{"id":35719,"date":"2021-09-30T13:00:07","date_gmt":"2021-09-30T13:00:07","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35719"},"modified":"2025-12-29T03:20:08","modified_gmt":"2025-12-29T03:20:08","slug":"o-fluxo-eletrico-e-a-lei-de-gauss","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/pt\/o-fluxo-eletrico-e-a-lei-de-gauss\/","title":{"rendered":"O Fluxo El\u00e9trico e a Lei de Gauss"},"content":{"rendered":"<style>\np, ul, ol{\ntext-align: justify;\n}\nh1{\ntext-align:center;\ntext-transform: uppercase;\n}\nh2{\ntext-align:center;\ntext-transform: uppercase;\nfont-size:24pt;\n}\nh3 { \n    text-align: center;\n    text-transform: uppercase;\n    font-size: 24px !important;\n}\n<\/style>\n<h1>O Fluxo El\u00e9trico e a Lei de Gauss<\/h1>\n<p>\nEm eletrost\u00e1tica, calcular o campo el\u00e9trico \u201cdo zero\u201d pode tornar-se muito custoso quando a geometria da distribui\u00e7\u00e3o de carga n\u00e3o \u00e9 trivial. O fluxo el\u00e9trico e a Lei de Gauss oferecem um caminho mais inteligente: em vez de lidar com integrais intermin\u00e1veis, escolhe-se uma superf\u00edcie fechada adequada e aproveita-se a simetria do sistema para obter resultados claros e verific\u00e1veis. Na pr\u00e1tica, isso se traduz em menos etapas, menos erros e maior controle conceitual sobre o que est\u00e1 sendo feito. Se voc\u00ea deseja passar de \u201csei a receita\u201d para \u201centendo o m\u00e9todo\u201d, aqui ver\u00e1 como Gauss transforma problemas que parecem pesados em solu\u00e7\u00f5es diretas, e quando realmente conv\u00e9m utiliz\u00e1-la.\n<\/p>\n<p style=\"text-align:center;\"><b>Objetivos de Aprendizagem<\/b><\/p>\n<ol>\n<li><strong>Explicar<\/strong> o funcionamento da lei de Gauss para o campo el\u00e9trico.<\/li>\n<li><strong>Utilizar<\/strong> a lei de Gauss para calcular campos el\u00e9tricos explorando as simetrias dos sistemas de coordenadas cartesianas, cil\u00edndricas e esf\u00e9ricas.<\/li>\n<li><strong>Relacionar<\/strong> as formas integral e diferencial por meio do Teorema da Diverg\u00eancia, identificando o que representa cada termo.<\/li>\n<li><strong>Contrastar<\/strong> a abordagem de Gauss com o c\u00e1lculo direto pela integral de Coulomb, explicando quando reduz a complexidade e quando n\u00e3o fornece uma solu\u00e7\u00e3o fechada.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><b><u>\u00cdNDICE DE CONTE\u00daDOS<\/u>:<\/b><br \/>\n<a href=\"#1\">Resolu\u00e7\u00e3o da eletrost\u00e1tica<\/a><br \/>\n<a href=\"#2\">Linhas de campo el\u00e9trico<\/a><br \/>\n<a href=\"#3\">Nota sobre a densidade das linhas de campo e sua representa\u00e7\u00e3o<\/a><br \/>\n<a href=\"#4\">Fluxo de Campo El\u00e9trico<\/a><br \/>\n<a href=\"#5\">Lei de Gauss<\/a><br \/>\n<a href=\"#6\">Problemas com simetria esf\u00e9rica<\/a><br \/>\n<a href=\"#7\">Mais Simetrias<\/a><br \/>\n<a href=\"#8\">Problemas com com simetria cil\u00edndrica e planar<\/a>\n<\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/b96XremJiKQ\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><br \/>\n<a name=\"1\"><\/a><\/p>\n<h2>Resolu\u00e7\u00e3o da eletrost\u00e1tica<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=128s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Com o que foi revisado at\u00e9 agora, temos que<\/span><\/strong><\/a> basta conhecer a forma do elemento de campo el\u00e9trico e sua distribui\u00e7\u00e3o no espa\u00e7o para determinar o campo el\u00e9trico total. Se tivermos uma distribui\u00e7\u00e3o volum\u00e9trica, ent\u00e3o<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\vec{E}(\\vec{r}) = \\int_V