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Potential energy of point charges

  1. Feb 27, 2009 #1
    I am puzzled by the solution to a problem in electrical potential. The situation involves three identical point charges (same charge, same mass) each one placed at a corner of a square. The charges are released simultaneously and the problem is to find the maximum speed attained by each of them.

    The published solution of the problem states that by symmetry each will end up with the same kinetic energy when the charges have moved to infinity. So the procedure is to calculate the potential energy of the configuration, divide that by 3, set that equal to one-half m v squared, and solve for v, where v is the speed of one charge.

    I am having trouble seeing the notion of “symmetry” here since the charges are not equidistant from each other. I also pose the argument that what if a fourth charge were to be included, one that would be orders of magnitude away from the original 3 and would consequently experience far less force due to the other particles. Would the individual charges each still have 1/4 the potential energy of the assembly of four? The only rationale I can muster is that the symmetry is not local symmetry but symmetry at infinity, and I have difficulty picturing that. Hope you can help me sort this out.
     
  2. jcsd
  3. Feb 27, 2009 #2
    I agree that the notion of symmetry seems a little... well, odd to me.

    It seems like the middle particle would end up with more than 1/3 of the total energy, whereas the two others would end up with the same value splitting the difference.

    In fact, I would argue that this can't possibly be right. Because this would certainly lead to different behavior then three particles arranged on the vertices of an equilateral triangle, and that would certainly give a 3-way split of energy.

    Perhaps you could simulate this numerically and look at asymptotic behavior to get a feel for what actually happens.
     
  4. Feb 28, 2009 #3
    When I simulate it numerically, it appears that the one particle ends up with around 1.061 times the individual energies of the other two. So you have 1-1.061-1 as ar as energy distribution goes.

    That's out to 10000 seconds. It was about the same for 1000 seconds; I can give you more data if you want.
     
  5. Feb 28, 2009 #4
    Thanks for your analysis - it seems to confirm my hunch. I was trying to do that myself. I know how to get v from F(t), but am unsure how to get v when I know F(r); I suspect a double integration. How did you do the analysis?
     
  6. Feb 28, 2009 #5
    When you do a numerical integration like I did, you don't have to get v = v(t) in closed form... that's the point of the approximation.

    You just use a scheme like:

    dx = vdt => delta(x) ~= v delta(t) => x' = x + v delta(t)
    dv = adt => delta(v) ~= a delta(t) => v' = v + a delta(t)
    a = F/m where F = f(x, v, t).

    This is notoriously easy to iterate on a computer. There are several methods, the simplest of which is Euler's method, though better schemes - such as Runge Kutta or more stable methods - exist.
     
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