Is the Net Force on a Charge at the Center of a 13-Sided Polygon Zero?

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aaaa202
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Homework Statement


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


Homework Equations


principle of superposition
symmetry

The Attempt at a Solution


I am not sure if the net force is 0 or not. Had the polygon been 12-sided the symmetry would be obvious, but I'm not sure in this case. Can anyone explain if the net force is 0 or not?
 
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hi aaaa202! :smile:
aaaa202 said:
13 equal charges are placed in the corners of a regular 13-sided polygon. What is the force on a test chrage at the center?

try it first for an equilateral triangle (3 equal charges) …

is that 0 ?​

then use the same method for 13 (you'll need a formula for summing a trig series)
 
Did the calculations for a triangle as you said, where i fixed the coordinates such that one of the charges sat on the y-axis. From there it was easy to see that Fres = 0, and I think that this must hold for any rotation of the coordinate system. After all I think the length of a vector is more or less defined to be preserved on a rotation in the euclidean coordinate system. However, I'm not sure, so can you confirm this? :/
And can you perhaps come up with a symmetry argument that makes it easy to see, that rotation of the coordinate system should not change the lFresl? :)
 
aaaa202 said:
Did the calculations for a triangle as you said, where i fixed the coordinates such that one of the charges sat on the y-axis. From there it was easy to see that Fres = 0

yes!

presumably you did cos0 + cos2π/3 + cos4π/3 = 1 - 1/2 - 1/2 = 0 ?

ok you now need to prove cos0 + cos2π/13 + cos4π/13 + … cos24π/13 = 0 :smile: