Charges in a polygon distribution

1. Mar 20, 2013

mathnerd15

1. The problem statement, all variables and given/known data
13 equal charges q are placed at the corners of a polygon and as on a clockface, what is the net force on a test charge at the center Q? I assume the charge at the center is positive and the charge on the polygon so they repel

2. Relevant equations
(1/4piEo)(qQ/r^2)

3. The attempt at a solution
it seems there is a net force in one direction where there is a lack of symmetry in the Fx, Fy. for the situation of q=3 I calculated that there is a net charge upwared of
(1/4piEo)(qQ/r^2)(1-2sin(pi/6))=0
with the equilateral triangle pointing upward, sin 30 degrees=1/2

but as n sides of the polygon approaches infinity even in an odd polygon this would resemble a smooth circle and the charges would cancel out with Fnet=0? take out one charge in the 13 sided polygon and the charge motion and electrical force is in the direction of the missing charge?

Last edited: Mar 20, 2013
2. Mar 20, 2013

epenguin

You don't need all of that calculation. All you'd need is that the force towards the centre is the same for each of the charges situated on the polygon corners. You might as well call it 1. The the problem is reduced to resolved components of these equal forces, an almost pure geometrical or trigonometrical problem.

We know what you mean by not symmetrical but it is not true there is no symmetry.

Try it for 3 charges (disposed at 120° tp each other, say one at the bottom) and look at the horizontal sum of forces - easy - and then look at the vertical sum - not difficult. Then see if concolsions generalise.

Yes.

3. Mar 20, 2013

mathnerd15

so for a circle with n=3, with the point at the bottom, Ey=(1-2sin(pi/6))=0 and Ex=0 since we rotated one point directly in -y direction

Last edited: Mar 20, 2013
4. Mar 20, 2013

epenguin

OK and what is sin(π/6) ?

5. Mar 20, 2013

mathnerd15

sin pi/6=sin 30degrees=1/2 so the y components go to zero...I'm not sure it's exactly 30 degrees when you divide the angle?

Last edited: Mar 20, 2013
6. Mar 20, 2013

epenguin

Thought they might. It only remains to generalise.

And maybe you don't need to make a big meal of that either - maybe even the above is too complicated.

How about if what you did at the bottom node for the horizontal net force you did for each node in turn?