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Question

# Two identical particles of mass m carry a charge Q each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed υ. The closest distance of approach will be

A
14πϵ0 Q2mv
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B
14πϵ0,4Q2mv2
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C
14πϵ0,2Q2mv2
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D
14πϵ0,3Q2mv2
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Solution

## The correct option is B 14πϵ0,4Q2mv2 At the closest distance of approach the two particles must have the same velocities applying COE ki+vi=kf+vf12mv2+kQ2∞=12mv21+12mv21+kQ2x⇒kQ2x=12mv2−mv21 ⋯(1) From COM, mv=mv1+mv1 ⇒v1=v2 ⋯(2) (2) in (1) gives kQ2x=mv24⇒x=14πϵ0,4Q2mv2

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