Integral mathematical part of physics question

  • #1
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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:

Answers and Replies

  • #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
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but i dont have q in the result
only 2pi*r *density
 
  • #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.
 
  • #6
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i got the idea 2pi r is q so its ok
thanks
 
  • #7
Cyosis
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I just hope q isn't supposed to be charge.
 
  • #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.)
 
  • #9
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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)
 
  • #10
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I just hope q isn't supposed to be charge.
q is the total charge.
whats the problem?
 
  • #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!
 
  • #12
tiny-tim
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charge density

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"
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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