  Dan Fullerton

Gauss's Law

Slide Duration:

Section 1: Electricity
Electric Charge & Coulomb's Law

30m 48s

Intro
0:00
Objective
0:15
Electric Charges
0:50
Matter is Made Up of Atoms
0:52
Most Atoms are Neutral
1:02
Ions
1:11
Coulomb
1:18
Elementary Charge
1:34
Law of Conservation of Charge
2:03
Example 1
2:39
Example 2
3:42
Conductors and Insulators
4:41
Conductors Allow Electric Charges to Move Freely
4:43
Insulators Do Not Allow Electric Charges to Move Freely
4:50
Resistivity
4:58
Charging by Conduction
5:32
Conduction
5:37
Balloon Example
5:40
Charged Conductor
6:14
Example 3
6:28
The Electroscope
7:16
Charging by Induction
7:57
Bring Positive Rod Near Electroscope
8:08
Ground the Electroscope
8:27
Sever Ground Path and Remove Positive Rod
9:07
Example 4
9:39
Polarization and Electric Dipole Moment
11:46
Polarization
11:54
Electric Dipole Moment
12:05
Coulomb's Law
12:38
Electrostatic Force, Also Known as Coulombic Force
12:48
How Force of Attraction or Repulsion Determined
12:55
Formula
13:08
Coulomb's Law: Vector Form
14:18
Example 5
16:05
Example 6
18:25
Example 7
19:14
Example 8
23:21
Electric Fields

1h 19m 22s

Intro
0:00
Objectives
0:09
Electric Fields
1:33
Property of Space That Allows a Charged Object to Feel a Force
1:40
Detect the Presence of an Electric Field
1:51
Electric Field Strength Vector
2:03
Direction of the Electric Field Vector
2:21
Example 1
3:00
Visualizing the Electric Field
4:13
Electric Field Lines
4:56
E Field Due to a Point Charge
7:19
Derived from the Definition of the Electric Field and Coulomb's Law
7:24
Finding the Electric Field Due to Multiple Point Charges
8:37
Comparing Electricity to Gravity
8:51
Force
8:54
Field Strength
9:09
Constant
9:19
Charge Units vs. Mass Units
9:35
Attracts vs. Repel
9:44
Example 2
10:06
Example 3
17:25
Example 4
24:29
Example 5
25:23
Charge Densities
26:09
Linear Charge Density
26:26
Surface Charge Density
26:30
Volume Charge Density
26:47
Example 6
27:26
Example 7
37:07
Example 8
50:13
Example 9
54:01
Example 10
1:03:10
Example 11
1:13:58
Gauss's Law

52m 53s

Intro
0:00
Objectives
0:07
Electric Flux
1:16
Amount of Electric Field Penetrating a Surface
1:19
Symbol
1:23
Point Charge Inside a Hollow Sphere
4:31
Place a Point Charge Inside a Hollow Sphere of Radius R
4:39
Determine the Flux Through the Sphere
5:09
Gauss's Law
8:39
Total Flux
8:59
Gauss's Law
9:10
Example 1
9:53
Example 2
17:28
Example 3
22:37
Example 4
25:40
Example 5
30:49
Example 6
45:06
Electric Potential & Electric Potential Energy

1h 14m 3s

Intro
0:00
Objectives
0:08
Electric Potential Energy
0:58
Gravitational Potential Energy
1:02
Electric Potential Energy
1:11
Electric Potential
1:19
Example 1
1:59
Example 2
3:08
The Electron-Volt
4:02
Electronvolt
4:16
1 eV is the Amount of Work Done in Moving an Elementary Charge Through a Potential Difference of 1 Volt
4:26
Conversion Ratio
4:41
Example 3
4:52
Equipotential Lines
5:35
Topographic Maps
5:36
Lines Connecting Points of Equal Electrical Potential
5:47
Always Cross Electrical Field Lines at Right Angles
5:57
Gradient of Potential Increases As Equipotential Lines Get Closer
6:02
Electric Field Points from High to Low Potential
6:27
Drawing Equipotential Lines
6:49
E Potential Energy Due to a Point Charge
8:20
Electric Force from Electric Potential Energy
11:59
E Potential Due to a Point Charge
13:07
Example 4
14:42
Example 5
15:59
Finding Electric Field From Electric Potential
19:06
Example 6
23:41
Example 7
25:08
Example 8
26:33
Example 9
29:01
Example 10
31:26
Example 11
43:23
Example 12
51:51
Example 13
58:12
Electric Potential Due to Continuous Charge Distributions

