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 equilibirum, the stable equilibrium , the distance of the particle is

The potential energy of a particle in a force field is U = A/(r^2) -

1.	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 equilibirum, the stable equilibrium , the distance of the particle is
Answer» `B/(2A)` `(2A)/(B)` `A/B` `(B)/A` Solution :Here , `U = A/(r^2) = B/r` For equilibrium, `(DU)/(dr) = 0` `:. - (2A)/(r^3) + B/(r^2) = 0 "or " (2A)/(r^3) = (B)/(r^2) "or " r = (2A)/(B)` For stable equilibrium, `(dU)/(dr) = 0` `:. - (2A)/(r^3) = B/(r^2) = 0 " or " (2A)/(r^3) = B/(r^2) " or " r = (2A)/(B)` for stable equilibrium, `(d^2U)/(dr^2) > 0` `(d^2 U)/(dr^2) = (6A)/(r^4) - (2B)/(r^3)` `(d^2U)/(dr^2)\|_(r = (2A//B)) = (6AB^4)/(16A^4) = (2B^4)/(8A^3) = (B^4)/(8A^3) > 0`. So for stable EQUILIBIRIUM, the DISTANCE of the particle is `(2A)/(B)`.