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Suppose I have a uniformly charged ring. What I want to know is that if a charged particle, constrained to move only in the plane of ring and initially placed at the centre of the ring when displaced slightly from the centre, , leads to change in potential energy or not. I've tried to find the electric field at a general point , but it turns out to be an ugly integral of the form $\int{\sqrt{1- k sin^2(x)}} dx$, which I could not simplify.

Although what I feel is that no change of potential energy takes place. I'm driving this analogously from the fact that a charged particle placed in a shell experiences no force ( if I take a cross section of that shell which has the charged particle in plane with it. The left cutout portion exerts equal forces ). Is this correct?

Thanks in advance.

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    $\begingroup$ Out of curiosity, would you expect the analogy to extend to 1D? That is, if you had equal charges at $(+R,0,0)$ and $(-R,0,0)$ would you expect the line segment connecting the charges to have constant potential. If not, where do you think the analogy with the 3D case breaks down? $\endgroup$
    – Brian Moths
    Commented Sep 24, 2017 at 15:51
  • $\begingroup$ Well it surely breaks for the 3d case ( a shell) but not for 1d case. In the 3d case the charge is proportional to the solid angle subtended ( which leads to zero field inside the shell ). I'm using the analogy of plane angle with this. I'm pretty sure the concept wouldn't apply in 1d case... $\endgroup$
    – user150098
    Commented Sep 24, 2017 at 16:00
  • $\begingroup$ You should be able to identify a "small" parameter in your integral, make a series in powers of this small paramter and integerate term by term to any order you want. $\endgroup$
    – ZeroTheHero
    Commented Sep 24, 2017 at 16:07
  • $\begingroup$ Why place the charge in the center of the ring? Why not place the charge close to the ring but in the plane of the ring? $\endgroup$
    – Cinaed Simson
    Commented Apr 24, 2019 at 11:26
  • $\begingroup$ You might fine it easier to calculate the potential as a function of position on a radius. $\endgroup$
    – R.W. Bird
    Commented Jan 21, 2020 at 19:48

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A Hint !! Your integral might be described in terms of elliptic functions by doing some changes in variables, namely elliptic function of second kind: $$ E(k)=\int_0^{\pi/2} \sqrt{1-k \sin^2 (x)} dx $$

See Abramowitz Handbook for details. You can then use any CAS (Mathematica for example) to plot the result.

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