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Homework Help: Two thin rods on the x-axis

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Two thin rods of length L lie along the x-axis, one between x = a and x = a + L, and the other between
    x = –a and x = –a – L. Each rod has positive charge Q distributed uniformly along its length.
    a. Calculate the electric field as a function of x produced by the rod on the left hand side
    (from –a – L to –a) at points along the positive x-axis.
    b. Determine the magnitude of the force that one rod exerts on the other.


    2. Relevant equations
    E=∫[dq/4πε_naught*r^2] where dq=λdx for a line of charge


    3. The attempt at a solution
    Well since for part a, they only want the electric field in the positive x-axis I came up with:

    E=∫from -a to -a-L of [λdx/4πε_naught*(-a-L+x)^2

    giving me:

    E=L/2a+L(2a+2L) when the integral is evaluated from -a to -a-L. - I'm aware I need to multiply my constants back in but for right now, I'm just worried about the integral.

    Am I on the right track?
     
  2. jcsd
  3. Oct 12, 2011 #2

    SammyS

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    Please read and follow the rules of these Forums. Wait 24 hours before bumping.

    Regarding your integral: E=∫from -a to -a-L of [λdx/4πε_naught*(-a-L+x)^2 ,

    What does x represent?

    You seem to have used x for two different things in the same expression.
     
  4. Oct 12, 2011 #3
    x is suppose to represent where you are on the x axis.

    -regarding the bump, sorry it won't happen again.
     
  5. Oct 12, 2011 #4

    SammyS

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    But you have (λ dx) in the integral, so in that sense, you're using x as the integration variable.

    Use something like x0 (or some variable like u or s), for the place at which you are finding E. You then need to express the distance from x0 to the point x on the rod, over which you are integrating. --- r = x0 - x .
     
  6. Oct 12, 2011 #5
    Sorry, I'm not following. The way I see it, when you take all the constants out you are just integrating dx/(-a-L+x)^2
     
  7. Oct 12, 2011 #6

    SammyS

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    Yes, but that's not what you should be integrating.

    (-a-L+x) is the distance between point x, and point (a + L) on the x axis.

    What would the integral be if you were finding E at a fixed location, such as at x = 1.357 meter, on the x-axis ?
     
  8. Oct 13, 2011 #7
    Okay, I think I know what you mean.

    E=∫kdq/r^2 let dq=λdl and r =(a+x), the distance from the end of the rod to a +x value
    E=λk/(a+x)^2∫dl -integrating from -a-L to -a


    E=λk/(a+x)^2[(-a)-(-a-L)]

    E=λkL/(a+x)^2

    How's that look?

    edit: I decided to use k rather than 1 over 4pi epsilon naught
     
  9. Oct 13, 2011 #8

    SammyS

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    No. r should be the distance between the element of charge, dq, and x. So, r = x - l.
     
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