Question

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 :

• A. $\frac{Qq}{4\pi {ϵ}_0{R}^2}$
• B. $0$
• C. $\frac{Qq}{4{\pi }^2{ϵ}_0{R}^2}$
• D. $\frac{Qq}{8{\pi }^2{ϵ}_0{R}^2}$

Answer 1

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

Let the linear charge density of the ring be given by $\lambda$

That is, $\lambda =\frac{Q}{2\pi R}$

where Q is the charge on the ring and R is its radius.

Now, the force acting between a small section $dx$ of the ring and the charge $q$ in the center will be,

F $=\frac{kq\lambda dx}{R^2}$

If $\theta$ be the angle subtended at the centre by $dx$, the force acting will be, 2T sin $\frac{\theta }{2}=T\theta$ ( $\theta$ is very small )

Thus we get $\frac{dx}{R}=\theta$

Thus force should be equal to $\frac{kq\lambda \theta }{R}$

$T=\frac{kq\lambda }{R}$

Substituting, the value of $\lambda$ we get,

T $=\frac{kQq}{2\pi R^2}=\frac{Qq}{8{\pi }^2{ϵ}_0{R}^2}$