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Field due to semicircle, cut up into pieces

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the electric field due to a half circle of radius R and total charge Q. (a) First estimate by dividing the charge into five equal parts. (b) Repeat for 9 equal parts, (c) Using integral calculus solve exactly.


    2. Relevant equations

    E = kQ/r^2

    3. The attempt at a solution
    Cant remember too much from doing these before...for a start I guess...

    a) break into 5 parts....

    Q = Q/5

    Just need some help getting going!

    Thanks in advance!
     
  2. jcsd
  3. May 7, 2014 #2

    SammyS

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    Are you finding the field at the center of the circle, or at some general point ?
     
  4. May 7, 2014 #3
    Doesn't specify...That is the entire question...hmm based on the level of what the class would be at, I would say at some general point
     
  5. May 7, 2014 #4

    haruspex

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    You need to decide what the parts look like and how you will approximate the field due to each (as point charges, presumably).
     
  6. May 8, 2014 #5
    Alright so I've been working on this...and understand that this question want the field at the center of the circle

    So here's what I've been doing...

    Untitled.png

    Total Charge = Q

    Charge per unit length = Q/L

    ---------------------------------------------------------------------------------------------

    For Part A....Q = Q/5 and L would be a half circle (pi * r) divided into 5 parts...so (5 pi * r)

    So we would have λ = .2Q / (5 pi * r)

    The charge on each slice will be dQ = λRdθ

    So for a small portion of the electric field we would have

    dE = kdQ / r^2 = (kλ / r) dθ


    Now the components of the field will be

    E_x = dEcosθ
    E_y = dEsinθ

    So for the total field we will have

    E_x = ∫(kλ / r)cosθ dθ = (kλ / r) ∫ cosθdθ = (kλ / r)sinθ evaluated from 0 to pi
    And
    E_y = ∫(kλ / r)sinθ dθ = (kλ / r) ∫ sinθdθ = -(kλ / r)cosθ evaluated from 0 to pi

    --------------------------------------------------------------------------------------------

    The x-components will cancel out....and the y-components will come out to (2kλ / r) right?

    So would that be it? Knowing λ = .2Q /(5 pi * r) ?

    Sorry for not knowing how to make that look nice...not too veteran here just yet :P

    ---------------------------------------------------------------------------------------------

    Can anyone let me know if I went completely off...or something is right :) thanks again!
     
  7. May 8, 2014 #6

    haruspex

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    What's 5 pi * r? That's two-and-a-half times around the circle!
    You seem to have confused two approaches. The cutting into 5 parts approach should be an approximation that avoids calculus. I would assume you are meant to represent each of the parts as a point charge. λ=Q/L, where L is the total length of the arc, πr, in all cases.
    Your exact calculus solution looks right (except at the end where you plugged in the wrong value for λ).
     
  8. May 9, 2014 #7
    Oh wow was that a mistake...what got me confused on that was I was looking at it as

    (1/5) (.2Q / (pi * r)) which then lead me to the (5 pi * r)

    alright so then it SHOULD be ,if wanting to use decimals here, (.2pi * r)

    Well that's interesting...now the (.2)'s cancel and we only have the Q/(pi *r)? Is that correct?
     
  9. May 9, 2014 #8

    haruspex

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    Yes, but as I posted, for the exact (calculus) method there was no point in going through that 0.2 stuff in the first place.
    You are asked to use two methods. In one you cut it up into 5 or 9 pieces and apply some simple approximation - no calculus. In the other, you keep it as a single piece and apply the calculus method you posted.
     
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