Question

Find the value of r for equilibrium position?

Answer 1

Answer :

Potential energy in a field is given by

$\dag\:\boxed{\bf{U=-\dfrac{A}{r^6}+\dfrac{B}{r^{11}}}}$

We have to find the value of r at equilibrium position$.$

◈ We know that in equilibrium position, net force acts on the particle = zero.

For conservative force, $\bf{F=-\dfrac{dU}{dr}}$

$:\implies\sf\:-\dfrac{dU}{dr}=0$

$:\implies\sf\:-{\huge(}-\dfrac{A}{r^6}+\dfrac{B}{r^{11}}{\huge{)}}\:\dfrac{1}{dr}=0$

$:\implies\sf\:-(-Ar^{-6}+Br^{-11})\:\dfrac{1}{dr}=0$

$:\implies\sf\:\dfrac{Ar^{-6}}{dr}-\dfrac{Br^{-11}}{dr}=0$

$:\implies\sf\:(-6Ar^{-7})-(-11Br^{-12})=0$

$:\implies\sf\:\dfrac{11B}{r^{12}}=\dfrac{6A}{r^{7}}$

$:\implies\sf\:\dfrac{r^{12}}{r^7}=\dfrac{11B}{6A}$

$:\implies\sf\:r^{5}=\dfrac{11B}{6A}$

$:\implies\boxed{\bf{\purple{r={\huge(}\dfrac{11B}{6A}{\huge)}^{\frac{1}{5}}}}}$

Cheers!