1h 1m 28s

Intro
0:00
Objectives
0:10
Potential Due to a Charged Ring
0:27
Potential Due to a Uniformly Charged Desk
3:38
Potential Due to a Spherical Shell of Charge
11:21
Potential Due to a Uniform Solid Sphere
14:50
Example 1
23:08
Example 2
30:43
Example 3
41:58
Example 4
51:41
Conductors

20m 35s

Intro
0:00
Objectives
0:08
Charges in a Conductor
0:32
Charge is Free to Move Until the
0:36
All Charge Resides at Surface
2:18
Field Lines are Perpendicular to Surface
2:34
Electric Field at the Surface of a Conductor
3:04
Looking at Just the Outer Surface
3:08
Large Electric Field Where You Have the Largest Charge Density
3:59
Hollow Conductors
4:22
Draw Hollow Conductor and Gaussian Surface
4:36
Applying Gaussian Law
4:53
Any Hollow Conductor Has Zero Electric Field in Its Interior
5:24
5:35
Electric Field and Potential Due to a Conducting Sphere
6:03
Example 1
7:31
Example 2
12:39
Capacitors

41m 23s

Intro
0:00
Objectives
0:08
What is a Capacitor?
0:42
Electric Device Used to Store Electrical Energy
0:44
Place Opposite Charges on Each Plate
1:10
Develop a Potential Difference Across the Plates
1:14
Energy is Stored in the Electric Field Between the Plates
1:17
Capacitance
1:22
Ratio of the Charge Separated on the Plates of a Capacitor to the Potential Difference Between the Plates
1:25
Units of Capacitance
1:32
1:37
Formula
1:52
Calculating Capacitance
1:59
Assume Charge on Each Conductor
2:05
Find the Electric Field
2:11
Calculate V by Integrating the Electric Field
2:21
Utilize C=Q/V to Solve for Capitance
2:33
Example 1
2:44
Example 2
5:30
Example 3
10:46
Energy Stored in a Capacitor
15:25
Work is Done Charging a Capacitor
15:28
Solve For That
15:55
Field Energy Density
18:09
Amount of Energy Stored Between the Plates of a Capacitor
18:11
Example
18:25
Dielectrics
20:44
Insulating Materials Place Between Plates of Capacitor to Increase The Devices' Capacitance
20:47
Electric Field is Weakened
21:00
The Greater the Amount of Polarization The Greater the Reduction in Electric Field Strength
21:58
Dielectric Constant (K)
22:30
Formula
23:00
Net Electric Field
23:35
Key Take Away Point
23:50
Example 4
24:00
Example 5
25:50
Example 6
26:50
Example 7
28:53
Example 8
30:57
Example 9
32:55
Example 10
34:59
Example 11
37:35
Example 12
39:57
Section 2: Current Electricity
Current & Resistance

17m 59s

Intro
0:00
Objectives
0:08
Electric Current
0:44
Flow Rate of Electric Charge
0:45
Amperes
0:49
Positive Current Flow
1:01
Current Formula
1:19
Drift Velocity
1:35
Constant Thermal Motion
1:39
Net Electron Flow
1:43
When Electric Field is Applied
1:49
Electron Drift Velocity
1:55
Derivation of Current Flow
2:12
Apply Electric Field E
2:20
Define N as the Volume Density of Charge Carriers
2:27
Current Density
4:33
Current Per Area
4:36
Formula
4:44
Resistance
5:14
Ratio of the Potential Drop Across an Object to the Current Flowing Through the Object
5:19
5:23
Resistance of a Wire
6:05
Depends on Resistivity
6:09
Resistivity Relates to the Ability of a Material to Resist the Flow of Electrons
6:25
Refining Ohm's Law
7:22
Conversion of Electric Energy to Thermal Energy
8:23
Example 1
9:54
Example 2
10:54
Example 3
11:26
Example 4
14:41
Example 5
15:24
Circuits I: Series Circuits

