Question 199 of 311

A uniformly charged ring of radius $3a$ and total charge $q$ is placed in $x‐y$ plane centred at origin. A point charge $q$ is moving towards the ring along the $z-$ axis and has speed $v$ at $z=4a$ . The minimum value of $v$ such that it crosses the origin is:

A$\sqrt{\dfrac{2}{m}}{\left(\dfrac{1}{15}\dfrac{{q}^{2}}{4\pi {ϵ}_{0}a}\right)}^{1/2}$B$\sqrt{\dfrac{2}{m}}{\left(\dfrac{4}{15}\dfrac{{q}^{2}}{4\pi {ϵ}_{0}a}\right)}^{1/2}$C$\sqrt{\dfrac{2}{m}}{\left(\dfrac{1}{5}\dfrac{{q}^{2}}{4\pi {ϵ}_{0}a}\right)}^{1/2}$D$\sqrt{\dfrac{2}{m}}{\left(\dfrac{2}{15}\dfrac{{q}^{2}}{4\pi {ϵ}_{0}a}\right)}^{1/2}$

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Answer: (D) $\sqrt{\dfrac{2}{m}}{\left(\dfrac{2}{15}\dfrac{{q}^{2}}{4\pi {ϵ}_{0}a}\right)}^{1/2}$

From the conservation of energy,
$\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{{q}^{2}}{\sqrt{{\left(3a\right)}^{2}+{\left(4a\right)}^{2}}}+\dfrac{1}{2}m{v}^{2}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{{q}^{2}}{\left(3a\right)}$
$v=\sqrt{\dfrac{{q}^{2}}{4\pi {\epsilon }_{0}a}\times \dfrac{2}{15}\times \dfrac{2}{m}}$
$=\sqrt{\dfrac{2}{m}}{\left(\dfrac{2}{15}\dfrac{{q}^{2}}{4\pi {\epsilon }_{0}a}\right)}^{1/2}$