Gauss's Law: Difference between revisions

Latest revision as of 18:54, 14 April 2024

Claimed by James Swope (Fall 2020), Edited by Nabiha Ahsan (Fall 2021), Edited by Tyler Morgan(Fall 2023), Edited by Anshu Dendukuri(Spring 2024)

Gauss's Law is a very powerful law that spans a diverse array of fields, with applications in physics, mathematics, chemistry, and engineering, among others. Along with James Maxwell's other three equations, Gauss's Law forms the foundation of classical electrodynamics.

The Main Idea

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Three-dimensional Gaussian surface

Picture a three-dimensional solid object in space that does not have any charges around it. Suppose that you discover an electric field is directed radially outward from all faces of the object. You then measure the magnitude of the electric field at each face. Without being able to open the object and determine its charge distribution, how might you determine the charge inside the object?

This simple situation illustrates one basic application of Gauss's Law: Using the measured electric field flowing through the faces of a closed surface to determine the charge that lies inside.

A Mathematical Model

Gauss's Law elegantly relates the net charge enclosed within a Gaussian surface to the patterns of electric field that flow over its faces (Electric Flux). A Gaussian surface is any surface belonging to a closed three-dimensional object. In words, the law states that the net electric flux outside a surface is equal to the ratio of the total enclosed charge inside the surface to the permittivity of free space. Symbolically, this is written as the area integral

[math]\displaystyle{ \oint \vec E \cdot \hat{n} ~dA = \frac{\sum Q_{inside}}{\varepsilon_{0}} }[/math]

where [math]\displaystyle{ \vec E }[/math] is the net electric field acting through the surface, [math]\displaystyle{ \hat{n} }[/math] is the unit normal vector perpendicular to the face of the surface, and [math]\displaystyle{ dA }[/math] is the surface area of the face being examined.

From the definition of Electric Flux, Gauss's Law may also be written equivalently as

[math]\displaystyle{ \Phi_{el} = \frac{Q_{enclosed}}{\varepsilon_{0}} }[/math]

Relation to Coulomb's Law

Additionally, Coulomb's Law and Gauss's Law are innately connected. Coulomb's Law relates charge to electric field. Gauss's Law relates charge to electric flux. Both laws, relate charge to an electrical property. For clarification, the electric flux is a quantity that is equal to the product of the perpendicular component of E-field and the area of the closed surface. Not only do these two laws look similar at the surface, but using one law it is quite easy to derive the other law.

Relationship Between Charge and Flux

We know from the previous introduction of flux that there is a relationship between the electric flux on a closed surface and the charges inside the surface. However, we would like to figure out what that particular factor is.

We start by considering a point charge of [math]\displaystyle{ +Q }[/math] enclosed by an imaginary spherical shell.

Everywhere on the imaginary shell, the electric field produced is parallel to the normal unit vector. Cos(0) = 1, so the dot product is simply the magnitude of the electric field. Recall that the surface area of the imaginary sphere is [math]\displaystyle{ 4 \pi r^{2} }[/math]. With Coulomb's Law and the surface area of a sphere, we get that electric flux is equal to [math]\displaystyle{ \frac{+Q}{\varepsilon_{0}} }[/math]. This implies that the factor is [math]\displaystyle{ \frac{1}{\varepsilon_{0}} }[/math].

Note that if the point charge was negative, the electric field would still be parallel, just opposite direction. Cos(180) = -1 so the dot product would be -1 times the magnitude of the electric field.

To understand this law more fully, one can look at the diagram below. As can be seen on the left side of this diagram, change in flux equals electric field multiplied by change in area. This is the basis for the equation ф=∮E da. When you integrate the change of A, you get the flux.

Image Taken from Hyperphysics

To more clearly state it, in integral form, the formula for this Law is the electric flux equals the total charge contained by a closed surface, divided by the permittivity ([math]\displaystyle{ \varepsilon_{0} = 8.854187817...×10^{-12} ~F*m^{-1} }[/math]).

Additionally, Gauss's law can be written in differential form.

