# Electric charge due to a Charged ring

I was studying the electric field produced by a ring when I came across this.

RESNICK|HALLIDAY|WALKER pg : 640

$$\pmb {Integrating}$$. Because we must sum a huge number of these components,each small,we set up an integral that moves along the ring , from element to element,from a starting point (call it s = 0) through the full circumference ( s = 2$$\pi$$R). Only the quantity s varies as we go through the elements ; the other symbols in $$Eq.22-14$$ remain the same , so we move them outside the integral.We find$$E=\int dE\text{ cos }\theta = \frac {z\lambda}{4\pi \varepsilon_0 (z^2 + R^2)^\text{3/2}}\int_0^{2\pi R} ds$$ $$\qquad\qquad\qquad\qquad=\frac {z\lambda(2\pi R)}{4\pi \varepsilon_0(z^2 + R^2)^{3/2}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{(22-15)}$$

I do not understand where the $$2\pi R$$ comes from in $$eq (22-15)$$

why does the integral of $$ds$$ from $$0$$ to $$2\pi R$$give me $$2\pi R$$?

• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. Apr 20 '20 at 9:49

As the paragraph mentions, $$s$$ denotes the distance around the ring over which you take infinitesimally thin slices. So starting from distance $$0$$ around the circle to one full revolution around it, you travel a distance of $$2 \pi r$$.
We know that $$s=R\theta$$. $$I=\int_0^{2\pi R}ds$$ The differential of $$s$$ is $$ds=Rd\theta$$. If $$s=2\pi R=R\theta$$, then $$\theta=2\pi$$, and if it is $$0$$, then the angle is also $$0$$. Therefore: $$I=\int_0^{2\pi R}ds=\int_0^{2\pi}Rd\theta=2\pi R$$