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Finite Charge Distribution

  1. Feb 25, 2005 #1


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    Finite Number of Charges around a regular Polygon

    I know this is probably an easy one, but I'm just unsure...

    There are 13 equal charges, q, that are placed at the corners of a regular 13-sided polygon.

    a) What is the Force on a test charge Q at the centre?

    What I've been donig is simply saying that all but three of the charges will cancel with the remaining three charges having a Y shape. then here, the total force F =2Fcos(theta) -F and then calculated theta as 2pi/13 (F is the force of one point charge).

    Is this right? it seems very complicated for what at first glance seems like a simple question.

    b) If one of the 13 charges is removed, what is the force on Q?

    In this case I have: F=Fcos[theta] - F. Again, this just doesn't "feel" right.

    Can anyone help?
    Last edited: Feb 26, 2005
  2. jcsd
  3. Feb 26, 2005 #2
    use symmetry in your part a, the answer need no calculation at all...
    remove a positive charge is same as place a negative charge on top of that positive charge... after you get part a, just add a negative charge and you'll get the answer, again, no calculation is needed
  4. Feb 26, 2005 #3


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    The resultant of all forces will be zero because of symmetry. Think: if there is a nonzero resultant force, and you turn the whole polygon by 2pi/13 the resultant force vector would turn with the same angle. But the configuration is the same as before, you can not see any difference if the "first" charge is upmost or the "second" one. That means that the resultant should be the same as before. What is the vector that stays the same if you turn it?

    You can remove a charge by putting an opposite one at the same place. You know the resultant of all 13 equal charges, now you have an additional charge, and the forces add up...

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