A uniformly charged ring of radius \(3a\) and total charge \(q\) is placed in \(xy-\) plane centered 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

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(D) \(\sqrt {\frac{2}{m}} {\left( {\frac{2}{15}\frac{{{q^2}}}{{4\pi { \in _0}a}}} \right)^{1/2}}\)

Detailed explanation

\(\mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{f}}\)

\(\frac{\mathrm{kq}^{2}}{\sqrt{16 \mathrm{a}^{2}+9 \mathrm{a}^{2}}}+\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{kq}^{2}}{3 \mathrm{a}}\)

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

\(v=\sqrt{\frac{4 k q^{2}}{15 m a}}\)

Same subject

From JEE Mains