Two charged beads with a third bead in equilibrium

[putongren]

I understood the situation intuitively and conceptually concerning the equilibrium issue, but couldn't wrap my head around how the mathematics dictate whether the equilibrium is stable or not.

 

[kuruman]

I understood the situation intuitively and conceptually concerning the equilibrium issue, but couldn't wrap my head around how the mathematics dictate whether the equilibrium is stable or not.

Mathematically, one evaluates the second derivative of the potential energy at the equilibrium position.

• If the result is positive, the equilibrium is stable.
• If the result is negative, the equilibrium is unstable.

 

[haruspex]

Mathematically, one evaluates the second derivative of the potential energy at the equilibrium position.

• If the result is positive, the equilibrium is stable.
• If the result is negative, the equilibrium is unstable.

I'm not sure whether @putongren's difficulty is with the mathematical formulation or why it is correct.
At the equilibrium point, the first derivative is zero; i.e. the gradient is horizontal. If the second derivative is positive then the gradient is increasing as x increases. Since the gradient is zero at the equilibrium point, that means it is negative to the left of the point and positive to the right. Hence it is in the bottom of a dip.

There is also the possibility that the second derivative is also zero. In this case, you just have to keep taking more derivatives until you reach the first that is nonzero there.
Having reached it, it matters whether you had to differentiate an odd or even number of times. If an even number, positive indicates a minimum and negative indicates a maximum, as in the simple case discussed above.
But if an odd number of differentiations then it is neither a minimum nor a maximum. Instead, it is a "saddle", a kind of inflexion point.

E.g. consider $y=x^3$. The first nonzero derivative at x=0 is the third, making zero an inflexion. But for $y=x^4$ the first nonzero derivative at x=0 is the fourth, where it is 24, making it a minimum.