29m 8s

Intro
0:00
Objectives
0:08
Ohm's Law Revisited
0:39
Relates Resistance, Potential Difference, and Current Flow
0:39
Formula
0:44
Example 1
1:09
Example 2
1:44
Example 3
2:15
Example 4
2:56
Electrical Power
3:26
Transfer of Energy Into Different Types
3:28
Light Bulb
3:37
Television
3:41
Example 5
3:49
Example 6
4:27
Example 7
5:12
Electrical Circuits
5:42
Closed-Loop Path Which Current Can Flow
5:43
Typically Comprised of Electrical Devices
5:52
Conventional Current Flows from High Potential to Low Potential
6:04
Circuit Schematics
6:26
Three-dimensional Electrical Circuits
6:37
Source of Potential Difference Required for Current to Flow
7:29
Complete Conducting Paths
7:42
Current Only Flows in Complete Paths
7:43
Left Image
7:46
Right Image
7:56
Voltmeters
8:25
Measure the Potential Difference Between Two Points in a Circuit
8:29
Can Remove Voltmeter from Circuit Without Breaking the Circuit
8:47
Very High Resistance
8:53
Ammeters
9:31
Measure the Current Flowing Through an Element of a Circuit
9:32
Very Low Resistance
9:46
Put Ammeter in Correctly
10:00
Example 8
10:24
Example 9
11:39
Example 10
12:59
Example 11
13:16
Series Circuits
13:46
Single Current Path
13:49
Removal of Any Circuit Element Causes an Open Circuit
13:54
Kirchhoff's Laws
15:48
Utilized in Analyzing Circuits
15:54
Kirchhoff's Current Law
15:58
Junction Rule
16:02
Kirchhoff's Voltage Law
16:30
Loop Rule
16:49
Example 12
16:58
Example 13
17:32
Basic Series Circuit Analysis
18:36
Example 14
22:06
Example 15
22:29
Example 16
24:02
Example 17
26:47
Circuits II: Parallel Circuits

39m 9s

Intro
0:00
Objectives
0:16
Parallel Circuits
0:38
Multiple Current Paths
0:40
Removal of a Circuit Element May Allow Other Branches of the Circuit to Continue Operating
0:44
Draw a Simple Parallel Circuit
1:02
Basic Parallel Circuit Analysis
3:06
Example 1
5:58
Example 2
8:14
Example 3
9:05
Example 4
11:56
Combination Series-Parallel Circuits
14:08
Circuit Doesn't Have to be Completely Serial or Parallel
14:10
Look for Portions of the Circuit With Parallel Elements
14:15
Lead to Systems of Equations to Solve
14:42
Analysis of a Combination Circuit
14:51
Example 5
20:23
Batteries
28:49
Electromotive Force
28:50
Pump for Charge
29:04
Ideal Batteries Have No Resistance
29:10
Real Batteries and Internal Resistance
29:20
Terminal Voltage in Real Batteries
29:33
Ideal Battery
29:50
Real Battery
30:25
Example 6
31:10
Example 7
33:23
Example 8
35:49
Example 9
38:43

