The potential energy of a particle in a certain field is given by $U = \frac{a}{r^{2}} - \frac{b}{r},$ where a and b are positive constants and r is the distance from the centre of the field. The distance of particle in the stable equilibrium position is

\frac{a}{b}

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- \frac{a}{b}

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\frac{2 a}{b}

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- \frac{2 a}{b}

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Solution

The correct option is C $\frac{2 a}{b}$
$U = \frac{a}{r^{2}} - \frac{b}{r}$
$F = \frac{d u}{d r} = - \frac{2 a}{r^{3}} + \frac{b}{r^{2}}$
For equilibrium
$F = 0 \Rightarrow - \frac{2 a}{r^{3}} + \frac{b}{r^{2}} = 0$
$- \frac{2 a}{r^{3}} + \frac{b}{r^{2}} \Rightarrow r = \frac{2 a}{b}$

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