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Magnitude of electric field by a curved rod.

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data
    http://i39.tinypic.com/qsp1rr.jpg
    Determine the magnitude of the electric field at point P.

    2. Relevant equations
    lambba=(q/l)
    Length of arc = (radius)(radians)

    There's another equation with a sin60 in it that gives the answer, but I forgot what it is...

    3. The attempt at a solution

    I first tried getting lambda by diving the charge 8e-6C by the length of the arc, .418m, then put that in this equation:
    [(lambda)/(4piEo)]int(1/r^s) but that didn't give me the right answer, which is 1.49e6 N/C upwards.

    I know the answer and most of the information, I just don't know what equation to apply to it.
     
  2. jcsd
  3. Feb 23, 2009 #2

    Delphi51

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    Is the charged part supposed to be circular in shape? And of uniform charge density per unit length? If not, you can't do it!

    You must consider a small element of length on the arc - dL,
    with charge dQ = lambda*dL.
    Sketch the electric field vector due to this charge at the point P. Write an expression for its magnitude. Use symmetry to decide what component of this dE will not be cancelled out by a similar dQ at the opposite angle, and write an expression for this component - with a sine or cos of an angle in it. Finally, integrate over the length of the charge to sum up all the dE contributions to the total E.
     
  4. Feb 23, 2009 #3
    http://i42.tinypic.com/s4nddw.jpg

    I attempted this first but got the wrong answer and veered off to a wrong direction trying weird equations, but with your help I tried it again and got the right answer.

    Thanks for the help.
     
  5. Feb 23, 2009 #4

    Delphi51

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    That isn't right - say dQ is at angle A. Then you'll have a sin(A) in your integral and A varies so it can't be taken through the integral sign.
     
  6. Feb 23, 2009 #5
    Erm, sin(A) isn't in the integral at all?

    Does that mean the sin 60 shouldn't be in there at all?

    Should I treat it as a straight rod then?
     
  7. Feb 23, 2009 #6

    Delphi51

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    The sin(A) IS in the integral. You can't take it out. You must integrate from -60 to +60 degrees and all that.
     
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