The potential energy U of a particle in a field varies with position r as `U = a/(r^2) - b/r`
where a and b are positive constants. Find the positions, where the particle:will be in stable equilibrium.

Explanation

If no force acts on a particie, it will be in stable equilibrium, i.e., F=0 or
`F = - (dU)/(dr) = 0`
`implies - d/(dr) (a/(r^2) - b/r) = 0`
`implies (2a)/(r^3) - b/(r^2) = 0 implies r = r_0 = (2a)/b`
Stable equilibrium corresponds to minimum potential energy,
i.e., `(dU)/(dr) = 0` and `(d^2 U)/(dr^2) gt 0` . If `r = r_0 = (2a)/(b), (dU)/(dr) = 0`, then U can be minimum or maximum.
If `(d^2 U)/(dr^2) > 0 ` at `r = r_0 U` will be minimum
Now, `(dU)/(dr) = -(2a)/(r^3) + b/(r^2)`
`therefore = (d^2 U)/(dr^2) = (6a)/(r^4) - (2b)/(r^3)`
At `r = r_0 = (2a)/b`
`((d^2 U)/(dr^2))_(at r = r_0) = 6a xx (b/(2a))^4 - 2b (b/(2a))^3`
`= 3/8 (b^4)/(a^3) - (b^4)/(4a^3) = (b^4)/(8a^3)`, which is positive.
Note: For stable equilibrium U is minimum and for unstable equilibrium, U is maximum.
For stable equilibrium, `(dU)/(dr) = 0 ` and `(d^2 U)/(dr^2) > 0`
For unstable equilibrium, `(dU)/(dr) = 0` and `(d^2 U)/(dr^2) lt 0`