{"id":30966,"date":"2021-08-31T13:00:42","date_gmt":"2021-08-31T13:00:42","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=30966"},"modified":"2025-01-02T19:30:27","modified_gmt":"2025-01-02T19:30:27","slug":"loi-de-coulomb-et-la-force-electrostatique","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/fr\/loi-de-coulomb-et-la-force-electrostatique\/","title":{"rendered":"Loi de Coulomb et la Force \u00c9lectrostatique"},"content":{"rendered":"<style>\n    p, ul, ol {\n        text-align: justify;\n    }\n    h1, h2 {\n        text-align: center;\n    }\n<\/style>\n<h1>Loi de Coulomb et la Force \u00c9lectrostatique<\/h1>\n<p style=\"text-align:center;\">\nLa \u00abLoi de Coulomb et la Force \u00c9lectrostatique\u00bb n&#8217;a pas seulement \u00e9largi notre compr\u00e9hension des forces \u00e9lectriques, mais elle a aussi g\u00e9n\u00e9r\u00e9 des anecdotes inattendues. Benjamin Franklin, lors d&#8217;une exp\u00e9rience visant \u00e0 \u00e9tourdir et \u00e0 cuire une dinde avec de l&#8217;\u00e9lectricit\u00e9, est devenu lui-m\u00eame le sujet de l&#8217;exp\u00e9rience : une d\u00e9charge l&#8217;a laiss\u00e9 \u00e9tourdi avec les cheveux h\u00e9riss\u00e9s, illustrant comme dans la r\u00e9alit\u00e9 les lignes du champ \u00e9lectrique. C&#8217;est maintenant \u00e0 notre tour d&#8217;\u00e9tudier les forces \u00e9lectriques.\n<\/p>\n<p style=\"text-align:center;\"><strong>Objectifs d&#8217;apprentissage :<\/strong><br \/>\n\u00c0 la fin de cette le\u00e7on, l&#8217;\u00e9tudiant sera capable de :<\/p>\n<ol>\n<li><strong>Mod\u00e9liser<\/strong> des ph\u00e9nom\u00e8nes \u00e9lectriques en utilisant le principe de superposition pour calculer la force r\u00e9sultante sur une charge de test.<\/li>\n<li><strong>Simplifier<\/strong> l&#8217;\u00e9tude des forces \u00e9lectriques en la limitant au cas \u00e9lectrostatique.<\/li>\n<li><strong>Appliquer<\/strong> la Loi de Coulomb pour d\u00e9terminer la force entre deux charges dans diverses situations.<\/li>\n<li><strong>Analyser<\/strong> des syst\u00e8mes centr\u00e9s sur la source de charge en utilisant une formulation simplifi\u00e9e de la Loi de Coulomb.<\/li>\n<li><strong>R\u00e9soudre<\/strong> des probl\u00e8mes pratiques li\u00e9s aux distributions de charges.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><strong><u>TABLE DES MATI\u00c8RES :<\/u><\/strong><br \/>\n<a href=\"#1\">Le principe de superposition<\/a><br \/>\n<a href=\"#2\">La simplification \u00e9lectrostatique<\/a><br \/>\n<a href=\"#3\">La Loi de Coulomb<\/a><br \/>\n<a href=\"#4\">Loi de Coulomb pour les syst\u00e8mes centr\u00e9s sur la source de charges<\/a><br \/>\n<a href=\"#5\">Exercices<\/a>\n<\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/RV3Uzorlkr4\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Il est maintenant temps de commencer \u00e0 mod\u00e9liser math\u00e9matiquement ces ph\u00e9nom\u00e8nes, et pour cela, nous introduirons la <strong>Loi de Coulomb.<\/strong> Mais avant, il est n\u00e9cessaire d&#8217;expliquer quelques points : le principe de superposition et la simplification \u00e9lectrostatique.<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Le principe de superposition<\/h2>\n<p>Le probl\u00e8me fondamental de l&#8217;\u00e9lectrodynamique consiste \u00e0 d\u00e9terminer la force qu&#8217;un \u00abnuage\u00bb de charges \u00e9lectriques <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1<\/span><\/span>, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_2<\/span><\/span>, <span class=\"katex-eq\" data-katex-display=\"false\">\\cdots<\/span> exerce sur une charge de test <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_0<\/span><\/span>, lorsque la position de chacune d&#8217;entre elles est une fonction connue du temps. En g\u00e9n\u00e9ral, les sources de charges ainsi que la charge de test sont en mouvement relatif.<\/p>\n<p>La solution de ce probl\u00e8me est facilit\u00e9e par le principe de superposition, qui nous dit que l&#8217;interaction de la charge de test avec une source est compl\u00e8tement ind\u00e9pendante de l&#8217;interaction avec les autres sources de charges. Cela signifie qu&#8217;il est toujours possible de d\u00e9terminer la force <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_1<\/span><\/span> produite par la charge <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1<\/span><\/span>, la force <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_2<\/span><\/span> produite par <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_2<\/span><\/span>, et ainsi de suite, pour finalement additionner et obtenir la force totale :<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_{tot} = \\displaystyle \\sum_{i}\\vec{F}_i<\/span><\/span><\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>La simplification \u00e9lectrostatique<\/h2>\n<p>S&#8217;il suffit d&#8217;additionner les forces, on pourrait dire qu&#8217;il suffit d&#8217;indiquer l&#8217;\u00e9quation qui d\u00e9crit la force que chaque source de charge exerce sur la charge de test et le probl\u00e8me serait r\u00e9solu ; cependant, le probl\u00e8me n&#8217;est pas aussi simple. Le probl\u00e8me r\u00e9side dans le fait que la force d\u00e9pend non seulement de la distance et de la magnitude des charges, mais aussi de la vitesse et de l&#8217;acc\u00e9l\u00e9ration relatives de chaque particule. De plus, \u00abl&#8217;information \u00e9lectrique\u00bb sur les changements de position, de vitesse et d&#8217;acc\u00e9l\u00e9ration de chaque particule voyage \u00e0 la vitesse de la lumi\u00e8re, ce qui signifie qu&#8217;elle met un certain temps pour atteindre la charge de test et produire son effet.<\/p>\n<p>Ainsi, dans le but de simplifier notre \u00e9tude pour l&#8217;instant, nous nous limiterons au cas <strong>\u00e9lectrostatique<\/strong>, c&#8217;est-\u00e0-dire que toutes les sources de charges resteront stationnaires, seule la charge de test pourra se d\u00e9placer ; et c&#8217;est dans ce contexte que la <strong>Loi de Coulomb<\/strong> \u00e9merge.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>La Loi de Coulomb<\/h2>\n<p>Supposons que nous avons une charge de test <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_0<\/span><\/span>, situ\u00e9e \u00e0 la position <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{r}<\/span><\/span>, et une source de charge <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span>, situ\u00e9e \u00e0 la position <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{r}^\\prime<\/span><\/span>. Quelle sera la force <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_{q \\to q_0}(\\vec{r})<\/span><\/span> exerc\u00e9e par la source de charge sur la charge de test ? La r\u00e9ponse \u00e0 cette question est donn\u00e9e par la <strong>Loi de Coulomb<\/strong>, qui s&#8217;exprime par la formule suivante :<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}_{q \\to q_0} (\\vec{r}) =\\displaystyle \\frac{1}{4\\pi \\epsilon_0}  \\frac{q  q_0 }{\\|\\vec{r} - \\vec{r}^\\prime \\|^2} \\frac{\\vec{r} - \\vec{r}^\\prime}{\\|\\vec{r} - \\vec{r}^\\prime\\|}<\/span><\/span><\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-amszqS6iLRk\/YSwCV4TCIZI\/AAAAAAAAFec\/XoTDZm0i-6oPZY4KuGv0xOEbrsYCfJM4wCLcBGAsYHQ\/s0\/fuerzaelectrica.PNG\" width=\"492\" height=\"390\" alt=\"Force \u00c9lectrique\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-amszqS6iLRk\/YSwCV4TCIZI\/AAAAAAAAFec\/XoTDZm0i-6oPZY4KuGv0xOEbrsYCfJM4wCLcBGAsYHQ\/s0\/fuerzaelectrica.PNG\" width=\"492\" height=\"390\" alt=\"Force \u00c9lectrique\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p>La loi de Coulomb ne r\u00e9sume pas seulement la loi des signes pour la force \u00e9lectrostatique, mais \u00e9tablit \u00e9galement que la force entre des charges \u00e9lectriques est inversement proportionnelle au carr\u00e9 de la distance qui les s\u00e9pare.<\/p>\n<p>La constante <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon_0<\/span><\/span> est appel\u00e9e <strong>constante de permittivit\u00e9 \u00e9lectrique du vide.<\/strong> Sa valeur dans le syst\u00e8me international est :<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\epsilon_0 = 8,85 \\cdot 10^{-12} \\left[ \\frac{C^2}{N\\cdot m^2}\\right] <\/span><\/span><\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Loi de Coulomb pour les syst\u00e8mes centr\u00e9s sur la source de charges<\/h2>\n<p>La <strong>loi de Coulomb<\/strong> peut \u00eatre exprim\u00e9e de mani\u00e8re plus simple si nous pla\u00e7ons l&#8217;observateur au niveau de la source de charges, c&#8217;est-\u00e0-dire en posant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{r}^\\prime = \\vec{0}<\/span><\/span>. Ainsi, la formule devient :<\/p>\n<p style=\"text-align: center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\vec{F}_{q \\to q_0} (\\vec{r}) = \\frac{1}{4\\pi \\epsilon_0}  \\frac{q  q_0 }{\\|\\vec{r}\\|^2}   \\frac{\\vec{r} }{\\|\\vec{r} \\|} = \\frac{1}{4\\pi \\epsilon_0}  \\frac{q   q_0 }{\\|\\vec{r}\\|^2} \\hat{r}<\/span><\/span><\/p>\n<p>O\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\hat{r}=\\vec{r}\/\\|\\vec{r}\\|<\/span><\/span> est le vecteur unitaire dirig\u00e9 de la source vers la charge de test.