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The potential energy of a particle in a ...

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

A

`B//2A`

B

`2A//B`

C

`A//B`

D

`B//A`

Text Solution

Verified by Experts

The correct Answer is:
B
`U=A/r^2-B/r,F=(-dU)/(dx)=-[-2A/r^3+B/r^2]`
At equilibrium -
`F=0implies (2A)/r^3 -B/r^2 =0 implies 1/r^2[(2A)/r-B]=0`
`r=(2A)/B`
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