{"id":284,"date":"2018-04-23T07:42:28","date_gmt":"2018-04-23T07:42:28","guid":{"rendered":"http:\/\/www.milq.info\/?page_id=284"},"modified":"2026-08-08T08:22:55","modified_gmt":"2026-08-08T06:22:55","slug":"m41_mittelung_des_coulomb-potentials","status":"publish","type":"page","link":"https:\/\/www.milq.info\/en\/m41_mittelung_des_coulomb-potentials\/","title":{"rendered":"Averaging the Coulomb potential"},"content":{"rendered":"
<\/div>

<\/p>\n

Averaging the Coulomb Potential over a Sphere of Radius R<\/h3>\n

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\".<\/p>\n

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\".<\/p>\n

\"\overline{W} = \dfrac{1}{V} \cdot \displaystyle\int \overline{W} (\vec{x}) d^3x\"<\/p>\n

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:<\/p>\n

\"\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\"<\/p>\n

Evaluating the integral: \"\displaystyle\int_{0}^{R} r dr = \dfrac{1}{2} R^2\". This gives the mean energy<\/p>\n

\"\overline{W} = - \dfrac{e^2}{4\pi\epsilon_0} \cdot \dfrac{4\pi}{V} \cdot \dfrac{1}{2} R^2\".<\/p>\n

Finally, \"V = \dfrac{4\pi}{3} R^3\" must be substituted. The result is:<\/p>\n

\"\overline{W} = - \dfrac{e^2}{4\pi\epsilon_0} \cdot \dfrac{4}{\dfrac{4\pi}{3} R^3} \cdot \dfrac{1}{2} R^2\"<\/p>\n

or<\/p>\n

\"\overline{W} = - \dfrac{3}{2} \cdot \dfrac{e^2}{4\pi\epsilon_0R} = + \dfrac{3}{2} E\".<\/p>\n

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.<\/p>\n

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.<\/p>\n

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)\".<\/p>\n

\"\"<\/div>\n

The mean value is now calculated over all spherical shells, weighted by the volume of each shell:<\/p>\n

\"\overline{W} = \dfrac{1}{V} \displaystyle\sum\limits_{\text{all shells}} W (\Delta V_n) \cdot \Delta V_n\".<\/p>\n

Writing out the Coulomb potential and \"\Delta V = 4 \pi r^2 \Delta r\" gives:<\/p>\n

\"\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\".<\/p>\n

Pulling the constant terms out of the sum and cancelling \"r\":<\/p>\n

\"\overline{W} = - \dfrac{4\pi}{V} \dfrac{e^2}{4\pi\epsilon_0} \displaystyle\sum\limits_{\text{all r-values}} r \cdot \Delta r\".<\/p>\n

In the limit \"\Delta r \to 0\", this is exactly the definition of the integral:<\/p>\n

\"\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\".<\/p>\n

Substituting \"V = \dfrac{4\pi}{3} R^3\":<\/p>\n

\"\overline{W} = - \dfrac{e^2}{4\pi\epsilon_0} \dfrac{4\pi}{\dfrac{4\pi}{3}R^3} \cdot \dfrac{1}{2} R^2\".<\/p>\n

Cancelling finally leads to the result:<\/p>\n

\"\overline{W} = - \dfrac{3}{2} \dfrac{e^2}{4\pi\epsilon_0R}\".<\/p>\n

In this way, the use of volume integrals can be avoided.<\/p>\n

Back<\/a><\/p>","protected":false},"excerpt":{"rendered":"

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 over a sphere of radius . Inside the potential well, the potential has a constant value . The potential well approximates the Coulomb potential best if is… Continue reading Averaging the Coulomb potential<\/span><\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-284","page","type-page","status-publish","hentry","without-featured-image"],"yoast_head":"\nAveraging the Coulomb potential - milq<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.milq.info\/en\/m41_mittelung_des_coulomb-potentials\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Averaging the Coulomb potential - milq\" \/>\n<meta property=\"og:description\" content=\"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 over a sphere of radius . 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Inside the potential well, the potential has a constant value . 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