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 ]

Step-by-step solution

From conservation in mechanical energy: - Δ P . E . = Δ K . E . 1 4 π ε 0 1 × 10 6 2 1 10 - 3 - 1 9 × 10 - 3 = 1 2 × 4 × 10 - 6 × V 2 9 × 10 9 × 10 - 12 8 9 × 10 - 3 = 2 × 10 - 6 × V 2 V 2 = 4 × 10 6 V = 2 × 10 3   m   s - 1