1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
    -wisky40
     
  5. Jul 21, 2004 #4
    I'm going to give you some hints more:

    [tex]dF=E\lambdadx_2[/tex]

    [tex]F=\int_{-a}^aE\lambdadx_2[/tex]

    and finally

    [tex]\lambda^2=\frac{q^2}{4a^2}[/tex]

    -wisky40
     
  6. Jul 21, 2004 #5
    I'm sorry my Latex still bad

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

    -wisky40
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electric Field/Force btw rods
Loading...