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 center of the field. For stable equilibrium, the distance of the particle is :

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$\frac{B}{2 A}$

$\frac{2 A}{B}$

$\frac{B}{A}$

$\frac{A}{B}$

answer is D.

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Detailed Solution

$F = - \frac{dU}{dr} = - \frac{2 A}{r^{3}} + \frac{B}{r^{2}} = 0$ (for equilibrium)
So, $r = \frac{2 A}{B}$
Also at $r = \frac{2 A}{B}, \frac{d^{2} U}{{dr}^{2}} = + ve$
So stable equilibrium.

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