In this equation, ρ is electric charge density, ∇*E is divergence of electric field, and [math]\displaystyle{ \varepsilon_{0} = 8.854187817...×10^{-12} ~F*m^{-1} }[/math].

Below is a brief summary of what has been covered.

A Computational Model

An extremely rudimentary explanation of Gauss's Law is: https://www.youtube.com/watch?v=f2Cccp6XBUY

To visualization of Gauss's law with computational model: https://www.youtube.com/watch?v=i5N36mWsdGo

Experimental demonstration of Gauss's law: https://www.youtube.com/watch?v=r3w4FAUOLuM

Visual explaination of Gauss Divergence Theorem(Khan Academy): https://www.youtube.com/watch?v=XyiQ2dwJHXE

To view the applications of Gauss's law in a coding setting (Python GLowScript): https://trinket.io/glowscript/f618920f61

Examples

Easy Examples

Easy Example 1

Looking at the images on the right hand side of the screen, we find an easy example in which we will find the electric field in a uniformly charged plate (+Q).

1. Recall Gauss's Formula:

2. Now look carefully at the diagram below and go through each of the surfaces one by one. Ask yourself, in what direction does the normal vector point?

3. Therefore, in the sides where there is 90 degree angle made with E, the flux will just be zero!

4. In the only case where the flux is not zero is in the sides A and B of the box (which are the same in magnitude). We compute the flux as follows:

Remember, when doing Gauss's problems, always think about you n vector before diving into the formula.

Easy Example 2

Consider a point charge [math]\displaystyle{ Q = +2 \, \text{μC} }[/math] enclosed by a spherical Gaussian surface with a radius of [math]\displaystyle{ r = 10 \, \text{cm} }[/math]. Calculate the electric flux through the surface.

- Solution:**

Using Gauss's Law, the electric flux [math]\displaystyle{ ( \Phi_{el} ) }[/math] through a closed surface is given by [math]\displaystyle{ \Phi_{el} = \frac{Q_{enclosed}}{\varepsilon_{0}} }[/math]. For a point charge enclosed by a spherical surface, the electric flux is simply [math]\displaystyle{ \frac{Q}{\varepsilon_{0}} }[/math]. Substituting the given values:

[math]\displaystyle{ \Phi_{el} = \frac{2 \times 10^{-6} \, \text{C}}{8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2} }[/math]

Intermediate Examples

Example 1

The example below shows how to determine the net charge located inside a box and is fundamental to understanding Gauss's Law. It demonstrates the idea that to calculate the net charge inside a box, we have to start off by calculating the fluxes acting on all surfaces on the box by using the Gauss's Law.

Example 2

The example below shows how to calculate the net charge enclosed by a box.

Example 3

In order to apply Gauss's Law, it is important to be certain you are working with a closed surface, then set electric flux equal to the internal field divided by the permittivity ([math]\displaystyle{ \varepsilon_{0} = 8.854187817...×10^{-12} ~F*m^{-1} }[/math]). An example of this Law being applied can be found below.

Example 4

In this example, the Qenc is equal to zero. This was determined by adding up all of the fluxes through every surface. This can also be determined by seeing that the same fluxes travel through the same areas of the tent. The equation Ф=∮B da=0 helps explain this when considering that Ф is the same in both equations.

Hard Examples

Example 1

Inside a non-uniformly charged sphere with a volume charge density [math]\displaystyle{ \rho(r) = \rho_0 r^2 }[/math] for [math]\displaystyle{ 0 \leq r \leq R }[/math], where [math]\displaystyle{ \rho_0 }[/math] is a constant and R is the radius of the sphere. Find the electric field at a distance r from the center, where r is between 0 and R, using Gauss's Law.

Solution:

To find the electric field inside the sphere, create a Gaussian surface and take advantage of the spherical symmetry. The charge enclosed within a Gaussian sphere of radius [math]\displaystyle{ r }[/math] is [math]\displaystyle{ \frac{4}{3}\pi \rho_0 r^3 }[/math]. Gauss's Law then gives [math]\displaystyle{ E = \frac{\rho_0 r}{3\varepsilon_0} }[/math].

