The potential energy of a particle in a force field is:
`U = (A)/(r^(2)) - (B)/(r )`,. Where `A` and `B` are positive
constants and `r` is the distance of particle from the centre of the field. For stable equilibrium the distance of the particle is

Options

A. `B/2A`
B. `2A/B`
C. `A/B`
D. `B/A`

Answer

`(2)`

Explanation

Here, `u = A/(r^(2)) - B/r`
For equilibrium, `(du)/(dr) = 0`
`therefore (- 2A)/(r^(3) + B/(r^(2)) or r = (2A)/B`
For stable equilibrium, `(d^(2)u)/(dr^(2)) gt 0`
`(d^(2)u)/(dr^(2)) = (6A)/(r^(4)) - (2B)/(r^(3))`
` (d^(2)u)/(dr^(2))|_(r)=((2A)/B) = (6AB)^(4))/(16 A^(4)) - (2B^(4))/(8A^(3)) = (B^(4))/(8A^(3)) gt 0` So for stable equilibrium, the distance of the particle is `(2A)/(B)`