Averaging the Coulomb potential

Averaging the Coulomb Potential over a Sphere of Radius R

A more exact method of adjusting the depth of the potential well consists in averaging the Coulomb potential $W$ over a sphere of radius $R$ .

Inside the potential well, the potential has a constant value $W_0$ . The potential well approximates the Coulomb potential best if $W_0$ is chosen so that it represents the “average depth” of the Coulomb potential. To this end, one calculates the mean value $\overline{W}$ of the Coulomb potential over a sphere of radius $R$ .

$\overline{W} = \dfrac{1}{V} \cdot \displaystyle\int \overline{W} (\vec{x}) d^3x$

Here $V = \dfrac{4\pi}{3} R^3$ is the volume of the sphere. Substituting $W (\vec{x}) = - \dfrac{e^2}{4\pi\epsilon_0r}$ and writing out the volume element gives:

$\overline{W} = - \dfrac{1}{V} \cdot \dfrac{e^2}{4\pi\epsilon_0} \displaystyle\int_{0}^{R} \dfrac{1}{r} \cdot 4\pir^2 dr$

Evaluating the integral: $\displaystyle\int_{0}^{R} r dr = \dfrac{1}{2} R^2$ . This gives the mean energy

$\overline{W} = - \dfrac{e^2}{4\pi\epsilon_0} \cdot \dfrac{4\pi}{V} \cdot \dfrac{1}{2} R^2$ .

Finally, $V = \dfrac{4\pi}{3} R^3$ must be substituted. The result is:

$\overline{W} = - \dfrac{e^2}{4\pi\epsilon_0} \cdot \dfrac{4}{\dfrac{4\pi}{3} R^3} \cdot \dfrac{1}{2} R^2$

$\overline{W} = - \dfrac{3}{2} \cdot \dfrac{e^2}{4\pi\epsilon_0R} = + \dfrac{3}{2} E$ .

With this relation, $W_0 = \overline{W} = \dfrac{3}{2} E$ , the depth of the potential well (i.e. the position of the bottom of the potential well below the zero level) is fixed.

The mathematics used here is somewhat complicated, since it involves a three-dimensional volume integral. In order to use the method presented here in the classroom, this elaborate mathematics has to be avoided. One possible way of doing so is shown below.

The sphere is divided into individual spherical shells of thickness $\Delta r$ (see figure). For small $\Delta r$ , the volume of a spherical shell is $\Delta V = \text{(surface area of the sphere)} \times \text{(thickness)} = 4 \pi r^2 \Delta r$ . Since the Coulomb potential is spherically symmetric, it is constant within each spherical shell (provided $\Delta r$ is small enough) and has the value $W (\Delta V)$ .

The mean value is now calculated over all spherical shells, weighted by the volume of each shell:

$\overline{W} = \dfrac{1}{V} \displaystyle\sum\limits_{\text{all shells}} W (\Delta V_n) \cdot \Delta V_n$ .

Writing out the Coulomb potential and $\Delta V = 4 \pi r^2 \Delta r$ gives:

$\overline{W} = \dfrac{1}{V} \displaystyle\sum\limits_{\text{all r-values}} \biggl( - \dfrac{e^2}{4\pi\epsilon_0r} \biggr) \cdot 4\pir^2 \cdot \Delta r$ .

Pulling the constant terms out of the sum and cancelling $r$ :

$\overline{W} = - \dfrac{4\pi}{V} \dfrac{e^2}{4\pi\epsilon_0} \displaystyle\sum\limits_{\text{all r-values}} r \cdot \Delta r$ .

In the limit $\Delta r \to 0$ , this is exactly the definition of the integral:

$\overline{W} = - \dfrac{e^2}{4\pi\epsilon_0} \dfrac{4\pi}{V} \displaystyle\int_{0}^{R} r dr = - \dfrac{e^2}{4\pi\epsilon_0} \dfrac{4\pi}{V} \dfrac{1}{2} R^2$ .

Substituting $V = \dfrac{4\pi}{3} R^3$ :

$\overline{W} = - \dfrac{e^2}{4\pi\epsilon_0} \dfrac{4\pi}{\dfrac{4\pi}{3}R^3} \cdot \dfrac{1}{2} R^2$ .

Cancelling finally leads to the result:

$\overline{W} = - \dfrac{3}{2} \dfrac{e^2}{4\pi\epsilon_0R}$ .

In this way, the use of volume integrals can be avoided.

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