d\\vec{E}(\\vec{r})= \\int_V \\frac{\\rho(\\vec{r}^\\prime)}{4\\pi\\epsilon_0}\\frac{\\vec{r}-\\vec{r}^\\prime}{\\|\\vec{r}-\\vec{r}^\\prime\\|^3}dV<\/span>\n<p>onde <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\rho(\\vec{r}^\\prime)<\/span><\/span> \u00e9 a densidade volum\u00e9trica de carga. No caso de termos uma densidade superficial ou linear de carga, substituiremos <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> por <span class=\"katex-eq\" data-katex-display=\"false\">\\sigma<\/span> ou <span class=\"katex-eq\" data-katex-display=\"false\">\\lambda<\/span>, respectivamente. A partir deste ponto, o que determina se podemos ou n\u00e3o encontrar o campo el\u00e9trico \u00e9 se conseguimos ou n\u00e3o resolver a integral.<\/p>\n<p>Embora formular o problema costume ser direto, mais cedo ou mais tarde descobriremos que nem sempre \u00e9 f\u00e1cil avali\u00e1-lo. De fato, grande parte do estudo da eletrost\u00e1tica consiste em desenvolver estrat\u00e9gias que permitam evitar o c\u00e1lculo de integrais desnecessariamente complicadas. Muitas dessas simplifica\u00e7\u00f5es prov\u00eam da an\u00e1lise vetorial, em particular do uso da diverg\u00eancia.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Linhas de campo el\u00e9trico<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=235s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Antes de introduzir a an\u00e1lise vetorial<\/span><\/strong><\/a> em nosso estudo da eletrost\u00e1tica, apresentaremos algumas ideias que ajudar\u00e3o a tornar o tema um pouco mais intuitivo. Refiro-me \u00e0s <strong>linhas de campo el\u00e9trico<\/strong>.<\/p>\n<p>Comecemos pelo mais simples: o campo el\u00e9trico de uma carga pontual localizada na origem do sistema de coordenadas. Ele \u00e9 da forma<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\vec{E}(\\vec{r}) = \\frac{1}{4\\pi\\epsilon_0}\\frac{q}{\\|\\vec{r}\\|^2}\\hat{r}<\/span>\n<p>Isso nos permite representar o campo el\u00e9trico no espa\u00e7o como um conjunto de \u201csetas\u201d cuja dire\u00e7\u00e3o e magnitude descrevem a dire\u00e7\u00e3o e a intensidade do campo el\u00e9trico em cada ponto.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEhlkOkwoYFrKYwfidmawFFb60AlBumv8u2irJnN87xYJnTfY7h2U1HL3Hzfh7kQZHhyctM7r70IfEXZkh0faUjZrGr3H_h6xrjeED-GqV39v_t2OzPD0zD9sjlm9t_twIEuaJcte9qhEGIKH3Bzn7a_AZ1rCh53DFunFLiXm09JalAMYQNAjKjv3oOosA\" width=\"400\" height=\"300\" alt=\"Campo el\u00e9trico de uma carga pontual em forma de vetores\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEhlkOkwoYFrKYwfidmawFFb60AlBumv8u2irJnN87xYJnTfY7h2U1HL3Hzfh7kQZHhyctM7r70IfEXZkh0faUjZrGr3H_h6xrjeED-GqV39v_t2OzPD0zD9sjlm9t_twIEuaJcte9qhEGIKH3Bzn7a_AZ1rCh53DFunFLiXm09JalAMYQNAjKjv3oOosA\" width=\"400\" height=\"300\" alt=\"Campo el\u00e9trico de uma carga pontual em forma de vetores\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p>Dado que a intensidade do campo el\u00e9trico decai com o quadrado da dist\u00e2ncia \u00e0 origem, os vetores tornam-se cada vez menores \u00e0 medida que nos afastamos. Al\u00e9m disso, eles apontam radialmente para fora a partir da carga.<\/p>\n<p>Essa representa\u00e7\u00e3o \u00e9 \u00fatil, mas existe outra ainda mais informativa: \u201cconectar o cont\u00ednuo de setas\u201d para formar um campo de linhas. Dessa forma, n\u00e3o ser\u00e1 o comprimento das setas que indicar\u00e1 a intensidade do campo el\u00e9trico, mas sim a \u201cdensidade de linhas de campo\u201d no diagrama.