Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Work Power Energy

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 particles is :

Options:

A/B

B/A

B/2A

2A/B

Correct Answer:

2A/B

Explanation:
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.