# Coulomb force on the center of a hexagon?

Six point charges $q$ are at the corners of a regular hexagon that has sides of length $a$. What is the force on another charge $Q$ which is located in the center of the hexagon? What is the force on $Q$ if we remove one charge $q$ from one corner?

I know I'm supposed to use Coulombs law and that the force on $Q$ is the sum of all other forces from the six point charges $q$. But how do I compute the forces on each of the six points?

Thanks much.

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Think about the symmetry of the problem. More specifically, look if some forces cancel out in the centre of the hexagon. –  Frédéric Grosshans Oct 2 '13 at 14:46
Hmm I calculated the locations of the charges and then the force on each point (is this correct for one charge at location (a,0) the force is F= kQ*q/a^2*vector(a,0) ?). I think all of them cancel out. So the force in the center would be 0 right? –  blondy Oct 2 '13 at 14:54
Correct, and that answers the first part of the question. For the second part note that some, but not all, of the five forces cancel out. –  John Rennie Oct 2 '13 at 15:07
You can also simplify the latter question by saying that the final force would be the hexagon-6-charge force (that indeed happens to cancel out itself), plus that of a single $-q$ charge at one of the corners. –  Nicolas Oct 2 '13 at 15:37