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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 equilibirum, the 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
Here , `U = A/(r^2) = B/r`
For equilibrium, `(dU)/(dr) = 0`
`:. - (2A)/(r^3) + B/(r^2) = 0 " or " (2A)/(r^3) = (B)/(r^2) " or " r = (2A)/(B)`
For stable equilibrium, `(dU)/(dr) = 0`
`:. - (2A)/(r^3) = B/(r^2) = 0 " or " (2A)/(r^3) = B/(r^2) " or " r = (2A)/(B)`
for stable equilibrium, `(d^2U)/(dr^2) > 0`
`(d^2 U)/(dr^2) = (6A)/(r^4) - (2B)/(r^3)`
`(d^2U)/(dr^2)|_(r = (2A//B)) = (6AB^4)/(16A^4) = (2B^4)/(8A^3) = (B^4)/(8A^3) > 0`.
So for stable equilibirium, the distance of the particle is `(2A)/(B)`.
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