# B Electric potential problem

1. Apr 26, 2017

### Henry Shi

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:
The internal energy of the system is conserved. Therefore, we have:
KE1+PE1=KE2+PE2
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=kq1q2/R
5,000,000=kq1q2/R
5,000,000=(9*10-9)(10*10-6)(50*10-6) / R
Solving for R, I got R=9*10-9m

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

What did I do wrong?

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2. Apr 26, 2017

### Staff: Mentor

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.
There is another mistake in this step.