# Electric Potential Energy problem

1. Apr 10, 2005

### futron

"Two particles each have a mass of 6.1*10-3 kg. One has a charge of +5.0*10-6 C, and the other has a charge of -5.0*10-6 C. They are initially held at rest at a distance of 0.80 m apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-half its initial value?"

What equations would I need to use in order to solve this problem? I tried using EPE/q=Va-Vb, but the voltage cancels out, so I'm unsure as to where to go from there. Thanks.

~Futron

2. Apr 10, 2005

### whozum

Use kinetic and potential energy relationships. Initially they both have no kinetic energy, but a certain amount of potential energy. When d = d/2 they have a certain about of potential and kinetic energy. Find the potential energy in both cases, any potential energy lost would have been translated into kinetic energy.

3. Apr 10, 2005

### futron

Alright, I tried using EPEi=EPEf+(1/2)mv^2 where EPE=q0*(Va-Vb), but that doesn't seem to work. Any ideas?

4. Apr 10, 2005

### whozum

$EPE_f-EPE_i = mv^2/2, and EPE = q(V_b-V_a)$ are the amended versions of your equations. Find the potential with $U = qE\Delta x$

This will take you in circles though.

The force between the charges is given by coulombs law. Find the difference in potential energy at d = r and d = r/2.

5. Apr 11, 2005

### futron

So if I have $EPE_a-EPE_b=q_0(V_A-V_B)$ where $V=kQ/r$, how would I then find the difference between the two when the distance is halved?

6. Apr 11, 2005

### whozum

$U = qE\Delta x$

The field for point charges is $E = \frac{kq}{r^2}$

$$\Delta x_i = 0.8m, \Delta x_f = 0.4m$$

7. Apr 11, 2005

### futron

Thanks, but I'm still not getting the correct answer. Which charge should I use for $q$ in $EPE=qE\Delta x$, and once I get that, would it simply be a matter of solving $EPE_f-EPE_i=(1/2)mv^2$?

8. Apr 12, 2005

### whozum

Can you show some work? In your first post you said the voltages cancel out which isnt correct, this would only happen if $V_b = v_a$ which isnt true.

Show me where your messing up, all these techniques should give the same answer, if you prefer voltages we can use that.

9. Apr 12, 2005

### futron

Alright, I tried using EPEf-EPEi=(1/2)mv^2 where EPEf=(5x10^-6)((9*10^9)*(-5*10^-6)/0.4)) and EPEi=(5*10^-6)x((9x10^9)(-5x10^-6)/0.8)), but that gives 5.43m/s, which is not the correct answer.

10. Apr 12, 2005

### GCT

I think that the potential energy state is $$U=k_{e}q_1q_2/r$$, you'll need to find $$\Delta U=k_eq_1q_2(1/r_{initial}~-1/r_{final})$$, this will equal the negative value of the change in kinetic energy $$=-mv^{2}_{final}-0$$

11. Apr 12, 2005

### futron

Thanks! I was jumbling too many equations together at once but that made it much clearer.