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Magnitude of charge on parallel plates

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A small plastic ball of mass 7.51 × 10-3 kg and charge +0.123http://edugen.wileyplus.com/edugen/courses/crs6407/art/qb/qu/c18/lower_mu.gifC is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.0o with respect to the vertical. The area of each plate is 0.02109 m2. What is the magnitude of the charge on each plate?

    2. Relevant equations
    electric field= q/EoA
    electric force= qE

    3. The attempt at a solution
    I have found the electric field and the force from above equations, I just get lost on how to go from here.
     
  2. jcsd
  3. Feb 13, 2016 #2

    TSny

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    Welcome to PF!
    You've done the hard work! Hopefully you've studies how the charge on the plates of a parallel plate capacitor is related to the electric field between the plates.
     
  4. Feb 16, 2016 #3
    are you talking about the capacitance? That's the next chapter and isn't related to this problem
     
  5. Feb 16, 2016 #4

    TSny

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    Look at the first equation you wrote under the section "Relevant equations" in your first post.
     
  6. Feb 16, 2016 #5
    yes but how is the mass and the angle come into play?
     
  7. Feb 16, 2016 #6

    TSny

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    Those come into play in finding the electric field, which you said that you already did.
     
  8. Feb 16, 2016 #7
    by using the equation I gave, which only uses the area and charge,
     
  9. Feb 16, 2016 #8

    cnh1995

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    You took q as the charge on the mass while the formula Q/εA uses the charge on the plates of the capacitor, which is unknown.
    I believe an FBD would be helpful here.
     
  10. Feb 16, 2016 #9
    This is the picture provided.
     

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  11. Feb 16, 2016 #10

    cnh1995

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    What are the forces acting on it? Can you write the equations of forces along horizontal and vertical directions? For that you'll need mass and the angle..
     
  12. Feb 16, 2016 #11
    the forces would be tension, weight of the ball and electric force.
    so since the ball is at equilibrium the net forces are equal.
    so if tension is pulling up and gravity down, tension is also pulling left and the electric force right.
    so if I find the tension would this be equal to the electric force?
     
  13. Feb 16, 2016 #12

    cnh1995

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    No. Tension is pulling the ball at an angle 30 degrees with the vertical. Electrical force is horizontal and weight of the ball is downward. The net force is 0. How will you write the equations for vertical and horizontal components of the forces?
     
  14. Feb 16, 2016 #13
    Tsin 30= mg
     
  15. Feb 16, 2016 #14

    cnh1995

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    30 degrees angle is with the vertical.. It should be Tsin60=mg...
    What about horizontal forces?
     
  16. Feb 16, 2016 #15
    electric force is pulling to the right and tension is also pulling to the left slightly but im unsure how I would make that an equation
     
  17. Feb 16, 2016 #16

    cnh1995

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    What is the horizontal component of tension? Equate it to electrical force.
     
  18. Feb 16, 2016 #17
    Tsin(theta)= electric force?
     
  19. Feb 16, 2016 #18
    sorry t cos(theta)
     
  20. Feb 16, 2016 #19

    cnh1995

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    Right..theta=60 degrees.. You just resolved tension into two components which balance the horizontal electrical force and downward weight. Now, what will you do to get rid of T from both the equations?
     
  21. Feb 16, 2016 #20
    well you can solve for T from the first equation since you know the mass and gravity. then plug it into the 2nd equation.
     
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