Example 2

A long, straight wire carries a non-uniform linear charge density [math]\displaystyle{ \lambda(z) = \lambda_0 e^{-az} }[/math], where [math]\displaystyle{ \lambda_0 }[/math] and a are constants, and z is the distance along the wire. Determine the electric field at a distance r from the wire using Gauss's Law.

Solution:

Set up a Gaussian cylindrical surface around the wire to take advantage of the cylindrical symmetry. The total charge enclosed within the Gaussian surface is found by integrating the linear charge density along the length of the cylinder. Gauss's Law then gives [math]\displaystyle{ E = \frac{\lambda_0}{2\pi\varepsilon_0 r} \int_{-\infty}^{\infty} e^{-az} \, dz }[/math]. Solving this integral yields the expression for the electric field as a function of r.

Connectedness

Gauss's Law can be found in many areas of science, technology, engineering, and mathematics. In fact, Gauss's Law and other Maxwell Equations form a basis for electrodynamics. They are the fundamental core of this field of study. Gauss's Law is also connected to the Gaussian/Normal distribution, which is a continuous probability distribution that is characterized by a bell-shaped curve. This distribution is found throughout statistics and probability and is used everyday by people in STEM fields.

As Industrial and Systems Engineering majors, we deal with many statistical distributions. A very important fundamental distribution is the Gaussian distribution or the Normal Distribution. With inference, the Gaussian Distribution comes up in confidence intervals for single statistics. With comparison inference, like finding the pairwise difference between statistics, generally we use other distributions such as the T-Distribution. However, the T-Distribution approximates the Gaussian distribution with degrees of freedom greater than 29.

As civil and environmental engineering majors, we also deal with the Gaussian/Normal Distribution in our fields. As an example, the speed data of traffic on a highway is said to follow the normal distribution.

Additional to many engineering majors. Gauss's Law has applications in many different STEM job fields. Engineering majors learn about the different Gaussian laws in college and apply them in there jobs. Millions of people are applying his laws to the world and will continue to do so. Gauss has had a huge impact on the world and will continue to have a big impact on the future. Gauss has done great work with his work.

History

Gauss's Law was first 'discovered' in 1773, by the great mathematician Joseph-Louis Lagrange. It was further developed by Carl Friedrich Gauss in 1813.

Gauss was a prolific German mathematician and physicist. He made significant contributions to a variety of fields in mathematics and physics. In fact his contributions are so great that he is referred to as the "greatest mathematician since antiquity" and the "foremost of mathematicians".

study.com

Gauss's Law: Difference between revisions

Latest revision as of 18:54, 14 April 2024

Contents

The Main Idea

A Mathematical Model

Relation to Coulomb's Law

Relationship Between Charge and Flux

A Computational Model

Examples

Easy Examples

Easy Example 1

Easy Example 2

Intermediate Examples

Example 1

Example 2

Example 3

Example 4

Hard Examples

Example 1

Solution:

Example 2

Solution:

