The potential energy of particle in a force field is U = A / r^2 − B / r , where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is ?

The potential energy of particle in a force field is  \( U = \frac{A}{r^2} - \frac{B}{r} \) , where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is ?

(a) B/2A

(b) 2A/B

(c) A/B

(d) B/A

Answer :  (b) 2A/B

\( \text{Stable equilibrium} \rightarrow \text{at } U_{\min} \ (\text{minimum potential energy}) \)

\( U = \frac{A}{r^2} - \frac{B}{r} \)

\( \frac{dU}{dr} = A(-2)r^{-3} - B(-1)r^{-2} \)

\( = \frac{-2A}{r^3} + \frac{B}{r^2} = 0 \)

\( \frac{-2A}{r^3} + \frac{B}{r^2} = 0 \)

\( \Rightarrow \frac{2A}{r} = B \)

\( r = \frac{2A}{B} \)