34m 3s

Intro
0:00
Objectives
0:17
Capacitors in Parallel
0:51
Store Charge on Plates
0:52
Can Be Replaced with an Equivalent Capacitor
0:56
Capacitors in Series
1:12
Must Be the Same
1:13
Can Be Replaced with an Equivalent Capacitor
1:15
RC Circuits
1:30
Comprised of a Source of Potential Difference, a Resistor Network, and Capacitor
1:31
RC Circuits from the Steady-State Perspective
1:37
Key to Understanding RC Circuit Performance
1:48
Charging an RC Circuit
2:08
Discharging an RC Circuit
6:18
The Time Constant
8:49
Time Constant
8:58
By 5 Time Constant
9:19
Example 1
9:45
Example 2
13:27
Example 3
16:35
Example 4
18:03
Example 5
19:39
Example 6
26:14
RC Circuits: Transient Analysis

1h 1m 7s

Intro
0:00
Objectives
0:13
Charging an RC Circuit
1:11
Basic RC Circuit
1:15
Graph of Current Circuit
1:29
Graph of Charge
2:17
Graph of Voltage
2:34
Mathematically Describe the Charts
2:56
Discharging an RC Circuit
13:29
Graph of Current
13:47
Graph of Charge
14:08
Graph of Voltage
14:15
Mathematically Describe the Charts
14:30
The Time Constant
20:03
Time Constant
20:04
By 5 Time Constant
20:14
Example 1
20:39
Example 2
28:53
Example 3
27:02
Example 4
44:29
Example 5
55:24
Section 3: Magnetism
Magnets

8m 38s

Intro
0:00
Objectives
0:08
Magnetism
0:35
Force Caused by Moving Charges
0:36
Dipoles
0:40
Like Poles Repel, Opposite Poles Attract
0:53
Magnetic Domains
0:58
Random Domains
1:04
Net Magnetic Field
1:26
Example 1
1:40
Magnetic Fields
2:03
Magnetic Field Strength
2:04
Magnets are Polarized
2:16
Magnetic Field Lines
2:53
Show the Direction the North Pole of a Magnet Would Tend to Point if Placed on The Field
2:54
Direction
3:25
Magnetic Flux
3:41
The Compass
4:05
Earth is a Giant Magnet
4:07
Earth's Magnetic North Pole
4:10
Compass Lines Up with the Net Magnetic Field
4:48
Magnetic Permeability
5:00
Ratio of the magnetic Field Strength Induced in a Material to the Magnetic Field Strength of the Inducing Field
5:01
Free Space
5:13
Permeability of Matter
5:41
Highly Magnetic Materials
5:47
Magnetic Dipole Moment
5:54
The Force That a Magnet Can Exert on Moving Charges
5:59
Relative Strength of a Magnet
6:04
Example 2
6:26
Example 3
6:52
Example 4
7:32
Example 5
7:57
Moving Charges In Magnetic Fields

29m 7s

Intro
0:00
Objectives
0:08
Magnetic Fields
0:57
Vector Quantity
0:59
Tesla
1:08
Gauss
1:14
Forces on Moving Charges
1:30
Magnetic Force is Always Perpendicular to the Charged Objects Velocity
1:31
Magnetic Force Formula
2:04
Magnitude of That
2:20
Image
2:29
Direction of the Magnetic Force
3:54
Right-Hand Rule
3:57
Electron of Negative Charge
4:04
Example 1
4:51
Example 2
6:58
Path of Charged Particles in B Fields
8:07
Magnetic Force Cannot Perform Work on a Moving Charge
8:08
Magnetic Force Can Change Its Direction
8:11
Total Force on a Moving Charged Particle
9:40
E Field
9:50
B Field
9:54
Lorentz Force
9:57
Velocity Selector
10:33
Charged Particle in Crosses E and B Fields Can Undergo Constant Velocity Motion
10:37
Particle Can Travel Through the Selector Without Any Deflection
10:49
Mass Spectrometer
12:21
Magnetic Fields Accelerate Moving Charges So That They Travel in a Circle
12:26
Used to Determine the Mass of An Unknown Particle
12:32
Example 3
13:11
Example 4
15:01
Example 5
16:44
Example 6
17:33
Example 7
19:12
Example 8
19:50
Example 9
24:02
Example 10
25:21
Forces on Current-Carrying Wires

