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Potential of a Finite Rod -Mastering physics

  1. Feb 11, 2005 #1
    Hi,
    Im having a little problem doing a question in mastering physics, I was wondering if anyone can help.

    here is the problem:
    A finite rod of length L has total charge q, distributed uniformly along its length. The rod lies on the x -axis and is centered at the origin. Thus one endpoint is located at (-L/2,0), and the other is located at (L/2,0). Define the electric potential to be zero at an infinite distance away from the rod. Throughout this problem, you may use the constant k in place of the expression 1/4*pi*epsilon_0.

    What is Va, the electric potential at point A (see the figure), located a distance d above the midpoint of the rod on the y axis?
    Express your answer in terms of k, q, d, and L.

    Here is what I have so far:

    r =sqrt(d^2+x^2)
    dV= k*(q/(2*(L/2)))*(dx/(sqrt(d^2+x^2)))
    V=K*(q/2*(L/2)))*(ln*((sqrt(d^2+x^2)+a)/(sqrt(d^2+x^2)-a)) *I got this from book*

    my problem is putting it all together, using k,q , d and L. especially the (ln*((sqrt(d^2+x^2)+a)/(sqrt(d^2+x^2)-a)) part, which does not want x in the answer. Im really lost.

    Thanks.
     
    Last edited: Feb 11, 2005
  2. jcsd
  3. Feb 11, 2005 #2

    StatusX

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    Are you sure you copied that formula for V correctly? It's not V, it's the antiderivative of dV/dx, and to get V you must evaluate it at the endpoints. In other words, after doing the indefinite integral of dV/dx with respect to x, you get what the book gives as V. But V is actually the definite integral over the portion of x where there is charge (-L/2, L/2). Also, where did 'a' come from?

    Edit: actually, that isn't the antiderivative of dv/dx. I don't know what that is.
     
    Last edited: Feb 11, 2005
  4. Feb 12, 2005 #3

    Galileo

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    The expression for dV looks correct.

    [tex]V=\int_{-L/2}^{L/2}dV=\frac{kq}{L}\int_{-L/2}^{L/2}\frac{dx}{\sqrt{x^2+d^2}}[/tex]

    Just evaluate it and put in the right limits (not 'a' or 'x').
     
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