The potential energy of particle in a force field is U = A/r² – b/r, where A and B

⇦

⇨

The potential energy of particle in a force field is U = A/r² – 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:

Options

(a) B / 2A
(b) 2A / B
(c) A / B
(d) B / A

Correct Answer:

2A / B

Explanation:

for equilibrium dU/ dr = 0
-2A/r³ + B/r² = 0 r = 2A / B
for stable equilibrium d²U /dr² should be positive for the value of r.
here d²U / dr² = 6A/r⁴ – 2B/r³ is +ve value for r = 2A / B So.

Related Questions:

1. The radii of circular orbits of two satellites A and B of the earth, are 4R and R
2. The magnetic flux linked with a coil at any instant t is given by ?, find emf
3. A Gaussian surface in the cylinder of cross-section πa² and length L is immersed
4. An electric dipole of dipole moment p is placed in a uniform external electric field
5. How much work is required to carry a 6 μC charge from the negative terminal

Topics: Electrostatics (146)
Subject: Physics (2479)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends