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Energy of charges in a square

  1. May 28, 2006 #1
    I'm not sure that I really understand this question:

    How much energy is needed to place four positive charges, each of magnitude +5.0mC, at the vertices of a square of side 2.5cm?


    what I was thinking is that V = kQ/r

    And since all the charges are equal, and the same distance apart V1=V2=V3=V4 = (9x10^9)(5x10^-3C)/2.5x10^-2m = 1.8x10^9V

    Vnet = V1 +V2 + V3 + V4 = 7.2x10^9V

    However I haven't taken into account the diaganols - do I need to?
     
  2. jcsd
  3. May 28, 2006 #2

    Doc Al

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    Staff: Mentor

    What you want to add is potential energy, not potential. The energy needed to bring two identical charges from infinity to a distance r apart is:
    [tex]U = kQ^2/r[/tex]

    Hint: Imagine the four charges being brought from infinity, one at a time.
     
  4. May 28, 2006 #3
    Ok I think I got it I added the potenial energy between each of the 6 charge sets and got 4.9x10^7J, hopefully that sounds about right.
    I think I was getting things confused with the second part of the question:

    Choose one way of assembling the charges and calculate the potential at each empty vertex as this set of charges is assembled. Clearly descrive the order of assembly.

    Would I be correct to use V =kQ/r for this part of the question?
     
    Last edited: May 28, 2006
  5. May 28, 2006 #4

    Hootenanny

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    Yes, you should sum all of the individual potentials, for example;

    [tex]V_{total} = \frac{kQ_{1}}{r_{1}} + ... + \frac{kQ_{n}}{r_{n}}[/tex]

    Where rn is the distance from the charge Qn to the empty vertex.

    ~H
     
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