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Semicircular Wire (Couloumb's law)

  1. Oct 20, 2012 #1
    Semicircular Wire
    A positively charged wire is bent into a semicircle of radius R, as shown in Figure 2.15.4.


    Screenshot from 2012-10-20 18:28:32.png


    The total charge on the semicircle is Q. However, the charge per unit length along the
    semicircle is non-uniform and given by

    λ = λo θ cos .

    (a) What is the relationship betweenλo , R and Q?

    (b) If a charge q is placed at the origin, what is the total force on the charge?



    Attempts:

    a)

    dq = λdl = λRdθ

    dq = λo.cosθ.R.dθ

    Q = ∫(-pi/2 to pi/2) [λo.cosθ.R.dθ]

    Q = 2λoR



    b)

    Force in x axis = ((Ke.q.λo.cosθ.R.dθ)/(R^2))*sinθ integrating that in order to dθ, the result is zero.

    Force in y axis = ((Ke.q.λo.cosθ.R.dθ)/(R^2))*cosθ integrating that in order to dθ, the result is (Ke.q.λo/R)*(0.5*pi)



    Is that correct??
     
    Last edited: Oct 20, 2012
  2. jcsd
  3. Oct 20, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    :biggrin: Bingo! Good work.
     
  4. Oct 21, 2012 #3
    To make your professor happy, you should solve for λ0 in terms of Q, and substitute that into your final answer.
     
  5. Oct 22, 2012 #4

    lol ok thanks for the help!
     
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