Where Is the Point of Equilibrium for Charges in a Cube?

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Homework Help Overview

The problem involves determining the point of stable equilibrium for charges located at the corners of a cube. The original poster considers the center of the cube as a potential equilibrium point but questions its validity based on the implications of Laplace's equation regarding potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of symmetry and Laplace's equation on the potential energy at the center of the cube. There is a consideration of the nature of equilibrium points, including the concept of saddle points and Earnshaw's theorem.

Discussion Status

The discussion is exploring various interpretations of equilibrium in the context of electrostatic forces. Some participants have provided insights into the nature of potential at the center and the implications of theoretical principles, while others are seeking clarification on specific concepts.

Contextual Notes

Earnshaw's theorem is referenced, indicating constraints on the equilibrium configurations of point charges. There is also mention of difficulties in visualizing the potential in three dimensions.

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This is an interesting problem!
Consider a cube with charges 'q' situated at its corners(a total of 8). Determine the point of stable equilibrium in this cube.
My Answer:
An off hand guess would be the point at the center of the cube, as though a positive charge at the center is suspended in midair.
A point of stable equilibrium is a point of local minimum in the potential energy. Here the potetial energy of 'q' is 'V'. But, we know that Laplace's equation do not allow a local minima for V.
So where would be the point of equilibrium be located?
 
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This seems like a silly question to ask. The force does vanish at the center by symmetry, but as you point out, it is a simple consequence of Laplace's equation that the potential can have no local maximum or local minimum in free space. I think this is a trick question.
 
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That's indeed an interesting question. It's clear the gradient of V is zero at the center, but as there cannot be a local maxima or minima it must be a saddle point. It is obvious now, but I must admit it got me thinking for awhile.

I've took the liberty to plot the potential in Maple. I can't do it in 3D though.
I've took all the charges and constants to equal unity. The origin is taken in the center of the cube with the axes perpendicular to the faces. The image shows the potential in a small region about the center in the xy-plane, so at z=0.
 

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Hi Galileo, thanks for your reply (I'm unable to open your attachment though:frown:). I don't understand what exactly a "saddle point" is. However, I found a theorem which can be applied here.

Earnshaw's theorem states that a collection of point charges cannot be maintained in an equilibrium configuration solely by the electrostatic interaction of the charges. This was first stated by Samuel Earnshaw in 1842. It is usually referenced to magnetic fields, but originally applied to electrostatic fields, and, in fact, applies to any classical inverse-square law force or combination of forces (such as magnetic, electric, and gravitational fields).
 
http://en.wikipedia.org/wiki/Saddle_point"
For an easier example, consider two charges of +q at the points (a,a,a) and (-a,-a,-a). Now, if I place another charge of +Q at the origin, what is the nature of equilibirium?
If the charge Q is displaced along the line joining (a,a,a) and (-a,-a,-a), there will be a restoring force and therefore there is a stable equilibirium in this direction. But, if the charge Q is displaced in any other direction, it will go off to infinity. So the equilibirium in this direction is unstable. This is an example of a saddle point.
 
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