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 isA. `A//B`B. `B//A`C. `B//2A`D. `2A//B`

The potential energy of a particle in a force field is: `U = (A)/(r^(2

1.	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 isA. `A//B`B. `B//A`C. `B//2A`D. `2A//B`
Answer» Correct Answer - D For Stable equilibrium, `F=(dU)/(dr)=0` we get `r=2A//B`

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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 isA. `A//B`B. `B//A`C. `B//2A`D. `2A//B`

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