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Homework Help: Charge on a point P from a uniform rod

  1. Apr 3, 2015 #1
    Consider a total charge of q = 3.50 multiply.gif 10^-9 C spread uniformly over a thin rod of length L = 0.8 m as shown. Point P is a distance H = 0.6 m away from the midpoint of the rod. Find the magnitude of the electric field at point P.

    Relevant equations:

    [tex] E = k \int \lambda dx/r^2 [/tex] where
    [tex] \lambda = q/L [/tex] and
    [tex] r = \sqrt{x^2 + H^2} [/tex]

    Attempt at solution

    [tex] E = k \int_{-.4}^{.4} \lambda dx/r^2 [/tex]
    [tex] E = k \int_{-.4}^{.4} q dx/(L)(x^2 + H^2) [/tex]
    [tex] E = (kq/L) \int_{-.4}^{.4} dx/(x^2 + H^2) [/tex]
    [tex] E = (kq/L) \arctan(x/H) |^.4 _{-.4} [/tex]

    plugging in k = 9 x 10^9, q = 3.5 x 10^-9, L = .8, H = .6 I get 46.3 N/C but this is incorrect.
  2. jcsd
  3. Apr 3, 2015 #2


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    Gold Member

    Your very first equation is wrong. The field from a point charge is a vector, and the net field from the distribution of charge is the integral of that vector.
    In the present case, there is a symmetry which allows you to figure out the direction of the resulting vector straight away. This means you only need to consider the component in that direction, reducing it to the integral of a scalar, but the integrand will be different.
  4. Apr 3, 2015 #3


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    In addition to haruspex' post: there is a factor H missing where you evaluate the integral.
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