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Semi-infinite non-conducting rod

  1. Mar 7, 2013 #1
    1. The problem statement, all variables and given/known data
    A "semi-infinite" nonconducting rod has a uniform linear charge density λ. Show that the electric field E at point P a distance R above one end of the rod makes an angle of 45° with the rod and that this result is independent of the distance R.



    2. Relevant equations
    $$\vec{E}=\int \frac{k\lambda }{r^2}dx$$



    3. The attempt at a solution
    I have found that $$\vec{E_x}=\frac{-k\lambda}{\sqrt{R^2+x^2}}$$ and $$\vec{E_y}=\frac{k\lambda x}{R\sqrt{R^2+x^2}}$$

    I have tried to show that, if Ex and Ey make a 45° angle, then they must be equal, but that leads nowhere. I have tried simply taking the arctan of Ey/Ex, but that leads to -x/R, which doesn't make sense. I have also tried evaluating the integrals at infinity and then using L'Hospital's rule, but it doesn't work for Ey.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 7, 2013 #2

    TSny

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    Gold Member

    After integrating, did you evaluate at the limits of integration? ##x## should not appear in the results.
     
  4. Mar 7, 2013 #3
    Oh, woops. I got $$\frac{k\lambda}{R}$$ for both. Of course the ratio of these is 1, so the angle is 45 degrees. Thanks.
     
  5. Mar 7, 2013 #4

    TSny

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    Looks good!
     
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