Calculating Velocity of Identical Charges with Fixed Distance: 3877 m/s

[TrippingSunwise]

First of all I've read the sticky and I am not trying to get somebody to do my homework. I have been working on this problem for the last two weeks and have the piles of paper to prove it. Here is my problem: Two identical 7.5u-c charges start from rest and are initially spaced 5.5 cm from each other. If we fix the one charge, and let the other go free, how fast will it be moving when the charges are 30 cm apart? Assume the charges have mass 10^-6 kg. The solution is 3877 m/s.

Roads I've been down:
v^2= vay^2 + 2ay(y-y0)
= 0 + 2ay(24.5)
v= sqrt(2a(24.5))
= 2632 m/s

mv^2= KQ^2/r
v^2 = KQ^2/rm
v=sqrt((9*10^9)(7.5 X 10^-6)^2)/(0.055m)(1*10^-6))

 

[StatusX]

What is the change in potential energy from the first point to the second (it goes down)? All of this is transformed into kinetic energy.

 

[TrippingSunwise]

I've been able to solve this problem for a large distance between the two charges.. in that case the potential difference is zero so I was able to get this:

deltaK + deltaU = 0
1/2mv^2 + 1/2mv^2 - 0 + Q(0-V) = 0 or
mv^2 = Q(kQ/r) = kQ^2/r
(1.0 * 10^-6kg)v^2 = (9 * 10^9)*(7.5*10^-6)/(0.055m)
v = 3.0 * 10^3 m/s

I'm not sure what the potential difference is at this distance.

 

[StatusX]

It's just V1-V2, or kQ^2/r1 - kQ^2/r2.

 

[TrippingSunwise]

I apologize for my daftness but that does not yield the velocity when they are 30 cm apart.

 

[StatusX]

Is your equation:

1/2 mv^2 = kq^2 (1/(5.5cm) - 1/(30 cm)) ?

If so, check your math, because that's right.

 

[TrippingSunwise]

Oh Thank You! Brain malfunction .. too long looking at the same problem.
Very much appreciated! I can't believe the incredibly quick response. I will highly recommend this site. All the best!