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Continuous Distribution of Charges Problem

  1. Feb 24, 2017 #1
    1. The problem statement, all variables and given/known data

    Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in the figure (Figure 1) .

    Find an expression for the electric field E⃗ at the center of the semicircle.
    Hint: A small piece of arc length Δs spans a small angle Δθ=Δs /R, where R is the radius.
    Express your answer in terms of the variables Q, L, unit vectors i^, j^, and appropriate constants.

    upload_2017-2-24_20-18-25.png
    2. Relevant equations


    3. The attempt at a solution
    upload_2017-2-24_20-21-28.png

    I drew a picture and established a coordinate system. Then I drew the vector of the electric fields at two point charges at point P. Then I saw that the y components of the field will cancel because of the symmetry. I then wrote what the field should be in terms of x. I'm not sure what to do now or how to get to a point where I could integrate.
     
  2. jcsd
  3. Feb 24, 2017 #2

    blue_leaf77

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    You mean ##x## as a coordinate? That's not necessary since you are asked to calculate the field on a fixed coordinate which is the center of the ring.
    After defining ##\lambda## to be the charge linear density ##Q/L = dQ/ds## you can integrate the last equation in your work.
     
  4. Feb 24, 2017 #3
    OK I substituted the linear charge density and integrated from 0 to L. Somethings still wrong though.
     
  5. Feb 24, 2017 #4
     
  6. Feb 24, 2017 #5
    Here's what I did:

    E = ∫dE = ∫(K*dQ*cosθ)/r^2 = ∫(K*λ*dS*cosθ)/r^2) = (Kλ)/r^2 ∫cos(S/r)ds = (Kλ/r) ∫ sin(S/r) ] 0 to L = (Kλ/r)[sin(L/r)-1]
     
  7. Feb 24, 2017 #6

    blue_leaf77

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    Leave ##\theta## as it is and use ##ds = rd\theta## and integrate with the angle. So that it will be
    $$
    \int \frac{k\lambda}{r^2} \cos\theta \ r\ d\theta
    $$
    Upon doing the integration, you may just integrate the upper/lower half of the semicircle owing to the symmetry.
     
  8. Feb 25, 2017 #7
    So I did (Kλ/r) ∫ cosθ dθ and got (Kλ/r) sinθ ] -pi/2 -> pi/2

    So my final answer is 2Kλ/r = (2KQ) / (RL)

    But my homework keeps saying "The correct answer does not depend on: KQ, RL."
     
  9. Feb 25, 2017 #8

    kuruman

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    Q and L are the given quantities, other than the appropriate constants. R is not given. Can you express R in terms of L?
     
  10. Feb 25, 2017 #9
    So I changed my answer to 2KQπ/L^2 and it worked. Thanks!
     
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