Integration of $dq$ with a ring charge?

Question

Considering the following scenario:

When calculating the electric field on the axis of a ring charge, you end up with:

And following the integration:

I am trying to understand the idea of $\int dq$ and how it evaluates to $Q$. I understand that integrating an infinitesimal like how $\int dx = x + C$ works, but I don't quite understand the idea of integration of these $dq$ terms with the ring being of finite $2(\pi)(a)$ in circumference. Why are we able to do this integral and know that it's integrating exactly once around the ring (summing an infinite amount of infinitesimal charges on each point of the circumference), not an infinite amount of times around the ring, as it doesn't have limits?

Source: physics.udel.edu

Answer 1

Let us assume that the ring is uniformly charged with the line charge density $\lambda$. An infinitesimal element of the ring is $r\times d\theta$. Therefore, the total charge on the ring is $\int_0^{2\pi }\lambda r d\theta$ which is $\int dq$, both of which are equal to $Q$. As you can see,integration is done as $\theta$ is varied between $0$ and $2 \pi$.

Answer 2

Think of it like this-

Integration is basically summing up.Also, every single charge on the ring is equidistant from the axis of the ring for a particular value of x, so you are integrating a constant function. Here, you can forget about the actual position of the charges on the ring as they all provide the same field irrespective of where they are on the ring. You are not going round the ring, instead you are adding up individual charges and hence the ∫dq= Q i.e sum of charges = total charge present on the ring. Simply think in terms of addition of charges.