In a conservative force field we can find the radial component of force from the potential energy function by using `F = -(dU)/(dr)`. Here, a positive force means repulsion and a negative force means attraction. From the given potential energy function U(r ) we can find the equilibrium position where force is zero. We can also find the ionisation energy which is the work done to move the particle from a certain position to infinity.
Let us consider a case in which a particle is bound to a certain point at a distance r from the centre of the force. The potential energy of the particle is : `U(r )=(A)/(r^(2))-(B)/(r )` where r is the distance from the centre of the force and A andB are positive constants. Answer the following questions.
The equilibrium is

Options

A. stable
B. unstable
C. neutral
D. cannot be predicted

Answer

To determine the type of equilibrium for the given potential energy function \( U(r) = \frac{A}{r^2} - \frac{B}{r} \), we will follow these steps:

### Step 1: Find the Force from the Potential Energy Function
The force \( F \) in a conservative force field can be calculated using the formula:
\[ F = -\frac{dU}{dr} \]
Given \( U(r) = \frac{A}{r^2} - \frac{B}{r} \), we need to differentiate \( U \) with respect to \( r \).

### Step 2: Differentiate the Potential Energy Function
Calculating the derivative:
\[ \frac{dU}{dr} = \frac{d}{dr}\left(\frac{A}{r^2}\right) - \frac{d}{dr}\left(\frac{B}{r}\right) \]
Using the power rule:
\[ \frac{d}{dr}\left(\frac{A}{r^2}\right) = -\frac{2A}{r^3} \]
\[ \frac{d}{dr}\left(\frac{B}{r}\right) = -\frac{B}{r^2} \]
Thus,
\[ \frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2} \]

### Step 3: Substitute into the Force Equation
Now substituting back into the force equation:
\[ F = -\left(-\frac{2A}{r^3} + \frac{B}{r^2}\right) = \frac{2A}{r^3} - \frac{B}{r^2} \]

### Step 4: Find the Equilibrium Position
At equilibrium, the force \( F \) must be zero:
\[ \frac{2A}{r^3} - \frac{B}{r^2} = 0 \]
Rearranging gives:
\[ \frac{2A}{r^3} = \frac{B}{r^2} \]
Multiplying both sides by \( r^3 \):
\[ 2A = Br \]
Thus, the equilibrium position \( r \) is:
\[ r = \frac{2A}{B} \]

### Step 5: Determine the Stability of the Equilibrium
To determine if this equilibrium is stable, we need to analyze the second derivative of the potential energy function \( U(r) \).

### Step 6: Calculate the Second Derivative
We need to find \( \frac{d^2U}{dr^2} \):
\[ \frac{d^2U}{dr^2} = \frac{d}{dr}\left(-\frac{2A}{r^3} + \frac{B}{r^2}\right) \]
Calculating the derivatives:
\[ \frac{d}{dr}\left(-\frac{2A}{r^3}\right) = 6\frac{A}{r^4} \]
\[ \frac{d}{dr}\left(\frac{B}{r^2}\right) = -2\frac{B}{r^3} \]
Thus,
\[ \frac{d^2U}{dr^2} = 6\frac{A}{r^4} + 2\frac{B}{r^3} \]

### Step 7: Evaluate the Second Derivative at Equilibrium Position
Substituting \( r = \frac{2A}{B} \) into the second derivative:
\[ \frac{d^2U}{dr^2}\bigg|_{r=\frac{2A}{B}} = 6\frac{A}{\left(\frac{2A}{B}\right)^4} + 2\frac{B}{\left(\frac{2A}{B}\right)^3} \]
This will yield positive values since \( A \) and \( B \) are positive constants.

### Conclusion
Since \( \frac{d^2U}{dr^2} > 0 \) at the equilibrium position, the equilibrium is stable.

### Final Answer
The equilibrium is stable.