{"id":35715,"date":"2021-09-30T13:00:44","date_gmt":"2021-09-30T13:00:44","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35715"},"modified":"2025-12-29T03:17:58","modified_gmt":"2025-12-29T03:17:58","slug":"electric-flux-and-gausss-law","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/en\/electric-flux-and-gausss-law\/","title":{"rendered":"Electric Flux and Gauss\u2019s Law"},"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>Electric Flux and Gauss\u2019s Law<\/h1>\n<p>\nIn electrostatics, calculating the electric field \u201cfrom scratch\u201d can become very costly when the geometry of the charge distribution is not trivial. Electric flux and Gauss\u2019s law offer a more efficient route: instead of struggling with endless integrals, you choose an appropriate closed surface and exploit the symmetry of the system to obtain clean and verifiable results. In practice, this translates into fewer steps, fewer errors, and greater conceptual control over what you are doing. If you want to move from \u201cI know the recipe\u201d to \u201cI understand the method,\u201d here you will see how Gauss transforms problems that appear cumbersome into direct solutions, and when it is actually convenient to use it.\n<\/p>\n<p style=\"text-align:center;\"><b>Learning Objectives<\/b><\/p>\n<ol>\n<li><strong>Explain<\/strong> the operation of Gauss\u2019s law for the electric field.<\/li>\n<li><strong>Use<\/strong> Gauss\u2019s law to calculate electric fields by exploiting the symmetries of Cartesian, cylindrical, and spherical coordinates.<\/li>\n<li><strong>Relate<\/strong> the integral and differential forms through the Divergence Theorem, identifying what each term represents.<\/li>\n<li><strong>Contrast<\/strong> the Gauss approach with direct calculation via Coulomb\u2019s integral, explaining when it reduces complexity and when it does not yield a closed-form solution.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><b><u>CONTENT INDEX<\/u>:<\/b><br \/>\n<a href=\"#1\">Resolution of electrostatics<\/a><br \/>\n<a href=\"#2\">Electric field lines<\/a><br \/>\n<a href=\"#3\">Note on the density of field lines and their representation<\/a><br \/>\n<a href=\"#4\">Electric Field Flux<\/a><br \/>\n<a href=\"#5\">Gauss\u2019s Law<\/a><br \/>\n<a href=\"#6\">Problems with spherical symmetry<\/a><br \/>\n<a href=\"#7\">More Symmetries<\/a><br \/>\n<a href=\"#8\">Problems with cylindrical and planar symmetry<\/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>Resolution of Electrostatics<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=128s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">From what has been reviewed so far, we have that<\/span><\/strong><\/a> it is sufficient to know the form of the electric field element and its distribution in space in order to determine the total electric field. If we have a volumetric distribution, then<\/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>where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\rho(\\vec{r}^\\prime)<\/span><\/span> is the volumetric charge density. In the case of a surface or linear charge density, we will replace <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">\\sigma<\/span> or <span class=\"katex-eq\" data-katex-display=\"false\">\\lambda<\/span>, respectively. From this point onward, what determines whether or not we can find the electric field is whether or not we are able to evaluate the integral.<\/p>\n<p>Although formulating the problem is usually straightforward, sooner or later we will discover that it is not always easy to evaluate it. In fact, a large part of the study of electrostatics consists of developing strategies that allow us to avoid the calculation of unnecessarily complicated integrals. Many of these simplifications arise from vector analysis, in particular from the use of the divergence.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Electric Field Lines<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=235s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Before introducing vector analysis<\/span><\/strong><\/a> into our study of electrostatics, we will present some ideas that will help make the topic somewhat more intuitive. I am referring to <strong>electric field lines<\/strong>.<\/p>\n<p>Let us begin with the simplest case: the electric field of a point charge located at the origin of coordinates. This has the form<\/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>This allows us to represent the electric field in space as a set of \u201carrows\u201d whose direction and magnitude describe the direction and intensity of the electric field at each point.<\/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=\"Electric field of a point charge represented as vectors\" 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=\"Electric field of a point charge represented as vectors\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p>Since the intensity of the electric field decreases with the square of the distance from the origin, the vectors become progressively smaller as we move away. In addition, they point radially outward from the charge.<\/p>\n<p>This representation is useful, but there is another that is even more informative: \u201cconnecting the continuum of arrows\u201d to form a field of lines. In this way, it is no longer the length of the arrows that indicates the intensity of the electric field, but rather the \u201cdensity of field lines\u201d in the diagram.<\/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>Note on the Density of Field Lines and Their Representation<\/h3>\n<p>Before continuing, it is worth noting a detail regarding the diagram of electric field lines. This type of representation is not entirely faithful when drawn on a plane (2D). In a 2D drawing, if we consider a circle of radius <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span>, the total number of lines is distributed along the circumference, so that the linear density is<\/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>This decreases with respect to <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> and not with respect to <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r^2<\/span><\/span>, as would be expected for the intensity of the electric field. However, if we interpret the model in three dimensions (like a hedgehog), then the total number of lines would be divided by the surface of a sphere<\/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>and this does decrease with respect to <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">r^2<\/span><\/span>. In other words, although the representation of field lines is usually carried out in two dimensions, what is actually intended is to synthesize a three-dimensional situation. We simply do not have three-dimensional paper to draw it: we represent in 2D what we wish to communicate in 3D.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Electric Field Flux<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=665s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">When we ask about the number<\/span><\/strong><\/a> of electric field lines that cross a given surface, the answer is given by the flux of the electric field through that surface. Thus, the electric flux of a field <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{E}<\/span><\/span> through a surface <span class=\"katex-eq\" data-katex-display=\"false\">S<\/span> is defined as<\/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>We should not be misled by the intuitive notion of the \u201cnumber of electric field lines crossing a surface.\u201d Recall that this number of lines (or line density) is a way of representing the intensity of the electric field. Therefore, the electric flux that we calculate is a scalar quantity associated with the intensity of the electric field passing through the surface <span class=\"katex-eq\" data-katex-display=\"false\">S<\/span>.<\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h3>Gauss\u2019s Law<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=758s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Since the intensity of the electric field<\/span><\/strong><\/a> is proportional to the electric charge, we should be able to express the electric flux through a surface that encloses a certain charge as a quantity proportional to the enclosed charge. In fact, it is not difficult to show that this is the case. Consider the following figure:<\/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=\"Electric flux through a closed surface\" 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=\"Electric flux through a closed surface\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p>From this, we obtain that:<\/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>In summary, we obtain:<\/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>This is <strong>Gauss\u2019s Law for the electric field in its integral form<\/strong>, and it shows a proportionality relationship between the electric flux through a closed surface and the enclosed charge. Note that it has been presented in its \u201cintegral form\u201d to emphasize that a differential form also exists, which is obtained by using Gauss\u2019s Divergence Theorem in the context of vector analysis.<\/p>\n<div style=\"background-color: #c0ffc0; padding: 20px;\">\n<h4>Gauss\u2019s Divergence Theorem<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=b96XremJiKQ&amp;t=1007s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">If <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{F}<\/span><\/span> is a differentiable vector field<\/span><\/strong><\/a> and <span class=\"katex-eq\" data-katex-display=\"false\">S<\/span> is a closed surface enclosing a volume <span class=\"katex-eq\" data-katex-display=\"false\">V<\/span>, then the following holds:<\/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;\">Applying the divergence theorem<\/span><\/strong><\/a> to the flux of the electric field over the closed surface <span class=\"katex-eq\" data-katex-display=\"false\">S<\/span>, we obtain<\/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>On the other hand, we also have<\/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>From these last two equations, we finally obtain<\/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>This is <strong>Gauss\u2019s Law for the electric field in its differential form.