• # question_answer 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 A)  B/A          B)  B/2AC)  2A/B         D)  A/B

Force experienced by the particle in field, $F=-\left( \frac{dU}{dr} \right)=-\left( \frac{-2A}{{{r}^{3}}}+\frac{B}{{{r}^{2}}} \right)$ At equilibrium $F=0\,F=-\left( \frac{-2A}{{{r}^{3}}}+\frac{B}{{{r}^{2}}} \right)=0$ $\Rightarrow$$r=\frac{2A}{B}$