Question

Potential energy of a system of particles is $U=\frac{\alpha }{3r^3}-\frac{\beta }{2r^2}$, where $r$ is distance between the particles. Here $\alpha$ and $\beta$ are positive constants. Which of the following are correct for the system?

• A. Equilibrium separation between the particles is $\frac{\alpha }{\beta }$
• B. For $r=\frac{\alpha }{\beta }$, the equilibrium is unstable
• C. For $r=\frac{\alpha }{\beta }$, the equilibrium is neutral.
• D. Work required to slowly move the particles to infinite separation from initial equilibrium position is $\frac{\beta }{\alpha }$

Answer 1

The correct option is A Equilibrium separation between the particles is $\frac{\alpha }{\beta }$

Given: $U=\frac{\alpha }{3r^3}-\frac{\beta }{2r^2}$
For equilibrium condition, $F=0$
$\Rightarrow -\frac{dU}{dr}=0$
$\Rightarrow -[\frac{-3\alpha }{3r^4}-\frac{-2\beta }{2r^3}]=0$
$\Rightarrow \frac{\alpha }{r^4}-\frac{\beta }{r^3}=0$
$\Rightarrow r_{\text{eq}}=\frac{\alpha }{\beta }$

Now,
$\frac{d^{2}U}{d{r}^2}=\frac{+4\alpha }{r^5}-\frac{3\beta }{r^4}$
At $r=\frac{\alpha }{\beta }$,
$\frac{d^{2}U}{d{r}^2}=\frac{4\alpha {\beta }^5}{{\alpha }^5}-\frac{3\beta {\beta }^4}{{\alpha }^4}=\frac{{\beta }^5}{{\alpha }^4}$
Since $\frac{d^{2}U}{d{r}^2}>0$
$⟹$stable equilibrium (as P.E is minimum)

Work required to slowly move the particle from $r=\frac{\alpha }{\beta }$ to $\infty$
$=U_{\infty }-U_{r=\frac{\alpha }{\beta }}$
$=0-\frac{-{\beta }^3}{6{\alpha }^2}$
$=\frac{{\beta }^3}{6{\alpha }^2}$

$\therefore$ Option $A$ is correct.