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Homework Help: Point charge help

  1. Oct 1, 2004 #1
    Question: Three point charges, which are initially at infinity, are placed at the corners of an equilateral triangle with sides d. Two of the point charges have a charge of q. If zero net work is required to place the three charges at the corners of the triangle, what must the value be of the third charge?

    What I did:
    q = 2 of the charges
    z = the third charge
    d = the final distance between each charge

    [tex] 0 = Uelec_q + Uelec_q + Uelec_z [/tex]

    [tex] 0 = kq(\frac{-1}{d}) + kq(\frac{-1}{d}) + kz(\frac{-1}{d})[/tex]

    [tex] 0 = \frac{-2kq}{d} - \frac{kz}{d} [/tex]

    [tex] -2q = z [/tex]

    The correct answer is [tex] \frac{-q}{2} = z [/tex]
    Can anyone see what I did wrong? I've tried this problem a few times and have gotten the same answer each time.
  2. jcsd
  3. Oct 2, 2004 #2


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    Homework Helper

    No work is needed to bring in the first charge (q).
    The second charge (q) feels the field of the first one. The work needed to place it at a distance d from the first charge is
    [tex]W_1 = k(\frac{q^2}{d})[/tex].
    The third charge feels the force from both charges already present.The potential at the third corner of the triangle is
    So the work done when the charge z is brought in from infinity is z times this potential,
    [tex]W_2 = 2k(\frac{qz}{d})[/tex].
    And the total work is zero.

  4. Oct 2, 2004 #3
    Ahh. I see where I messed up now. I wasn't taking into account the second charge in the [tex] Uelec [/tex] formula and was leaving out the [tex] Uelec [/tex] of the other 'z' on each. I was looking at it as all points moving together at the same time instead of each one moving individually.

    Thanks a bunch! :biggrin:
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