Electric Potential Energy problem

[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

 

[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.

 

[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?

 

[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.

 

[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?

 

[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

 

[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?

 

[whozum]

Can you show some work? In your first post you said the voltages cancel out which isn't correct, this would only happen if V_b = v_a which isn't true.

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

 

[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.

 

[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

 

[futron]

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