Connectedness

History

See also

Further Reading

External links

References

Navigation menu

@@ Line 1: / Line 1: @@
-'''Claimed by Charu Thomas (SPRING 2017)'''
+'''Claimed by James Swope (Fall 2020), Edited by Nabiha Ahsan (Fall 2021), Edited by Tyler Morgan(Fall 2023), Edited by Anshu Dendukuri(Spring 2024)'''
+<!--I apologize for my apparent lack of contribution. I had hoped to create a great deal of new content (specifically a computational model and a few hard examples), however, I ended up running out of time. I was also unable to finish tidying up/rewriting the article. Please don't judge me too critically.
-Gauss's Law is an equation relating the charges and electric flux. It states that net electric flux outside a surface is equal to the charge inside the surface over the constant permittivity of space (ε0 = 8.854187817...×10−12 F⋅m−1). Coulomb's Law and Gauss's Law are intimately connected because Coulomb's Law relates the charge to electric field. Gauss's Law relates electric flux, a quantity that is equal to the product of the perpendicular component of E-field and Area of closed surface.
+TO DO:
-It is one of Maxwell's Equations, formulated by Carl Friedrich Gauss, a notorious German mathematician.
+   Re-write for consistent tone/style
+   Provide more examples (easy/hard)
+   Add diagrams
+   Add brief descriptions to each external link/source
+   Create computational model-->
+Gauss's Law is a very powerful law that spans a diverse array of fields, with applications in physics, mathematics, chemistry, and engineering, among others. Along with [[James Maxwell]]'s other three equations, Gauss's Law forms the foundation of classical electrodynamics.
 ==The Main Idea==
+[[File:Gaussian Box Ex.jpg|thumb|Three-dimensional Gaussian surface]]
+Picture a three-dimensional solid object in space that does not have any charges around it. Suppose that you discover an electric field is directed radially outward from all faces of the object. You then measure the magnitude of the electric field at each face. Without being able to open the object and determine its charge distribution, how might you determine the charge inside the object?
+This simple situation illustrates one basic application of Gauss's Law: Using the measured electric field flowing through the faces of a closed surface to determine the charge that lies inside.
+===A Mathematical Model===
+Gauss's Law elegantly relates the net charge enclosed within a ''Gaussian surface'' to the patterns of electric field that flow over its faces ([[Electric Flux]]). A Gaussian surface is any surface belonging to a closed three-dimensional object. <!--Include picture of examples of surfaces that ARE Gaussian surfaces and surfaces that are NOT Gaussian surfaces-->
+In words, the law states that the net electric flux outside a surface is equal to the ratio of the total enclosed charge inside the surface to the permittivity of free space. Symbolically, this is written as the area integral
+<math>\oint \vec E \cdot \hat{n} ~dA = \frac{\sum Q_{inside}}{\varepsilon_{0}}</math>
+where <math> \vec E </math> is the net electric field acting through the surface, <math> \hat{n}</math> is the unit normal vector perpendicular to the face of the surface, and <math> dA </math> is the surface area of the face being examined.
+<!--"Returning to the example mentioned above, [provide a worked example calculating the charge inside the box, given values of E and the dimensions of the box]"-->
+From the definition of [[Electric Flux]], Gauss's Law may also be written equivalently as
+<math> \Phi_{el} = \frac{Q_{enclosed}}{\varepsilon_{0}}</math>
+====Relation to Coulomb's Law====
+Additionally, Coulomb's Law and Gauss's Law are innately connected. Coulomb's Law relates charge to electric field. Gauss's Law relates charge to electric flux. Both laws, relate charge to an electrical property. For clarification, the electric flux is a quantity that is equal to the product of the perpendicular component of E-field and the area of the closed surface. Not only do these two laws look similar at the surface, but using one law it is quite easy to derive the other law.
+====Relationship Between Charge and Flux====
 We know from the previous introduction of flux that there is a relationship between the electric flux on a closed surface and the charges inside the surface. However, we would like to figure out what that particular factor is.
