Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook