question_answer

A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed u at z = 4a. The minimum value of u such that it crosses the origin is :      [JEE Main 10-4-2019 Morning]

A) \[\sqrt{\frac{2}{m}}{{\left( \frac{1}{15}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]

B) \[\sqrt{\frac{2}{m}}{{\left( \frac{2}{15}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]

C) \[\sqrt{\frac{2}{m}}{{\left( \frac{4}{15}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]

D) \[\sqrt{\frac{2}{m}}{{\left( \frac{1}{5}\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a} \right)}^{1/2}}\]

Correct Answer: B

Solution :

[b] \[{{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}}\]

\[\frac{k{{q}^{2}}}{\sqrt{16{{a}^{2}}+9{{a}^{2}}}}+\frac{1}{2}m{{\text{v}}^{2}}=\frac{k{{q}^{2}}}{3a}\]

\[\frac{1}{2}m{{\text{v}}^{2}}=\frac{k{{q}^{2}}}{a}\left( \frac{1}{3}-\frac{1}{5} \right)=\frac{2k{{q}^{2}}}{15a}\]