# Insulating Spheres in Electric Fields

1. Oct 12, 2008

### faceoclock

Hi, I'd like to ask the good people of this forum for some help.
Here's a problem I've been working on for a while, and I'm seriously at my wit's end. I guess there's something I'm missing here...

1. The problem statement, all variables and given/known data
Two insulating spheres have radii r1 and r2, masses m1 and m2, and uniformly distributed charges -q1 and q2. They are released from rest when their centers are separated by a distance d. How fast is each moving when they collide? Suggestion: Consider conservation of energy and of linear momentum.

2. Relevant equations
I thought these were relevant:
Momentum=mv
Kinetic energy = 1/2(mv^2)
$$\Delta$$U = -q$$\int$$E dr

3. The attempt at a solution
First I solved for the potential energy that this system gains when the two spheres are moved apart:
$$\Delta$$U = q1$$\int^{d}_{d-r1-r2}$$E dr = k(q1)(q2)($$\frac{1}{d-r1-r2}$$ - 1/d)

I figured this is the amount of energy the spheres would have when they collide, so...
$$\Delta$$U = $$\frac{1}{2}$$(m1)v$$^{2}_{1}$$ + $$\frac{1}{2}$$(m2)v$$^{2}_{2}$$

From conservation of momentum, v2 = (m1/m2)v1 so subbing that into the above equation I got:
$$\Delta$$U = $$\frac{1}{2}$$m1v$$^{2}_{1}$$ + $$\frac{1}{2}$$$$\frac{m^{2}_{1}}{m_{2}}$$v$$^{2}_{1}$$

So then I solved for v1 to get:

v1 = $$\sqrt{\frac{2kq_{1}q_{2}((1/(d-r1-r2)-(1/d))}{m_{1}+\frac{m^{2}_{1}}{m_{2}}}}$$

And v2 can be figured out the same way. However, I know for a fact this isn't the right answer.

In closing
I'm don't really know what I did wrong, but I suspect it's because I treated the two spheres as point charges, and I'm not sure if I'm justified in doing that.

2. Oct 12, 2008

### nasu

They move from a distance d to a distance r1+r2.
What is d-r1-r2?
Look at your potential energy again. You don't need any integral. Just take the difference between the final energy and initial energy. I would treat them as point charges.

3. Oct 13, 2008

### faceoclock

Yup lol, that would do it. Thanks a lot for your help Nasu!