Question

A circular ring of radius $R$ with uniform positive charge density $\lambda$ per unit length is fixed in the Y-Z plane with its centre at the origin $O$. A particle of mass $m$ and positive charge $q$ is projected from the point $P(\sqrt{3}R,0,0)$ on the positive X-axis directly towards $O$, with initial velocity $v$. The smallest value of the speed $v$ such that the particle does not return to $P$ is $\sqrt{\frac{\lambda q}{x{\epsilon }_{0}m}}$. Find $x$.

Answer 1

Ring and Charge particle has a positive charge so both will repel each other.

But when Velocity of projection of charge q is given in such a way that it just cross the center of the ring then it will never come back because of repulsion force.

Potential at distance $r$ from circumference of curve is $V=\frac{k\lambda 2\pi R}{r}$

Initially, $r^2=(\sqrt{3}R{)}^2+R^2\Rightarrow r=2R$

finally $r=R$ (particle at center)

From conservation of energy:

Loss in KE $=$ Gain in Electrical potential energy.

so, $\frac{m(v_1^2-v_2^2)}{2}=\frac{k\lambda 2\pi R}{R}-\frac{k\lambda 2\pi R}{2R}$

$\frac{m(v_1^2-0)}{2}=\frac{k\lambda 2\pi }{1}-\frac{k\lambda 2\pi }{2}$

$\frac{m{v}_1^2}{2}=\frac{k\lambda 2\pi }{2}$ $K=\frac{1}{4\pi {ϵ}_0}$

$\Rightarrow v=\sqrt{\frac{\lambda q}{2{ϵ}_{o}m}}$

On comparing from question $x=2$