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

To find the distance at which the potential energy results in stable equilibrium, we need to analyze the given potential energy function:

U = \frac{A}{r^2} - \frac{B}{r}

In physics, for a particle to be in stable equilibrium in a potential field, the first derivative of the potential energy with respect to r must be zero, and the second derivative must be positive. Let's apply these conditions step-by-step:

1. Calculate the first derivative of U with respect to r and set it to zero to find the critical points:

   \frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2} = 0

   To solve for r, multiply the entire expression by r^3 to clear the fractions:

   -2A + Br = 0

   Solve for r:

   Br = 2A

   r = \frac{2A}{B}

2. Calculate the second derivative of U to determine the nature of the critical point:

   \frac{d^2U}{dr^2} = \frac{6A}{r^4} - \frac{2B}{r^3}

   Substitute r = \frac{2A}{B} into the second derivative:

   \frac{d^2U}{dr^2} = \frac{6A}{\left(\frac{2A}{B}\right)^4} - \frac{2B}{\left(\frac{2A}{B}\right)^3}

   Simplify the expression to ensure it is positive:

   If you simplify, you will find that the expression is positive, confirming that r = \frac{2A}{B} is indeed a point of stable equilibrium.

Therefore, for stable equilibrium, the correct distance of the particle is \frac{2A}{B}.