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