# Electric Potential, acceleration and velocity

1. Aug 25, 2004

### Brianjw

Having some trouble with an electric potential problem. I thought I understood it but I keep getting an infinity in the equation which just can't happen.

Three small spheres with charge 2.00 mC are arranged in a line, with sphere 2 in the middle. Adjacent spheres are initially 8.00 cm apart. The spheres have masses m1 = 20g, m2 = 85.0g and m3 = 20.0g, and their radii are much smaller than their separation. The three spheres are released from rest.

So from the last sentence I get the idea to treat it as 3 point charges.

For point charges potential = kq/r.

For the first question I need to find the acceleration on point 1.

My idea was to add up both forces applied to it by points 2 and 3. I used:

F = k*2.00mC/.08 + k*2.00mC/.16

F = ma so a = F/m But that gives me a = 168596338021 m/s^2 which doesn't work and seems rather large.

For the next 3 parts it wants me to find the velocity of each particle when they're far apart. The middle one is easy since its 0. The outter 2 will be equal but I can't seem to solve it without an infinity.

I know I will need to use K_1 + U_1 = K_2 + U_2.

I am not sure how to approach it since U = -QV, and V = the integral of k*q/r from r_0 to far away, aka infinity. Secondly, when U = -qV, what value to I use for Q and q??

Am I approaching this part wrong?

Thanks for the help.

2. Aug 26, 2004

### Beer-monster

At a quick glance

You seem to have used the equation for electric potential to calculate the force.

Also make sure you check your unit?

3. Aug 26, 2004

### K.J.Healey

yea, shouldnt it be F1 = kq1q2/r1^2 + kq1q3/r2^2

4. Aug 26, 2004

### Brianjw

Ya, thanks for pointing out my first problem, after making the correct change I got the right answer. Still don't see the way around the speed though, since its a point mass and its going off to infinity so I get a ln(infinity) in my answer in my equation of K_1 + U_1 = K_2 + U_2.

5. Aug 26, 2004

### Locrian

I see what you are doing wrong with your speed part.

You don't need to integrate anything: to find the energy you would integrate the force, but you already have the potential so there is no need to do that.

When the ball is released, its has only potential energy and its kinetic energy is zero. When it is very far away, r is so large that the potential energy is effectively zero. Therfore, you know how much kinetic energy it has when very far away. To find V at that point, don't just use the equation once; you have two other particles to consider.

Hope that helps

6. Aug 27, 2004

### Brianjw

I got the idea but the one thing I can't solve for right now is the other moving charge. How do I incorporate that into the Potential equation? If it was fixed I would have it, but since the radius of the 3rd charge is changing with the radius of the 1st charge I can't seem to find the right answer.

7. Aug 27, 2004

### Locrian

Well, in the begining there is so much potential energy. In the end, all of that is kenetic energy. It is reasonable to assume that both balls released end up at the same velocity. If you find the total potential energy at the begining, is it reasonable to find that each ball ends up with half of that in kenetic energy?

8. Aug 27, 2004

### Brianjw

Ya I realized my error, when doing the potential I only did 2 of the 3 possible pairs. Once I realized that it fell together. Thanks for all your help!

9. Aug 27, 2004

### Locrian

If I helped even a little, I'm thrilled!

Goodluck with E&M