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Homework Help: Help with Electric Potential

  1. Mar 24, 2005 #1
    There is a plastic rod with a length L and a uniform positive charge lying on the x axis. With V = 0 at infinity find the electric potential at point P on the axis at distance d from one end of the rod. I got an answer but I have no clue if it is right or not. This is what I know:

    [tex] dv = \frac{k dq}{x} \ \ \ \ \ \lambda = \frac{q}{x} [/tex]

    [tex] dv = \frac{k\lambda dx}{x} \ \ \ \ \ dq = \lambda dx [/tex]

    [tex] \int dv = k\lambda \int \frac{dx}{x} [/tex] Upper = d + L Lower = d

    [tex] V = k\lambda(ln(d+L) - ln(d)) [/tex]

    Is there anything I am missing?

    The next problem is similar but [tex] \lambda = cx [/tex] where c is some positive constant. If I just substitute in cx then the x cancles leaving dx which gives x afer the integral. The answer contains a ln so I am missing something. Any help is appreciated.
  2. jcsd
  3. Mar 24, 2005 #2
    Seems right to me. For part ii, V = KCL. Answer does not have an ln.
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