Question

See attachment below

Answer 1

this is a nice problem. do you know how to start?

Answer 2

In equilibrium, the cord is stretched in the direction of resultant force of G =mg and F = q E, where E stands for the electric field strength of the ring on the axis in distance x from the plane of the ring

Answer 3

how about you @nincompoop ??

Answer 4

Using the triangle similarity, one can write
\[\frac{ x }{ R } = \frac{ E q }{ mg }\]

Answer 5

For the calculation of the electric field strength let us divide the ring to n identical parts, so as every part carries the charge \(Q/n\)

Answer 6

The electric field strength magnitude of one part of the ring is given by
\[\Delta E = \frac{ Q }{ 4 \pi \epsilon_0 l^2 n}\]

Answer 7

Hint:
here, for symmetry reason, the electric force is directed along the z-axis, and the magnitude of its x-component, is:

\[\Large {F_x} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Qq}}{{{l^3}}}\sqrt {{l^2} - {R^2}} \]

Answer 8

now in order to find l, we have to reire that the subsequent condition is checked:
\[\Large mg = {F_x}\tan \theta \]
where \[\theta \] is sulike in the figure below: