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2 spheres connected by wire, find tension

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data
    2 conducting spheres of radius r=0.5cm are connected by a long thin piece of conducting wire of length L=2metres. A charge of 60 micro Coulombs is put onto one of the spheres, find the tension T in the wire. Assume the charge is evenly distributed along the surface of the two spheres.


    2. Relevant equations
    [tex]F=k_{e}\frac{Qq}{r^2}[/tex]
    [tex]Q=30\mu C[/tex]
    Q is 30 micro Coulombs because I assumed that the charge would distribute evenly onto the surface of both spheres.
    3. The attempt at a solution
    This question was in the Gauss Law chapter of my book however I didn't use Gauss Law for this one.
    I said
    [tex]F_{12}=k_{e}\frac{Q^2}{L^2}[/tex]
    [tex]F_{21}=-F_{12}[/tex]
    So according to me each sphere is "pulling" the rope in opposite directions to produce a combined force of
    [tex]F=2F_{12}[/tex]
    The tension in the wire must balance this force otherwise it will break
    So
    [tex]F_{wire} = F=4N[/tex]
    However the answer gives 2 Newtons which I cant convince myself is right because each sphere is contributing its own force on the string.
     
  2. jcsd
  3. Aug 27, 2010 #2

    Doc Al

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    Staff: Mentor

    You are misunderstanding how tension is created. To create a tension F in a rope, for example, both ends must be pulled with the force F. The tension in the rope--or wire--is the force with which it pulls. The rope pulls each sphere with a force F, thus the tension is F, not 2F.
     
  4. Aug 27, 2010 #3
    OK thanks for that. I suppose one way of looking at it would be to consider the forces on one of the spheres, there is only the electric force and the tension, for the sphere to remain where it is these forces would have to be equal.

    How about the case when the force is different on each end? What would be the tension then?

    I always thought of tension as a mass suspended by a string with the string pulling up with the same force as the mass is pulling down.
     
  5. Aug 27, 2010 #4

    Doc Al

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    Staff: Mentor

    Sounds good.

    If the external forces on the spheres were different, you'd have to calculate the tension based on the net force and acceleration. But there would be a single tension throughout the rope, assuming it's massless.

    Good, that works. Realize that whatever's connected to the other end of the rope--the ceiling, perhaps--is also pulling up with the same force.
     
  6. Aug 27, 2010 #5
    Thanks for clearing that up, the bit at the end was what I was missing - the ceiling is pulling up with the same force.
     
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