I am having trouble understanding this problem in Electric Potential. (Source: OpenStax, Chapter 19)
Now consider another system of two point charges. One has a mass of 1000 kg and a charge of 50.0 µC, and is initially stationary. The other has a mass of 1.00 kg, a charge of 10.0 µC, and is initially traveling directly at the first point charge at 10.0 m/s from very far away. What will be the closest approach of these two objects to each other?

Here is my attempt:

Spoiler

The internal energy of the system is conserved. Therefore, we have:
KE_{1}+PE_{1}=KE_{2}+PE_{2}
The potential energy at time 1 is essentially 0, and the kinetic energy at time 2 is 0. PE=qV. Therefore, the equation becomes:
0.5(1)(10)^{2} + 0 = 0 + (10*10^{-6})V
50=(10^{-5})V
5,000,000=V
This number is implausible. I also used V=kq_{1}q_{2}/R
5,000,000=kq_{1}q_{2}/R
5,000,000=(9*10^{-9})(10*10^{-6})(50*10^{-6}) / R
Solving for R, I got R=9*10^{-9}m

However, the correct answer was R=9*10^{-2} m, or 9 cm.

Both will move a bit at the time of closest approach, but that is a small correction, not the 7 orders of magnitude difference you got.

Why?

You used V as voltage before. Now you use it as energy and set it equal. That doesn't work. If you would have worked with units, you would have spotted the problem immediately.