Earnshaw's Theorem and a photon

In summary, a hollow sphere charged with positive charge and a proton placed in the cavity will result in the proton staying stably in the center of the sphere due to the zero electric field present there. This configuration does not follow the assumptions of Earnshaw's Theorem, which states that systems with fixed field strength obeying inverse square law will not allow for stable levitation. The electric field inside the sphere is zero due to it acting as a Faraday cage, and the presence of a single free proton inside will result in image surface charges rearranging to provide a quasistable equilibrium. This behavior may change if the proton is replaced with a sphere attached by many magnets, but the stability of the levitation remains uncertain.
  • #1
Harmony
203
0
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.
 
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  • #2
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.
 
  • #3
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...
 
  • #4
CompuChip said:
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.
 
  • #5
I think the electric field will be zero everywhere inside the sphere because it acts like a Faraday cage(try googling this).
 
  • #6
A.T. said:
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 :-p
 
  • #7
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.
 

FAQ: Earnshaw's Theorem and a photon

1. What is Earnshaw's Theorem?

Earnshaw's Theorem is a fundamental concept in physics that states that it is impossible to achieve stable equilibrium for a system of charged particles solely through the use of electric or magnetic forces.

2. How does Earnshaw's Theorem apply to a photon?

Earnshaw's Theorem applies to a photon because, as a massless particle, a photon cannot be held in a stable equilibrium by any type of force. This is due to the fact that photons do not experience electric or magnetic forces, as they are electrically neutral.

3. What does Earnshaw's Theorem mean for the behavior of photons?

Earnshaw's Theorem means that photons are always in motion and cannot be trapped or contained by any external forces. This is why photons are able to travel vast distances through space without any hindrance or deviation from their path.

4. Can Earnshaw's Theorem be violated for photons?

No, Earnshaw's Theorem is a fundamental law of physics and cannot be violated. It applies to all charged particles, including photons, and has been extensively tested and proven through experiments and mathematical models.

5. How does Earnshaw's Theorem impact our understanding of electromagnetism?

Earnshaw's Theorem is a crucial concept in electromagnetism as it helps us understand the limitations of electric and magnetic fields in holding particles in a stable equilibrium. It also has practical applications in fields such as particle physics and engineering, where the behavior of charged particles is of great importance.

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