For more information, please see full course syllabus of AP Physics C/Electricity and Magnetism

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For more information, please see full course syllabus of AP Physics C/Electricity and Magnetism

For more information, please see full course syllabus of AP Physics C/Electricity and Magnetism

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

### Gauss's Law

- Electric Flux: If a uniform electric field E is perpendicular to a plane of area A, then the electric flux through the area is defined as the product EA. If E makes an angle theta with the normal (perpendicular) to the plane, then the flux is given by EAcos(theta).
- In general, given an electric field E in a certain region of space, and a closed surface, we define the electric flux through the closed surface as a surface integral of the dot product of E and da, where da is a vector with a magnitude equal to the area of the surface element da and whose direction is the outward normal to the surface element da, and E is the electric field at the position of the element da.
- Gauss’s law: The electric flux through any closed surface is equal to Q_enc / epsilon_0, where Q_enc is the charge enclosed within the closed surface, and epsilon_0 = 8.854 x 10^-12 in SI units.

### Gauss's Law

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Electric Field Lines
- Electic Flux: Constant E
- Field Lines Equally Spaced
- Area Perpendicular To Field Lines
- Electric Flux
- Area Perpendicular to Electric Lines
- Tilt the Area
- Flux of E Through Area
- Electric Flux: General Case
- Perpendicular at Different Directions
- Electric Field Given On a Patch
- Magnitude of Field
- Direction is Outward Normal
- Flux Through Patch
- Example
- Gauss's Law: Charge Outside
- Gauss's Law: Charge Enclosed
- Example 1: Flux Through Square
- Example 2: Flux Through Cube
- Example 3: Flux Through Pyramid

- Intro 0:00
- Electric Field Lines 0:11
- Magnitude of Field
- Unit Area and Unit Lines
- Number of Lines Passing Through the Unit
- Electic Flux: Constant E 6:51
- Field Lines Equally Spaced
- Area Perpendicular To Field Lines
- Electric Flux
- Area Perpendicular to Electric Lines
- Tilt the Area
- Flux of E Through Area
- Electric Flux: General Case 20:46
- Perpendicular at Different Directions
- Electric Field Given On a Patch
- Magnitude of Field
- Direction is Outward Normal
- Flux Through Patch
- Example 36:09
- Electric Field in Whole Space
- Sphere of Radius 'r'
- Flux Through Sphere
- Gauss's Law: Charge Outside 46:02
- Flux Through Radius Phase is Zero
- Outward normal 'n'
- Gauss's Law: Charge Enclosed 1:00:30
- Drawing Cones
- Example 1: Flux Through Square
- Example 2: Flux Through Cube
- Example 3: Flux Through Pyramid

0 answers

Post by Jamal Tischler on February 10, 2016

I did solve the integral for the first extra example, I wrote 4 pages..

The most important thing about that integral is to realise that da=dy*dz.

And then calculate the double integral.

0 answers

Post by Marcus Suzuki on June 8, 2014

This professor may be slow, but has an amazing amount of knowledge. He explains things very clear.

1 answer

Last reply by: spencer frame

Mon Dec 17, 2012 4:52 PM

Post by Herve Gnidehoue on December 24, 2011

In the case of the sphere, the magnetic flux turned out to be Q/Epsilon O because the area of the sphere is " accidentally" 4 pi R ^2 . The R^2 and the 4 pi later cancel. If the area was a complicated figure, wouldn't the flux be a function of something else besides Q ? Awesome professor.

0 answers

Post by Hamza albobaz on October 25, 2011

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