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Homework Help: Electric Field/Force btw rods

  1. Jul 19, 2004 #1
    Identical thin rods of length (2a) carry equal charges +Q uniformly distributed along their lengths.

    The rods lie on (along) the x axis with their *centers* separated by a distance b > 2a.

    (Left rod from x=-a to x=a, right rod from x=b-a to x=b+a)

    Show that the magnitude of the force exerted by the left rod on the right one is

    [tex]F = k_e\frac{Q^2}{4a^2} \ln \frac{b^2}{b^2-4a^2}[/tex]

    Thanks very much for answering :-)
    Last edited: Jul 19, 2004
  2. jcsd
  3. Jul 21, 2004 #2
    if you want you can use this ideas: [tex]q=\lambda2a[/tex] or for a small piece
    [tex]dq=\lambdadx[/tex] in this case I'm not going to write the electric field as vector because is in the same direction. -->
    [tex]E=K\lambda\int_{-a}^a\frac{dx_1}{r^2}[/tex]and [tex]r=x_1+x_2+b[/tex] because x_1 is the variable on the first rod , x_2 on the other and b is the ditance between them. After this you can integrate with respect dx_2 and take again the interval from a to -a and you'll get the result your looking for.
  4. Jul 21, 2004 #3
    I'm sorry my Latex is not good.However, dq=(lambda)dx and make these substitutions
    dr=dx_1 , x_2+b= constants for the first integral. I hope this hepls.
  5. Jul 21, 2004 #4
    I'm going to give you some hints more:



    and finally


  6. Jul 21, 2004 #5
    I'm sorry my Latex still bad

    F=int of E(lambda)dx_2 ,from -a to a

  7. Jul 21, 2004 #6
    This reminds me of a problem where I had to find the gravitational force on one rod due to the other. This was a problem in my Calc. book. The easiest way to solve this is to find the force on a point by the rod and then extending the point into a rod.
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