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Potential Energy of a System of charged particles

  1. Jan 19, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=54869&stc=1&d=1358636005.jpg


    2. Relevant equations
    V = kQ/r
    PE = Vq

    3. The attempt at a solution
    Tried a lot of ways... get an order of magnitude less than the answer and also off by a bit too.

    Please help. I think you are supposed to take V at each location then multiply by q since this is PE. I carry everything in vector form but my numbers are confusing. Attraction is supposed to be negative PE. Well we have attraction/ repulsion both working on a particle. It gets attracted and repelled to a certain direction. Is this contributing negative or positive potential energy?

    I've had a lot of partial insights but no complete solution. Thanks
     

    Attached Files:

  2. jcsd
  3. Jan 19, 2013 #2

    SammyS

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    V due to what? Multiply by which q ?
    What do you get for V at the location of Q1, for example?
     
  4. Jan 19, 2013 #3
    V at Q1 should be..
    = -15900i - 15900j + 4440i - 8880j

    = -11460i -24780j
    = 27301r N*m/C

    I mean q at that location.

    So |PE| would be qV = 27301* 2.8e-6 = 0.0764 J
     
  5. Jan 19, 2013 #4

    haruspex

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    Voltage is a scalar.
     
  6. Jan 19, 2013 #5
    Then how do i figure out the total voltage from two different particles?
    I can only think of taking a scalar value between one charge and the point Q1,
    and doing a scalar addition of that voltage with the voltage between the other and point Q2. Do i simply do this for all the particles to get a scalar total?
     
    Last edited: Jan 19, 2013
  7. Jan 19, 2013 #6

    haruspex

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    The voltage at one charge due to the other two is simply the sum of the voltages due to each of those two. But be careful re the total PE of a system with more than two charges. I would suggest imagining one charge moving away to infinity, then the other.
     
  8. Jan 19, 2013 #7
    your visualization seems to be the case. Thank you so much. Quite clear now.
     
    Last edited: Jan 19, 2013
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