In free space, a particle A of charge 1 μ C is held fixed at point P . Another particle B of the same charge and mass 4 μ g is kept at a distance of 1 m m from P . If B is released, then its velocity

Question

In free space, a particle A of charge 1 μ C is held fixed at point P . Another particle B of the same charge and mass 4 μ 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: [Take 1 4 π ϵ 0 = 9 × 10 9 N m 2 C - 2 ]

TestHub · Accepted Answer

From conservation in mechanical energy:
$- Δ P . E . = Δ K . E .$
$\frac{1}{4 π ε_{0}} {(1 \times 10^{6})}^{2} [\frac{1}{10^{- 3}} - \frac{1}{9 \times 10^{- 3}}] = \frac{1}{2} \times 4 \times 10^{- 6} \times V^{2}$
$9 \times 10^{9} \times 10^{- 12} [\frac{8}{9 \times 10^{- 3}}] = 2 \times 10^{- 6} \times V^{2}$
$V^{2} = 4 \times 10^{6}$
$V = 2 \times 10^{3} m s^{- 1}$

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