What is the net force on a test charge at the centre of a tridecagon?

AI Thread Summary
The net force on a test charge at the center of a regular tridecagon, with 13 equal charges at its corners, is not zero due to the odd number of charges. While symmetry arguments apply to regular polygons with an even number of vertices, they do not hold for an odd number, as charges cannot cancel in pairs. Therefore, the electric field at the center is not zero, and the test charge experiences a net force. A careful analysis of the charge distribution is necessary to understand the resulting electric field. The conclusion is that the lack of symmetry leads to a non-zero net force on the test charge.
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Homework Statement



13 equal charges, q, are situated at the corners of a regular tridecagon. What is the net force on a test charge, Q, situated at the centre.


Homework Equations



F=k(qQ/r^2) r^

(r^=r vector/r)


The Attempt at a Solution



The net force will be zero due to the symetry of the tridecagon.
 
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What symmetry? Symmetry arguments work in regular polygons when you have an even number of vertices and therefore an even number of charges. Then the electric field at the centre is zero because contributions from charges cancel in pairs. Here you have an odd number of charges. They cannot cancel in pairs. Is the field at the centre still zero, though? You need to structure your argument more carefully.
 
Last edited:
No. There is no centre of symmetry.

ehild
 

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