1. Oct 3, 2010

### thetest

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

My reasoning is that when the charge stops that means that the force of friction equals the electric force. So I find the distance by setting the two equations equal to each other and solve for x which gives (keq1q2/(mgk))^(1/2), however what about the initial distance that they were set at? Do I add it to this value or subtract it or is it already included within the value of x that I solved for? Any help would be greatly appreciated. Thanks!

Last edited: Oct 3, 2010
2. Oct 3, 2010

### willem2

That won't work. When the force of friction equals the electric force, the net force is 0, so the object will keep on moving.
Use conservation of energy. The energy to move an object a certain distance with a constant friction is easy to find.

3. Oct 3, 2010

### thetest

How can we use the conservation of energy? Are you talking about the equation K1+U1+Wother= K2 + U2? But even if we use that we are not provided a velocity nor does the charge change in height so we can't use potential?

4. Oct 3, 2010

### willem2

You don't need a velocity. The velocity of the charge is 0 at the start and at the end, so the initial and final kinetic energies are 0. This leaves only the electrical potential energy and the work that friction does on the charge.

5. Oct 3, 2010

### thetest

Do we solve for x

Last edited: Oct 3, 2010
6. Oct 3, 2010

### willem2

kmg is the force of friction, not the work done

7. Oct 3, 2010

### thetest

Then we solve for x to find the distance right?

Last edited: Oct 3, 2010
8. Oct 3, 2010

### willem2

Yes! Don't confuse the different k's tough.

9. Oct 3, 2010

### thetest

Awesome thank you! I do have a question though, when I solve for x, I get xq1q2/l0 - (4*pi*e0)(kmgx^2) = q1q2. Is there a way to simplify this?

10. Oct 3, 2010

### willem2

The work done by friction should be $kmg(x - l_0)$ since the charge starts at $l_0$

11. Oct 3, 2010

### thetest

Oh okay and then we use the quadratic equation to solve for x? It is just the + value right?

12. Oct 3, 2010

### willem2

I'm not sure. x should be positive and there should be only 1 positive solution.

13. Oct 3, 2010

### thetest

Oh okay I found the positive solution. Are you sure this is the way to solve the problem because my professor says that the answers should not be very complicated? Thanks again though!