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Integral mathematical part of physics question

  1. May 20, 2009 #1
    i got a ring a small part of that ring is dq
    then the whole ring is q

    but i cant get this result in the integral
    [tex]
    \int_{0}^{2\pi}dq=2\pi-0=2\pi
    [/tex]
    i should get the result "q"
     
    Last edited: May 20, 2009
  2. jcsd
  3. May 20, 2009 #2

    Kurdt

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    If you're going to do it like that then dq will actually be equal to r*d(theta).
     
  4. May 20, 2009 #3
    thanks
     
  5. May 20, 2009 #4
    but i dont have q in the result
    only 2pi*r *density
     
  6. May 20, 2009 #5

    Kurdt

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    Can you write out the question in full then we might have a clue as to what you're aiming for here.
     
  7. May 20, 2009 #6
    i got the idea 2pi r is q so its ok
    thanks
     
  8. May 20, 2009 #7

    Cyosis

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    I just hope q isn't supposed to be charge.
     
  9. May 20, 2009 #8

    dx

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    Is q the circumference of the ring or the total charge on the ring? If it's the total charge on the ring, then q = 2πrλ where λ is the charge density per unit length of the circumference. (Assuming of course that the charge density is uniform.)
     
  10. May 20, 2009 #9
    dq = lambda·r·d(theta)
    Where lambda is the charge density (charge/distance in this case since it's lineal density).
    So by integrating you get 2pi·r·lambda = Q (length·density = charge)
     
  11. May 20, 2009 #10
    q is the total charge.
    whats the problem?
     
  12. May 20, 2009 #11

    Cyosis

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    The problem is that [itex]2 \pi r \neq q[/itex]. Just check the units, on the left side you have meters on the right side Coulombs, it does not match. A few people have already shown you how to do it correctly though so read those posts!
     
  13. May 20, 2009 #12

    tiny-tim

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    charge density

    Hi transgalactic! :smile:

    (have you been away?)

    You need to integrate the charge density, not just the charge …

    (and technically you integrate from 0 to 2π, not dq … how can q = 2π?)

    so you integrate qdθ/2π, which is the charge on the tiny arc dθ …

    giving ∫0 qdθ/2π = q :wink:
     
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