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Net Electric Field Intensity please help

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data
    A rod of length l with a uniform charge per unit length lambda is placed a distance d from the origin along the x axis. A similar rod with the same charge is placed along the y axis. Determine the net electric field intensity at the origin.


    2. Relevant equations
    Electric Field Intensity is the same as electric field, right? So you would use E=kq/r^2 (substituting d in for r?)

    lambda = Q/l


    3. The attempt at a solution
    I am under the impression the net electric field will be zero, as the charge on the x axis should cancel out the charge on the y axis, right? Or since the rods are d away from the origin, and not actually on the origin, does this automatically mean the field is zero?
     
    Last edited: Sep 28, 2008
  2. jcsd
  3. Sep 28, 2008 #2

    Doc Al

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    That equation is for the field from a point charge, not a line of charge. Hint: Break the line into segments and add up each of their contributions to the field. You'll have to integrate.
    No. If the line charges where on opposite sides of the origin, then you could conclude that their fields would cancel out at the origin. But that's not the case here.

    You'll need to find the field contribution from each line charge and then add them up. Don't forget that electric field is a vector.
     
  4. Sep 28, 2008 #3
    The fields on the x and y are the same though right? So I find one, and can multiply it by 2?
     
  5. Sep 28, 2008 #4
    So dE = kdq/r^2 ?? r^2 being l^2 + d^2 ?

    So dE=(klambda*dl)/(l^2 + d^2)
     
  6. Sep 28, 2008 #5

    Doc Al

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    The magnitudes are the same, but they have different directions. (So you can't just multiply by 2 to get their total.)
     
  7. Sep 28, 2008 #6
    So the components for the x and y axis are the same in magnitude, but opposite direction? So once you get the components for the one rod, you are able to just change the direction to find the other?
     
  8. Sep 28, 2008 #7
    I feel so confused. Right now I have for the rod on the y axis:

    Ex = k*lambda*x*(1/x^2)(sinA)(limit A1-A2)

    Ex = (k*lambda)/x *{[(l+d)/SQRT(l^2+d^2)] - -(l+d)/SQRT(l^2+d^2)]
    Ex = [2k*lambda*(l+d)]/[x * SQRT(l^2+d^2]
     
  9. Sep 28, 2008 #8

    Doc Al

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    Can we assume that this rod is parallel to the x axis (actually, directly along the x-axis), not perpendicular to it?
     
  10. Sep 28, 2008 #9

    The picture which I did not include (sorry about that), shows both rods directly on their respective axis
     
  11. Sep 28, 2008 #10

    Doc Al

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    Good, as I had imagined it. The differential field element from a segment dx, ignoring signs, would be given by:

    [tex]dE = \frac{k \lambda}{x^2} dx[/tex]

    Take the integral over the proper limits, from x = d to d + l.
     
  12. Sep 28, 2008 #11
    Thanks very much. I have never done intergration before, so I am not really sure what it means and how to do it. Do you take out the k*lambda/x^2 first?
     
  13. Sep 28, 2008 #12

    Doc Al

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    The k*lambda is a constant, so it just tags along. What's the integral (anti-derivative) of 1/x^2?
     
  14. Sep 28, 2008 #13
    I have done derivatives before, but never antiderivatives before and am not sure. Just giving a wild guess would be, x ?
     
  15. Sep 29, 2008 #14

    Defennder

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    An anti-derivative of a function f(x) is simply a function g(x) such that when it is differentiated, it gives you f(x). So in this case, the question becomes: What must you differentiate in order to give you 1/x^2?
     
  16. Sep 29, 2008 #15

    Doc Al

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    There's nothing wrong with guessing an antiderivative, but then you must immediately check if your guess was correct by taking the derivative. Try it.
     
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