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Homework Help: Calculate the Electric Potential

  1. Nov 4, 2003 #1
    What am I doing wrong here?

    Here's the problem:
    For the arrangement descripbed in the previous problem (see attachment), calculate the electric potential at point B that lies on the perpendicular bisector of the rod a distance b above the x axis.

    [lamb] = [alpha] x where [alpha] is a constant.

    The correct answer is
    V = -(k[alpha]L/2)ln(([squ] [(L^2/4) + b^2)] - L/2)/[squ] [(L^2/4) + b^2)] - L/2))

    I'm not getting the same answer.

    So far I've got the following: [the] = [<]'a' on the diagram.

    V = kq/r
    x' = btan[the]
    dx' = bsec^2[the]d[the]
    x = L/2 + x'
    r = bsec[the]

    dq = [lamb]d(L/2 + x') = [lamb]dx'
    dq = [alpha](L/2 + x')dx' = [alpha](L/2 + btan[the])bsec^2[the]d[the]

    so dV = k[alpha](L/2 + btan[the])bsec^2[the]d[the]/bsec[the]

    Taking the integral of both sides from -[the] to +[the] doesn't yield the correct result.

    I'd appreciate it if someone could point out where I went wrong. I have a feeling the problem's in [lamb] = dq/dL = dq(L/2 + x').

  2. jcsd
  3. Nov 4, 2003 #2
    I forgot the attachment. Here it is.

    Attached Files:

  4. Nov 4, 2003 #3
    How is that rod charged?

    You say
    dq = &lambda; dx' with
    &lambda; = &alpha; x.

    This would mean that the rod is NOT uniformly charged. Are you sure this is what the problem says?

    Normally, problems like this have a uniformly charged rod, meaning
    sth. like dq = &lambda; dx', where &lambda; is a constant.
  5. Nov 4, 2003 #4
    The rod is not uniformly charged. [lamb] varies with x.
  6. Nov 5, 2003 #5
    Small typo in the answer. Should be:

    -(k&alpha;L/2)ln{( sqrt[(L2/4) + b2)] + L/2)/ (sqrt[(L2/4) + b2] - L/2)}
    (the answer you posted equals 0)

    Don't set it up in terms of the angle.
    You want V = -k&int;dq/r
    dq = &Lambda;dx and &Lambda; = &alpha;x
    dq = &alpha;xdx
    r = sqrt(b2 + (x - L/2)2)

    so your integral is

    V = -k&alpha;&int; (xdx/(sqrt(b2 + (x - L/2)2)) from x=0 to x=L

    It actually does work out. Happy integrating. :smile:
    Last edited: Nov 5, 2003
  7. Nov 5, 2003 #6
    Thanks again gnome.
  8. Nov 5, 2003 #7
    That integral is a royal pain!!!

    I'm curious. What was wrong with the integral I was originally working with other than the fact that it might be easier integrating with respect to 'x' rather than the angle?
  9. Nov 5, 2003 #8
    Actually, nothing. Shoulda done it your way.

    dV = k&alpha;(L/2 + btan&theta;)bsec^2&theta;d&theta;/bsec&theta;
    dV = k&alpha;(L/2 + btan&theta;)sec&theta;d&theta;
    V = k&alpha;L/2 &int;sec&theta;d&theta; + k&alpha;b &int;tan&theta;sec&theta;d&theta;
    V = k&alpha;L/2 {ln|sec&theta; + tan&theta;|}{from -&theta; to &theta;} + k&alpha;b{sec&theta;}{from -&theta; to &theta;}
    The way you have defined theta, up at the top there, sec(-&theta;) = sec(&theta;) = (sqrt[L2 + 4b2])/(2b) so the second term of the integral drops out completely.

    tan&theta; = L/(2b)
    tan(-&theta;) = -L/(2b)

    so the integral, evaluated from -&theta; to &theta;, becomes:
    (k&alpha;L/2)(ln{(sqrt[L2 + 4b2])/(2b) + L/(2b)} - ln{(sqrt[L2 + 4b2])/(2b) - L/(2b)})

    Play around with that a little & you get exactly the same answer.
  10. Nov 5, 2003 #9

    It's a relief to know I set the integral up correctly. A stupid mistake on my part evaluating the integral turned into something it's not, so I thought I was going about it the wrong way.
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