-We start by considering a point charge of +Q enclosed by an imaginary spherical shell.
+We start by considering a point charge of <math>+Q</math> enclosed by an imaginary spherical shell.
 [[File:gaussv2212.jpg]]
-Everywhere on the imaginary shell, the electric field produced is parallel to the normal unit vector. Cos(0) = 1, so the dot product evaluates to the magnitude of the electric field. Recall that the surface area of the imaginary sphere is 4*pi*r^2. With Coulomb's Law and the surface area of a sphere, we get that electric flux is equal to +Q/ε0. This implies that the factor is 1/ε0.
+Everywhere on the imaginary shell, the electric field produced is parallel to the normal unit vector. Cos(0) = 1, so the dot product is simply the magnitude of the electric field. Recall that the surface area of the imaginary sphere is <math>4 \pi r^{2}</math>. With Coulomb's Law and the surface area of a sphere, we get that electric flux is equal to <math>\frac{+Q}{\varepsilon_{0}}</math>. This implies that the factor is <math>\frac{1}{\varepsilon_{0}}</math>.
 Note that if the point charge was negative, the electric field would still be parallel, just opposite direction. Cos(180) = -1 so the dot product would be -1 times the magnitude of the electric field.
-The idea of Gauss's Law is that the electric flux out of a closed surface is equivalent to the charge enclosed, divided by the permittivity.
-There is a near identical law to this law, known as Gauss's law for Magnetism.  The variation found is that magnetic fields are used instead of electric fields in the calculations.  Also, Gauss's Law for Gravity is very similar as well.  To state it again, the electric flux passing through a closed surface is the same as the charge enclosed, divided by permittivity of the surface. This implies that the electric flux is proportional to the total charge enclosed. Any closed surface can be have Gauss's Law applied to it.  For symmetrically shaped objects, Gauss's Law greatly simplifies calculation of electric field enclosed by surface.
+To understand this law more fully, one can look at the diagram below.  As can be seen on the left side of this diagram, change in flux equals electric field multiplied by change in area. This is the basis for the equation ф=∮E da. When you integrate the change of A, you get the flux.
+Image Taken from Hyperphysics
+[[File:Gaulaw.gif]]
+To more clearly state it, in integral form, the formula for this Law is the electric flux equals the total charge contained by a closed surface, divided by the permittivity (<math>\varepsilon_{0} = 8.854187817...×10^{-12} ~F*m^{-1}</math>).
+[[File:Adc2dff3156800a39ef0a9df76a7d868.png]]
+Additionally, Gauss's law can be written in differential form.
+[[File:Screen Shot 2018-04-18 at 5.31.54 PM.png]]
+In this equation, ρ is electric charge density, ∇*E is divergence of electric field, and <math>\varepsilon_{0} = 8.854187817...×10^{-12} ~F*m^{-1}</math>.
+Below is a brief summary of what has been covered.
-===A Mathematical Model===
+[[File:Gauss_14.jpg]]
-A very helpful and clear summary of this Law can be found in the diagram below.  As can be seen on the left side of this diagram, change in flux equals electric field multiplied by change in area.
+==A Computational Model==
-Image Taken from Hyperphysics
+An extremely rudimentary explanation of Gauss's Law is: https://www.youtube.com/watch?v=f2Cccp6XBUY
-[[File:Gaulaw.gif]]
+To visualization of Gauss's law with computational model: https://www.youtube.com/watch?v=i5N36mWsdGo
+Experimental demonstration of Gauss's law: https://www.youtube.com/watch?v=r3w4FAUOLuM
-To more clearly state it, the formula for this Law is the electric flux equals the total charge contained by a closed surface, divided by the permittivity (epsilon naught: ε0 = 8.854187817...×10−12 F⋅m−1).
+Visual explaination of Gauss Divergence Theorem(Khan Academy): https://www.youtube.com/watch?v=XyiQ2dwJHXE
-[[File:Adc2dff3156800a39ef0a9df76a7d868.png]]
+To view the applications of Gauss's law in a coding setting (Python GLowScript): https://trinket.io/glowscript/f618920f61
 ==Examples==
-In order to apply Gauss's Law, it is important to be certain you are working with a closed surface, then set electric flux equal to the internal field divided by the permittivity (epsilon naught: ε0 = 8.854187817...×10−12 F⋅m−1).  An example of this Law being applied can be found below.
+===Easy Examples===
+====Easy Example 1====
