# Electric potential and conservations of energy

1. Jan 29, 2012

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. Jan 29, 2012

### Delphi51

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. Jan 29, 2012

i did that in step 2 where the radius is √2(d^2)

4. Jan 29, 2012

### Delphi51

Oops, I didn't sort that out. So you ended up with U = (2.707*kq²/(d²), did you?

5. Jan 29, 2012

i got U=Kq^2/√(2*d^2)

6. Jan 29, 2012

### Delphi51

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. Jan 29, 2012

yea that's what i got for the sum of the total energy. and i equated that to 2mv^2

8. Jan 29, 2012

### Froster78

are you using 2mv^2 because there are four spheres? I think what you calculated is the PE for one sphere right?

9. Jan 29, 2012

That is correct

10. Jan 29, 2012

### Delphi51

Okay, so we agree on U = (2.707*kq²/(d²) ?
Isn't it ½mv² rather than 2mv² ?

11. Jan 29, 2012

It'll be Utotal=0.5mv^2(1+1+1+1) since there's four spheres and it will simplify to 2mv^2

12. Jan 29, 2012

how'd you get 2.7*kq^2 on the numerator?

13. Jan 29, 2012

### Froster78

You're looking for the velocity of only one sphere, so shouldn't be 1/2mv^2

14. Jan 29, 2012

so dosen't that mean i have to dive the total energy by 4 also?

15. Jan 29, 2012

### Froster78

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. Jan 29, 2012

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. Jan 29, 2012

I also tried what you suggested but it's still incorrect

18. Jan 29, 2012

### Delphi51

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. Jan 29, 2012

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. Jan 29, 2012