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

The potential energy of a particle in a conservative field is U =a/r^3 - b/r^2 where a and b are positive constants and r is the distance of particle from the centre of field. For equilibrium, the value of r is

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.

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, r is the distance from the centre of the field. Find: (a) the value of r_0 corresponding to the equilibrium position of the particle, examine where this position is steady, (b) the maximum magnitude of the attraction force, draw the plots U(r) and F_r(r) (the projections of the force on the radius vector r).

The potential energy function of a particle is given by U(r)=A/(2r^(2))-B/(3r) , where A and B are constant and r is the radial distance from the centre of the force. Choose the correct option (s)