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Force on charges placed at vertices of a square.

  1. Jul 15, 2013 #1
    Hello. This isn't a homework, but rather my own self-study of a textbook (Franklin's Classical Electromagnetism), so if this is an inappropriate place for these types of questions let me know.

    1. The problem statement, all variables and given/known data

    Four equal charges q are placed at the vertices of a square of side L. What is the magnitude of force on one of the charges?

    Further, if the charges are released from rest in this configuration use this force to find the velocity a long time after the charges are released.

    2. Relevant equations

    Coulomb's law extended to many charges:

    [itex]\mathbf{F}=q\sum_i\frac{q_i(\mathbf{r}-\mathbf{r}_i)}{|\mathbf{r}-\mathbf{r}_i|^3}[/itex]

    3. The attempt at a solution

    In order to find the magnitude of the force I simply carried out the sum and used Pythagoras' theorem on the result. It is possible from inspection of the square to determine the vector quantities describing the distances from the single charge to the other three. For definiteness I am considering the force on the upper right-hand charge.

    So,

    [itex]\mathbf{F}=q\Big\{\frac{qL\mathbf{i}}{L^3}+\frac{qL \mathbf{j}}{L^3}+\frac{q(L\mathbf{i}+L\mathbf{j})}{(\sqrt{2}L)^3}\Big\}\\
    =\Big(\frac{q}{L}\Big)^2\Big(1+\frac{1}{(\sqrt{2})^3}\Big)( \mathbf{i}+\mathbf{j})[/itex]

    Then the magnitude of this force is

    [itex]|\mathbf{F}|=\sqrt{2\Big(\frac{q}{L}\Big)^4\Big(1+ \frac{1}{2^{\frac{3}{2}}}\Big)^2}\\
    =\Big(\frac{q}{L}\Big)^2\Big(\frac 12 +\sqrt{2}\Big)[/itex]

    So assuming that is correct I continue, and here is where my trouble begins. Intuitively, the charges accelerate away from one another along the diagonals of the square. The force diminishes over time and so eventually the charges approach some final velocity.

    I begin with Newton's #2:

    [itex]\mathbf{F}=\Big(\frac{q}{L}\Big)^2\Big(1+\frac{1}{(\sqrt{2})^3}\Big)( \mathbf{i}+\mathbf{j})\\
    =m\frac{d^2\mathbf{r}}{dt^2}[/itex]

    The motion is identical in the x and y directions, so it suffices to consider motion in, say, the x direction.

    [itex]\frac{d^2x}{dt^2}=\Big(\frac{q^2}{mL^2}\Big)\Big(1+\frac{1}{(\sqrt{2})^3}\Big)[/itex]

    This is where I start to feel a little shaky in my logic. I want to say that L is now the initial separation of the charges along a certain direction. That is, taking the upper right-hand charge, L describes the initial separation between it and the charge to its left. Then I can integrate once to find the velocity. However this leads to a linear relationship between the velocity and time, which means that the velocity increases without bound.

    Any thoughts?

    Thanks in advance.

    P.S. I can't figure out why some of the LaTeX code is not formatting properly. I hope it's clear regardless.
     
    Last edited: Jul 15, 2013
  2. jcsd
  3. Jul 15, 2013 #2

    tiny-tim

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    Hello TheShrike! :smile:
    you have dv/dt = K/x2

    how did you get a linear relation out of that? :confused:
    you need to put in some spaces

    the forum software inserts spaces every so often if you don't, and if they come in the middle of a codeword, it ruins the code! :wink:
     
  4. Jul 15, 2013 #3
    Oh, I was considering the L as a constant, so that if I integrate I'd get

    [itex]v(t)=\Big(\frac{q}{l}\Big)^2\Big(1+\frac{1}{(\sqrt{2})^3}\Big)t+A[/itex]

    with A an arbitrary constant. Although putting in the initial condition would lead to A=L.

    Using the interpretation you've given I'd have no idea how to solve it. Wouldn't it then be a non-linear differential equation? :eek:

    I did consider that way, but I suspected I was over-complicating things.

    Ah, thanks! Sorted now.
     
  5. Jul 15, 2013 #4

    tiny-tim

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    Hello TheShrike! :smile:
    The trick is to write dv/dt = dv/dx dx/dt = v dv/dx :wink:
     
  6. Jul 15, 2013 #5
    The center of the square doesn't move, so it's easiest to measure distance relative to the center of the square. At any time, the distance of each charge from the center of the square is L√2/2. The charges move along the diagonals, so that [tex]a=\frac{\sqrt{2}}{2}\frac{d^2L}{dt^2}[/tex].
     
  7. Jul 17, 2013 #6

    rude man

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    Or, you can forget forces and use potentials.

    The first charge is emplaced with zero energy. The 2nd, 3rd and 4the charges have what potential? Then, when the charges are released, what must their total kinetic energies be?

    EDIT: sorry, the above pertains to just part b of the question, and moreover you're supposed to integrate the force. You can use the above alternative approach as an easy doule-check on your answer to part b.
     
  8. Jul 28, 2013 #7
    Alright, I've given a few of these a try. Taking tiny-tim's suggestion first:

    I'm going to have the differential equation:

    [itex]v\frac{dv}{dx}= \frac{q^2}{mx^2}\Big(1+\frac{1}{\sqrt{2}}\Big)[/itex]

    Let those constant terms be collapsed into the constant A.

    Then, by separation of variables, the ODE can be solved to give:

    [itex]v=-\frac{A}{x}+L[/itex]

    where L is the arbitrary constant determined by the initial condition of the separation between the charges.

    I'm not quite sure how to regain v in terms of t after this. I thought about integrating v to get x in terms of t and then differentiating again, but I'm not sure how to carry out that integration either. (Starting to feel stupid...)

    Alternatively, with rude man's approach:

    I can integrate the force, using [itex]\mathbf{F}=-\nabla\phi[/itex]. This gets:

    [itex]\phi=q^2(1+\frac{1}{2^{\frac{3}{2}}})(\frac{1}{x}+\frac{1}{y})+A[/itex]

    where A is the usual arbitrary constant. To do this I had to assume that [itex]L^2=x^2=y^2[/itex] from the original equation for the force. I'm not sure if I can determine the constant.

    There is potentially a small point of confusion here, as it seems like I've selected an origin, however I just used distances by inspection of the problem in the original derivation. However, looking at the information it looks like the origin is at the bottom leftmost charge, remembering that I am considering the effects on the right uppermost charge.

    Alright, now I assume that enough time has passed that all potential energy has been converted to kinetic energy. Thus,

    [itex]v=\sqrt{2m\phi}[/itex]

    with the constant A still undetermined.

    How does that sound so far?

    P.S. Sorry for taking so long. I get distracted easily sometimes (Dammit, Skyrim!)

    P.P.S. Is there a quick way of wrapping Tex tags around words, similar to how it can be done for Italics or Bold?
     
  9. Jul 28, 2013 #8

    rude man

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    I didn't have integrating of forces in mind.

    My way is to compute the potential energy of the system. To do this, take two charges at a time, calculate their mutual potential energy, and add all the potential energies (six of them).

    When all charges reach infinity, what happened to their potential energy?

    Realize that the mass m of each charge is involved in answering the second part of the question. Ignore gravitational attraction among the charges to reach your answer (happy, mfb?).
     
  10. Jul 28, 2013 #9

    tiny-tim

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    Hi TheShrike! :smile:
    nooo, the LHS is wrong: the integral of v dv/dx is not v :wink:
    Just put v= dx/dt …

    then dx/something = dt, and integrate. :smile:
     
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