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

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Correct answer: D. r / 4

At closest distance r its whole KE is converted into PE (1)/(2) m v^2=(1)/(4 π _0) (Q · q)/(r) r=(1)/(4 π _0) (Q · q)/(m v^2) In next case, r^ =(1)/(4 π _0) (Q q)/(m(2 v)^2) r^ =(1)/(4)((1)/(4 π _0) · (Q q)/(m v^2)) r^ =r / 4

A charged particle ' q ' is shot with speed v towards another fixed...

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