17m 52s

Intro
0:00
Objectives
0:08
Forces on Current-Carrying Wires
0:42
Moving Charges in Magnetic Fields Experience Forces
0:45
Current in a Wire is Just Flow of Charges
0:49
Direction of Force Given by RHR
4:04
Example 1
4:22
Electric Motors
5:59
Example 2
8:14
Example 3
8:53
Example 4
10:09
Example 5
11:04
Example 6
12:03
Magnetic Fields Due to Current-Carrying Wires

24m 43s

Intro
0:00
Objectives
0:08
Force on a Current-Carrying Wire
0:38
Magnetic Fields Cause a Force on Moving Charges
0:40
Current Carrying Wires
0:44
How to Find the Force
0:55
Direction Given by the Right Hand Rule
1:04
Example 1
1:17
Example 2
2:26
Magnetic Field Due to a Current-Carrying Wire
4:20
Moving Charges Create Magnetic Fields
4:24
Current-Carrying Wires Carry Moving Charges
4:27
Right Hand Rule
4:32
Multiple Wires
4:51
Current-Carrying Wires Can Exert Forces Upon Each Other
4:58
First Right Hand Rule
5:15
Example 3
6:46
Force Between Parallel Current Carrying Wires
8:01
Right Hand Rules to Determine Force Between Parallel Current Carrying Wires
8:03
Find Magnetic Field Due to First Wire, Then Find Direction of Force on 2nd Wire
8:08
Example
8:20
Gauss's Law for Magnetism
9:26
Example 4
10:35
Example 5
12:57
Example 6
14:19
Example 7
16:50
Example 8
18:15
Example 9
18:43
The Biot-Savart Law

21m 50s

Intro
0:00
Objectives
0:07
Biot-Savart Law
0:24
Brute Force Method
0:49
Draw It Out
0:54
Diagram
1:35
Example 1
3:43
Example 2
7:02
Example 3
14:31
Ampere's Law

26m 31s

Intro
0:00
Objectives
0:07
Ampere's Law
0:27
Finds the Magnetic Field Due to Current Flowing in a Wire in Situations of Planar and Cylindrical Symmetry
0:30
Formula
0:40
Example
1:00
Example 1
2:19
Example 2
4:08
Example 3
6:23
Example 4
8:06
Example 5
11:43
Example 6
13:40
Example 7
17:54
Magnetic Flux

7m 24s

Intro
0:00
Objectives
0:07
Magnetic Flux
0:31
Amount of Magnetic Field Penetrating a Surface
0:32
Webers
0:42
Flux
1:07
Total Magnetic Flux
1:27
Magnetic Flux Through Closed Surfaces
1:51
Gauss's Law for Magnetism
2:20
Total Flux Magnetic Flux Through Any Closed Surface is Zero
2:23
Formula
2:45
Example 1
3:02
Example 2
4:26

1h 4m 33s

Intro
0:00
Objectives
0:08
0:44
0:46
Direction of the Induced Current is Given by Lenz's Law
1:09
Formula
1:15
Lenz's Law
1:49
Lenz's Law
2:14
Lenz's Law
2:16
Example
2:30
Applying Lenz's Law
4:09
If B is Increasing
4:13
If B is Decreasing
4:30
Maxwell's Equations
4:55
Gauss's Law
4:59
Gauss's Law for Magnetism
5:16
Ampere's Law
5:26
5:39
Example 1
6:14
Example 2
9:36
Example 3
11:12
Example 4
19:33
Example 5
26:06
Example 6
31:55
Example 7
42:32
Example 8
48:08
Example 9
55:50
Section 4: Inductance, RL Circuits, and LC Circuits
Inductance

6m 41s

Intro
0:00
Objectives
0:08
Self Inductance
0:25
Ability of a Circuit to Oppose the Magnetic Flux That is Produced by the Circuit Itself
0:27
Changing Magnetic Field Creates an Induced EMF That Fights the Change
0:37
Henrys
0:44
Function of the Circuit's Geometry
0:53
Calculating Self Inductance
1:10
Example 1
3:40
Example 2
5:23
RL Circuits

