Solve the Electric Field distance z above a circular loop of radius r. The charge/length = $\lambda$

The arc-length is 2$\pi$r. So the smallest portion of the circle is 2$\pi r \delta \theta$ and charge is therefore \begin{align} q&=2\pi r \delta \theta*\lambda \\ R&=\sqrt{r^2+z^2}= \text{constant} \\ E&= \frac {1}{4 \pi \epsilon _o}\frac {q}{R^2}=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \delta \theta*\lambda}{\sqrt{r^2+z^2}}\end{align}

And we only need the z component.

$$E =\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \delta \theta*\lambda}{\sqrt{r^2+z^2}}\sin(\theta)=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \delta \theta*\lambda}{\sqrt{r^2+z^2}} \frac {z}{\sqrt{z^2+r^2}}$$ and everything is constant except for $\delta \theta$ $$E=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \lambda}{\sqrt{r^2+z^2}} \frac {z}{\sqrt{z^2+r^2}}\int_0^{2\pi}{\delta \theta}$$ So I thought the correct answer must be: $$E=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \lambda z}{[r^2+z^2]^{3/2}} \cdot 2\pi$$ But the correct answer does not multiply by 2$\pi$

Correct: $$E=\frac {1}{4 \pi \epsilon _o}\frac {2\pi r \lambda z}{[r^2+z^2]^{3/2}}$$

Why was I wrong? where did I slip up?



The length element should be $r d\theta$ not $2\pi r d\theta$. So the charge element is $$dq=\lambda r d\theta$$ but not $$dq=\lambda 2\pi r d\theta.$$

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  • $\begingroup$ Why is the length element not equal to the arclength? $\endgroup$ – Jess L Oct 24 '16 at 20:49
  • $\begingroup$ The arclength is $rd\theta$. Or put this way, the circumference is $2\pi r$. But the portion you have is just $d\theta/2\pi$ of it. $\endgroup$ – velut luna Oct 24 '16 at 20:51

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