Two small beads having positive charges 3q and q are fixed

[richyw]

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 is free to slide on the rod. At what point is the third bead in equilibrium?

Explain whether it can be in stable equilibrium

Homework Equations

$$\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$$

The Attempt at a Solution

I have already found the equilibrium point. My trouble is explaining whether it can be in stable equilibrium. The answer says as long as it is a positive charge it can be. Is this because if it were a negative particle, any deviation from equilibrium and it would be accelerated into one of the two charges, whereas if it were positive it would oscillate until it reached equilibrium?

This makes sense in my head but I don't really know how to "show" it.

 

[Hypersphere]

I have already found the equilibrium point. My trouble is explaining whether it can be in stable equilibrium. The answer says as long as it is a positive charge it can be. Is this because if it were a negative particle, any deviation from equilibrium and it would be accelerated into one of the two charges, whereas if it were positive it would oscillate until it reached equilibrium?

This makes sense in my head but I don't really know how to "show" it.

That's basically the reason, yes. In the stable case, the force will push it back towards the equilibrium, and not away from it.

Another way to think about it is that the force on the particle is related to a potential through $\mathbf{F} = - \nabla V \left( x \right)$. If the equilibrium point corresponds to a potential minimum, the equilibrium will be stable, whereas a maximum gives an unstable equilibrium (e.g. a ball rolling down a hill, or being stuck in a valley). And, of course, you can determine if it's a maximum or minimum by considering the second derivative of the potential function, the sign of which will depend on the sign of the charge.

 

[richyw]

ah thank you. that makes perfect sense now!