Electrostatic containment (impossible!)

  • Thread starter LHarriger
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  • #1
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I just came across Earnshaw's Theorem which states:

A charged particle cannot be held in stable equilibrium by electrostatic forces alone.

As an example it said that equal and fixed point charges at the corners of a cube could not hold stationary a point charge at the center.

There was no proof provided, but after a little thought it occured to me that Poisson's Equation for the potential would reduce to Laplace's Equation:

[itex] \nabla^{2}V=-\frac{\rho}{\epsilon_{0}} \Longrightarrow \nabla^{2}V=0[/itex]

since there is no charge anywhere but at the corners.
Because Laplace's Equation does not allow solutions with local maxima or minima then it would be impossible to find a a stable configuration.

Mathematically this seems fine. However, visually it seems that if I were to lightly tap the charge at the center then it would feel a repulsion pushing it back to the center (causing it to then oscillate about the center). Clearly this is wrong. Does anyone know of a physical way to imagine the instability?
 
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Answers and Replies

  • #2
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Why don't you just calculate by hand the (direction of the) force that will apply to the charge, once you've shifted it slightly to the side?
 
  • #3
Claude Bile
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You can get points of unstable equilibrium, but these points are unsuitable for trapping. A point of unstable equilibrium is like a ball sitting on a hill. In the direct centre of the hill, it is stable, however once you nudge the ball slightly, it rolls off - which is what I suspect cesiumfrog is alluding to.

Consider too empty space, where the force on a charged particle is zero everywhere - you would not regard that particle as being trapped.

To trap a particle, you need more than a central point of equilibrium, you need the surrounding region of space to push the particle back into the point of equilibrium should it get displaced.

Claude.
 
  • #4
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Why don't you just calculate by hand the (direction of the) force that will apply to the charge, once you've shifted it slightly to the side?
I have a general equation for the potential at any point inside the cube but it is really ugly:
[itex]\frac{q}{4\pi\epsilon_{0}}\sum{\frac{1}{r_i}}[/itex]
where
[itex]r_i=\sqrt{(a+x)^{2}+(a+y)^{2}+(a+z)^{2}} ; 1 \leq i \leq 8[/itex]
and the sign ordering preceding the x, y, and z is different for each r
ie: +++, ++-, +-+, +--, -++, -+-, --+, ---
The gradient of this still needs to be taken to get first E and then the force. Considering how messy this already is, I dont see how looking at the equation of force will help much. In any case I am after a conceptual picture, the math part I am OK with.
You can get points of unstable equilibrium... A point of unstable equilibrium is like a ball sitting on a hill. In the direct centre of the hill, it is stable, however once you nudge the ball slightly, it rolls off - which is what I suspect cesiumfrog is alluding to.
I was also alluding to this when I said that Laplace's Equation has no maxima or minima. Indeed the potential energy of a charge distribution is simply qV(r) and the standard method for calculating stability is two take the second derivative and check to see if it is less than zero. However, in our case taking the second derivative just gives us Laplace's Equation. Since it is equal to zero this stability test usually does not work. (I could always expand the potential to higher order terms and check these, but like I said, I am looking for a conceptual answer.)

Edit: Ignore that last thing in parenthses, a better statement would be to reiterate that although the usual stability test fails we already know by theorem that the Laplace Equation admits no solutions with maxima or minima.
 
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  • #5
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Pardon my ignorance, but what is "stable equilibrium" for charged particles? Or unstable equilibrium for that matter? Is this about motion?

- Bryan
 
  • #6
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Equilibrium in this context means a spot with zero force. For example, if you have a wire cube holding equal negative charges in each corner, then an electron is kind-of in "equilibrium" if you leave it in the exact centre, because by symmetry the electric forces in each direction cancel.

Stable equilibrium means that if the electron shifts a fraction in any direction (as will inevitably happen in real life) then the forces in that new position will push it back to where it was.

In the case of our electron, if it moves slightly closer to one of the faces of the cube, then the forces there will actually push it even further from the centre (making the centre an unstable equilibrium point). I don't know how to explain it more conceptually than this.

At first you might think that as it moves toward one face it would be rejected toward the centre again, because the repulsive force from each of those four nearer charges (individually) is stronger than each of the oppositely directed forces from the four individual charges on the back face. But those "restoring" forces are directed almost in opposition to each other, whilst the forces from the other charges are now directed more in the direction of the perturbation; if you do the math (which is fairly easy, since by symmetry you only have to consider two of the corner charges, and you can take the limit where the perturbation is small so as to simplify fractions) you see that the net force only accentuates the perturbation further.
 
  • #7
jtbell
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For a more real-life example of stable versus unstable equilibrium, imagine a spherical bowl with a smooth surface inside and out. First turn the bowl so the "inside" faces upwards, and put a marble at the bottom. If you put the marble at just the right place, the net force on it is zero, so it is in equilibrium. Furthermore, if you push the marble slightly away from that position, gravity pulls it back, so it is a stable equilibrium.

Now turn the bowl upside down, and put the marble at the very top. If you put the marble in just the right place, again there is no net force on it, so it is in equilibrium again. However, if you push it slightly away from that position, gravity pulls it further away, and it never comes back, so it is an unstable equilibrium.
 
  • #8
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At first you might think that as it moves toward one face it would be rejected toward the centre again, because the repulsive force from each of those four nearer charges (individually) is stronger than each of the oppositely directed forces from the four individual charges on the back face. But those "restoring" forces are directed almost in opposition to each other, whilst the forces from the other charges are now directed more in the direction of the perturbation
That makes sense, thanks cesium that was the kind of explanation I was after.
if you do the math (which is fairly easy, since by symmetry you only have to consider two of the corner charges, and you can take the limit where the perturbation is small so as to simplify fractions) you see that the net force only accentuates the perturbation further.
So Taylor expand just two of the above terms (one from each face) and look at those to see how the forces work out?
 

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