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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 $$x=2$$

Physics
NCERT
Standard XII

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