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Electric Potential/Energy question

  1. Nov 23, 2003 #1
    Here's the question: "How much work must be done to bring three electrons from a great distance apart to within 1.0 x 10^-10 m from each other?"

    Here's what I did :

    .(work to bring one to another)

    Work = PE + KE
    = PE + 0
    =q(V final - V initial) ; V initial = 0
    =qV final
    =q(KQ/r)
    = (1.6 x 10^-19)((9 x 10^9 x 1.6 x 10^-19)/(1.0 x 10^-10))
    = 2.3 x 10^-18 J

    .(work to bring the other one)

    Work = q(KQ/r)
    = (1.6 x 10^-19)((9 x 10^9 x 1.6 x 10^-19)/(2 x 1.0 x 10^-10))
    = 1.2 x 10^-18 J

    . I then add them together to get the total work done

    (2.3 x 10^-18 J) + (1.2 x 10^-18 J)
    = 3.4 x 10^-18 J

    Now, is that right? Feel free to correct me.
     
  2. jcsd
  3. Nov 23, 2003 #2
    I may be interpreting the question differently from you. I interpret it to mean that the three electrons are brought into the configuration of an equilateral triangle, with each edge of the triangle r=1 angstrom in length.

    If so, then you can compute the final potential energy of one of the pairs of charges with ke2/r, and the final total potential energy of the three charges is three times that (energy from three pairs of charges).
     
  4. Nov 23, 2003 #3
    I see. I'm visualizing the question as three electrons coming together in a line. Bleh, the book didn't provide a diagram. However, I think you're right. Thanks for guiding me in the right direction.
     
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