+Looking at the images on the right hand side of the screen, we find an easy example in which we will find the electric field in a uniformly charged plate (+Q).
+. Recall Gauss's Formula:
+[[File:Gauss for.jpg|400px]]
+. Now look carefully at the diagram below and go through each of the surfaces one by one. Ask yourself, in what direction does the normal vector point?
+[[File:Gauss ex.jpg|400px]]
+. Therefore, in the sides where there is 90 degree angle made with E, the flux will just be zero!
+[[File:Gauss fl.jpg|400px]]
+. In the only case where the flux is not zero is in the sides A and B of the box (which are the same in magnitude). We compute the flux as follows:
+[[File:Gauss ans.jpg|600px]]
+Remember, when doing Gauss's problems, always think about you n vector before diving into the formula.
+====Easy Example 2====
+Consider a point charge <math> Q = +2 \, \text{μC} </math> enclosed by a spherical Gaussian surface with a radius of <math> r = 10 \, \text{cm} </math>. Calculate the electric flux through the surface.
+**Solution:**
+Using Gauss's Law, the electric flux <math>( \Phi_{el} )</math> through a closed surface is given by <math> \Phi_{el} = \frac{Q_{enclosed}}{\varepsilon_{0}} </math>. For a point charge enclosed by a spherical surface, the electric flux is simply <math> \frac{Q}{\varepsilon_{0}} </math>. Substituting the given values:
+<math> \Phi_{el} = \frac{2 \times 10^{-6} \, \text{C}}{8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2} </math>
+===Intermediate Examples===
+====Example 1====
+The example below shows how to determine the net charge located inside a box and is fundamental to understanding Gauss's Law. It demonstrates the idea that to calculate the net charge inside a box, we have to start off by calculating the fluxes acting on all surfaces on the box by using the Gauss's Law.
+[[File:Wiki resource.jpg|600px]]
+====Example 2====
+The example below shows how to calculate the net charge enclosed by a box.
+[[File:Wiki resource2.jpg|600px]]
+====Example 3====
+In order to apply Gauss's Law, it is important to be certain you are working with a closed surface, then set electric flux equal to the internal field divided by the permittivity (<math>\varepsilon_{0} = 8.854187817...×10^{-12} ~F*m^{-1}</math>).  An example of this Law being applied can be found below.
+[[File:Gauss_law3.png|600px]]
+====Example 4====
+In this example, the Qenc is equal to zero. This was determined by adding up all of the fluxes through every surface. This can also be determined by seeing that the same fluxes travel through the same areas of the tent. The equation Ф=∮B da=0 helps explain this when considering that Ф is the same in both equations.
+[[File:Gauss's_Law1234.jpeg|600px]]
+===Hard Examples===
+====Example 1====
-[[File:Gauss_law3.png]]
+Inside a non-uniformly charged sphere with a volume charge density <math>\rho(r) = \rho_0 r^2 </math> for <math>0 \leq r \leq R </math>, where  <math>\rho_0</math>  is a constant and  R  is the radius of the sphere. Find the electric field at a distance  r  from the center, where  r is between 0 and  R, using Gauss's Law.
-Below is a further example of Gauss's Law with explanation.
+=====Solution:=====
+To find the electric field inside the sphere, create a Gaussian surface and take advantage of the spherical symmetry. The charge enclosed within a Gaussian sphere of radius  <math>r</math>  is  <math>\frac{4}{3}\pi \rho_0 r^3 </math>. Gauss's Law then gives  <math>E = \frac{\rho_0 r}{3\varepsilon_0} </math>.
-[[File:Gauss_14.jpg]]
+====Example 2====
-This example shows how to find the electric field in a sphere using Gauss's Law.
+A long, straight wire carries a non-uniform linear charge density <math> \lambda(z) = \lambda_0 e^{-az} </math>, where  <math>\lambda_0</math>  and  a  are constants, and  z  is the distance along the wire. Determine the electric field at a distance r  from the wire using Gauss's Law.
-[[File:gauss's_sphere.jpg]]
+=====Solution:=====
+Set up a Gaussian cylindrical surface around the wire to take advantage of the cylindrical symmetry. The total charge enclosed within the Gaussian surface is found by integrating the linear charge density along the length of the cylinder. Gauss's Law then gives  <math>E = \frac{\lambda_0}{2\pi\varepsilon_0 r} \int_{-\infty}^{\infty} e^{-az} \, dz </math>. Solving this integral yields the expression for the electric field as a function of  r.
 ==Connectedness==