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEgebGCEuU6Akvc8M-p8m0FjK-a1AGtq4H7A2fAuy1r6M08uAnWQRYJfHIRtcRAvGt3CQZHCI7EwCHv3os55aZpef1KDTHFDiS2Sf8nvyXH_ctiildMSeSK-suC7al5kbGmFReywKsEJh1GVsHDtTqRShAyiJZFQER2fKav4wOcn9z8q7zjmjk3T07UHTQ\" width=\"400\" height=\"300\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEgebGCEuU6Akvc8M-p8m0FjK-a1AGtq4H7A2fAuy1r6M08uAnWQRYJfHIRtcRAvGt3CQZHCI7EwCHv3os55aZpef1KDTHFDiS2Sf8nvyXH_ctiildMSeSK-suC7al5kbGmFReywKsEJh1GVsHDtTqRShAyiJZFQER2fKav4wOcn9z8q7zjmjk3T07UHTQ\" width=\"400\" height=\"300\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><br \/>\n<a name=\"3\"><\/a><\/p>\n<h3>Nota sobre a densidade das linhas de campo e sua representa\u00e7\u00e3o<\/h3>\n<p>Antes de continuar, conv\u00e9m observar um detalhe sobre o diagrama de linhas de campo el\u00e9trico. Esse tipo de representa\u00e7\u00e3o n\u00e3o \u00e9 totalmente fiel quando desenhado em um plano (2D). Em um desenho 2D, se considerarmos um c\u00edrculo de raio <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span>, o n\u00famero total de linhas distribui-se ao longo do per\u00edmetro da circunfer\u00eancia, de modo que a densidade linear \u00e9<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{n}{2\\pi r}<\/span>\n<p>Isso decai em rela\u00e7\u00e3o a <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> e n\u00e3o em rela\u00e7\u00e3o a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r^2<\/span><\/span>, como se espera que a intensidade do campo el\u00e9trico o fa\u00e7a. No entanto, se interpretarmos o modelo em tr\u00eas dimens\u00f5es (como um ouri\u00e7o), ent\u00e3o o n\u00famero total de linhas ficaria dividido pela superf\u00edcie de uma esfera<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{n}{4\\pi r^2}<\/span>\n<p>e isso, sim, decai em rela\u00e7\u00e3o a <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r^2<\/span><\/span>. Em outras palavras, embora a representa\u00e7\u00e3o das linhas de campo seja habitualmente feita em 2 dimens\u00f5es, o que se pretende sintetizar \u00e9, na realidade, uma situa\u00e7\u00e3o em 3 dimens\u00f5es. Simplesmente n\u00e3o dispomos de papel em tr\u00eas dimens\u00f5es para desenh\u00e1-la: representamos em 2D aquilo que queremos comunicar em 3D.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Fluxo de Campo El\u00e9trico<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=665s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Quando nos perguntamos pelo n\u00famero<\/span><\/strong><\/a> de linhas de campo el\u00e9trico que atravessam uma determinada superf\u00edcie, a resposta \u00e9 dada pelo fluxo do campo el\u00e9trico sobre essa superf\u00edcie. Assim, define-se o fluxo el\u00e9trico de um campo <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{E}<\/span><\/span> sobre uma superf\u00edcie <span class=\"katex-eq\" data-katex-display=\"false\">S<\/span> como<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Phi_{\\vec{E},S} =\\displaystyle \\int_S \\vec{E}\\cdot d\\vec{S}<\/span>\n<p>N\u00e3o devemos nos deixar enganar pela no\u00e7\u00e3o intuitiva de \u201cn\u00famero de linhas de campo el\u00e9trico que atravessam uma superf\u00edcie\u201d. Lembremos que esse n\u00famero de linhas (ou densidade de linhas) \u00e9 uma forma de representar a intensidade do campo el\u00e9trico. Portanto, o fluxo el\u00e9trico que calculamos \u00e9 uma grandeza escalar associada \u00e0 intensidade do campo el\u00e9trico que atravessa a superf\u00edcie <span class=\"katex-eq\" data-katex-display=\"false\">S<\/span>.