  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 $$P\left( \sqrt { 3 } R,0,0 \right)$$ 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 $$\sqrt { \cfrac { \lambda q }{ x{ \varepsilon }_{ 0 }m } }$$. Find $$x$$.

Solution

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 $$V=\dfrac{k\lambda2\pi R}{r}$$Initially, $$r^2=(\sqrt 3R)^2+R^2\Rightarrow r=2R$$finally $$r=R$$ (particle at center)From conservation of energy:Loss in KE $$=$$ Gain in Electrical potential energy.so, $$\dfrac{m(v_1^2-v_2^2)}{2}=\dfrac{k\lambda2\pi R}{R}-\dfrac{k\lambda2\pi R}{2R}$$$$\dfrac{m(v_1^2-0)}{2}=\dfrac{k\lambda2\pi }{1}-\dfrac{k\lambda2\pi }{2}$$$$\dfrac{mv_1^2}{2}=\dfrac{k\lambda2\pi }{2}$$             $$K=\dfrac1{4 \pi \epsilon_0}$$$$\Rightarrow v=\sqrt{\dfrac{\lambda q}{2\epsilon_o m}}$$On comparing from question $$x=2$$PhysicsNCERTStandard XII

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