42m 17s

Intro
0:00
Objectives
0:11
Inductors in Circuits
0:49
Inductor Opposes Current Flow and Acts Like an Open Circuit When Circuit is First Turned On
0:52
Inductor Keeps Current Going and Acts as a Short
1:04
If the Battery is Removed After a Long Time
1:16
Resister Dissipates Power, Current Will Decay
1:36
Current in RL Circuits
2:00
Define the Diagram
2:03
Mathematically Solve
3:07
Voltage in RL Circuits
7:51
Voltage Formula
7:52
Solve
8:17
Rate of Change of Current in RL Circuits
9:42
Current and Voltage Graphs
10:54
Current Graph
10:57
Voltage Graph
11:34
Example 1
12:25
Example 2
23:44
Example 3
34:44
LC Circuits

9m 47s

Intro
0:00
Objectives
0:08
LC Circuits
0:30
Assume Capacitor is Fully Charged When Circuit is First Turned On
0:38
Interplay of Capacitor and Inductor Creates an Oscillating System
0:42
Charge in LC Circuit
0:57
Current and Potential in LC Circuits
7:14
Graphs of LC Circuits
8:27
Section 5: Maxwell's Equations
Maxwell's Equations

3m 38s

Intro
0:00
Objectives
0:07
Maxwell's Equations
0:19
Gauss's Law
0:20
Gauss's Law for Magnetism
0:44
1:00
Ampere's Law
1:18
Revising Ampere's Law
1:49
Allows Us to Calculate the Magnetic Field Due to an Electric Current
1:50
Changing Electric Field Produces a Magnetic Field
1:58
Conduction Current
2:33
Displacement Current
2:44
Maxwell's Equations (Complete)
2:58
Section 6: Sample AP Exams
1998 AP Practice Exam: Multiple Choice Questions

32m 33s

Intro
0:00
0:11
Multiple Choice 36
0:36
Multiple Choice 37
2:07
Multiple Choice 38
2:53
Multiple Choice 39
3:32
Multiple Choice 40
4:37
Multiple Choice 41
4:43
Multiple Choice 42
5:22
Multiple Choice 43
6:00
Multiple Choice 44
8:09
Multiple Choice 45
8:27
Multiple Choice 46
9:03
Multiple Choice 47
9:30
Multiple Choice 48
10:19
Multiple Choice 49
10:47
Multiple Choice 50
12:25
Multiple Choice 51
13:10
Multiple Choice 52
15:06
Multiple Choice 53
16:01
Multiple Choice 54
16:44
Multiple Choice 55
17:10
Multiple Choice 56
19:08
Multiple Choice 57
20:39
Multiple Choice 58
22:24
Multiple Choice 59
22:52
Multiple Choice 60
23:34
Multiple Choice 61
24:09
Multiple Choice 62
24:40
Multiple Choice 63
25:06
Multiple Choice 64
26:07
Multiple Choice 65
27:26
Multiple Choice 66
28:32
Multiple Choice 67
29:14
Multiple Choice 68
29:41
Multiple Choice 69
31:23
Multiple Choice 70
31:49
1998 AP Practice Exam: Free Response Questions

29m 55s

Intro
0:00
0:14
Free Response 1
0:22
Free Response 2
10:04
Free Response 3
16:22
Bookmark & Share Embed

## Copy & Paste this embed code into your website’s HTML

Please ensure that your website editor is in text mode when you paste the code.
(In Wordpress, the mode button is on the top right corner.)
×
• - Allow users to view the embedded video in full-size.

