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Charges, conservation of energy

  1. Jul 26, 2008 #1

    Four particles, two with charge +1.0 x 10^-6 C and two with charge -1.0 x 10^-6 C, are placed at the corners of a square of side 1.0 m. Each has mass 1.0 mg, and they are released from rest. Find the time taken for the square of charges to collapse to zero size.

    The particles are placed with like charges on the corners diagonal to each other, like so:

    + -
    - +

    2. Relevant equations

    Potential energy for a pair of point charges:

    [tex]U = \frac{kQ_{1}Q_{2}}{d}[/tex]

    3. The attempt at a solution

    The charges move along the diagonals of the square toward the center. When a side of the square decreases in length by an amount ds, each charge moves an amount sqrt(2)*ds toward the center. The time taken for it to do so is dt = [sqrt(2)*ds]/v, where v is the velocity of the charge at its current position.

    Electric force is conservative and no outside forces are acting, so energy is conserved. So I can determine the velocity in terms of the current length of the side of the square using conservation of energy. I determine the electric potential energy of the system when the charges are first released by adding pairs:

    [tex]U = -\frac{kQ^{2}}{d} - \frac{kQ^{2}}{d} + \frac{kQ^{2}}{\sqrt{2}d} - \frac{kQ^{2}}{d} - \frac{kQ^{2}}{d} + \frac{kQ^{2}}{\sqrt{2}d}[/tex]

    where k = 9 x 10^9 N*m^2/C^2, Q = 1.0 x 10^-6 C, d = 1.0 m, so

    U = -0.02327 J. K = 0 because the charges start at rest.

    At side length a with a < d,

    [tex]U = \frac{-0.02327 N*m^{2}}{a}, K = 4(0.5mv^{2}) = 2mv^{2}[/tex]

    Setting the states equal, plugging in given values, and solving for v gives:

    [tex]v = 108 \sqrt{\frac{1 - a}{a}}[/tex]

    Next, I plug this into the expression for dt to get the differential time, but I use a = 1 - s in the velocity to put it in terms of the charge's displacement from the side of the original square. Adding them all up gives the integral:

    [tex]t = \frac{\sqrt{2}}{108}\int\sqrt{\frac{1-s}{s}}ds[/tex]

    evaluated from s = 0 m to s = 0.5 m (center of the square, where the sides collapse to zero).

    Admittedly, my calculus skills are rusty and I forgot how to evaluate that integral (help?), but I used Mathematica and got a value of 1.2854. This gave me:

    t = 0.01311 * 1.2854 = 0.017 s

    I have no idea whether or not I'm going about this correctly. Can anyone point out any mistakes I've made? Thanks.
  2. jcsd
  3. Jul 26, 2008 #2
    Nobody? Is there anything unclear in how I've presented my attempt at a solution? I really wish textbooks would come with answers to all problems rather than only a select few. It would make self-study much easier.
  4. Jul 27, 2008 #3


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    Homework Helper

    I think your technique is all right. I didn't verify it numerically, though since I don't have access to Mathematica and the antiderivative given by the online integrator is really intimidating to manipulate. And I don't see anything wrong with using Mathematica, that integral is really hard to solve.
  5. Jul 27, 2008 #4
    Thanks for the response. The antiderivative given by Mathematica had an arctan in it, so I was thinking trigonometric substitution was to be used somewhere along the line, but all of the techniques I tried failed. It's good to hear that someone else finds it a difficult integral and that I haven't necessarily forgotten all of my calculus!

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