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
Answer & explanation
Correct answer: option 4
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 $T=\frac{Q q}{8 \pi^2 \in_0 R^2}$