The potential energy of a particle in a force field is $U = \frac{A}{r^{2}} - \frac{B}{r}$ 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
1)A/B
2)B/A
3)B/2A
4)2A/B

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Solution

Potential energy is given as; $U = \frac{A}{r^{2}} - \frac{B}{r}$
For condition of stable equilibrium force should be zero or change in potentia energy will be zero.
$F = - \frac{d U}{d r} = 0$
$F = \frac{d \frac{A}{r^{2}}}{d r} - \frac{d \frac{B}{r}}{d r}$
$0 = - 2 A r^{- 3} - (- B r^{- 2})$

$- 2 A r^{- 3} = (- B r^{- 2})$
$\frac{2 A}{B} = r$
So, For condition of stable equilibrium diustance should be,
$\frac{2 A}{B} = r$

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