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Homework Help: Electric potential (rod)

  1. Mar 14, 2009 #1
    [SOLVED] Electric potential

    1. The problem statement, all variables and given/known data

    The charge on the rod of the figure (length 2l, center at the origin) has a nonuniform linear charge distribution, λ = ax.
    Determine the potential V at:

    (a) points along the y-axis.
    (b) points along the x-axis. (Assume x > l)

    (express all answers in terms of a, x, l, ε0 and appropriate constants)

    2. Relevant equations

    dQ = λdx
    dV = dQ/(4*pi*ε0*r)

    3. The attempt at a solution

    For part a, V = 0 because
    dV = dQ/[4*pi*ε0*(x2 + y2)1/2] dx with limits of integration from -l to l.

    For part b, I'm having a really hard time determining the limits of integration and what "r" in dV is (ie, is it x or is it r, a segment of the rod?). I tried a lot of things, none of which produced the correct answer. Right now, I have

    Let k = 1/(4*pi*ε0)
    I'm treating "x" as a fixed distance from the rod, while calling r a segment or distance along the rod starting at x - l.

    dV = [k(ar)]/(x - l + r) dr with integration limits from x - l to x + l (one end of the rod to the other)

    With change of variables,
    u = x - l + r
    dx = du
    r = u - x + l
    integration limits become 2x - 2l to 2x

    dV = [ka(u - x + l)] /u du
    = ka (1 + (l - x)/u) du

    V = ka(u + (l - x)ln(u))
    V = ka(2x - (2x - 2l) + (l - x)ln(2x) - (l - x)ln(2x - 2l))
    V = ka(2l + (l - x)ln(2x / 2x - 2l))
    V = ka(2l + (l - x)ln(x / (x - l)))

    I have a feeling that this is wrong, especially because I still don't completely understand what I'm supposed to integrate along, etc. Can someone explain how to go about solving this problem and point out what I am doing incorrectly?

    Thank you!
    Last edited: Mar 15, 2009
  2. jcsd
  3. Mar 14, 2009 #2


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    Along x, aren't you integrating in r at any point x > l from the distance x - l the closest point to x + l the farthest distance?
  4. Mar 14, 2009 #3
    umm, I was until the change of variables part--unless I don't need a change of variables and integrated the wrong expression?
  5. Mar 15, 2009 #4


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    I guess what I was trying to suggest was that you have an r for some point X > l such that r(x) = X - x. You would be taking the integral of this I was thinking from x = l to x = -l

    dV = k*a*x/r(x)*dx = k*a*x/(X - x)*dx
  6. Mar 15, 2009 #5
    Oh! Okay, that makes sense (where X is treated like it's fixed but r(x) is still the distance from X to x, some point on the rod). Thank you!
    Last edited: Mar 15, 2009
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