Question

The potential energy of a 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. $\frac{B}{2A}$
• B. $\frac{2A}{B}$
• C. $\frac{A}{B}$
• D. $\frac{B}{A}$

Answer 1

The Correct Option is B

For Stable equilibrium, $F =-\frac{ d U }{ dr}=0$
we get $r =2 A / B$