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

B

Explanation

The potential energy of particle in a force field is
`U=(A)/(r^(2))-(B)/(r)`
`(dU)/(dr)=-(2A)/(r^(3))+(B)/(r^(2))`
For equilibrium, `(dU)/(dr)=0` or`-(2A)/(r^(3))+(B)/(r^(2))=0`
or `r=(2A)/(B)`
The equilibrium will be stable, if `(d^(2)U)/(dr^(2))=+` for this value of `r` .
Now, `(d^(2)U)/(dr^(2))=(d)/(dr)(-(2A)/(r^(3))+(B)/(r^(2)))=((6A)/(r^(4))-(2B)/(r^(3)))`, which is positive for `r=2A//B`
Hence choice (b) is correct.