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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

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The correct Answer is:
B
(b) Given, the potential energy of a particle in a force field
`U=(A)/(r^(2))-(B)/(r )`
For stable equilibrium, `F=-(dU)/(dr)=0`
`o=-(2A)/(r^(3))+(B)/(r^(2))" or "(2A)/(r )=B`
The distance of particle from the centre of the field
`r=(2A)/(B)`
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