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Electric potential and conservations of energy

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data
    The four 2.23 g spheres shown below have q+ve=15.0 nC and d=3.28 cm. The spheres are released simultaneously and allowed to move away from each other. What is the speed of each sphere when they are very far apart?

    http://capa.physics.mcmaster.ca/figures/kn/Graph29/kn-pic2978.png [Broken]
    2. Relevant equations
    U=Kqq/r
    U=0.5mv^2


    3. The attempt at a solution
    1. 2U=k(q^2)/d (potential energy between particles along d)
    2. U=k(q^2)/√(2(d^2) (potential energy of the particles across form each other)
    3.U(total)=2mv^2 (assuming the particles would have the same velocity)

    i then solved for velocity but i didn't get the right answer
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 29, 2012 #2

    Delphi51

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    2U=k(q^2)/d would be the PE due to q's position near two other q's. But there is additional PE due to its position near the third q on the diagonal. Try adding that in.
     
  4. Jan 29, 2012 #3
    i did that in step 2 where the radius is √2(d^2)
     
  5. Jan 29, 2012 #4

    Delphi51

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    Oops, I didn't sort that out. So you ended up with U = (2.707*kq²/(d²), did you?
     
  6. Jan 29, 2012 #5
    i got U=Kq^2/√(2*d^2)
     
  7. Jan 29, 2012 #6

    Delphi51

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    Don't we have total U = U due to q + U due to other q + U due to diagonal q ?
    U = kq²/d² + kq²/d² + kq²/(√2d²)
     
  8. Jan 29, 2012 #7
    yea that's what i got for the sum of the total energy. and i equated that to 2mv^2
     
  9. Jan 29, 2012 #8
    are you using 2mv^2 because there are four spheres? I think what you calculated is the PE for one sphere right?
     
  10. Jan 29, 2012 #9
    That is correct
     
  11. Jan 29, 2012 #10

    Delphi51

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    Okay, so we agree on U = (2.707*kq²/(d²) ?
    Isn't it ½mv² rather than 2mv² ?
     
  12. Jan 29, 2012 #11
    It'll be Utotal=0.5mv^2(1+1+1+1) since there's four spheres and it will simplify to 2mv^2
     
  13. Jan 29, 2012 #12
    how'd you get 2.7*kq^2 on the numerator?
     
  14. Jan 29, 2012 #13
    You're looking for the velocity of only one sphere, so shouldn't be 1/2mv^2
     
  15. Jan 29, 2012 #14
    so dosen't that mean i have to dive the total energy by 4 also?
     
  16. Jan 29, 2012 #15
    No, you have already found the PE for just one sphere due to the other 3. I haven't checked numbers but your method looks sound.
     
  17. Jan 29, 2012 #16
    when i do it... my answer keeps coming up to 0.193 m/s so it would be nice if someone can confirm this :D
     
  18. Jan 29, 2012 #17
    I also tried what you suggested but it's still incorrect
     
  19. Jan 29, 2012 #18

    Delphi51

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    1 + 1 + 1/root 2 = 2.707 is what I get.
    My final answer for the speed is a little over 2 m/s.
    Maybe show your work so we can critique. I often make mistakes in calcs now.
     
  20. Jan 29, 2012 #19
    U=(8.98e9)(15e-9)^2 /0.0328
    =6.16e-5
    =2(U)
    =1.23e-4
    U(diagonal)=(8.98e9)(15e-9)^2/(d*sqrt(2))
    =4.35e-5
    Utotal=(1.23e-4)+(4.35e-5)
    =(1.67e-4)

    1.67e-4=2(2.23e-3 kg)v^2
    v=0.193 m/s
     
  21. Jan 29, 2012 #20
    the velocity of 2 m/s is also incorrect sadly ;(
     
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