A ring has charge Q and radius R. If a charge q is placed at its centre then the increase in tension in the ring is

$\frac{Q q}{8 \pi^2 \varepsilon_0 R^2}$

Consider a small element AB, $\theta$ is very small. Then

$A B=R(2 \theta)$

Change on AB is $d Q=\frac{Q}{2 \pi R}(2 R \theta)=\frac{Q \theta}{\pi}$

$2 T \sin \theta=\frac{d Q \cdot q}{4 \pi \in_0 R^2}=\frac{Q q \theta}{4 \pi^2 \in_0 R^2}$

$2 T \theta=\frac{Q q \theta}{4 \pi^2 \in_0 R^2}$  or  $T=\frac{Q q}{8 \pi^2 \in_0 R^2}$