<\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h3>Lei de Gauss<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=758s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Dado que a intensidade do campo el\u00e9trico<\/span><\/strong><\/a> \u00e9 proporcional \u00e0 carga el\u00e9trica, dever\u00edamos ser capazes de expressar o fluxo el\u00e9trico sobre uma superf\u00edcie que envolve certa carga como uma quantidade proporcional \u00e0 carga envolvida. De fato, n\u00e3o \u00e9 dif\u00edcil demonstrar que isso ocorre. Consideremos a seguinte figura:<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEgGcCL8WVnhwXmxDkhsW5W31AyJiEsJDsVZZDNm1kQ-MREYYaaBvYb7CBkGSCkfPgiNbDGFP-R4LHr_9pH6ijy0Ji7m1VgzO2pjJwjFDOqAd61VGMJfb4CDfmGyn9uacon7VcpXlB9cd7ZltDUEc3fhDQ86PuKqQb7kN-JuNgGxInlRKiyY91nU2zHfIg\" width=\"500\" height=\"400\" alt=\"Fluxo el\u00e9trico em uma superf\u00edcie fechada\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEgGcCL8WVnhwXmxDkhsW5W31AyJiEsJDsVZZDNm1kQ-MREYYaaBvYb7CBkGSCkfPgiNbDGFP-R4LHr_9pH6ijy0Ji7m1VgzO2pjJwjFDOqAd61VGMJfb4CDfmGyn9uacon7VcpXlB9cd7ZltDUEc3fhDQ86PuKqQb7kN-JuNgGxInlRKiyY91nU2zHfIg\" width=\"500\" height=\"400\" alt=\"Fluxo el\u00e9trico em uma superf\u00edcie fechada\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p>A partir disso, tem-se que:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\n\\displaystyle \\oint_S \\vec{E}\\cdot d\\vec{S} &amp;= \\displaystyle \\oint_S \\left(\\frac{1}{4\\pi\\epsilon_0} \\frac{q_{enc}}{\\|\\vec{r}\\|^2}\\hat{r} \\right)\\cdot d\\vec{S} \\\\ \\\\\n\n&amp; = \\displaystyle \\frac{q_{enc}}{4\\pi\\epsilon_0} \\oint_S \\frac{\\hat{r}}{\\|\\vec{r}\\|^2}\\cdot d\\vec{S} \\\\ \\\\\n\n&amp; = \\displaystyle \\frac{q_{enc}}{4\\pi\\epsilon_0} \\underbrace{\\oint_S d{\\Omega}}_{= 4\\pi} = \\frac{q_{enc}}{\\epsilon_0}\n\n\\end{array}\n\n<\/span>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEjLQhqvLJZMuQHfXHj_WYfbajP9PYwVdNgs4eVflg_jQAJSFu5czNfBgMBTWOWXCE5Tx3-DYwrs8eNpOuJoflvQYbUwpl3BG4BaZxJdnJirqRPsbZM00TfnzyGQvuAimfenB3GUYnEJdZDh2xiXWX5ftu0bN-UYH3G4rydnrnBqEpKDNnNXgdpi5EP81w\" width=\"400\" height=\"300\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEjLQhqvLJZMuQHfXHj_WYfbajP9PYwVdNgs4eVflg_jQAJSFu5czNfBgMBTWOWXCE5Tx3-DYwrs8eNpOuJoflvQYbUwpl3BG4BaZxJdnJirqRPsbZM00TfnzyGQvuAimfenB3GUYnEJdZDh2xiXWX5ftu0bN-UYH3G4rydnrnBqEpKDNnNXgdpi5EP81w\" width=\"400\" height=\"300\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p>Em s\u00edntese, obtemos:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\color{blue}{\\oint_S \\vec{E}\\cdot d\\vec{S} = \\frac{q_{enc}}{\\epsilon_0}}<\/span>\n<p>Esta \u00e9 a <strong>Lei de Gauss para o campo el\u00e9trico em sua forma integral<\/strong>, e mostra uma rela\u00e7\u00e3o de proporcionalidade entre o fluxo el\u00e9trico atrav\u00e9s de uma superf\u00edcie fechada e a carga envolvida. Note-se que ela foi apresentada em sua \u201cforma integral\u201d para enfatizar que tamb\u00e9m existe uma forma diferencial, a qual \u00e9 obtida utilizando o Teorema da Diverg\u00eancia de Gauss no contexto da an\u00e1lise vetorial.<\/p>\n<div style=\"background-color: #c0ffc0; padding: 20px;\">\n<h4>Teorema da Diverg\u00eancia de Gauss<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=1007s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Se <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}<\/span><\/span> \u00e9 um campo vetorial diferenci\u00e1vel<\/span><\/strong><\/a> e <span class=\"katex-eq\" data-katex-display=\"false\">S<\/span> \u00e9 uma superf\u00edcie fechada que envolve um volume <span class=\"katex-eq\" data-katex-display=\"false\">V<\/span>, ent\u00e3o vale que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\oint_S\\vec{F}\\cdot d\\vec{S} = \\int_V (\\vec{\\nabla}\\cdot \\vec{F})dV<\/span>\n<\/div>\n<p>&nbsp;<\/p>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=1055s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Aplicando o teorema da diverg\u00eancia<\/span><\/strong><\/a> ao fluxo do campo el\u00e9trico sobre a superf\u00edcie fechada <span class=\"katex-eq\" data-katex-display=\"false\">S<\/span>, tem-se que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\oint_S\\vec{E}\\cdot d\\vec{S} = \\int_V (\\vec{\\nabla}\\cdot\\vec{E})dV = \\frac{q_{enc}}{\\epsilon_0}<\/span>\n<p>Por outro lado, tamb\u00e9m se tem<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{q_{enc}}{\\epsilon_0} = \\int_V \\frac{\\rho}{\\epsilon_0} dV<\/span>\n<p>A partir dessas duas \u00faltimas equa\u00e7\u00f5es, obt\u00e9m-se finalmente que<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\vec{\\nabla}\\cdot\\vec{E} = \\frac{\\rho}{\\epsilon_0}}<\/span>\n<p>Esta \u00e9 a <strong>Lei de Gauss para o campo el\u00e9trico em sua forma diferencial.<\/strong><\/p>\n<p>Agora podemos recorrer \u00e0 Lei de Gauss para aproveitar melhor as simetrias geom\u00e9tricas de alguns problemas e simplificar, em grande medida, o c\u00e1lculo das integrais que conduzem ao campo el\u00e9trico.<\/p>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Problemas com simetria esf\u00e9rica<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/04itEuVNDN4\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<ol>\n<li>Encontrar o campo el\u00e9trico a uma dist\u00e2ncia <span class=\"katex-eq\" data-katex-display=\"false\">z<\/span> do centro de uma superf\u00edcie esf\u00e9rica de raio <span class=\"katex-eq\" data-katex-display=\"false\">R<\/span> que possui uma densidade de carga uniforme <span class=\"katex-eq\" data-katex-display=\"false\">\\sigma<\/span>. Analise ambos os casos: quando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">z\\lt R<\/span><\/span> e quando <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">z\\geq R<\/span><\/span>.<\/li>\n<li>Realize a mesma an\u00e1lise do exerc\u00edcio anterior, mas agora considerando uma esfera maci\u00e7a e uniformemente carregada, com densidade volum\u00e9trica <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span>. Em seguida, construa um gr\u00e1fico de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{E}\\|<\/span><\/span> em fun\u00e7\u00e3o de <span class=\"katex-eq\" data-katex-display=\"false\">z<\/span>.<\/li>\n<li>Suponhamos que o campo el\u00e9trico, a uma dist\u00e2ncia <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> da origem do sistema de coordenadas, seja <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{E}=kr^2\\hat{r}<\/span><\/span>, com <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> constante. Encontre a densidade de carga <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> associada a esse campo.<\/li>\n<\/ol>\n<p><a name=\"7\"><\/a><\/p>\n<h2>Mais Simetrias<\/h3>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/6eMaax9orAo\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=6eMaax9orAo&amp;t=122s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">A lei de Gauss \u00e9 sempre verdadeira<\/span><\/strong><\/a>, mas nem sempre \u00e9 \u00fatil. Nos exemplos anteriores, se <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> n\u00e3o fosse uniforme, se n\u00e3o tiv\u00e9ssemos simetria esf\u00e9rica, ou se fosse escolhida outra forma para a superf\u00edcie gaussiana, continuaria sendo verdade que o fluxo el\u00e9trico \u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_{enc}\/\\epsilon_0<\/span><\/span>, mas o campo el\u00e9trico n\u00e3o teria por que ser constante nem estar orientado na mesma dire\u00e7\u00e3o que o elemento <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d\\vec{S}<\/span><\/span>; e, sem essas condi\u00e7\u00f5es, n\u00e3o podemos extrair <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{E}\\|<\/span><\/span> da integral.