# Integral mathematical part of physics question

transgalactic
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:

Staff Emeritus
Gold Member
If you're going to do it like that then dq will actually be equal to r*d(theta).

transgalactic
thanks

transgalactic
but i don't have q in the result
only 2pi*r *density

Staff Emeritus
Gold Member
Can you write out the question in full then we might have a clue as to what you're aiming for here.

transgalactic
i got the idea 2pi r is q so its ok
thanks

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

Homework Helper
Gold Member
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.)

springo
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)

transgalactic
I just hope q isn't supposed to be charge.

q is the total charge.
whats the problem?

Homework Helper
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!

Homework Helper
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 