Question

The potential energy of a particle in a certain field has the form $U=\frac{a}{r^2}-\frac{b}{r},$ where $a$ and $b$ are postive constants, $r$ is the distance from the centre of the field.

• A. the value of $r_0$ corresponding to the equilibrium position of the particle is $\frac{a}{b}$
• B. Particle is in unstable equiltibrium.
• C. the minimum attractive force is $-\frac{b^2}{27a^3}$
• D. the value of $r_0$ corresponding to the equilibrium position of the particle is $\frac{2a}{b}$

Answer 1

Answer: (C), (D)

The correct options are
C the minimum attractive force is $-\frac{b^2}{27a^3}$
D the value of $r_0$ corresponding to the equilibrium position of the particle is $\frac{2a}{b}$

$U=\frac{a}{r^2}-\frac{b}{r}$

$F=-\frac{dU}{dr}$

$U=a{r}^{-2}-b{r}^{-1}$

$-\frac{dU}{dr}=-\left[a\right(-2)r^{-2-1}-b(-1\left)r^{-1-1}\right)]$

$=-[-2a{r}^{-3}+b{r}^{-2}]$

$F=\frac{2a}{r^3}-\frac{b}{r^2}$
At equilibrium,
$F=0$
Hence,
$\frac{2a}{r_0^3}-\frac{b}{r_0^2}=0$

$\Rightarrow \frac{1}{r_0^2}(\frac{2a}{r_0}-b)=0$

$\Rightarrow \frac{2a}{r_0}-b=0$

$\Rightarrow \frac{2a}{r_0}=b\Rightarrow r_0=\frac{2a}{b}$
Hence, option (d) is correct

For option (b)

$U=\frac{a}{r^2}-\frac{b}{r}=a{r}^{-2}-b{r}^{-1}$

$\frac{dU}{dr}=a(-2)r^{-2-1}-b(-1)r^{-1-1}$

$=-2a{r}^{-3}+b{r}^{-2}$

$\frac{d^{2}U}{d{r}^2}=-2a(-3)(r{)}^{-4}+(-2)b{r}^{-3}$

$\frac{d^{2}U}{d{r}^2}=\frac{6a}{r^4}-\frac{2b}{r^3}$

at $r_0=\frac{2a}{b}$

$=\frac{6a}{(2a{)}^4}\times b^4-\frac{2b}{(2a{)}^3}\times b^3$

$=\frac{6a{b}^4}{16a^4}-\frac{2b^4}{8a^3}$

$=\frac{b^4}{a^3}(\frac{3}{8}-\frac{1}{4})$

$=\frac{b^4}{a^3}\left(\frac{3-2}{8}\right)=\frac{1}{8}\frac{b^4}{a^3}$

This is positive value, so it is in stable equilibrium.Hence option (b) is incorrect.


For option (c)
We have $F=\frac{2a}{r^3}-\frac{b}{r^2}$
for maximum or minimum value of force,
$\frac{dF}{dr}=0$

$\begin{array}{cc}\frac{dF}{dr}& =2a(-3)r^{-4}-(-2)b{r}^{-3}=0\\ & =-\frac{6a}{r^4}+\frac{2b}{r^3}=0\end{array}$

$⟹r_m=\frac{3a}{b}$

$\frac{d^{2}F}{dr}=\frac{d}{dr}(-6a{r}^{-4}+2b{r}^{-3})=-6(-4)a{r}^{-5}+2(-3)b{r}^{-4}$

$=\frac{6}{r^4}(\frac{4a}{r}-b)$

$\frac{4a}{r}-b$ is positive for $r_m=\frac{3a}{b}$ which says it is a local minima.
We have $r_m=\frac{3a}{b}$. So,

$\begin{array}{cc}{F}_{r_m}& =\frac{2a}{r_0^3}-\frac{b}{b_0^2}\\ & =\frac{2a\times b^3}{(3a{)}^3}-\frac{b}{(3a{)}^2}\times b^2\\ & =\frac{2a{b}^3}{27a^3}-\frac{b^3}{9a^2}\end{array}$

$=\frac{-b}{27a^3}$
this tells us that it is an attractive force.

For detailed solution watch the next video.