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Finding dipole moment / Electric Field Problem

  1. Jan 31, 2016 #1
    1. The problem statement, all variables and given/known data
    Problem 3 from the attached photo

    2. Relevant equations
    p = qd

    3. The attempt at a solution
    I know the magnitude of the dipole moment is p = qd, where d is the distance between the 2 oppositely charged particles. I know the direction of the dipole moment vector is from - to +. My first thought is there's no charge separation, and hence a dipole moment equals zero. Something tells me this isn't the case.
     

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  2. jcsd
  3. Jan 31, 2016 #2

    haruspex

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    I don't know why you think there is no charge separation. Some charges are R√2 from the nearest opposite charge. Only where the two semicircles meet is there no separation.
     
  4. Jan 31, 2016 #3
    Something like p = ∫dp = ∫∫dq⋅dr, where 0 ≤ r ≤ R ?
     
  5. Jan 31, 2016 #4

    haruspex

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    The charges are along arcs, not distributed over semicircular areas, so I don't see how you get a double integral.
     
  6. Jan 31, 2016 #5
    What if I setup like this (attached), using:
    p = ∫dp = q∫2y = 2q∫(R2-x2)0.5dx from R to -R? Don't I need to differentialize Q somehow?
     

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  7. Jan 31, 2016 #6

    haruspex

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    That integral should work, but you will find it easier in polar coordinates. I don't know what you mean by "differentialize Q".
     
  8. Jan 31, 2016 #7
    So, using...
    Picture2.jpg
    ...would we use d = charge separation = 2R, not d=R?

    So, p = ∫dp = ∫d(dq) = ∫(2R)dq = ∫(2R)[λRdθ]

    Where λ=Q/s, so Q = λs, or dq=λds=λRdθ, resulting in...

    So, p = 2λR2∫dθ from pi to -pi?

    This isn't feeling right since nothing is mentioned about charge density...??
     
  9. Jan 31, 2016 #8

    haruspex

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    No, the integral you have now won't work. The dipole moment is a vector. Integration is a sum, so when you integrate vectors you need that sum to reflect vector addition. Your diagram above pairs charges in such a way that the dipoles willpoint in different directions.
    Your previous way of pairing them was fine, and so was the integral you wrote down. You just need to rewrite that integral in polar form.
     
  10. Jan 31, 2016 #9
    That was my attempt at re-writing in polar form... which apparently is incorrect. I'm staring at...
    Picture1.jpg
    If I try x=Rcosθ, y=Rsinθ
    1. p = ∫dp
    2. = ∫(Q)(2y)
    3. = 2Q∫Rsinθdθ
    4. = -2QRcosθ
    5. eval'd from 0 to pi?
    For #2, don't I need a variable of integration? If I use dy, rather than y, I'd get a sine term in #4, which would result in an answer of zero - which I'm pretty sure isn't correct?
    Also, still doesn't seem like I'm capturing the charge, Q, correctly via this method?
     
  11. Jan 31, 2016 #10
    Unless I integrate from π/2 to -π/2... ?
     
  12. Jan 31, 2016 #11

    haruspex

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    ρρ
    In line 2, Q is the charge on a small element of arc, ρRdθ.
    In line 3, you have correctly left the dθ inside the integral, so the Q outside is now ρR.
    Other than that, your steps 1 to 5 are correct. It does not give zero.
     
  13. Jan 31, 2016 #12
    okay - thank you *very* much for all your help!
     
  14. Jan 31, 2016 #13

    haruspex

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    You are welcome. What was your final answer? Bear in mind that you were given the total charge on each arc, not the charge density.
     
  15. Jan 31, 2016 #14
    Right. I assume I'd use:
    1. ρ= Q/s
    2. = Q/(2πR)/2, or
    3. ρ = Q/(πR)?
     
  16. Jan 31, 2016 #15

    haruspex

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    Good.
     
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