Electric potential and conservations of energy

  • Thread starter maiad
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  • #1
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Homework Statement


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]

Homework Equations


U=Kqq/r
U=0.5mv^2


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
 
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Answers and Replies

  • #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.
 
  • #3
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i did that in step 2 where the radius is √2(d^2)
 
  • #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?
 
  • #5
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i got U=Kq^2/√(2*d^2)
 
  • #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)
 
  • #7
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yea that's what i got for the sum of the total energy. and i equated that to 2mv^2
 
  • #8
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are you using 2mv^2 because there are four spheres? I think what you calculated is the PE for one sphere right?
 
  • #9
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That is correct
 
  • #10
Delphi51
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Okay, so we agree on U = (2.707*kq²/(d²) ?
Isn't it ½mv² rather than 2mv² ?
 
  • #11
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It'll be Utotal=0.5mv^2(1+1+1+1) since there's four spheres and it will simplify to 2mv^2
 
  • #12
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how'd you get 2.7*kq^2 on the numerator?
 
  • #13
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You're looking for the velocity of only one sphere, so shouldn't be 1/2mv^2
 
  • #14
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so dosen't that mean i have to dive the total energy by 4 also?
 
  • #15
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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.
 
  • #16
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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
 
  • #17
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I also tried what you suggested but it's still incorrect
 
  • #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.
 
  • #19
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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
 
  • #20
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the velocity of 2 m/s is also incorrect sadly ;(
 
  • #21
Delphi51
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U=(8.98e9)(15e-9)^2 /0.0328
=6.16e-5
=2(U)
I don't understand why you say U = 2U.
Why did you leave out the factor of 2.707 in finding U?
Oh, you did the three parts separately instead of just multiplying by 2.707. Or did you? Check to see if the total works out to U = 2.707*kq²/d.

Someone pointed out that this kq²/d is the energy of the ONE charge, but you are still using Ek = 2mv² !

I found a mistake in my work; now getting an answer about half way between 1 and 2 m/s.
 

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