2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Electrostatics & Capacitance

JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Critical Thinking

  • question_answer
    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                                                                                                            [AIEEE 2004]

    A)                    r

    B)                                      2r

    C)                    r/2                                    

    D)            r/4

    Correct Answer: D

    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. \[\frac{1}{2}m{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{qQ}{r}\] Therefore the distance of closest approach is given by \[r=\frac{qQ}{4\pi {{\varepsilon }_{0}}}.\frac{2}{m{{v}^{2}}}\] Þ \[r\propto \frac{1}{{{v}^{2}}}\] Hence if v is doubled, r becomes one fourth.


You need to login to perform this action.
You will be redirected in 3 sec spinner