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Homework Help: Closest distance of approach

  1. Jul 11, 2012 #1

    AGNuke

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    A positive charge +Q is fixed at a point A on line AC. Another positively charged particle of mass m and charge +q is projected from a point B with velocity u. The point C is at large distance from A and B is situated at distance d perpendicular from point C from AC

    Find the minimum distance of approach of +q towards +Q during motion.




    Take [itex]Qq = 4\pi \varepsilon _{0}[/itex] and [itex]d=\sqrt{2}-1[/itex]



    I tried at an instance where the velocity component directed towards +Q becomes zero. But I can't do anything about the perpendicular component of velocity as how it will increase the distance of particle from line AC.
     
  2. jcsd
  3. Jul 11, 2012 #2

    tiny-tim

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    Hi AGNuke! :smile:
    Is that the complete question? :confused:

    C (and the distance d) seems to have no relevance unless the initial velocity u is parallel to AC.

    If so, use conservation of angular momentum (because … ?) :wink:
     
  4. Jul 11, 2012 #3

    AGNuke

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    Yeah. I got it. I conserved the angular momentum as well as mechanical energy (to get the velocity at closest distance of approach out of business) and got my answers.

    Thanks a lot. :smile:
     
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