How Is the Radial Component of an Electric Field Proven to Cancel Out in a Ring?

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SUMMARY

The discussion focuses on the mathematical proof of the cancellation of the radial component of the electric field generated by a uniformly charged ring with radius r, observed at a distance z along its axis. The radial electric field element is expressed as dE_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda d\theta}{(r^2+z^2)}\hat{\textbf{r}}. The cancellation occurs due to symmetry when integrating over the full angle of the ring, leading to E_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda}{(r^2+z^2)}\int_0^{2\pi} \hat{\textbf{r}} d\theta, which results in zero for the radial components.

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rick_2009
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This is about the electric field of a ring with radius r, at a distance z from center, along the axis of the ring. The ring carries a uniform line charge [tex]\lambda[/tex]. We always say that the radial component of the field cancels out due to symmetry. Can somebody tell how to prove it mathematically (using cylindrical coordinate system only)?

[tex]dE_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda d\theta}{(r^2+z^2)}\hat{\textbf{r}}[/tex]
[tex]E_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda}{(r^2+z^2)}\int_0^{2\pi} \hat{\textbf{r}} d\theta[/tex]

?
 
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First thing you need to do is convert the directional vector that is local to the current element to a directional vector at the point of observation.

So the electric field element points in the r-hat direction, convert that to cylindrical coordinates and then also provide the appropriate translation since you will be observing it at a point offset from the origin of the coordinate system of the electric field element.
 
I don't get your answer
 

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