Equilibrium? What does this mean?

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The discussion revolves around the stability of equilibrium for a charged bead sliding on a rod between two fixed charged beads. The user derived an equation to find the position of equilibrium, resulting in x = 0.951m. The book suggests that equilibrium can be stable if the charge is positive, but the user questions whether a negative charge would also achieve stability, citing the cancellation of forces on both sides. The analogy of a marble at rest on a hill versus in a valley is used to illustrate different types of equilibrium stability. The key takeaway is the distinction between stable and unstable equilibrium in the context of charged particles.
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Homework Statement



Two small beads having positive charges 3q and q are fixed at the opposite ends of a horizontal insulating rod, extending from the origin to the point D, a third small charged bead is free to slide on the rod at a distance x away from the origin (but shorter than D). Can the equilibrium be stable?

Now I broke it down to

0 = 3d2 - 6xd +2x2

Solving x = 0.951

The book said that it can be stable if it is positive. But wouldn't a negative charge also worked since

k\frac{3q*-q_{x}}{x^2} = k\frac{q*-q_{x}}{(d-x)^2}

Negative and both sides and hence they cancel?
 
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Consider the equilibrium situations for a marble at rest balanced at the top of a hill versus a marble at rest in a valley between hills. Are both stable equilibriums?
 
I forgot t mention d = 1.50m...
 
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