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 particles is :
Answer & explanation
Correct answer: option 4
F = -\(\frac{dU}{dr} = \frac{2A}{r^3} - \frac{B}{r^2} = 0\)
⇒ r = \(\frac{2A}{B}\)
At r =\(\frac{2A}{B} , \frac{d^2U}{dr^2}\) = +ve
Hence Its a case of stable stable equilibrium.