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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
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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
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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
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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
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1 Answer 1

For that kind of homework, you should always try to

  • simplify using symmetries (leading to questions such as "what cancels out", "what is equal", "what is uniform"...). For this a drawing is an absolute must-have.
  • use the fact that you can always decompose a seemingly complex structure into a superposition of different structures, since Maxwell's equations are linear.

before you calculate anything.

(I post this as an answer to a slightly generalized form of the question, since the specific answer should already be obvious from the comments. However this could be transformed into a comment as well.)

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