JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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
- A \(\sqrt {\frac{2}{m}} {\left( {\frac{1}{5}\frac{{{q^2}}}{{4\pi { \in _0}a}}} \right)^{1/2}}\)
- B \(\sqrt {\frac{2}{m}} {\left( {\frac{1}{15}\frac{{{q^2}}}{{4\pi { \in _0}a}}} \right)^{1/2}}\)
- C \(\sqrt {\frac{2}{m}} {\left( {\frac{4}{15}\frac{{{q^2}}}{{4\pi { \in _0}a}}} \right)^{1/2}}\)
- D \(\sqrt {\frac{2}{m}} {\left( {\frac{2}{15}\frac{{{q^2}}}{{4\pi { \in _0}a}}} \right)^{1/2}}\)
Answer & Solution
Correct Answer
(D) \(\sqrt {\frac{2}{m}} {\left( {\frac{2}{15}\frac{{{q^2}}}{{4\pi { \in _0}a}}} \right)^{1/2}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{f}}\) \(\frac{\mathrm{kq}^{2}}{\sqrt{16 \mathrm{a}^{2}+9 \mathrm{a}^{2}}}+\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{kq}^{2}}{3 \mathrm{a}}\)…
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