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Homework Help: Thin rod w/linear charge density, dipole moment

  1. Apr 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A thin rod of length 2L has a linear charge density that isλ0 at the left end but decreases linearly with distance going from left to right in such a way that the charge on the entire rod is zero.


    E = −kλ0/L(d/(L−d)−ln(d−L)+d/(L+d)+ln(L+d))

    for a point P that is distance d>l to the right of the center of the rod:

    How does this expression simplify to a familiar result for the limiting case d≫L? (Hint: The dipole moment of the rod has magnitude p=2λ0L2/3.) Express your answer in terms of some or all of the variables p, d, and Coulomb's constant k.

    2. Relevant equations
    p=q*d ?
    E = −kλ0/L(d/(L−d)−ln(d−L)+d/(L+d)+ln(L+d))

    3. The attempt at a solution
    I don't have a clear idea of where I'm going in this problem. I started with the limiting case, and narrowed down the E equation to

    E= (−kλ0/L)(1-ln(d)+1+ln(d))
    E= (−kλ0/L)(1+1)
    E= (−kλ0/L)(2)
    E= -2kλ0/L

    I don't think this is correct, because the answer has to be defined by no other variable than p, d, and k. Because of that, I know I have to use their equation for p, but I don't know how to use it. Frankly, I don't even know if my first step was correct.

    I could definitely use some guidance in the right direction right now.

    Thank you very much!
  2. jcsd
  3. Apr 18, 2016 #2


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    If you are given p, then it's not wise to overwrite that with something like p = q d. That d means something else than your d.
    $$p = \int x dq = \int_0^{2L} x\ \lambda\, dx = \lambda_0 \int_0^{2L} x \left (1-{x\over L}\right ) \, dx $$which gives you p (check !)​
    Where you are going in this exercise should be obvious: if they give you the dipole moment and mention d>>L, you are going towards an expression for the dipole field -- which you should write down for the given situation so you know where you are going.

    In that context you can't simply ignore L/d (and write d/(L-d) = 1 ), but have to write out the E expression in terms of L/d (which is a small number...)
    Last edited: Apr 18, 2016
  4. Apr 18, 2016 #3
    Okay, I've checked the integral for p, and it works, and I can appreciate how this condenses to a dipole field, since the rod and dipole would look the same given enough distance.

    So that means I'm trying to condense Erod to the form E=c*k*p/d3, c being a numerical multiplier, correct?
  5. Apr 19, 2016 #4


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    I should be flogged for giving all that away ...
  6. Apr 19, 2016 #5
    Well, I'm not going to give you a hard time about it. Thank you very much for your help!
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