0$$,
to this function in spherical coordinates (https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates), yielding
$$ \nabla \cdot {\bf E}(r) = \frac{1}{r^2} \frac{\delta (r A_r(r))}{\delta r} = 0.$$
This indicates that $A_r(r)$ i.e. the electric field has to be constant wrt r.
Now let's combine the result from section 1) that the electric field is zero at the origin, and from 3) that the electric field is constant to yield that the electric field has to vanish in the whole system.
To gain further insight, consider how Gauss law is connected to conservation of field lines, and consider that in spherical symmetry.
A: 
If I were to put a negative charge inside a negatively charged spherical shell, will it move to the center?


Electric field inside the shell due to the shell is zero (Gauss's Law), would that mean the charge inside the sphere faces no force?

The answer depends upon whether the spherical shell is conductive, and if not, whether or not the charges on the spherical shell are uniformly distributed.
If the charges on the spherical shell are uniformly distributed, then there will be no net electric field due to the shell anywhere within the shell.
However, if the spherical shell is conductive, and the charge within the cavity is not perfectly centered, the charge within the cavity of the shell will cause the charge on the shell to re-distribute itself. The resulting charge distribution on the shell will not be uniform. Consequently, the shell will exert a net force upon the charge in the cavity. The direction of the force will be away from the center, not toward it. That is, an uncentered charge in the cavity will want to move to the surface.
