# Homework Help: Electric Potential problem

1. Jan 18, 2005

### NeoKrypt

One particle has a mass of 3.00x10^-3 kg and a charge of +7.80 µC. A second particle has a mass of 6.00x10^-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.100 m, the speed of the 3.00x10^-3 kg particle is 130 m/s. Find the initial separation between the particles.

Using the information given from the 3x10^-3 kg particle, I tried to use enegry to solve the problem. EPEinitial=EPEfinal+KE I subsituted and eventually ended up with kq^2/r(initial)=kq^2/r(final)+1/2mv^2. I solved the equation for r(final) and I got an answer that made sense, but it was not the right answer. Can anyone tell me what I am doing wrong?

2. Jan 18, 2005

### MathStudent

remember that with energy problems you have to keep in mind what the "system" is that you are evaluating. In this case, the system contains BOTH particles, so the right side of the equation should contain two kinetic energy components (one for each particle).
The new question now is what is the speed of the other particle?

Last edited: Jan 18, 2005
3. Jan 18, 2005

### Gamma

If you take the system as a whole, net force is zero. So may be we can use the conservation of mometum to find the velocity of the other particle

4. Jan 18, 2005

### NeoKrypt

So if I add the other kenetic energy to the other side of the equation I get kq^2/r(initial)=kq^2/r(final)+1/2mV^2+1/2mv^2. Instead of one unknown in the equation I now have two, the velocity of the second particle and the initial distance between the particles? How do I go about getting the velocity of the second particle?