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).

Options

A. `2a/b`
B. `2a/b`
C. `-a/b`
D. `a/b`

Answer

`(1)`

Explanation

For stable equilibrium `(Du)/(dr) = 0 `
`U = (a)/(r^(2)) - (b) /(r)`
`(dU)/(dr) = a(-2)r^(3) + br^(-2) = 0`
`(2a)/(r^(3)) = (b)/(r^(2)) Rightarrow r = (2a)/b`