-Gauss's Law, as well as the other Maxwell Equations form a basis for electrodynamics. They are the fundamental core of this field of study.
+Gauss's Law can be found in many areas of science, technology, engineering, and mathematics. In fact, Gauss's Law and  other Maxwell Equations form a basis for electrodynamics. They are the fundamental core of this field of study. Gauss's Law is also connected to the Gaussian/Normal distribution, which is a continuous probability distribution that is characterized by a bell-shaped curve. This distribution is found throughout statistics and probability and is used everyday by people in STEM fields.
-Magnetostatics study is also closely related to Gauss's Law, but in particular Gauss's Law of Magnetism, which is very similar to Gauss's Law relating to electric fields.
+As Industrial and Systems Engineering majors, we deal with many statistical distributions. A very important fundamental distribution is the Gaussian distribution or the Normal Distribution. With inference, the Gaussian Distribution comes up in confidence intervals for single statistics. With comparison inference, like finding the pairwise difference between statistics, generally we use other distributions such as the T-Distribution. However, the T-Distribution approximates the Gaussian distribution with degrees of freedom greater than 29.
+As civil and environmental engineering majors, we also deal with the Gaussian/Normal Distribution in our fields. As an example, the speed data of traffic on a highway is said to follow the normal distribution.
-As an Industrial and Systems Engineering major, we deal with many statistical distributions. A very important fundamental distribution is the Gaussian distribution or the Normal Distribution. With inference, the Gaussian Distribution comes up in confidence intervals for single statistics. With comparison inference, like finding the pairwise difference between statistics, generally we use other distributions such as the T-Distribution. However, the T-Distribution approximates the Gaussian distribution with degrees of freedom greater than 29.
+Additional to many engineering majors. Gauss's Law has applications in many different STEM job fields. Engineering majors learn about the different Gaussian laws in college and apply them in there jobs. Millions of people are applying his laws to the world and will continue to do so. Gauss has had a huge impact on the world and will continue to have a big impact on the future. Gauss has done great work with his work.
 ==History==
-[[File:220px-Carl_Friedrich_Gauss_(C._A._Jensen).jpg]]
+[[File:220px-Carl_Friedrich_Gauss_(C._A._Jensen).jpg||Portrait of Gauss]]
+Gauss's Law was first 'discovered' in 1773, by the great mathematician Joseph-Louis Lagrange. It was further developed by Carl Friedrich Gauss in 1813.
-Carl Friedrich Gauss was a German Mathematician and Physicist who contributed notably to a wide variety of fields regarding mathematical and scientific study. He has been referred to as the "greatest mathematician since antiquity" and the "foremost of mathematicians".  He is considered one of the most impactful and influential contributors to the fields of Mathematics and Physics in history.
+Gauss was a prolific German mathematician and physicist. He made significant contributions to a variety of fields in mathematics and physics. In fact his contributions are so great that he is referred to as the "greatest mathematician since antiquity" and the "foremost of mathematicians".
 == See also ==
+===Further Reading===
 Gauss's Law is tied in closely with the other of Maxwell's equations that can be found here in the Physics Book.
@@ Line 76: / Line 190: @@
 http://physics.info/law-gauss/
+http://teacher.nsrl.rochester.edu/phy122/Lecture_Notes/Chapter24/Chapter24.html
 https://en.wikipedia.org/wiki/Gauss%27s_law
 https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss
+http://physicscatalyst.com/elec/guass_0.php
 ==References==
 http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html
+http://www.pas.rochester.edu/~stte/phy114S09/lectures/lect04.pdf
 https://web.iit.edu/sites/web/files/departments/academic-affairs/academic-resource-center/pdfs/Gauss_Law.pdf
+http://www.sciencedirect.com/science/article/pii/S2090447911000165
 spiff.rit.edu
@@ Line 92: / Line 214: @@
-[[Category:Which Category did you place this in?]]
+[[Category: Electric Field]]
+[[Category: Fields]]
+[[Category: Gauss's Law]]
+[[Category: Maxwell's Equations]]