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Net electric field of a charged arc

  1. Feb 17, 2017 #1
    1. The problem statement, all variables and given/known data
    So this was a problem worked in class by the professor in class.
    Find the net electric field at the origin due to the arcs
    upload_2017-2-17_14-44-14.png

    2. Relevant equations
    L=2πr/4
    λ=q/L
    E=kQ/r2

    3. The attempt at a solution
    So the professor gave the answer using the fromula
    Enet1(2sin45°)/(4πε0r1)+λ2(2sin45°)/(4πε0r2)+λ3(2sin45°)/(4πε0r3)

    I can understand the formula except for where the 2sin45° comes from. I mean the arc is a quarter circle of 90° in the second quadrant but I am unsure where this term comes from and how it would change if the parameters of the problem were to change.
     
  2. jcsd
  3. Feb 17, 2017 #2

    TSny

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    See if this diagram helps
    upload_2017-2-17_15-10-28.png
     
  4. Feb 21, 2017 #3
    So if U am understanding this correctly then it is the added sine values of the two angles created when drawing the vector for the field?
     
  5. Feb 21, 2017 #4

    TSny

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    I'm not following what you are saying here. In order to see why there is a factor of 2sin(45o) in the answer, you need to go through the derivation. This means setting up and evaluating the integral for the net electric field. Are you having trouble setting up the integral?
     
  6. Feb 23, 2017 #5
    Yes, I am having trouble coming up with the integral.
     
  7. Feb 23, 2017 #6

    TSny

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    OK. Show us your attempt at setting up the integral and we can go from there. It helps to first consider the direction of the net electric field.
     
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