Earnshaw's Theorem and a photon

[Harmony]

Imagine there is a hollow sphere charged with positive charge, distributed evenly. Then a proton is placed in the cavity. Would the proton levitate stably in the middle of the cavity?

By Earnshaw Theorem, any configuration of system with fixed field strength which obey inverse square law will not enable us to have stable levitation. (Since there are no local maxima or minima). But, in the above case, if the proton doesn't stay stably in the middle, where can it go? It is repelled in every direction isn't it?

Does the system above do not obey some of the assumption required by Earnshaws Theorem?

Thanks in advanced.

 

[CompuChip]

The proton will stay in the center of the sphere, because there the electric field is zero.

I don't know the Earnshaw Theorem, but I think it does not apply in this case because inside a uniformly charged sphere the electric field goes like E ~ r and not as E ~ 1/r^2.

 

[Harmony]

Thanks for the prompt reply. Yeah, I forgot that this is in uniform field, hence the field strength decrease by 1/r rather than 1/r^2.

Suppose if I substitute the proton with a sphere attached by many magnet with the same polarity facing up, and similarly, attach many magnet on the interior surface of the hollow sphere. How would the behaviour change? Would it still be the same? Or would the levitation lose its stability? I am not very sure whether the magnetic field strength decline by 1/r^2 or 1/r in this case...

 

[A.T.]

The proton will stay in the center of the sphere, because there the electric field is zero.

Isn't the electric field zero inside the entire cavity? That would mean, it is not really a 'stable' configuration, in the sense that the proton returns to the center, in case of a small perturbation.

 

[Dadface]

I think the electric field will be zero everywhere inside the sphere because it acts like a Faraday cage(try googling this).

 

[CompuChip]

Isn't the electric field zero inside the entire cavity? That would mean, it is not really a 'stable' configuration, in the sense that the proton returns to the center, in case of a small perturbation.

You are right, I was thinking solid spheres. If the sphere is hollow the field is zero (apply Gauss's law, no enclosed charge, etc). Still, that is not E ~ 1/r^2 or even E ~ 1/r

 

[Bob S]

If you have a hollow conducting sphere (without any apertures), there can be no electric fields within the metal (from the inside surface to the outside surface). This means that there is no net charge inside the hollow sphere. If there is charge on the inside surface of the hollow sphere, there must be free compensating charge inside the hollow sphere. Use Gauss' Law.

If you have a single free proton inside a hollow conducting sphere, the surface image charges will re-arrange to provide suitable image charge (surface image charge density doesn't have to be integer charges, just its integral over the entire 4 pi surface). If the proton is in the center, then it is in quasistable equilibrium. If the proton is off center, it will be accelerated to the wall by the rearranging image charge. Compare this to a free charge proximal to a conducting metal plate. There is an image surface charge whose arrangement is equivalent to a mirror charge behind the conducting plate. There is net attraction.