Question

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).

Answer 1

From the equation `F_r=-(dU)/(dr)` we get `F_r=[-(2a)/(r^3)+(b)/(r^2)]`
(a) we have at `r=r_0`, the particle is in equilibrium position. i.e. `F_r=0` so, `r_0=(2a)/(b)`
To check, whether the position is steady (the position of stable equilibrium), we have to satisfy
`(d^2U)/(dr^2)gt0`
We have `(d^2U)/(dr^2)=[(6a)/(r^4)-(2b)/(r^3)]`
Putting the value or `r=r_0=(2a)/(b)`, we get
`(d^2U)/(dr^2)=(b^4)/(8a^3)`, (as a and b are positive constant)
So, `(d^2U)/(dr^2)=(b^2)/(8a^3)gt0`,
which indicates that the potential energy of the system is minimum, hence this position is steady:
(b) We have `F_r=-(dU)/(dr)=[-(2a)/(r^3)+(b)/(r^2)]`
For F, to be maximum, `(dF_r)/(dr)=0`
So, `r=(3a)/(b)` and then `F_(r(max))=(-b^3)/(27a^2)`,
As `F_r` is negative, the force is attractive.