Equilibrium of Charged Beads on a Rod

AI Thread Summary
The discussion revolves around finding the equilibrium position of a third charged bead on a rod, with two fixed beads of charges 3q and q at opposite ends. Participants emphasize using Coulomb's law to analyze the forces acting on the third bead, noting that the net force must equal zero for equilibrium. There is uncertainty about the charge of the third bead, with suggestions that it may not matter for determining equilibrium but could affect stability. The conversation encourages applying the principles of force balance and charge interaction to solve the problem. Ultimately, the focus remains on using the correct equations and understanding the conditions for equilibrium.
sqenixs
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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?
 
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You just said this is an equilibrium problem. What is the condition for the forces on a particle in equilibrium?
 
they are equal and opposite of each other?
 
sqenixs said:
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?
 
Welcome to PF!

sqenixs said:
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! :smile:

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? :wink:
 
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.
 
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