<\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h2>Exercices<\/h2>\n<ol>\n<li>Douze charges ponctuelles de m\u00eame magnitude <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> sont plac\u00e9es aux coins d&#8217;un polygone r\u00e9gulier \u00e0 douze c\u00f4t\u00e9s (comme les chiffres d&#8217;une horloge). Quelle sera la force nette sur une charge ponctuelle <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> plac\u00e9e au centre ?<\/li>\n<li>Une des douze charges de l&#8217;exercice pr\u00e9c\u00e9dent est retir\u00e9e, supposons que c&#8217;est celle qui serait \u00e0 12 heures (si on l&#8217;imagine comme une horloge). Quelle force ressentira maintenant la charge ponctuelle <span class=\"katex-eq\" data-katex-display=\"false\">q<\/span> au centre ?<\/li>\n<li>\u00c9tendez le raisonnement des deux exercices pr\u00e9c\u00e9dents pour une distribution de <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> sources de charge r\u00e9parties sur un polygone r\u00e9gulier \u00e0 <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> c\u00f4t\u00e9s, avec une charge de test plac\u00e9e au centre.<\/li>\n<li>On dispose de trois charges ponctuelles : <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1=+3[nC]<\/span><\/span> plac\u00e9e \u00e0 la position <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(0;0)[mm]<\/span><\/span>, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_2=-5[nC]<\/span><\/span> plac\u00e9e \u00e0 la position <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(0,56;0)[mm]<\/span><\/span>, et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_3=+7[nC]<\/span><\/span> plac\u00e9e \u00e0 la position <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(1;1)[mm]<\/span><\/span>. Calculez la force totale exerc\u00e9e sur la charge <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_3<\/span><\/span>.<\/li>\n<li>Sur un axe se trouve une charge <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_1 = 3[C]<\/span><\/span>, et \u00e0 une distance de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">40[mm]<\/span><\/span>, une autre charge <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_2 = 7[C]<\/span><\/span>. Si une troisi\u00e8me charge est plac\u00e9e entre ces deux charges de mani\u00e8re \u00e0 ce que la somme des forces sur elle soit nulle, quelle sera la distance entre cette troisi\u00e8me charge et les deux autres ?<\/li>\n<li>Deux petites sph\u00e8res de cuivre, chacune ayant une masse de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0,040[kg]<\/span><\/span>, sont plac\u00e9es \u00e0 une distance de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2,0[m]<\/span><\/span>. En consid\u00e9rant que la masse molaire du cuivre est <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">63,5[g\/mol]<\/span><\/span> et que son num\u00e9ro atomique est 20, r\u00e9pondez aux questions suivantes :\n<ol>\n<li type=\"a\">Combien d&#8217;\u00e9lectrons contient chaque sph\u00e8re ?<\/li>\n<li type=\"a\">Combien d&#8217;\u00e9lectrons faut-il d\u00e9placer d&#8217;une sph\u00e8re \u00e0 l&#8217;autre pour produire une force d&#8217;attraction entre les sph\u00e8res d&#8217;environ <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">10^4[N]<\/span><\/span> ?<\/li>\n<li type=\"a\">Quelle fraction des \u00e9lectrons des sph\u00e8res cela repr\u00e9sente-t-il ?<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p><center><br \/>\n<iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/jxO34PM91Io\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><br \/>\n<br \/>\n<iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/EBOxmKPQe9U\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><br \/>\n<\/center><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Loi de Coulomb et la Force \u00c9lectrostatique La \u00abLoi de Coulomb et la Force \u00c9lectrostatique\u00bb n&#8217;a pas seulement \u00e9largi notre compr\u00e9hension des forces \u00e9lectriques, mais elle a aussi g\u00e9n\u00e9r\u00e9 des anecdotes inattendues. Benjamin Franklin, lors d&#8217;une exp\u00e9rience visant \u00e0 \u00e9tourdir et \u00e0 cuire une dinde avec de l&#8217;\u00e9lectricit\u00e9, est devenu lui-m\u00eame le sujet de l&#8217;exp\u00e9rience [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":30950,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":8,"footnotes":""},"categories":[722,647],"tags":[],"class_list":["post-30966","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-electromagnetisme","category-physique"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Loi de Coulomb et la Force \u00c9lectrostatique - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"La loi de Coulomb d\u00e9crit la force \u00e9lectrostatique entre deux charges : proportionnelle au produit des charges et 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