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Classical physics problem

  1. Aug 5, 2004 #1
    HI,
    I just have a simple question if you have a square array of charges of +1 where the charges lie on the vertices of the square(so the charges form a suare patten where 4 of them form a unit square). The way you can balance the charges is by placing a -1 charge in the center of each square or two -1/2 along the diagnols of each square thus making the system neutral.Now is it possible to use simple force balance to come up with a psoition where the -1/2 charges will lie.

    Anybody need more clarification tell me.
    thanks
     
  2. jcsd
  3. Aug 5, 2004 #2
    Yeah, it is possible.

    Step one is in focusing on a diagonal, and realizing that the forces due to the charges on the other diagonal cancel through symmetry (i.e., we can neglect them. So, the problem simpifies to this scheme:

    q-----------(-q/2)-------(-q/2)----------q

    For this system to be in equilibrium along the line shown, we require that a test charge q' placed at the centre of the line must experience no net force. This is only possible if the negative charges (-q/2) are equidistant from the centre of the line, say, at a distance (a/2) each side of the centre (so that the separation of each negative charge is a).

    By applying Coulomb's law to the test charge, it is then a straightforward step to obtain a relationship between a and the distance r between the positive charges. This turns out to be [tex]a=r/\sqrt{2}[/tex]; since, by geometry, [tex]r=\sqrt{2}[/tex], we have a=1.

    In other words, the negative charges form a square with sides of length 1/4 at the centre of the square of positive charges.
     
  4. Aug 5, 2004 #3
    thanks a further clarification

    That was very good analysis however i still need some more fundamental(i know i may sound lame but i am just beginning with real physics).
    How did you use coulombs law to find your relation. I got zero. Please elaborate.

    thanks
     
    Last edited: Aug 5, 2004
  5. Aug 5, 2004 #4
    your argument was erroneous

    no the argument that the two other chargejs produce forces would cancel out is erroneous. Infact the two charges have a components in the direction that you just drew. The force is oblique angle so it has one component that cancels but other one does contribute.
     
  6. Aug 5, 2004 #5

    turin

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    What do you mean by "simple?" Do you know trig? As dannyboy pointed out, there is a high degree of symmetry. However, as you noticed, you cannot use this symmetry to ignore the oblique sources.
     
  7. Aug 5, 2004 #6
    Yes fortunately i do and i do find it simple. From what i did if i considered oblique sources i was getting a complex solution. This is what is bothering me by the way.
     
  8. Aug 5, 2004 #7

    turin

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    ggups,
    Do you mean "complex" as in not "real-valued", or "complicated?" I expect the solution to be real-valued and complicated; however, I can't immediately rule out the complex-valued possibility by inspection. If the solution does indeed turn out to be not real-valued, then this result indicates that there is no way to arrange the four -q/2 charges to achieve balance.
     
  9. Aug 5, 2004 #8
    ya i mean mathematically a complex valued solution. Can you give it a try i just wanna check my math. By the way there are not four -q/2 charges but only two. There are four +1 charges located around the vertices of square and two -q/2 charges lie along one of the diagnol. I know that the solution is possible i just dont know how to get there through my math.
    thanks anyways
     
  10. Aug 5, 2004 #9

    turin

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    Well, wait a second. Now I'm not so convinced that a nontrivial solution is possible, because the situation has now become non-symmetrical. Also, consider this: If you have only two -q/2 charges that lie along the diagonal, and you already know that a -q charge at the exact center is a solution, then this seems to suggest that both -q/2 charges must lie at the center, thus returning to the original situation. Of course, given the quadratic nature of the force, I suppose there is another solution out there, in principle.
     
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