# 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 =?

## (1) $\frac{3 a}{b}$ (2) $\frac{2 b}{3 a}$ (3) $\frac{2 a}{3 b}$ (4) $\frac{3 b}{2 a}$

Sep 7, 2017

(4) $x = \frac{3 b}{2 a}$

#### Explanation:

You are looking for a point where the force on the system goes to zero.

$F \left(x\right) = - \frac{d}{\mathrm{dx}} U \left(x\right)$

$- \frac{d}{\mathrm{dx}} \left(\frac{a}{x} ^ 2 - \frac{b}{x} ^ 3\right) = \frac{- 3 b + 2 a x}{x} ^ 4$

By inspection notice that the numerator will be zero when both of those terms are equal. One of the terms includes an $x$. Determine when:

$3 b = 2 a x$

$\frac{3 b}{2 a} = x$

The functional form of $F \left(x\right)$ will look like this:
graph{(-3 + 2 x)/x^4 [-1, 5, -0.1, 0.2]}
However the particular magnitudes will change depending on the constants $a$ and $b$. Here I've set them both to $1$. The equation crosses zero at a point where $x = \frac{3}{2}$.

The graph of $U \left(x\right)$ looks like this:
graph{1/x^2 - 1/x^3 [-1, 5, -0.1, 0.2]}