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Question: Answered & Verified by Expert
A particle A of charge μC is held fixed at a point P in free space. Another particle B of same charge and mass 4 mg is kept at a distance of mm from P. If B is released then its velocity at a distance of mm from P is (Take 14πε0=9×109 Nm2C-2)
PhysicsElectrostaticsNEET
Options:
  • A 1.5×102 m/s
  • B 1.0  m/s
  • C 3.0×104 m/s
  • D 2.0×103 m/s
Solution:
2197 Upvotes Verified Answer
The correct answer is: 2.0×103 m/s

Loss in PE=Gain in KE

K×10-6×10-6110-3-19×10-3=12 mv2

9×109×10-6×10-610-3×89=12×4×10-6×V2

V=2×103 m/s

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