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

• A. $\frac{B}{2A}$
• B. $\frac{2A}{B}$
• C. $\frac{A}{B}$
• D. $\frac{B}{A}$

Answer 1

Answer: (B)

Here, $U=\frac{A}{r^2}-\frac{B}{r}$
For equilibrium, $\frac{dU}{dr}=0$
$\therefore -\frac{2A}{r^3}+\frac{B}{r^2}=0$
or $\frac{2A}{r^3}=\frac{B}{r^2}$
or $r=\frac{2A}{B}$
For stable equilibrium, $d^{2}U/d^2>0$
$\frac{d^{2}U}{d{r}^2}=\frac{6A}{r^4}-\frac{2B}{r^3}$
${\frac{d^{2}U}{d{r}^2}|}_{r=(2A/B)}$
$=\frac{6A{B}^4}{16A^4}-\frac{2B^4}{8A^3}=\frac{B^4}{8A^3}>0$
So for stable equilibrium, the distance of the particle is $2A/B$.