In free space, a particle A of charge 1,μ C is held fixed at a point P. Another particle B of the same charge and mass 4,μ g is kept at a distance of 1,mm from P. If B is released, then its velocity at a distance of 9,mm from P is [ Take 1 4π 0 = 9 × 109,N m2 C - 2 ]

JEE Mains · Physics · STD 12 - Electric potential and capacitance

In free space, a particle \(A\) of charge \(1\,\mu C\) is held fixed at a point \(P.\) Another particle \(B\) of the same charge and mass \(4\,\mu g\) is kept at a distance of \(1\,mm\) from \(P\). If \(B\) is released, then its velocity at a distance of \(9\,mm\) from \(P\) is [ Take \(\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N{m^2}{C^{ - 2}}\) ]

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Verified Solution

Correct Answer

(D) None of these

Detailed explanation

\({{\text{q}}_{\text{A}}} = 1\,\mu {\text{C}};\,{{\text{q}}_8} = 1\,\mu {\text{C}},\)

\({{\text{m}}_{\text{B}}} = 4 \times {10^{ - 9}}\,{\text{kg}},\quad {{\text{r}}_{{\text{AB}}}} = {10^{ - 3}}\,{\text{m}}\)

\(\frac{1}{2} M_{B}\, V^{2}=k q_{A} q_{B}\left\{\frac{1}{10^{-3}}-\frac{1}{9 \times 10^{-3}}\right\}\)

\(\frac{1}{2} 4 \times 10^{-9}\, \mathrm{V}^{2}=9 \times 10^{8} \times 10^{-6} \times 10^{-6} \times \frac{8}{9} \times 10^{3}\)

\(V^{2}=\frac{8}{2} \times 10^{9}=4 \times 10^{9}\)

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