<\/strong><\/p>\n<p>We can now make use of Gauss\u2019s Law to better exploit the geometric symmetries of certain problems and greatly simplify the calculation of the integrals that lead to the electric field.<\/p>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Problems with Spherical Symmetry<\/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>Find the electric field at a distance <span class=\"katex-eq\" data-katex-display=\"false\">z<\/span> from the center of a spherical surface of radius <span class=\"katex-eq\" data-katex-display=\"false\">R<\/span> that has a uniform surface charge density <span class=\"katex-eq\" data-katex-display=\"false\">\\sigma<\/span>. Analyze both cases: when <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">z\\lt R<\/span><\/span>, and when <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">z\\geq R<\/span><\/span>.<\/li>\n<li>Carry out the same analysis as in the previous exercise, but now considering a solid sphere uniformly charged with a volumetric charge density <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span>. Then construct a plot of <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{E}\\|<\/span><\/span> as a function of <span class=\"katex-eq\" data-katex-display=\"false\">z<\/span>.<\/li>\n<li>Suppose that the electric field, at a distance <span class=\"katex-eq\" data-katex-display=\"false\">r<\/span> from the origin of coordinates, is <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{E}=kr^2\\hat{r}<\/span><\/span>, with <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> constant. Find the charge density <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> associated with this field.<\/li>\n<\/ol>\n<p><a name=\"7\"><\/a><\/p>\n<h2>More Symmetries<\/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;\">Gauss\u2019s law is always true<\/span><\/strong><\/a>, but it is not always useful. In the previous examples, if <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> were not uniform, if we did not have spherical symmetry, or if a different shape were chosen for the Gaussian surface, it would still be true that the electric flux is <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q_{enc}\/\\epsilon_0<\/span><\/span>, but the electric field would not necessarily be constant nor oriented in the same direction as the element <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">d\\vec{S}<\/span><\/span>; and without these conditions we cannot extract <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{E}\\|<\/span><\/span> from the integral.<\/p>\n<p>Symmetry is crucial in the application of Gauss\u2019s Law to problem solving.<\/p>\n<p>There are many types of symmetries that we can exploit. Among all of them, the following three are the most frequent:<\/p>\n<ol>\n<li><strong>Spherical symmetry:<\/strong> The Gaussian surface is a concentric sphere.<\/li>\n<li><strong>Cylindrical symmetry:<\/strong> The Gaussian surface is a coaxial cylinder.<\/li>\n<li><strong>Planar symmetry:<\/strong> The Gaussian surface is a rectangular box.<\/li>\n<\/ol>\n<p><a name=\"8\"><\/a><\/p>\n<h3>Problems with cylindrical and planar symmetry<\/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;\">Consider an infinitely long<\/span><\/strong><\/a> straight cylindrical wire of radius <span class=\"katex-eq\" data-katex-display=\"false\">R<\/span>, charged with a charge density <span class=\"katex-eq\" data-katex-display=\"false\">\\rho<\/span> of the form\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>where <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> is a constant. Compute the electric field inside the cylinder.<\/li>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=6eMaax9orAo&amp;t=1895s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Find the electric field produced<\/span><\/strong><\/a> by an infinite plane endowed with a uniform surface charge density <span class=\"katex-eq\" data-katex-display=\"false\">\\sigma<\/span>.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Electric Flux and Gauss\u2019s Law In electrostatics, calculating the electric field \u201cfrom scratch\u201d can become very costly when the geometry of the charge distribution is not trivial. Electric flux and Gauss\u2019s law offer a more efficient route: instead of struggling with endless integrals, you choose an appropriate closed surface and exploit the symmetry of the [&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":6,"footnotes":""},"categories":[710,635],"tags":[],"class_list":["post-35715","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-electromagnetism","category-physics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Electric Flux and Gauss\u2019s Law - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"The electric flux measures how much \u201coutflow\u201d of the field passes through a surface; Gauss\u2019s Law relates it to the enclosed charge \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\/en\/electric-flux-and-gausss-law\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Electric Flux and Gauss\u2019s Law\" \/>\n<meta property=\"og:description\" content=\"The electric flux measures how much \u201coutflow\u201d of the field passes through a surface; 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