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Electric potential of a ring

  1. Dec 8, 2016 #1
    1. The problem statement, all variables and given/known data

    A ring of radius a is made from a charge wire with a uniform charge density λ.

    a) Calculate the electric potential due to the ring as a function of distance from its center along the axis of the ring passing through the center, perpendicular to its plane

    b) If a particle of mass m with a charge identical to that of the ring displaced ever so slightly from the ring's center what speed will the particle eventually attain after being repulsed by the ring.

    2. Relevant equations

    V = k∫(1/r)dq



    3. The attempt at a solution

    So for a I drew the ring here is my best redraw of it in ms paint below. After drawing it i used dq = λ dL than i plug that in to V = k∫(1/r)dq and I plug (a2 + d2)1/2 for r. getting me V = k∫(λdL)/(a2 + d2)1/2 getting me
    V = (kλL)/(a2 + d2)1/2 since L = 2πa I can replace it in the fuction. So I get V = (2kλπa)/(a2 + d2)1/2

    I have to clue weather or not this is right and have no idea how to begin part b.
     

    Attached Files:

  2. jcsd
  3. Dec 8, 2016 #2

    haruspex

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    You have part a correct.
    For part b, think about energy.
     
    Last edited: Dec 9, 2016
  4. Dec 9, 2016 #3
    So since V = U/Q I can set it as (2kλπa)/(a2 + d2)1/2 = U/Q since the Q is the same as the one the ring has I can replace Q as λL since L = 2πa, Q = λ2πa So U = (4kπ2λ2a2)/(a2 + d2)1/2 and U = KE = 1/2mv2, so v2 = (2kπ2λ2a2)/(m(a2 + d2)1/2)
    then I take the square of both sides and I get v = (πλa(2k)1/2)/(m(a2 + d2)1/2)1/2
     
  5. Dec 9, 2016 #4

    haruspex

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    Where d is....?
    I think there may be a factor of 2 error.
     
  6. Dec 9, 2016 #5
    Can you please elaborate on what you mean? I don't know if you are saying that is the speed at d or if you are asking where d is at.

    I see it now the (2k)1/2 is actually a (8k)1/2
     
  7. Dec 9, 2016 #6

    TSny

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    haruspex is pointing out that you never defined what the symbol d represents.
    Yes.
     
  8. Dec 9, 2016 #7

    haruspex

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    Right, but more importantly that d is not given as part of the question, so should not appear in the answer.
     
  9. Dec 9, 2016 #8
    Since d can't be in the solution should I be looking to relate d to a or am I just going about this the wrong way?
     
  10. Dec 9, 2016 #9

    haruspex

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  11. Dec 10, 2016 #10
    I understand the problem is saying that the particle is starting a distance ds, but i still don't understand how that helps me solve the problem. Is ds somehow related to d cause I just don't see it.
     
  12. Dec 10, 2016 #11

    haruspex

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    It means that you can treat d as zero initially.
     
  13. Dec 10, 2016 #12
    So does this mean the answer is just v = πλa(8k/ma)1/2
     
  14. Dec 10, 2016 #13

    haruspex

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    Yes, but you can simplify a bit more.
     
  15. Dec 10, 2016 #14
    v = 2πλ(2ak/m)1/2
     
  16. Dec 10, 2016 #15

    haruspex

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    Looks good.
     
  17. Dec 10, 2016 #16
    Thanks for all the help!
     
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