PlanetPhysics/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

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

The relationship between force and potential energy is

$$ \mathbf{F} = -\nabla U $$

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

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

Taking the derivative yields

"$ \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

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

If we further assume that the constant C is so much larger than x, the force will simplify. Rearranging to get $$ F = -\frac{A}{x^2} \left [ 1 + B e^{-x/c} -\frac{Bx}{C}e^{-x/c} \right ] $$

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

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

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

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