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Question: Answered & Verified by Expert
In free space, a particle A of charge 1 μC is held fixed at point P . Another particle B of the same charge and mass 4 μg 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  14πϵ0=9×109 N m2 C-2 ]
PhysicsElectrostaticsJEE MainJEE Main 2019 (10 Apr Shift 2)
Options:
  • A 1.0 m s-1
  • B 1.5×102 m s-1
  • C 2.0×103 m s-1
  • D 3.0×104 m s-1
Solution:
1017 Upvotes Verified Answer
The correct answer is: 2.0×103 m s-1
From conservation in mechanical energy:
-ΔP.E.=ΔK.E.
14πε01×1062110-3-19×10-3=12×4×10-6×V2
9×109×10-1289×10-3=2×10-6×V2
V2=4×106
V=2×103 m s-1

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