The potential energy of a particle in a force field is $U=\frac{A}{r^{2}}-\frac{B}{r}$ where $A$ and

Question

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 the particle from the centre of the field. For stable equilibrium, the distance of the particle is\

Professor PC Thomas Classes · Answer

(b) : Here, $U=\frac{A}{r^{2}}-\frac{B}{r}$ For equilibrium, $\frac{d U}{d r}=0$\ $	herefore \quad-\frac{2 A}{r^{3}}+\frac{B}{r^{2}}=0$ or $\frac{2 A}{r^{3}}=\frac{B}{r^{2}} \quad$ or $\quad r=\frac{2 A}{B}$\