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Electrostatics Basics -- 2 plastic balls hanging on threads attract each other

  1. Mar 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Two small plastic balls hang from threads of negligible mass. Each ball has a mass of
    0.14 g and a charge of magnitude q. The balls are attracted to each other and the threads attached to the balls make an angle of 20.0 degree. with the vertical. The balls are 2.05 cm apart.
    A. Find the magnitude of the electric fore acting on each ball
    B. The tension in each of the threads
    C The magnitude of the charge of the charge on the balls.
    2u6jyfb.jpg
    2. Relevant equations
    11bfec934e963740d4bed02c2062a4e7.png

    3. The attempt at a solution
    A. Since they attract each other. there has to be an attraction force between the two charges.
    I assumed that there is a force that acts against it in the horizontal direction. The cable has tension so I thought that the x-component of the force has to act against the attraction force. With that in mind I needed to find out the x-component of the tension, FT which would then equal the magnitude of the electric force.
    I am not sure if my thought process is right. I am also confused on how to calculate the the x component of the tension. Would it be: ∑Fx = FTx= mg sin(20) = ma (which is zero since it doesn't move) ?

    Since I couldn't figure out A, I was reluctant to continue.



     
  2. jcsd
  3. Mar 16, 2015 #2
    The horizontal component of the tension cannot incorporate g because gravity only acts vertically. The diagram suggests that you can make a right-angled triangle with the tension FT as the hypotenuse, giving you the correct expression for the horizontal component. This would not be zero as you suggest, however, since Newton's second law deals with resultant forces i.e. the F in F=ma refers to all the forces in one direction minus all the forces in the opposite direction. It's a matter of choice which direction is positive and which is negative, but usually the context can show you which arrangement is more convenient.

    Your general reasoning is correct though.
     
  4. Mar 16, 2015 #3
    Ah, ok. It makes sense that g is not part of the horizontal component of tension. I know that the tension would be FT = mg. To get the x-component, would I have to get FTsin(20) ?
    My sentence about the force summation in the x-direction was really bad sorry bout that. What I actually meant was that since the the balls are not moving, that there is no acceleration.
    So FTx - Fattraction force = ma (ma is zero i think )
    FTx = Fattraction force
     
  5. Mar 17, 2015 #4

    andrevdh

    User Avatar
    Homework Helper

    That is correct.
    Now you can also balance the force components in the y-direction.
     
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