>[!summary] Gauss's law is a tool for calculating the amount of flux penetrating an area. > **General equation:** ${\oint \vec{E}\cdot \vec{dA}} = \frac{Q_{enc}}{\epsilon_0}$ > **Flux:** $\Phi = \frac{Q_{enc}}{\epsilon_0}$ > **Enclosed charge:** $Q_{enc} = \int \lambda \cdot dV$ > **Inf sheet energy:** $E = \frac{\lambda}{2\epsilon_0 }$ > **Uniform charge line:** $E = \frac{\lambda }{2\pi r \epsilon_0}$ >[!info]+ Read Time **⏱ 5 mins** # Definition Gauss' law is a tool useful for finding the flow of [[Electric Fields|electric fields]] through a **closed surface**, usually imaginary. It's to describe the net electric [[Flux|flux]] per enclosed [[Charge|charge]]. The equation is denoted by the [[Closed Integral|closed integral]] to denote that it is integrated through a closed surface. The closed integral is through a [[Gaussian Surfaces|Gaussian surface]] to define the little parts of area $\vec{dA}$ $ \Phi=\oint \vec{E} \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_0} $ Although always valid, Gauss's law is best used in highly symmetric cases because it relies on symmetry arguments to assume the electric field is always constant along the area. > [!note] The $Q_{enc}$ is the charge enclosed. If we had a object with [[Charge Density|charge density]] integration is required to find $Q_{enc}$ # Deriving Gauss's Law from Coulomb's Law >[!warning] Assumptions To derive Gauss' law from [[Coulomb's Law| Coulomb's law]], we need to assume the following: >- From [[Coulomb's Law|coulombs law]] the electric field for a point charge is $E = \frac{q}{4\pi \epsilon R^2}$ >- The Gaussian surface is a sphere with area $4\pi R^2$ >- Assume dA and the electric field will always be parallel to each other >- $\Phi = \oint \vec{E} \cdot \vec{dA}$ (from [[Flux|flux]]) ![[gau_3.png|500]] [^1] >[!note] Explanation Example of a Gaussian surface around a point charge a distance R away. From flux, we know this measurement is true: $\begin{align*} \Phi &= \oint \vec{E} \cdot d\vec{A} \\ &= \oint \frac{q}{4\pi \varepsilon_0 R^2} \, dA \\ &= \frac{q}{4\pi \varepsilon_0 R^2} \oint dA \\ &= \frac{q}{4\pi \varepsilon_0 R^2} \cdot 4\pi R^2 \\ &= \frac{q}{\varepsilon_0} \\ \Rightarrow\quad \Phi &= \frac{q}{\varepsilon_0} \\ \Phi &= \oint \vec{E} \cdot \vec{dA} \\ \oint \vec{E} \cdot \vec{dA} &= \frac{q}{\epsilon_0} \end{align*} $ # Electric Field for A Infinite Sheet >[!warning] Assumptions If we have an infinite sheet and want to find an equation for the electric field. The simplest approach to this is through gauss's law. > We are going to assume the following: >- There is a **uniform** charge density over the inf sheet ($\lambda$) >- Our Gaussian surface is a **cylinder** (So the electric field is always constant) >- The **electric field out the inf sheet is postive** ![[gau_1.png]] [^2] >[!note] Explanation Electric field of a infinite sheet with a Gaussiann surface over a part of it. First we want to find $Q_{enc}$ $\begin{array}{c} Q_{enc}= \int \lambda \cdot dA = \lambda \cdot A \\ \text{Note because the charge density is uniform} \\ \text{We just needed to intergrate over area} \end{array}$ Then we choose a Gaussian surface of a cylinder so the electric is constant $\begin{array}{c} \oint E\cdot dA = \frac{Q_{enc}}{\epsilon_0} \\ E \oint dA = \frac{Q_{enc}}{\epsilon_0} \\\\ \text{If we imagine intergrating over this} \\ \text{Than there is an electric field point out on both sides, refer to photo above} \\ \\ 2EA = \frac{\lambda A}{\epsilon_0} \\ E = \frac{\lambda}{2\epsilon_0 } \end{array}$ # Electric Field of A Uniform Charged Line >[!warning] Assumptions If we have an uniform charge line, that has a uniform charge density we can find the electric field the same way using gauss law. > Were going to assume that following: >- **Uniform charge density** ($L$) >- $Q_{enc} = \lambda L$ >- Were going to use a cylinder Gaussian surface with an area ($A$) of $2\pi r L$ ![[gau_2.png]] [^3] >[!note] Explanation Uniforms charged line with a gaussian surface over it. Since we already assumed $Q_{enc}$ were just going to integrate to find the electric field using our known cylinder Guassian area. $\begin{array}{c} \oint \vec{E} \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_0} \\ \text{Assume a cylinder as gassian surface} \\ E \oint dA = \frac{Q_{enc}}{\epsilon_0} \\ E A = \frac{Q_{enc}}{\epsilon_0} \\ E\cdot 2\pi rL = \frac{\lambda L}{\epsilon_0} \\ E = \frac{\lambda }{2\pi r \epsilon_0} \end{array}$ # Resources <iframe width="560" height="315" src="https://www.youtube.com/embed/htjxMwHKuyI?si=uVauU1nDgHvnRW4N" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe> <iframe width="560" height="315" src="https://www.youtube.com/embed/f2Cccp6XBUY?si=WkfGhB4BGB7SiMoa" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe> --- > 💡 Found this concept helpful? [Star Math & Matter on GitHub](https://github.com/rajeevphysics/Obsidan-MathMatter) to support more intuitive science breakdowns like this. --- [^1]: Taken from https://tikz.net/electric_field_sphere/ by Izaak Neutelings (Februari, 2020) [^2]: Taken from https://tikz.net/electric_field_plane/ by Izaak Neutelings (November, 2018) [^3]: Taken from https://tikz.net/electric_field_rod/ by Izaak Neutelings (July 2018)