The potential energy of a system of two particles is given by $U\left(x\right)=\frac{a}{x^2}-\frac{b}{x}$ . Find the minimum potential energy of the system, where $x$ is the distance of separation and $a,b$ are positive constants.

The correct option is A $-\frac{b^2}{4a}$
Given,

$U\left(x\right)=\frac{a}{x^2}-\frac{b}{x}.........\left(1\right)$

$\Rightarrow \frac{dU\left(x\right)}{dx}=\frac{d}{dx}[\frac{a}{x^2}-\frac{b}{x}]=[\frac{-2a}{x^3}+\frac{b}{x^2}]$

W know that, extrema occurs at $\frac{dU\left(x\right)}{dx}=0$

$\Rightarrow (-\frac{2a}{x^3}+\frac{b}{x^2})=0$

$\Rightarrow \frac{2a}{x^3}=\frac{b}{x^2}$ $\Rightarrow x=\frac{2a}{b}$

As we know,

For maxima: $\frac{d^{2}U}{d{x}^2}<0$

For minima: $\frac{d^{2}U}{d{x}^2}>0$

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

$\Rightarrow {\left[\frac{d^{2}U}{d{x}^2}\right]}_{x=\frac{2a}{b}}=\frac{6a}{{\left(\frac{2a}{b}\right)}^4}-\frac{2b}{{\left(\frac{2a}{b}\right)}^3}$

$\Rightarrow {\left[\frac{d^{2}U}{d{x}^2}\right]}_{x=\frac{2a}{b}}=\frac{b^4}{8a^3}>0[\because \text{a,b are positive constants}]$

$\therefore U(x{)}_{min}=\frac{a}{{\left(\frac{2a}{b}\right)}^2}-\frac{b}{\left(\frac{2a}{b}\right)}=-\frac{b^2}{4a}$

Hence, $\left(A\right)$ is the correct answer.
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