Question

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

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

Answer 1

Answer: (D)

$T$ is tension in the ring.
Electrostatic force in radial direction will balance the tension component in that direction
$2T\sin \theta =\frac{qQ(2\theta R)}{\frac{2\pi R}{4\pi {\epsilon }_0{R}^2}}<br/>$
angle $\theta$ is very small then $\sin \theta =\theta$


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