# Coulomb's law problem

1. Dec 18, 2012

### spaghetti3451

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Coulomb's law
Principle of Superposition

3. 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 realise that the forces will add up to zero based on the orientation of the charges.

2. Dec 18, 2012

### 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.

3. Dec 18, 2012

### PeterO

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.

4. Dec 18, 2012

### Staff: Mentor

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.

5. Jan 5, 2013

6. Jan 5, 2013

### spaghetti3451

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

7. Jan 5, 2013

### spaghetti3451

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

8. Jan 5, 2013

### Staff: Mentor

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.

9. Jan 5, 2013

### haruspex

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.