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Charge Distrubution evenly on Arc (Radius R)

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data

    A charge Q is distributed evenly on a wire bent into an arc of radius R, as shown in
    the figure below.What is the mathematical expression that describes the electric field
    at the center of the arc (point P indicated) as a function of the angle θ? Sketch
    a graph of the electric field as a function of θ for 0 < θ < 180.

    I added the figure for the question as an attachment.



    3. The attempt at a solution

    lambda=Q/pi R

    dE= kdQ/R^2
    dE= (kdQ/R^2) cos θ

    dQ=lambda dl
    dl= Rd theta
    dQ=lambda R dθ

    dE=(k[lambda R d θ]/R^2)cos θ


    E=(k lambda R cos θ/R^2)d θ(from pi/2 to -pi/2)
    E=k lambda/R cos θ dθ
    E=k lambda/Rsin θ
    E=k lambda/R[sin(pi/2)-sin(-pi/2)]
    E=k lambda/2R
    E=k(Q/piR)/2R=2kQ/piR^2
    E= 2kq/piR^2

    Is this right? I used K for 1/4pi(E) to make it easier to type.
    Also i am kind of lost on what is expected for the sketch.
     

    Attached Files:

  2. jcsd
  3. Oct 10, 2011 #2

    lightgrav

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    looks good ... might make more sense if the E-field component was upward (to balance gravity, if you put a charged item in the center, say)
     
  4. Oct 10, 2011 #3

    SammyS

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    Why are you integrating from -π/2 to π/2 ?

    Integrate from -θ0 to θ0 or similar.
     
  5. Oct 10, 2011 #4
    i didn't notice that. i was integrating from the wrong limits. Thank you for pointing that out. I am still confused on the sketch of the electric field as a function of theta if anyone could point me in the right direction with this it would be much appreciated.
     
  6. Oct 10, 2011 #5

    SammyS

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    Finish working out the solution before you can do the plot.
     
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