Physics-
General
Easy

Question

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

  1. r    
  2. 2r    
  3. r/2    
  4. r/4    

The correct answer is: r/4


    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. fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q Q over denominator r end fraction
    Therefore the distance of closest approach is given by
    r equals fraction numerator q Q over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 2 over denominator m v to the power of 2 end exponent end fraction Þ r proportional to fraction numerator 1 over denominator v to the power of 2 end exponent end fraction
    Hence if v is doubled, r becomes one fourth.

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