Question

A ring has charge $Q$, radius $R$ and a charge $q$ is placed at the centre. The increase in tension in the ring is :

• A. $\frac{Qq}{4\pi {\epsilon }_0{R}^2}$
• B. $\frac{Qq}{4{\pi }^2{\epsilon }_0{R}^2}$
• C. $\frac{Qq}{8{\pi }^2{\epsilon }_0{R}^2}$
• D. zero

Answer 1

The correct option is C $\frac{Qq}{8{\pi }^2{\epsilon }_0{R}^2}$

T is tension in the ring.

Electrostatic force in radial direction will balance the tension component in that direction

$2Tsin\theta =\frac{qQ\left(2\theta R\right)}{\frac{2\pi R}{4\pi {\epsilon }_0{R}^2}}$

angle $\theta$ is very small then $\sin \theta =\theta$

$2T\theta =\frac{qQ2\theta }{8{\pi }^2{\epsilon }_0{R}^2}$

$T=\frac{qQ}{8{\pi }^2{\epsilon }_0{R}^2}$