Gauss's Flux Theorem

Claimed by Brenda Dang Claimed by Patrycja Kotowska (Fall 2016)

This is a page about Gauss's Flux Theorem. A work in progress by Jeff Patz

The Main Idea

Gauss's Flux Theorem is a way of relating charge distribution to its resulting electric field. Analyzing the direction and magnitude of the electric field present on a closed surface helps in understanding the total sum of the charge on the inside of the closed surface. The mathematical model used to related electric field and charge distribution displays that there is a proportional relationship between the amount of electric field passing through a surface and the amount of charge inside of the surface. This application of Gauss's Flux Theorem can be applied to a surface with any shape and size. Since Gauss's Law uses the superposition principle in order to find the sum of all of the charges inside of a closed surface, charges present on the outside of the surface contribute no net flux and produce no electric field on the surface of the object.

A Mathematical Model

In words, the electric flux of a closed surface is equal to the total charge enclosed in the closed surface over the constant epsilon naught. The electric flux of a closed surface is also equal to the surface integral of the electric field evaluated over the closed surface.

[math]\displaystyle{ \Phi_E = \frac{\Sigma q}{\varepsilon_0} = \oint_C E\bullet \hat{n} dA }[/math]

Where E is the electric field, dA is the infinitesimal area in the direction of the electric field, \hat{n} is the unit normal vector, q is the sum of the charges inside the closed surface, and the dot denotes a dot product.

For the special case of a constant electric field, the electric flux is equal to the electric field over the closed surface multiplied by the area and the cosine of the angle between the two vectors.

[math]\displaystyle{ \Phi_E = EAcos(\theta) }[/math]

Where E is the electric field, A is area of the surface, and /theta is the angle between the E and A

Examples

One Surface and Uniform Electric Field

A disk of radius 5 centimeters is in an area of uniform electric field with magnitude 400 Volts/Meter. The angle between the electric field and the disk is 35 degrees.

Using the simplified version of Gauss's Law because the electric field is uniform: [math]\displaystyle{ \Phi_E = EAcos(\theta) }[/math], fill out the known values, which in the case is all values needed.

[math]\displaystyle{ \Phi_E = (400)(\pi0.05^2)cos(35) }[/math]

<Bold textmath>\Phi_E = 2.573</math> Volt Meters

Multiple Surfaces and Uniform Electric field

An equilateral prism lies on top of a rectangular prism with dimensions l = 5cm, w = 7cm, h = 19cm, and is in an area of uniform electric field with magnitude 560 Volts/Meter perpendicular to the long sides of the rectangular prism.

Using the simplified version of Gauss's Law because the electric field is uniform: [math]\displaystyle{ \Phi_E = EAcos(\theta) }[/math]

First of all it is important to note that there are four surfaces that contribute to the net electric flux. two of the two long sides of the equilateral prism and two of the long sides of the rectangular prism. We know the angles between E and A because the top prism is an equilateral triangle, so all the side angles are 60 degrees. With this equation we can write:

[math]\displaystyle{ \Phi_E = E[(w*h)cos(120) + (w*h)cos(60) + (w*h)cos(180) + (w*h)cos(0)] }[/math]

[math]\displaystyle{ \Phi_E = 0 }[/math]

Non-Uniform Electric Field

A solid sphere of radius R has a charge +Q uniformly distributed throughout. Find the Electric field at locations r1<R, r2>R.

For this problem we can utilize the equality of [math]\displaystyle{ \frac{Q}{\varepsilon_0} = \oint_C E\bullet dA }[/math]

For r1<R, we know the line integral of dA is just the circumference of the gaussian surface we took, a sphere of radius r1.

[math]\displaystyle{ \frac{Q}{\varepsilon_0} = E\bullet 4\pi r_1^2 }[/math]

Since we're taking a portion of the sphere we need to find the portion of the charge that is inside of the Gaussian surface. This is as easy as a ratio of volumes.

[math]\displaystyle{ q = \frac{3Q}{4\pi r_1^3}*\frac{4\pi R^3}{3} }[/math]

[math]\displaystyle{ q = Q\frac{r_1^3}{R^3} }[/math]

[math]\displaystyle{ E = Q\frac{r_1}{4\pi R^3} }[/math]

Connectedness

Most of the applications of Gauss's Law are finding the electrical fields of different shapes. This is related to my major because as a EE knowing the electrical fields of different objects definitely comes in handy.

History

Gauss's Law was formulated by Fredreich Gauss in 1835 as a different way to derive Coulomb's Law, and remains as one of Maxwell's equations, the four equations that make up classical electrodynamics.

See also

External links

http://web.hep.uiuc.edu/home/serrede/P435/Lecture_Notes/A_Brief_History_of_Electromagnetism.pdf

http://ethw.org/Gauss'_Law

http://www.storyofmathematics.com/19th_gauss.html

References

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

http://www.physnet.org/modules/pdf_modules/m132.pdf