Question

A charge q uniformly distributed over a uniform thin metallic ring of radius r. Now a point charge Q is placed at the centre of this ring, then increment of the tension stretching the wire is

Answer 1

$\frac{Qq}{8\pi^2\varepsilon_0 r^2}$

Answer 2

The charge $q$ on the ring creates a linear charge density $\lambda = q/(2\pi r)$. A small element $dq = \lambda dl = \lambda r d\theta$ on the ring experiences an outward electrostatic force $dF_e = k Q dq / r^2 = (Q \lambda d\theta) / (4\pi\varepsilon_0 r)$ from the central charge $Q$. For the ring to be in equilibrium, this outward force on any segment must be balanced by the inward radial component of the tension $T$ in the ring. For a small segment subtending angle $d\theta$, the inward radial tension force is $2T \sin(d\theta/2) \approx T d\theta$. Equating forces, $T d\theta = dF_e$, which leads to $T = (Q \lambda) / (4\pi\varepsilon_0 r)$. Substituting $\lambda = q/(2\pi r)$ gives the increment in tension $T = Qq / (8\pi^2\varepsilon_0 r^2)$.