<\/p>\n<p>A simetria \u00e9 crucial na aplica\u00e7\u00e3o da Lei de Gauss na resolu\u00e7\u00e3o de problemas.<\/p>\n<p>Existem muitos tipos de simetrias que podemos aproveitar. Entre todas elas, as tr\u00eas seguintes s\u00e3o as mais frequentes:<\/p>\n<ol>\n<li><strong>Simetria esf\u00e9rica:<\/strong> A superf\u00edcie gaussiana \u00e9 uma esfera conc\u00eantrica.<\/li>\n<li><strong>Simetria cil\u00edndrica:<\/strong> A superf\u00edcie gaussiana \u00e9 um cilindro coaxial.<\/li>\n<li><strong>Simetria planar:<\/strong> A superf\u00edcie gaussiana \u00e9 uma caixa retangular.<\/li>\n<\/ol>\n<p><a name=\"8\"><\/a><\/p>\n<h3>Problemas com con simetria cil\u00edndrica e planar<\/h3>\n<ol>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=6eMaax9orAo&amp;t=630s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Considere um cabo cil\u00edndrico<\/span><\/strong><\/a> infinitamente longo, reto, de raio <span class=\"katex-eq\" data-katex-display=\"false\">R<\/span> e carregado com uma densidade de carga <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> da forma\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\rho(r) = \\left\\{\\begin{array}{lll}\n\nkr &amp; ; &amp; r\\lt R \\\\ \\\\\n\n0 &amp; ; &amp; R\\lt r \\\\ \\\\\n\n\\end{array}\\right.<\/span>\n<p>onde <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> \u00e9 uma constante. Calcule o campo el\u00e9trico no interior do cilindro.<\/li>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=6eMaax9orAo&amp;t=1895s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Encontrar o campo el\u00e9trico que produz<\/span><\/strong><\/a> um plano infinito provido de uma densidade uniforme de carga <span class=\"katex-eq\" data-katex-display=\"false\">\\sigma<\/span>.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>O Fluxo El\u00e9trico e a Lei de Gauss Em eletrost\u00e1tica, calcular o campo el\u00e9trico \u201cdo zero\u201d pode tornar-se muito custoso quando a geometria da distribui\u00e7\u00e3o de carga n\u00e3o \u00e9 trivial. O fluxo el\u00e9trico e a Lei de Gauss oferecem um caminho mais inteligente: em vez de lidar com integrais intermin\u00e1veis, escolhe-se uma superf\u00edcie fechada adequada [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":35708,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":1,"footnotes":""},"categories":[712,637],"tags":[],"class_list":["post-35719","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-eletromagnetismo","category-fisica-pt"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>O Fluxo El\u00e9trico e a Lei de Gauss - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"O fluxo el\u00e9trico mede quanta \u201csa\u00edda\u201d do campo atravessa uma superf\u00edcie; a Lei de Gauss o relaciona com a carga envolvida \u222eE\u00b7dA = Qenc\/\u03b50\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/toposuranos.com\/material\/pt\/o-fluxo-eletrico-e-a-lei-de-gauss\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"O Fluxo El\u00e9trico e a Lei de Gauss\" \/>\n<meta property=\"og:description\" content=\"O fluxo el\u00e9trico mede quanta \u201csa\u00edda\u201d do campo atravessa uma superf\u00edcie; 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