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Pleas help electric field

  1. Sep 14, 2005 #1
    pleas help!! electric field

    a semi infinite insulting rod carries a constant charge per unit length of lambda. SHow that he electric field at the point P makes an angle of 45 with the rod and that this result is independant of R

    [tex] q = \lambda x [/tex]
    [tex] dq = \lambda dx [/tex]

    for the Y direction poiting downward
    [tex] dE_{y} = \int_{0}^{\infty} \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{R^2 + x^2} \sin{\theta} [/tex]
    [tex] \sin{\theta} = \frac{x}{R^2 + x^2} [/tex]
    [tex] dE_{y} = \int_{0}^{\infty} \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda R dx}{(R^2 + x^2)^\frac{3}{2}} [/tex]
    [tex] dE_{y} = \frac{\lambda R}{4 \pi \epsilon_{0}} \int_{0}^{\infty} \frac{dx}{(R^2 + x^2)^\frac{3}{2}} [/tex]
    [tex] dE_{y} = \frac{\lambda R}{4 \pi \epsilon_{0}} [\frac{x}{R^2 \sqrt{R^2+x^2}}]_{0}^{\infty} [/tex]
    now im not sure about this integral and whether or not it converges...

    for the horizontal pointing left
    [tex] \frac{\lambda R}{4 \pi \epsilon_{0}} \int_{0}^{\infty} \frac{xdx}{(R^2 + x^2)^\frac{3}{2}} [/tex]
    same problem here

    please help!
     

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    Last edited: Sep 14, 2005
  2. jcsd
  3. Sep 15, 2005 #2

    quasar987

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    The integral is right and the limit to evaluate is almost trivial: pull x² out of the square root.
     
  4. Sep 15, 2005 #3
    ok for the Y part
    [tex] E_{y} = \frac{\lambda}{4 \pi \epsilon_{0}R} [/tex]
    i swhat i got the Y part. Am i right??

    for the X part
    i got the same thing.
    However the resultant for these two given a 45 degree angle... however it isn ot independat of R as in
    [tex] E = \sqrt{2} \frac{\lambda}{4 \pi \epsilon_{0}R} [/tex]
     
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