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Electric potential at center of semicircle

  1. Feb 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A total charge Q is uniformly distributed on a thread of length L. This thread forms a semicircle. what is the potential at the center? (assume V=0 at large distances)

    2. Relevant equations

    V = -∫Edl
    E = kQ/r^2
    3. The attempt at a solution
    V = - ∫Edr = -E∫dr = -Er = - kQr/r^2 = -kQ/r
    is this correct? I am not sure if the sign should be negative
     
  2. jcsd
  3. Feb 8, 2015 #2

    mfb

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    E is a vector and it is not constant. Finding the electric field is more complicated, it is sufficient to calculate the potential here but that has a different formula.
    Also, using r both for radius and for the length along the circle is confusing and leads to wrong results.
     
  4. Feb 8, 2015 #3
    hmmm so would it be correct to say
    V = ∫dV = k∫dQ/r = kQ/r because Q is constant?
     
  5. Feb 8, 2015 #4

    mfb

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    The result is right but you'll have to justify the steps you used.
     
  6. Feb 9, 2015 #5
    Does this mean i have to plug something in for dQ?
     
  7. Feb 9, 2015 #6

    mfb

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    The length should appear somewhere.

    Oh, and I guess the final result should be expressed in terms of L instead of r as well I guess.
     
  8. Feb 9, 2015 #7
    I think i can use arc length formula L = theta r to relate r and L. I am not sure how I can plug that in to the equation for V. I think this doesn't seem like it should be a tough problem but I must be missing something obvious =[
     
  9. Feb 9, 2015 #8

    mfb

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    This is not a tough problem. Yes you can use the arc length formula.

    If a charge Q is uniformly distributed over a length L, what is the charge per length?
     
  10. Feb 10, 2015 #9
    I think it would simply be Q/L so dQ in terms of the charge density would be dQ = λdL = (Q/L)dL

    V = k∫dQ/r = k∫(Q/L)dL/r = kQL/Lr = kQ/r ??
     
  11. Feb 11, 2015 #10

    mfb

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    Looks fine. Now you just have to express r in terms of L.
    The semicircle has a length of L.
     
  12. Feb 11, 2015 #11
    so since L = theta r then r = L/theta so the final step would be
    kQ/r = kQtheta/L
    but since they dont give me theta can I get away with doing this?
     
  13. Feb 11, 2015 #12

    Nathanael

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  14. Feb 11, 2015 #13
    assuming a semicircle is a half circle (forgive me if this is incorrect I am not 100 percent sure) then L would equal the perimeter or L = (pi/2)r^2
    so r = sqrt(2L/pi) so it would be kQ/(sqrt(2L/pi))
     
  15. Feb 11, 2015 #14

    Nathanael

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    Yes, semicircle just means half-circle

    I think you're mixing up the area and circumference formulas. On the left side you have units of length but on the right side you have units of length squared.
     
  16. Feb 12, 2015 #15
    oops sorry L = 2(pi/2)r = pi r
    so r = L/pi
    so kQ/r = kQpi/L
     
  17. Feb 12, 2015 #16

    mfb

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    Right.
     
  18. Feb 12, 2015 #17
    thanks guys
     
  19. Feb 12, 2015 #18
    i feel like i did something wrong. if my integral was k∫(Q/L)dL/r then wouldnt it be
    k∫(Q/L)dL/r = kQ/r∫dL/L(because Q and r are constant) = kQ/r (ln(L)) ??
     
  20. Feb 12, 2015 #19
    or If i put everything in terms of L instead of r from the begining:
    k∫((Q/L)dL)/(L/pi) = k∫piQdL/L^2 = kpiQ∫dL/L^2 = - kpiQ/L
    so it would be negative?
     
  21. Feb 12, 2015 #20

    mfb

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    If you use L like this, you have to integrate over a different variable, like L' (or x, or whatever). The L in Q/L does not change in the integral.
     
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