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Detector moving toward charged ring - rate of change of V

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data
    A thin ring (radius r = 1.41 cm) carries a charge Q = 8.57 pC distributed uniformly along its length. The ring lies in the y-z plane, so the axis through its center is the x-axis .

    A small detector is moving along the positive x-axis toward the ring at velocity v = -0.543i mm/s. When the detector reaches the point (5.91 cm,0), at what rate does it measure the potential is changing, in V/m? The sign will indicate whether it is increasing or decreasing.
    WP_20150917_004.jpg
    2. Relevant equations
    V = kQ / (x2+R2)1/2

    3. The attempt at a solution

    I honestly cannot even figure out now looking at my notes what I actually did as it was towards the morning when I did this :biggrin: but this is what I have:

    V = (Q/2πε0r2)*[(x2+r2)1/2-x]

    It seems I took dV/dt and got (Q/2πε0r2)* x/(x2+r2)1/2


    and then plugging in values I got - 0.409 :confused:
     
  2. jcsd
  3. Sep 17, 2015 #2

    haruspex

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    I suggest you start again with a clear head. Your 'relevant equation' is right, but I cannot see how you got from there to your next equation. We cannot point out where you went wrong if you do not post your working.
     
  4. Sep 18, 2015 #3

    rude man

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    Confusing question. The problem asks for rate of change of potential in V/m, suggesting what is sought is dV/dx evaluated at P. On the other hand, the velocity of P is also given, suggesting what is sought is dV/dt evaluated at P.
    Take your pick I guess ...
     
  5. Sep 18, 2015 #4

    mfb

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    They could also ask how the electric field (in V/m) is changing as function of time.
    Pick one, or calculate more than one.
     
  6. Sep 18, 2015 #5

    mfb

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    Yes, but then why did they give the speed? Something is odd, I just considered another option.
     
  7. Sep 18, 2015 #6

    rude man

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    Sorry, I thought I had deleted my post. Yes, agreed, as I say, a confusing question!
     
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