# Integration of $dq$ with a ring charge?

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

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 $2 \pi$.
• Ah this is very helpful, thanks!! How exactly did you go from the first integral with limits to the indefinite integral $dq$? Commented Jan 15, 2017 at 8:17
• the first integral is what they mean by $\int dq$. It is not an indefinite integral because you already know that you have to integrate over the ring. You have the limits. Commented Jan 15, 2017 at 8:25
• I'm sorry but I'm having trouble seeing how you got your first integral and how integrating that will get you $Q$. Commented Jan 17, 2017 at 5:47