1D example of the relation between force and potential energy

For a simple one dimensional example of the relationship between force and potential energy, assume that the potential energy of a particle is given by the equation

$\displaystyle U(x) = -\frac{A}{x} \left [ 1 + B e^{-x/c} \right]
$

The realtionship between force and potential energy is

$\displaystyle \mathbf{F} = -\nabla U$ (1)

For our 1D example, where the potential energy is dependent only on the position, $ x$

$\displaystyle F = -\frac{dU}{dx}$

Taking the derivative yields

$\displaystyle \frac{dU}{dx} = \frac{A}{x^2} \left [ 1 + B e^{-x/c} \right ] - \frac{A}{x} \left [ -\frac{B}{C}e^{-x/c} \right ]$

Therefore, the force on the particle is governed by the equation

$\displaystyle F = -\frac{A}{x^2} \left [ 1 + B e^{-x/c} \right ] + \frac{A}{x} \left [ -\frac{B}{C}e^{-x/c} \right ]$ (2)

If we further assume that the constant C is so much larger than x, the force will simplify. Rearraning to get

$\displaystyle F = -\frac{A}{x^2} \left [ 1 + B e^{-x/c} -\frac{Bx}{C}e^{-x/c} \right ]
$

Because $ x \ll C$, $ e^{-x/c} \Rightarrow e^0 = 1$, we get

$\displaystyle F = -\frac{A}{x^2} \left [ 1 + B - \frac{Bx}{C} \right ] $

Further simplifying $ \frac{Bx}{C} = 0$ gives the force under this assumption

$\displaystyle F = -\frac{A}{x^2} \left [ 1 + B \right ]$ (3)



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