The potential energy of a particle in a certain field has the form `U=(a//r^2)-(b//r)` , where a and b are positive constants and r is the distance from the centre of the field. Find the value of `r_0` corresponding to the equilibrium position of the particle, examine whether this position is stable.

Explanation

(a) `U(r) = a/r^(2) - b/r `
At equilibrium, F = -` (dU)/(dr) = 0`
`F = -(br - 2a)/(r^(3))`
Hence `br - 2a = 0` at equilibrium, ` r - r_(0) = 2a/b` corresponds to equilibrium.
(b) At stable equilbrium , the potential energy is minimum and at unstable equilbrium, it is maximum. From calculus, we known that for minimum value around a point ` r - r_(0)` the first derivative should be zero and the second derivative should be positive. For minimum potential energy,
`(dU)/(dr) = 0 and (d^(2)U)/(dr^(2)) gt at r = r_(0)`
At ` r = r_(0) = 2a/b`
`(d^(2)U)/(dr^(2)) = (6a - 2br_(0))/(r_(0)^(4)) = (2a)/(r_(0)^(4)) gt 0`
Hence the potential energy function `U(r)` has a minimum value at `r_(0) = (2a)/(b)` The system has a stable equilibrium at minimum potential energy state.