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E-field of a semi circle.

  1. Aug 1, 2009 #1
    A line charge of uniform density [tex]\lambda[/tex] forms a semi-circle of radius R0. Determine the magnitude and direction of the electric field intensity at the center of the semi-circle.

    I won't bother with uploading a picture I'm pretty sure you can picture this without.

    My trouble with this is I'm unsure whether I need to do a double integral.

    So here is my working.

    E = (1/4[tex]\pi[/tex][tex]\epsilon[/tex]) q/r2 r

    As the problem is symmetric only the cos([tex]\theta[/tex]) components add. So for a small piece of charge dq

    Ei = (1/4[tex]\pi[/tex][tex]\epsilon[/tex]) dq/R02 cos([tex]\theta[/tex])

    Now for this I have to sum up all the dq's but they are also all have a different angle. So do I do an integral for theta from -[tex]\pi[/tex]/2 to [tex]\pi[/tex]/2 ?

    Anyway the answer I get is [tex]\lambda[/tex]/2[tex]\epsilon[/tex]R0

    Any help with this would be greatly appreciated.

    P.S. I'm new to latex.
     
    Last edited: Aug 1, 2009
  2. jcsd
  3. Aug 2, 2009 #2

    diazona

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    Homework Helper

    First of all: no. Because it's a line of charge, i.e. a one-dimensional shape, you only need to do a single integral. (Think about it: if there were a double integral, what two variables would you integrate over?) If it were a two-dimensional surface, you might need to do a double integral; if it were a 3D volume, you might need a triple integral, etc.

    Actually dE, not Ei:
    [tex]\mathrm{d}\mathbd{E} = \frac{1}{4\pi\epsilon_0} \frac{\mathrm{d}q}{R_0^2}\cos\theta[/tex]
    but anyway:
    I think your answer is off by a factor of [itex]\pi[/itex] but the procedure is basically right.
     
  4. Aug 2, 2009 #3
    When I say double integral I mean in respect to all the dq's and all the d[tex]\theta[/tex]'s.

    When you sum up all the dq's to get the total charge isn't it [tex]\lambda[/tex][tex]\pi[/tex]Ro? And the integral of cos([tex]\theta[/tex]) from -pi/2 to pi/2 = 2?

    So when I put it all together I get

    E = (2/4[tex]\pi[/tex][tex]\epsilon[/tex])([tex]\lambda[/tex][tex]\pi[/tex]Ro/Ro2)

    And it cancels to what I said?
     
  5. Aug 2, 2009 #4

    rl.bhat

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    For a small element dL, charge dq = λ*dL and dL = R*dθ.
    So the electric field at the center dE = μο/4π*λR*cosθ*dθ/R^2
    Or dE = μο/4π*λ*cosθ*dθ/R. Τake the integration from 0 to π.
     
  6. Aug 2, 2009 #5
    isn't the integral of cos(θ)dθ from 0 to pi equal to 0?
     
  7. Aug 2, 2009 #6

    Doc Al

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    There's no double integral. You need to express dq in terms of the variable θ: dq = λrdθ
     
  8. Aug 2, 2009 #7
    Yeah I see that now... but rl.bhat said

    dE = μο/4π*λ*cosθ*dθ/R. Τake the integration from 0 to π.

    isnt that integral 0? (isn't the integral of cos(θ)dθ from 0 to pi equal to 0?)
     
  9. Aug 2, 2009 #8

    Doc Al

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    The integral should be from -π/2 to +π/2. (To avoid confusion, next time describe the orientation of the semicircle and how you define θ.)
     
  10. Aug 2, 2009 #9
    I figured that after the first post I should of put up a picture. Thanks for clearing that up Doc Al (like always :) ). All makes sense now!

    Thanks all!
     
  11. Aug 2, 2009 #10

    rl.bhat

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    Sorry. When you start with one end of the semicircle, the component of the field which add up is dEsinθ, rather dE*cosθ. Ιn that case the limit is 0 to pi.
     
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