# Charge/Field/Velocity Problem

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1. Oct 7, 2015

### Callix

1. The problem statement, all variables and given/known data

2. Relevant equations
F=kqQ/D^2=Ma
E=kQ/D^2
D=vt?

3. The attempt at a solution
So far, I've been able to label the forces on each ship through superposition. But that is all I currently able to understand. I know that my end solution is a speed/velocity, but I am not sure how to get there with the given (Q, t, D, and M)

Any help and direction would be greatly appreciated!

2. Oct 7, 2015

### Staff: Mentor

Have you considered a conservation law approach?

3. Oct 7, 2015

### Callix

I have not because we haven't discussed it. How might I use said approach?

4. Oct 7, 2015

### Staff: Mentor

Usually one considers the trade off between potential and kinetic energy. What form of potential energy is involved here?

5. Oct 7, 2015

### Callix

There is definitely potential energy between each ship.

6. Oct 7, 2015

### Staff: Mentor

Yes... what type and how much?

7. Oct 7, 2015

### Callix

Electrical Potential Energy
U12=kq1q2/D
U23=kq2q3/D
U13=kq1q3/D

8. Oct 7, 2015

### Staff: Mentor

What's the total PE of the system in terms of the variables that you were given?

9. Oct 7, 2015

### Callix

U(total) = 3* kq^2/D, right?

10. Oct 7, 2015

### Staff: Mentor

Looks good.

So how will that change as the ships all tend towards infinite separation? Where will it go?

11. Oct 7, 2015

### Callix

All the potential will simply turn into kinetic energy right?

12. Oct 7, 2015

### Staff: Mentor

That's right.

13. Oct 7, 2015

### Callix

Alright, so then 3* kq^2/D=1/2Mv^2 and just solve for v

14. Oct 7, 2015

### Staff: Mentor

Remember that the energy will be split three ways.

15. Oct 7, 2015

### Callix

So is the multiplying by 3 completely necessary if the charges and D are the same? Couldn't I just solve it as kq^2/D=1/2Mv^2

16. Oct 7, 2015

### Staff: Mentor

Sure. In your presented solution be sure to state why you are justified in doing it.

17. Oct 7, 2015

### Callix

Ah okay, I understand now! Thank you very much! :)