What is the force on a test charge at the center?

[spaghetti3451]

Homework Statement

13 equal charges are placed at the corners of a regular 13 sided polygon. What is the force on a test charge at the center?

Homework Equations

Coulomb's law
Principle of Superposition

The Attempt at a Solution

Coulomb's law gives an answer of 0, but it is cumbersome. I am looking for a much more elegant solution, where I do not actually calculate the individual forces and add them up, but somehow realize that the forces will add up to zero based on the orientation of the charges.

 

[haruspex]

Suppose there is a nonzero net force. Consider a vertex that lies with 90 degrees of that direction from the centre. The component of the net force towards that vertex is positive. By symmetry, it must have the same component towards every vertex. But there must also be a vertex between 90 degrees and 270 degrees from the net force direction, and the component of the net force towards it must be negative.

 

[PeterO]

Homework Statement

13 equal charges are placed at the corners of a regular 13 sided polygon. What is the force on a test charge at the center?

Homework Equations

Coulomb's law
Principle of Superposition

The Attempt at a Solution

Coulomb's law gives an answer of 0, but it is cumbersome. I am looking for a much more elegant solution, where I do not actually calculate the individual forces and add them up, but somehow realize that the forces will add up to zero based on the orientation of the charges.

If you draw a force diagram, you will see 13 forces radiating out, each angled at (360/13) degrees to its neighbour.
If you translate those forces, to join them head to tail, and do it in order, you will just form another regular 13 sided polygon - indicating that the net force is zero.

 

[Chestermiller]

All 13 forces point to the center of the polygon, and, by symmetry, none of them are preferred over any of the others. If there was one that is preferred, you would have to be able to identify which one it was. But, they are all identical, so you couldn't.

 

[spaghetti3451]

Suppose there is a nonzero net force. Consider a vertex that lies with 90 degrees of that direction from the centre. The component of the net force towards that vertex is positive.

I understand.

By symmetry, it must have the same component towards every vertex.

I don't see how! The non-zero force F will have two vertexes on its two sides. If the first vertex makes an angle θ with the direction of the force F, then the component of the force along that vertex is F cosθ, but then the component of the force along the other vertex must be F cos (360/13 - θ). Clearly, the two components are not the same.

 

[spaghetti3451]

If you draw a force diagram, you will see 13 forces radiating out, each angled at (360/13) degrees to its neighbour.
If you translate those forces, to join them head to tail, and do it in order, you will just form another regular 13 sided polygon - indicating that the net force is zero.

Thank you for your answer. Now, I'm hoping I can understand harispex and Chestermiller's methods as they are also very good.

 

[spaghetti3451]

All 13 forces point to the center of the polygon,

by symmetry, none of them are preferred over any of the others.

What is the symmetry we are talking about? And what does the word 'preferred' mean in this context?

 

[Chestermiller]

What is the symmetry we are talking about? And what does the word 'preferred' mean in this context?

If you put a charge in the center of the polygon, what direction would you expect it to move in? And if you could specify a direction that you would expect it to move in, there would be 12 other equally valid directions for it to move. This would eliminate all 13 directions as possibilities. You could do this exercise for all possible directions.

 

[haruspex]

I don't see how! The non-zero force F will have two vertexes on its two sides. If the first vertex makes an angle θ with the direction of the force F, then the component of the force along that vertex is F cosθ, but then the component of the force along the other vertex must be F cos (360/13 - θ). Clearly, the two components are not the same.

There is a symmetry between the vertices. Each vertex is as good as any other. If there is a reason for there being a net force towards one, the same reason must apply to all the others. But there cannot be a net force towards all of them.