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Question:

In free space, a particle A of charge 1µC is held fixed at a point P. Another particle B of the same charge and mass 4µg is kept at a distance of 1mm from P. If B is released, then its velocity at a distance of 9mm from P is: [Take 1/4πε₀ = 9 × 10⁹ Nm²/C²]

3.0 × 10⁴ m/s

1.0 m/s

1.5 × 10² m/s

2.0 × 10³ m/s

Solution:

The correct option is A
2.0 × 10³ m/s
WE = −[ΔU] = Ui − UF = 1/2mv²
U = kq₁q₂/r
(9 × 10⁹) × 10⁻¹² / 10⁻³ − (9 × 10⁹) × 10⁻¹² / 9 × 10⁻³ = 1/2 × (4 × 10⁻⁶)v²
v² = 4 × 10⁶
v = 2 × 10³ m/s