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JIPMER2017PhysicsElectrostatics

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

Options

  1. Ar
  2. B2 r
  3. Cr / 2
  4. Dr / 4

Correct answer

D. r / 4

Step-by-step solution

At closest distance r its whole KE is converted into PE 1 2 m v^2= 1 4 ₀ Q q r r= 1 4 ₀ Q q m v^2 In next case, r^ = 1 4 ₀ Q q m(2 v)^2 r^ = 1 4 ( 1 4 ₀ Q q m v^2 ) r^ =r / 4

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