# Homework Help: Find the electric potential energy of this system of charges

1. Feb 3, 2010

### matt72lsu

1. The problem statement, all variables and given/known data
A point charge q_1 = 86.5 uC is held fixed at the origin. A second point charge, with mass m = 5.90×10−2 kg and charge q2 = -2.95 uC, is placed at the location (0.323 m, 0)
1) Find the electric potential energy of this system of charges.
2)If the second charge is released from rest, what is its speed when it reaches the point (0.121 m, 0)?

2. Relevant equations

W = EPEa - EPEb

3. The attempt at a solutionI think the above equation is used in 1 but I'm not sure how to start it. For 2, I'm not sure how to figure out the problem using the x,y coordinates given. Any help would be appreciated

2. Feb 3, 2010

### dacruick

alright so what you are basically doing is bringing a charge from infinity to a point (0.323,0) The reason you have to think about it this way is that the only way that charge is in existance, but not part of the initial system's energy is if it is an infinite distance away. so what you want to do is take EPEi(i denoting infinity), and subtract from it EPEb.

For your second question think about it like gravity. You have the earth at the origin. you want to find the speed when something falls from 323 meters in the air to 121 meters in the air. So you are going to find the difference in potential energy. So the equation is the same. So you are going to have kq/0.121 - kq/0.323. that will give the kinetic energy gained by the charge. Throw that energy in the E = 0.5mv² and there you go. solve for v

3. Feb 3, 2010

### dacruick

oh and by the way EPEi is always 0 since r is infinity

4. Feb 4, 2010

### matt72lsu

ok so I'm still not exactly comprehending number 1. so basically we "start" at infinity and move to a point (.323), which in this case we find the difference (kq/.323- kq/0). so is this the EPE of the charges? there seems like there would be more to this problem.
for 2, i am getting a negative number when i subtract kq/.121 - kq/.323. thus i cant take a sq root. am i doing something wrong?