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Electric Potential of a Finite Rod

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    The figure shows a thin rod of length L and charge Q. It lies directly along the x axis with its center at the origin. Find an expression for the electric potential a distance x away from the center of the rod on the axis of the rod. (Give your answer in terms of x, L, Q and appropriate constants.)


    2. Relevant equations
    [tex]V=Q/(4\pi\epsilon_{0}x)[/tex]
    [tex]\lambda=Q/L[/tex]

    3. The attempt at a solution
    I know that I have to integrate each infinitesimal potential, but I am not sure what bounds to integrate over? Here is what I have done so far:

    [tex]dQ=\lambda*dx[/tex]
    [tex]dV=dQ/(4\pi\epsilon_{0}x)=\lambda*dx/(4\pi\epsilon_{0}x)[/tex]

    Now, when both sides are integrated the result is:

    [tex]V=\lambda/(4\pi\epsilon_{0})\int(1/x)dx[/tex]

    The integral becomes a natural log. I have tried with several combinations of lower boundaries (0, -L/2, +L/2) and upper boundaries (L/2, x) but none of my results are correct (I get instant feedback on whether my answer is correct).

    Thanks!
     
    Last edited: Mar 1, 2009
  2. jcsd
  3. Mar 1, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    Isn't the x of your integral the distance from the discrete element charges to the point X, so the range of the distances is x-L/2 and x-3L/2 for E field along the + x-axis?
     
  4. Mar 1, 2009 #3
    [tex]\left(\frac{1}{4{\cdot}{\pi}{\cdot}{\epsilon}_{0}}\right){\cdot}\left(\frac{Q}{L}\right){\cdot}{\ln}\left(\frac{2{\cdot}x+L}{2{\cdot}x-L}\right)[/tex]

    This is the correct answer! I'm not sure if what you said is correct but it set off a light bulb in my mind that made me realise that the integration variable had to run from x-L/2 to x+L/2! Thanks for your insight, and your welcome to the PF.
     
  5. Mar 1, 2009 #4

    LowlyPion

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    Homework Helper

    Right. I was sloppy in the 3/2 because that distance is x+L/2 to the other end. Glad you caught my error and it didn't mislead but rather served to possibly inspire.
     
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