Potential energy in a force field – Rankers Physics

Potential Energy & Equilibrium: Practice Problem & Solution

The potential energy of particle in a force field is \(U = \frac{A}{r} - \frac{B}{r^2}\) 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: (2012 Pre)
\(B/2A\)
\(2A/B\)
\(A/B\)
\(B/A\)

Solution Explained:

To solve this problem, we apply the core principles of Potential Energy & Equilibrium. Understanding the underlying formula is key to arriving at the correct answer below:

For equilibrium, the force is zero: \(F = -\frac{dU}{dr} = 0\). Given \(U = \frac{A}{r} - \frac{B}{r^2}\), if the question implicitly means \(U = \frac{A}{r^2} - \frac{B}{r}\), then \(\frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2}\). Setting \(\frac{dU}{dr} = 0\) gives \(\frac{2A}{r^3} = \frac{B}{r^2} \Rightarrow r = \frac{2A}{B}\). Checking stability, \(\frac{d^2U}{dr^2} = \frac{6A}{r^4} - \frac{2B}{r^3}\), which is positive at \(r = \frac{2A}{B}\).

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