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Electrostatics: force calculation

  1. Aug 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Is there a way to nicely compute the force between two configurations of point charges ?

    Here is an example. Two "squares" of charges (each separated by a rigid insulator of length a) are separated by a distance d, measured with respect to their centers. Instead of taking a sum over 2^4 = 16 terms, is it possible to "summarize" each square with something like an overall charge or an overall electric field?

    [tex]
    \begin{verbatim}
    e+ --a-- e-
    | |
    a a
    | |
    e+ --a-- e-
    \end{verbatim}
    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    Finding an "effective charge" for each square doesn't seem right (since the total charge is zero). Also, I can't find a way to get out of doing lots of algebra by using the electric field produced by one square and calculating the force on the other square from that, although it would seem that the vertical component of the field cancels along a line joining the centers. Is this right?
     
    Last edited: Aug 15, 2012
  2. jcsd
  3. Aug 15, 2012 #2

    Simon Bridge

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    In general? No.
    There needs to be some kind of helpful symmetry. You could represent the squares as two pairs of dipoles ... or study quadrapoles. But, from scratch? Doubt it.
     
  4. Aug 15, 2012 #3
    I didn't think so, either.

    To be honest, I was trying to solve problem 23-21 out of Giancoli, Physics for Scientists and Engineers, 2nd ed. The problem asks for the force between a molecule of thymine and a molecule of adenine in a DNA strand. Thymine and adenine are shown as a set of point charges in configurations with at least some symmetry. The problem seems tedious because there are 4*3=12 terms in the total force with like 10 of these terms having a different direction.

    Has anyone done this problem a different way?
     
  5. Aug 15, 2012 #4

    Simon Bridge

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    Ah - but chemists treat those sorts of things - +q on one side and -q on the other side - as dipoles. They are called "polar molecules". Much of the math is done for you ... just look for dipole-dipole interactions.

    It's a bit like with atomic bonding - you'd think you need to work out all the e-e interactions separately but it turns out you only need to worry about the valence shell.
     
  6. Aug 15, 2012 #5
    Thanks! I got the right order of magnitude at least but am not sure my reasoning is very sound.

    The thymine molecule has 4 charges indicated and the adenine has two. For the thymine, I found the net charge to be zero: (1+1-1-1)e = 0. The adenine's net charge would be (1+1-1)e = e. I calculated the center of charge for each molecule and then found the distance d between the centers.

    I treated the whole thing as a dipole like you suggested and gave each "side" a charge of
    q=e/2 (but of the opposite polarity) by just halving the net charge of the whole thing. Then I performed the following calculation:

    [tex]
    F \approx \frac{1}{4\pi \epsilon_0} \frac{q^2}{d^2} = C \times 10^{-10} \text { N}
    [/tex]

    My constant C underestimates the book's by a factor of 3 or so. Could I have made a mistake in my reasoning for the charge?
     
  7. Aug 16, 2012 #6

    Simon Bridge

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    Each molecule would be treated as a dipole due to the molecular charge separation. So thymine has 2e on one side and -2e on the other, separated by an effective distance.

    If there is a "correct" answer you can look up - then there will be a "correct" approximation associated with the question that Giancoli expected you to use. Re-read the chapter associated with the question.
     
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