The potential energy of a particle in a force field is:
`U = (A)/(r^(2)) - (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//A`
B. `B//2A`
C. `2A//B`
D. `A//B`

Explanation

To find the distance of a particle from the center of a force field for stable equilibrium, we start with the given potential energy function:

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

where \( A \) and \( B \) are positive constants, and \( r \) is the distance from the center of the field.

### Step 1: Find the expression for force
The force \( F \) acting on the particle can be derived from the potential energy \( U \) using the relation:

\[ F = -\frac{dU}{dr} \]

### Step 2: Differentiate the potential energy
Now, we differentiate \( U \) with respect to \( r \):

\[ \frac{dU}{dr} = \frac{d}{dr}\left(\frac{A}{r^2} - \frac{B}{r}\right) \]

Using the power rule for differentiation:

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

### Step 3: Set the force to zero for stable equilibrium
For stable equilibrium, the net force acting on the particle must be zero:

\[ F = -\frac{dU}{dr} = 0 \]

This leads us to:

\[ -\left(-\frac{2A}{r^3} + \frac{B}{r^2}\right) = 0 \]

Thus, we have:

\[ \frac{2A}{r^3} = \frac{B}{r^2} \]

### Step 4: Solve for \( r \)
Now, we can cross-multiply to solve for \( r \):

\[ 2A = Br \]

Rearranging gives:

\[ r = \frac{2A}{B} \]

### Conclusion
The distance of the particle from the center of the field for stable equilibrium is:

\[ r = \frac{2A}{B} \]