Question

The potential energy of a particle in a central field has the form $U(r)=\frac{1}{r^2}-\frac{1}{r}$, where ' $r^{\prime }$ is the distance form the centre of the field. The magnitude of the maximum attractive force in Newton is

• A. $\frac{1}{27}$
• B. $\frac{1}{9}$
• C. $\frac{1}{3}$
• D. 1

Answer 1

Answer: (A)

Given, equation of potential energy,
$U(r)=\frac{1}{r^2}-\frac{1}{r}$
As we know that, attractive force,
$F=-\frac{dU}{dr}$
$F=-\frac{d}{dr}\left(\frac{1}{r^2}-\frac{1}{r}\right)$
$=-\frac{d}{dr}\left(r^{-2}-r^{-1}\right)$
$=-\left(-2r^{-3}+r^{-2}\right)$
$=\left(2r^{-3}-r^{-2}\right)\dots \text{ (i) }$
Now, for $F_{\max }$ i.e. maximum value of force,
$\frac{dF}{dr}=0$
$\frac{d}{dr}\left(2r^{-3}-r^{-2}\right)=0$
$-6r^{-4}+2r^{-3}=0$
$2r^{-3}=6r^{-4}$
$\therefore 1=3r^{-1}\Rightarrow r=3$Substituting in Eq. (i), we get
$F=\frac{2}{r^3}-{\frac{1}{r^2}|}_{r=3}$
$=\frac{2}{3^2}-\frac{1}{3^2}$
$=\frac{2}{27}-\frac{1}{9}$
$=\frac{2}{27}-\frac{3}{27}$
$=-\frac{1}{27}$