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Find min velocity so that particle grazes the shell

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle of mass 1 kg and charge 1/3 μC is projected towards a non conducting fixed spherical shell having the same charge uniformly distributed on its surface. Find the minimum initial velocity of projection required if the particle just grazes the shell.
    1zv7vnr.jpg
    a)[itex]\sqrt{\frac{2}{3}}[/itex]
    b)[itex]2\sqrt{\frac{2}{3}}[/itex]
    c)[itex]\frac{2}{3}[/itex]
    d)none


    2. Relevant equations



    3. The attempt at a solution
    Since the charge is uniformly distributed over the shell, the potential at any point on the surface or inside the shell is KQ/R, where R is the radius. Therefore,
    [tex]\frac{1}{2}mu^2=\frac{KQq}{R}[/tex]
    where u is the initial velocity and Q and q are the charges of shell and particle respectively. Plugging the values, i get
    [tex]u=\sqrt{2}[/tex]

    But this is wrong, please help me proceed in the right direction.
     
  2. jcsd
  3. Jun 3, 2012 #2

    ehild

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    The velocity of the charged particle (strange particle it is, with 1 kg mass:smile:) need not be zero at the surface of the sphere: remember that not only the energy conserves in a central field but the angular momentum, too.

    ehild
     
  4. Jun 3, 2012 #3
    Why can't we assume that velocity becomes zero at the surface as we need to find the minimum velocity?
    About what point we will conserve angular momentum?
     
  5. Jun 3, 2012 #4

    ehild

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    As the angular momentum must be conserved.

    About the centre of the sphere, of course. It is a central force field!


    ehild
     
  6. Jun 3, 2012 #5
    I think i need to clear my concepts about angular momentum and central forces.

    Thanks, i think i have got the answer. :smile:
    Conserving angular momentum, i get:
    Vf=V/2

    Vf is final velocity and V is the initial velocity.
    Using conservation of Energy
    [tex]\frac{1}{2}mV^2=\frac{KQq}{R}+\frac{1}{8}mV^2[/tex]
    Solving, i get:
    [tex]V=2\sqrt{\frac{2}{3}}m/s[/tex]

    Thanks for the help!
     
  7. Jun 3, 2012 #6

    ehild

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    Excellent! :cool: (although you omitted the "f" subscript from the energy equation)

    ehild
     
  8. Jun 3, 2012 #7
    Thanks! :smile:
    Yes, i skipped the step and wrote 1/8mV^2.
     
  9. Jun 3, 2012 #8

    ehild

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    I see. Think of the supervisors, their mind is not so quick as yours: Do not skip the first step.

    ehild
     
  10. Jun 3, 2012 #9
    Haha, i can do it here because the supervisor is you. :smile:
     
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