A particle located in one dimensional potential field has potential energy function`U(x)=(a)/(x^(2))-(b)/(x^(3))`, where a and b are positive constants. The position of equilibrium corresponds to x equal to

Options

A. `(3a)/(2b)`
B. `(2b)/(3a)`
C. `(2a)/(3b)`
D. `(3b)/(2a)`

Answer

To find the position of equilibrium for a particle in a one-dimensional potential field with the potential energy function \( U(x) = \frac{a}{x^2} - \frac{b}{x^3} \), we need to follow these steps:

### Step-by-Step Solution:

1. **Understand the Condition for Equilibrium**:
The position of equilibrium corresponds to the point where the net force acting on the particle is zero. The force \( F \) can be derived from the potential energy \( U(x) \) using the relation:
\[ F = -\frac{dU}{dx} \]

2. **Differentiate the Potential Energy Function**:
We need to differentiate \( U(x) \) with respect to \( x \):
\[ U(x) = \frac{a}{x^2} - \frac{b}{x^3} \]
To find \( \frac{dU}{dx} \), we apply the power rule:
\[ \frac{dU}{dx} = -\frac{d}{dx}\left(\frac{a}{x^2}\right) + \frac{d}{dx}\left(\frac{b}{x^3}\right) \]
This gives:
\[ \frac{dU}{dx} = -\left(-2a x^{-3}\right) + \left(-3b x^{-4}\right) = \frac{2a}{x^3} - \frac{3b}{x^4} \]

3. **Set the Force to Zero**:
For equilibrium, we set the force \( F \) to zero:
\[ F = -\frac{dU}{dx} = 0 \]
This leads to:
\[ \frac{2a}{x^3} - \frac{3b}{x^4} = 0 \]

4. **Solve for \( x \)**:
Rearranging the equation gives:
\[ \frac{2a}{x^3} = \frac{3b}{x^4} \]
Cross-multiplying yields:
\[ 2a x^4 = 3b x^3 \]
Dividing both sides by \( x^3 \) (assuming \( x \neq 0 \)):
\[ 2a x = 3b \]
Solving for \( x \):
\[ x = \frac{3b}{2a} \]

5. **Conclusion**:
The position of equilibrium corresponds to:
\[ x = \frac{3b}{2a} \]

### Final Answer:
The position of equilibrium corresponds to \( x = \frac{3b}{2a} \).

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