NEET Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति NEET PYQ-Work Energy Power and Collision

The potential energy of a particle in a force field is \[U=\frac{A}{{{r}^{2}}}-\frac{A}{r},\] 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[AIPMT (S) 2012]

A) B/2A

B) 2A/B

C) A/B

D) B/A

Correct Answer: B

Solution :

Given, the potential energy of a particle in a force field \[U=\frac{A}{{{r}^{2}}}-\frac{B}{{{r}^{1}}}\]

For stable equilibrium, \[F=-\frac{dU}{dr}=0\]

\[0=-\frac{2A}{{{r}^{3}}}+\frac{B}{{{r}^{2}}}\] or \[\frac{2A}{r}=B\]

The distance of particle from the centre of the field \[r=\frac{2A}{B}\].