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Two balls carrying the same charge

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data
    The entire problem is in the attachment.


    2. Relevant equations
    F = k*(|q||q0|)/(r^2)



    3. The attempt at a solution
    F = 8.99e9(.450uC*.450uC)/(3cm^2) = 6.068e8

    This problem has me completely lost and Im turning to you guys as my last resort. I solved for F and have no idea what to do after this. Any sort of insight would be appreciated and I can provide formulas upon request. Thank you.
     

    Attached Files:

  2. jcsd
  3. Oct 17, 2012 #2

    Doc Al

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    Think in terms of energy, not force, for most of the questions.
     
  4. Oct 17, 2012 #3
    I could use a little more guidance than that. What kind of process should I follow to solve part A?
     
  5. Oct 17, 2012 #4

    Doc Al

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    Apply conservation of energy. What forms of energy are relevant here and how would you calculate them?
     
  6. Oct 17, 2012 #5
    PE = mgh?
     
  7. Oct 17, 2012 #6

    Doc Al

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    OK, that's gravitational PE.

    What other kinds of energy are involved here?
     
  8. Oct 17, 2012 #7
    Kinetic energy? KE = 1/2mv^2
     
  9. Oct 17, 2012 #8

    Doc Al

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    Yep, that's another.

    Keep going.
     
  10. Oct 17, 2012 #9
    I wouldn't know any others. A guess would be charge?
     
  11. Oct 17, 2012 #10

    Doc Al

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    Due to their electric field, those charges will have electric potential energy. Look it up!
     
  12. Oct 17, 2012 #11
    Electric Potential Energy: PE = k((q1*q1)/r)

    A pair of point charges separated by a distance
     
  13. Oct 17, 2012 #12
    You may also want to check the units you are using for r.
     
  14. Oct 17, 2012 #13
    Convert cm to m. 3cm = .03m
    And g to kg. 12g = .012kg

    Alright, next step?
     
  15. Oct 17, 2012 #14

    Doc Al

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    Good!

    Now see if you can come up with an expression for the total mechanical energy.
     
  16. Oct 17, 2012 #15
    TME = 1/2mv^2+mgh+k((q1*q1)/r)
     
    Last edited: Oct 17, 2012
  17. Oct 17, 2012 #16

    Doc Al

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    Good. But use the same distance r for both gravitational and electrostatic PE.
     
  18. Oct 17, 2012 #17
    So your saying that both h and r should be .03m? And I am solving for velocity?

    1/2(.012)(v^2) + (.012)(9.8)(.03) + (8.99e9(.45uC*.45uC)/(.03))?
     
    Last edited: Oct 17, 2012
  19. Oct 17, 2012 #18

    Doc Al

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    I would use r for both, where r is the distance between the centers of the balls.
    What's the velocity at the initial position? And at the highest position? (You're solving for that highest position.)

    You'll take advantage of the fact that the total mechanical energy remains constant. Evaluate it at the initial position.
     
  20. Oct 17, 2012 #19
    Okay so the r is going to be equal to .06m.

    The velocity at the initial position is zero, so I set v = 0?

    1/2(.012)(0) + (.012)(9.8)(.06) + (8.99e9(.45uC*.45uC)/(.06)) = .03739725
     
  21. Oct 17, 2012 #20

    Doc Al

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    Yes, at the initial position.
    Right.
    I didn't check your arithmetic, but that's the right idea. Now set up a general equation and solve for the value of r at the final position.
     
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