A charged particle q is shot towards another charged particle Q which is fixed, with a speed ν. It approaches Q upto a closest distance r and then returns. If q were given a speed 2ν , the closest distances of approach would be

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r/2

r/4

answer is D.

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Detailed Solution

Charge q will momentarily come to rest at a distance r from charge Q when all it's kinetic energy converted to potential energy i.e. 12mv2=14πε0.qQrTherefore the distance of closest approach is given by r=qQ4πε0.2mv2⇒r∝1v2

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