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.
Best Answer
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.