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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 $4 m g$ is kept at a distance of $1 m m$ from P. If B is released, then its velocity at a distance of $9 m m$ from P is: $[T a k e \frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}]$

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$1.0 m / s$

$3.0 \times 10^{4} m / s$

$2.0 \times 10^{3} m / s$

$1.5 \times 10^{2} m / s$

answer is C.

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Detailed Solution

$\frac{1}{2} {mv}^{2} = q (∆ V) = {kq}_{1} q_{2} [\frac{1}{d_{1}} - \frac{1}{d_{2}}]$

$v = \frac{8 \times 2 \times 10^{6}}{4} = 2 \times 10^{3} m / s$

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