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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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    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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