JEE Mains · Physics · STD 12 - 2. 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 \(\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N{m^2}{C^{ - 2}}\) ]
- A \(1.5\times 10^2\,m/s\)
- B \(2.0\times 10^3\,m/s\)
- C \(1.0\,m/s\)
- D None of these
Answer & Solution
Correct Answer
(D) None of these
Step-by-step Solution
Detailed explanation
\({{\text{q}}_{\text{A}}} = 1\,\mu {\text{C}};\,{{\text{q}}_8} = 1\,\mu {\text{C}},\) \({{\text{m}}_{\text{B}}} = 4 \times {10^{ - 9}}\,{\text{kg}},\quad {{\text{r}}_{{\text{AB}}}} = {10^{ - 3}}\,{\text{m}}\)…
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