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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

`a/b`

B

`b/a`

C

`(2a)/(b)`

D

`(2b)/(a)`

Text Solution

Verified by Experts

The correct Answer is:
C
`(dU)/(dr)=0,-(2a)/(r^3)+(b)/(r^2)=0,r=(2a)/(b)`
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Knowledge Check

  • 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
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    `(2A)/(B)`
    C
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    D
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