# Integral mathematical part of physics question

• #1
transgalactic
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i got a ring a small part of that ring is dq
then the whole ring is q

but i can't get this result in the integral
$$\int_{0}^{2\pi}dq=2\pi-0=2\pi$$
i should get the result "q"

Last edited:

## Answers and Replies

• #2
Staff Emeritus
Science Advisor
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If you're going to do it like that then dq will actually be equal to r*d(theta).

• #3
transgalactic
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0
thanks

• #4
transgalactic
1,395
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but i don't have q in the result
only 2pi*r *density

• #5
Staff Emeritus
Science Advisor
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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
transgalactic
1,395
0
i got the idea 2pi r is q so its ok
thanks

• #7
Homework Helper
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I just hope q isn't supposed to be charge.

• #8
Homework Helper
Gold Member
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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
springo
126
0
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
transgalactic
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I just hope q isn't supposed to be charge.

q is the total charge.
whats the problem?

• #11
Homework Helper
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The problem is that $2 \pi r \neq q$. 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
Science Advisor
Homework Helper
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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 can't get this result in the integral
$$\int_{0}^{2\pi}dq=2\pi-0=2\pi$$
i should get the result "q"

Hi transgalactic!

(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

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