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Please help with electric potential

  1. Nov 9, 2004 #1
    There is a rod with uniform charge distribution lambda . Point P is located directly above the rod at a distance d from the centre of the rod. Find the electric potential at point P

    Also in thee second part similar situation however one half of the rod has a negative charge distribution while the other is positivelambda.

    For the first part

    [tex] V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{r} [/tex] where r is the distance from the poin P to the infinitesimal point on the rod and x is the horizontal axsi upon whihc the rod sits

    so then [tex] r = \sqrt{d^2 + x^2} [/tex]
    and then it follows that

    [tex] V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{\sqrt{d^2 + x^2}} [/tex]

    For the second part

    wouldnt i get a zero answer because instead of 2 times the expression you'd get one of those expresions and then the negative of it which when you add them (right?) they cancel out?
  2. jcsd
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