{"id":209583,"date":"2024-10-21T09:09:17","date_gmt":"2024-10-21T09:09:17","guid":{"rendered":"https:\/\/www.electricity-magnetism.org\/?p=209583"},"modified":"2024-10-21T09:09:17","modified_gmt":"2024-10-21T09:09:17","slug":"comment-calculer-le-champ-electrique-dun-disque-charge-uniformement","status":"publish","type":"post","link":"https:\/\/www.electricity-magnetism.org\/fr\/comment-calculer-le-champ-electrique-dun-disque-charge-uniformement\/","title":{"rendered":"Comment calculer le champ \u00e9lectrique d&rsquo;un disque charg\u00e9 uniform\u00e9ment ?"},"content":{"rendered":"<p class=\"sidekick\">Calculer le champ \u00e9lectrique d&rsquo;un disque charg\u00e9 uniform\u00e9ment : guide pas-\u00e0-pas pour comprendre les principes de l&rsquo;ing\u00e9nierie thermique et leurs applications pratiques.<\/p>\n<h2>Comment calculer le champ \u00e9lectrique d&rsquo;un disque charg\u00e9 uniform\u00e9ment ?<\/h2>\n<p>Calculer le champ \u00e9lectrique d&rsquo;un disque charg\u00e9 uniform\u00e9ment est un probl\u00e8me classique en \u00e9lectrostatique. Pour une compr\u00e9hension claire, voici les \u00e9tapes d\u00e9taill\u00e9es qui vous aideront \u00e0 d\u00e9terminer le champ \u00e9lectrique \u00e0 une distance z au-dessus du centre du disque.<\/p>\n<h2>Hypoth\u00e8ses et Notations<\/h2>\n<ul>\n<li>Le disque a un rayon R.<\/li>\n<li>La densit\u00e9 de charge superficielle est uniforme et not\u00e9e \u03c3 (sigma), exprim\u00e9e en coulombs par m\u00e8tre carr\u00e9 (C\/m<sup>2<\/sup>).<\/li>\n<li>Nous cherchons le champ \u00e9lectrique \u00e0 une distance z au-dessus du centre du disque.<\/li>\n<\/ul>\n<h2>M\u00e9thode de Calcul<\/h2>\n<p>Pour r\u00e9soudre ce probl\u00e8me, nous utilisons la superposition des contributions de petits \u00e9l\u00e9ments de charge sur le disque.<\/p>\n<ol>\n<li>Consid\u00e9rons un \u00e9l\u00e9ment de surface infinit\u00e9simal dA sur le disque. Cet \u00e9l\u00e9ment peut \u00eatre d\u00e9crit en coordonn\u00e9es polaires (r, \u03b8) o\u00f9 r est la distance radiale et \u03b8 est l&rsquo;angle azimutal.<\/li>\n<li>La charge \u00e9l\u00e9mentaire dQ sur cet \u00e9l\u00e9ment est :\n<p>dQ = \u03c3 * dA = \u03c3 * r * dr * d\u03b8<\/p>\n<\/li>\n<li>La contribution au champ \u00e9lectrique, dE, \u00e0 une distance z au-dessus du centre due \u00e0 cette charge \u00e9l\u00e9mentaire est donn\u00e9e par la loi de Coulomb :\n<p>dE = \\(\\frac{k_e * dQ}{r^2 + z^2}\\)<\/p>\n<\/li>\n<li>L&rsquo;\u00e9l\u00e9ment de champ \u00e9lectrique dE a des composants \u00e0 la fois dans la direction z (verticale) et dans le plan. Cependant, gr\u00e2ce \u00e0 la sym\u00e9trie, les composants horizontaux s&rsquo;annulent, ne laissant que la composante verticale dE<sub>z<\/sub> qui contribue au champ total.\n<p>dE<sub>z<\/sub> = dE * \\(\\frac{z}{\\sqrt{r^2 + z^2}}\\) = \\(\\frac{k_e * dQ}{(r^2 + z^2)^{3\/2}} * z\\)<\/p>\n<\/li>\n<li>En int\u00e9grant dE<sub>z<\/sub> sur la surface du disque enti\u00e8re, nous obtenons :\n<p>E<sub>z<\/sub> = \\(\\int_{disk} \\frac{k_e * \\sigma * r * dr * d\u03b8 * z}{(r^2 + z^2)^{3\/2}}\\)<\/p>\n<\/li>\n<\/ol>\n<h3>Int\u00e9gration Finale<\/h3>\n<p>Pour effectuer cette int\u00e9gration, nous devons d&rsquo;abord int\u00e9grer sur \u03b8 de 0 \u00e0 2\u03c0, puis sur r de 0 \u00e0 R :<\/p>\n<p>E<sub>z<\/sub> = 2\u03c0\u03c3z \\(\\int_0^R \\frac{r * dr}{(r^2 + z^2)^{3\/2}}\\)<\/p>\n<p>Utilisons la substitution u = r^2 + z^2, donc du = 2r dr. Cela nous permet de r\u00e9\u00e9crire l&rsquo;int\u00e9grale :<\/p>\n<p>\\(E_z = 2\u03c0\u03c3z \\int_{z^2}^{R^2 + z^2} \\frac{du}{2u^{3\/2}}\\)<\/p>\n<p>Cette int\u00e9grale est \u00e9valu\u00e9e comme suit :<\/p>\n<p>\\(E_z = \u03c0\u03b1\u03c3 \\left[\\frac{-2}{u^{1\/2}}\\right]_{z^2}^{R^2 + z^2}\\)<\/p>\n<p>L&rsquo;\u00e9valuation de cette int\u00e9grale donne :<\/p>\n<p>\\(E_z = 2\u03c0\u03c3 \\left[1 &#8211; \\frac{z}{\\sqrt{R^2 + z^2}}\\right]\\)<\/p>\n<h2>Conclusion<\/h2>\n<p>Le champ \u00e9lectrique E<sub>z<\/sub> au-dessus du centre d&rsquo;un disque charg\u00e9 uniform\u00e9ment \u00e0 une distance z est donn\u00e9 par :<\/p>\n<p>\\(E_z = 2\u03c0\u03c3 \\left[1 &#8211; \\frac{z}{\\sqrt{R^2 + z^2}}\\right]\\)<\/p>\n<p>Ce r\u00e9sultat montre comment le champ \u00e9lectrique d\u00e9pend \u00e0 la fois de la densit\u00e9 de charge superficielle \u03c3, du rayon du disque R et de la distance z au-dessus du disque.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculer le champ \u00e9lectrique d&rsquo;un disque charg\u00e9 uniform\u00e9ment : guide pas-\u00e0-pas pour comprendre les principes de l&rsquo;ing\u00e9nierie thermique et leurs applications pratiques.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_generate-full-width-content":"","footnotes":""},"categories":[98],"tags":[],"class_list":["post-209583","post","type-post","status-publish","format-standard","hentry","category-des-questions","generate-columns","tablet-grid-50","mobile-grid-100","grid-parent","grid-50"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v17.9 - 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