Can someone explain Earnshaws Theory

  • Thread starter jonlg_uk
  • Start date
  • #1
141
0
I know Earnshaws Theory basically states that it is impossible to levitate a charged object using static electric field. But can someone explain it in simple lamens language what exactly going on on an atomic level and how it ties in with Gauss's law?

I look forward to hearing your replies
 
Last edited:

Answers and Replies

  • #2
5,601
40
I am not familiar with the theorem except for a brief read here...have you seen this explanation??

http://en.wikipedia.org/wiki/Earnshaw's_theorem#Explanation

or is there something about the explanation that doesn't make sense to you?

If you are questioning this:
Gauss's Law says that the divergence of any possible electric force field is zero in free space.
then I don't think anyone knows. It says electromagnetic radiation does not disappear nor originate in free space....the big bang might be an exception??? (I'm not sure.)

There is a hypothesis that gravity MIGHT be weak (relative to other forces) because it can "disappear" or "leak" into higher dimensions (parallel worlds) but insofar as is known other forces are restricted to three dimensions....It's possible that string theory or other theories with higher dimensions might offer some "understanding" but I've not seen any such explanation. But there are other theories which suggest gravity might transform from three to two dimensions at short scales....that was referenced in a recent paper here in the forums..maybe last week.
 
Last edited:
  • #3
141
0
Referring to the wiki link you sent me its this bit i dont understand. I need to broken down into lamens language :
Informally, the case of a point charge in an arbitrary static electric field is a simple consequence of Gauss's law. For a particle to be in a stable equilibrium, small perturbations ("pushes") on the particle in any direction should not break the equilibrium; the particle should "fall back" to its previous position. This means that the force field lines around the particle's equilibrium position should all point inwards, towards that position. If all of the surrounding field lines point towards the equilibrium point, then the divergence of the field at that point must be negative (i.e. that point acts as a sink). However, Gauss's Law says that the divergence of any possible electric force field is zero in free space.
 
  • #4
192
0
It is not that it is impossible to levitate an object. It is just impossible to make the levitation stable. An object in equilibrium will be unstable in at least one direction. A small displacement from equilibrium will grow and take the object further away. For instance, it is easy to see how an object can be stable in the vertical direction. If it is displaced up then the electrostatic force decreases and gravity will bring it down. Displace it down and electric force increases and pushes it up. But, if you displace it horizontally then the force points in the direction displaced and will push the object further in that direction, not back.

Now, you can create a combination of charges to create a saddle point, and place the object so that it is stable in the horizontal direction. However, this loses vertical stability. If the object is displaced vertically in the saddle point then the electric force actually increases now, instead of decreases, which means it will move further away from equilibrium.

Understanding why this is the case requires some knowledge of vector calculus. Like the article says, the divergence of the fields are zero in free space, which means the vectors can point toward each other in one direction but must point away from each other in some other direction.
 
  • #5
141
0
It is not that it is impossible to levitate an object. It is just impossible to make the levitation stable. An object in equilibrium will be unstable in at least one direction. A small displacement from equilibrium will grow and take the object further away. For instance, it is easy to see how an object can be stable in the vertical direction. If it is displaced up then the electrostatic force decreases and gravity will bring it down. Displace it down and electric force increases and pushes it up. But, if you displace it horizontally then the force points in the direction displaced and will push the object further in that direction, not back.
If it is displaced up then shouldnt the electrostatic force increase the closer the charged object comes to the source of electric field???

E=V/d=F/q

therefore

F=V*q/d

????
Or am I being stupid

Now, you can create a combination of charges to create a saddle point, and place the object so that it is stable in the horizontal direction. However, this loses vertical stability. If the object is displaced vertically in the saddle point then the electric force actually increases now, instead of decreases, which means it will move further away from equilibrium.
Why will it now reverse what is the reason??
I am guessing you are describing a penning trap now?
 
  • #6
192
0
I was speaking simply of a charge supported above other charges of the same sign.

a penning trap is quite a bit more complicated.
 
  • #7
141
0
Yes but why does it reverse direction when you have the sadle point situations as you described?
 
  • #8
141
0
I dont understand how earnshaws theory relates to gauss's law
 
  • #9
192
0
If you place two positive charges there will be a null point in between them where the fields are pointing toward the null, which is stable in that direction, but the field points away from the null in the other two direction. gauss's law states that the total flux of electric field through a closed surface is equal to the enclosed charge. So, if you construct a Gaussian surface around the null point then you can see that what goes in must come out since there is no spatial charge. The electric field cannot point toward the null from all directions as that would constitute net flux, and Q_enc != 0.

Take the case of a ring of charge: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c2

If you plot the z component as they have, there is a null at z=0 and it will increase in magnitude up to a max. So, the null point in the center of the ring is unstable in the z direction but stable in x-y. After the max the e-field starts decreasing again and so is stable in z, but now unstable in x-y.
 
  • #10
141
0
The electric field cannot point toward the null from all directions as that would constitute net flux, and Q_enc != 0.
I know that the electric field lines would be distributed uniformly around a point charge. But why does all the electric field lines pointing towards the null point eqaul Q_enc !=0 ??? (!= I take it that means NOT zero)
 
  • #11
192
0
If it points toward the null in all directions then there must be non-zero flux through the gaussian surface, which means Q is non-zero.
 
  • #12
141
0
If it points toward the null in all directions then there must be non-zero flux through the gaussian surface, which means Q is non-zero.
Sorry I must be being stupid, but I still dont understand that? There is something I am missing. Gauss's law states that the electric flux through any closed surface is proportional to the enclosed electric charge. So if all the lines of both postive charges pointed towards the null point then wouldnt you have a equal number of flux lines going in and a equal amount going out. Indicating a zero charge at the null point????????
 
  • #13
816
1
What is so hard to understand?
If you have a minimum in any field of this kind. Then the field vectors must all point inside there, so the charge that is trying to escape is being pushed back. This test charge is NOT part of the field calculation. So if you take a small volume around the charge and apply Gauss' law to the surface you find that another opposite charge must be trapped inside this volume with the first one, for arbitrarily small volumes.
Hence you cannot get a minimum without a charge in it, so you cannot levitate an electron without it falling onto a proton, we need quantum mechanics for that.
 
  • #14
141
0
What is so hard to understand?
So if you take a small volume around the charge and apply Gauss' law to the surface you find that another opposite charge must be trapped inside this volume with the first one, for arbitrarily small volumes.
Why must another opposite charge be trapped inside this volume?


Hence you cannot get a minimum without a charge in it, so you cannot levitate an electron without it falling onto a proton, we need quantum mechanics for that.
I dont understand that..
 

Related Threads on Can someone explain Earnshaws Theory

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
885
Replies
3
Views
2K
Replies
2
Views
1K
Replies
36
Views
4K
Replies
10
Views
26K
  • Last Post
Replies
24
Views
13K
Replies
10
Views
2K
Replies
7
Views
2K
Replies
2
Views
2K
Top