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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 $2\pi R$?

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  • $\begingroup$ 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. $\endgroup$
    – David Z
    Apr 20 '20 at 9:49
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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 $2 \pi r$.

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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: $$I=\int_0^{2\pi R}ds=\int_0^{2\pi}Rd\theta=2\pi R$$

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