# Help with solving the problem charge problem

1. Sep 12, 2011

### thomq

1. Two small beads having a positive charges 3q and q are fixed at the opposite ends of a horizontal, insulating rod, extending from the origin to point X. This length is the distance (d). A third small charged bead (Q) is free to slide on the rod. At what position is the third bead in equilibrium?

2. Coulomb's Law is used for this problem. F = kqq/d

F[1][/SUB] = kQ(3q)/x[2][/SUP]
F[2][/SUB] = kQq/(d-x)[2][/SUP]

F[1][/SUB] = F[2][/SUB]

3. I am attempting to solve for x.

3 (d-x)[2][/SUP] = (x)[2][/SUP]

I know I could make both side a square root so I have x one side, but I need help beyond this.

I also need to show all steps. Thanks!

2. Sep 12, 2011

### Delphi51

Welcome to PF, thom.
Looks good so far!
Recommend you expand the left side, collect like terms to get a quadratic equation.

I'm passing on a tip I find really helpful: don't bother with those sub, sup codes. Copy the ² symbol from the bottom of my post, or the complete set here:
https://www.physicsforums.com/blog.php?b=346 [Broken]

Last edited by a moderator: May 5, 2017
3. Sep 12, 2011

### susskind_leon

$$\frac{3}{x^2}=\frac{1}{(d-x)^2} \Rightarrow 3x^2-6dx+3d^2=x^2\Rightarrow 2x^2-6dx+3d^2=0\Rightarrow x^2-3dx+3/2d^2=0$$
$$x_{1,2}=\frac{3d}{2}\pm\sqrt{\frac{9d^2}{4}-\frac{3d^2}{2}}=\frac{3d}{2}\pm\sqrt{\frac{3d^2}{4}}=\frac{3\pm\sqrt{3}}{2}d$$
I hope I haven't made a mistake...

4. Sep 12, 2011

### Delphi51

That checks out when using the + sign. The other solution is larger than d, so it has to be extraneous.