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Question

# A point charge q having mass m is projected from a long distance with speed v towards another stationary particle of same mass and charge. The distance of closest approach of the particles is :

A

q22πε0mv2

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B

2q2πε0mv2

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C
q2πε0mv2
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D
Zero
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Solution

## The correct option is C q2πε0mv2Given, for the both point charges, mass=m charge=q For the first point charge initial velocity,vi=v For the second point charge initial velocity, ui=0 During minimum separation (closest approach) both particles will move with same velocity. (Let be u) Applying the energy conservation principle, Initial total energy=Final total energy P.Ei+K.Ei=P.Ef+K.Ef...(1) Here, Since the particle are at infinite distance so, initial potential energy, P.Ei=0 initial kinetic energy, K.Ei=12mv2 final potential energy when they are at rmin P.Ef=Kq2rmin final kinetic energy, K.Ef=12mu2+12mu2=mu2 Substituting the values in (1), we get ⇒0+12mv2=Kq2rmin+mu2 ⇒12mv2=Kq2rmin+mu2....(2) Applying momentum conservation: Initial momentum (P)i= Final momentum (P)f ⇒mv=mu+mu ⇒u=v2....(3) Using (2) in (3) we get, 12mv2=Kq2rmin+mv24 ⇒Kq2rmin=mv24 ⇒14πε0q2rmin=mv24,[K=14πε0] ∴rmin=q2πε0mv2 Hence option (c) is the correct answer.

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