Equilibrium of Charged Beads on a Rod

[sqenixs]

Homework Statement

Two small beads having positive charges 3q and q are fixed at opposite ends of a horizontal, insulating rod, extending from the origin to the point x = d. A third small, charged bead between the two is free to slide on the rod. At what position is the third bead in equilibrium?
Can it be in stable equilibrium?

Homework Equations

coulombs law - F = k (|q1| |q2|)/(r^2)

The Attempt at a Solution

I am not sure what to do because I do not know if the third bead is positively or negatively charged. Unless it doesn't matter?

 

[G01]

You just said this is an equilibrium problem. What is the condition for the forces on a particle in equilibrium?

 

[sqenixs]

they are equal and opposite of each other?

 

[G01]

they are equal and opposite of each other?

Well, I guess this is best formulated in terms of the net force. If there are equal and opposite forces on the particle, the the net force along the x-axis must be:

\Sigma F_x =0

There you go. Now, you should be able to use that equation to solve for q, by summing all the forces on the particle. (Remember to keep track of the signs of the forces) Can you take it from here?

 

[tiny-tim]

Welcome to PF!

coulombs law - F = k (|q1| |q2|)/(r^2)
…
I am not sure what to do because I do not know if the third bead is positively or negatively charged. Unless it doesn't matter?

Hi sqenixs! Welcome to PF!

Yes … use Coulomb's law, with r and d - r, and subtract.

(But isn't it q1 q2, not |q1| |q2|?)

As to the charge … maybe it doesn't matter … or maybe it affects the stability … who knows?

 

[ZeroVersion]

I think you can assume that one of the charges to be negative and the other is negative and it will give you the right answer..

and welcome to PF.