• ## Related Books 1 answer Last reply by: Professor Dan FullertonFri Feb 16, 2018 5:53 PMPost by Kevin Fleming on February 16, 2018Professor Fullerton,For example 5, question a, part ii, why did you equate the Q(enclosed) by the gaussian surface surrounding the outer radius of the metallic sphere to be +Q if, in part i, you said that the Q(enclosed) for the gaussian surface situated between the inner and outer surface was eqaual to 0. Wouldn't the +Q charge be neutralized by the -Q charge of the inner surface? Thanks 2 answersLast reply by: Sunanda EluriTue Sep 13, 2016 9:56 AMPost by Sunanda Eluri on September 13, 2016Hello sir, If the radius of the gaussian surface is doubled, will it affect the flux passing through the surface? 1 answer Last reply by: Professor Dan FullertonSat Sep 3, 2016 3:58 PMPost by isaac maingi on September 3, 2016when moving the 4pir^2 from the left side of the equation to the right side of the equation to solve for E, whey does the 4pir^2 go to the denominator instead of the numerator? 4 answers Last reply by: Professor Dan FullertonMon Feb 22, 2016 6:17 AMPost by Jim Tang on February 20, 2016Hm... Same thingIn Example 4, my intuition told me the caps were 0 since dA and E were at 90 degrees...I think that's what you meant by "cancelling out." In this case, it's not cancelling out between objects...but of simply cancelling out? 2 answersLast reply by: Jim TangSat Feb 20, 2016 9:02 PMPost by Jim Tang on February 20, 2016For Example 2, I think you have the wrong explanation for why the sides cancel out? My intuition tells me since cos(90)=0, the E will all be 0 along those sides. I can't see anything "cancelling out." 1 answer Last reply by: Professor Dan FullertonThu Feb 18, 2016 10:32 AMPost by Daniel Jansson on February 18, 2016Mr. Fullerton, I'd love some examples of Gauss law in differential form as well!/Danny (Engineering physics student) 1 answer Last reply by: Professor Dan FullertonSun Apr 19, 2015 3:02 PMPost by Patrick Jin on April 19, 2015Hi! Thank you for the lecture first of all. I have questions about the example 2. So when we use the cylinder shaped gaussian field for the infinite field, all the electric flux to the side cancel out. Would it be correct if I say that up and down flux cancels out too just like the flux on the side, but since we are dealing with an infinite plane ( which technically only has top and bottom part with no left and right sides ) we are only considering the magnitude of the up and down flux ? 1 answerLast reply by: Miras KarazhigitovSun Apr 5, 2015 7:52 PMPost by Miras Karazhigitov on April 5, 2015Hello, can you explain why in 5th example (1. a) ii ) the total charge is equal to +Q? 1 answer Last reply by: Professor Dan FullertonFri Apr 3, 2015 8:48 AMPost by Luvivia Chang on April 2, 2015hello professor, Can you explain why the flux through top and bottom are equal in the example 1 while the sum of the fluxes through left and right caps are zero in example 3? what is the diffeence between the 2 situations? 1 answer Last reply by: Professor Dan FullertonTue Feb 3, 2015 6:08 AMPost by Arjun Srivatsa on February 3, 2015Why does the electric flux of the sides of a cyclinder cancel out by symmetry in Example 2? 1 answer Last reply by: Professor Dan FullertonFri Nov 14, 2014 6:06 AMPost by QuangNguyen VoHuynh on November 14, 2014Also, I would like to know the reason why the electric filed in the conductor is zero? Is the reason because the total flux is zero? 1 answer Last reply by: Professor Dan FullertonFri Nov 14, 2014 6:05 AMPost by QuangNguyen VoHuynh on November 14, 2014Can you explain me the reason why we must use the cylinder when finding the electric filed in infinite plane using Gauss's Law?

### Gauss's Law

• Electric flux is the amount of electric field penetrating a surface.
• Normals to closed surfaces point from the inside to the outside.
• The total flux through a closed surface is positive is there is more flux from inside to outside than outside to inside.
• Gauss’s Law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of the material enclosed.
• Gauss’s Law is useful for finding the electric field due to charge distributions for cases of spherical, cylindrical, and planar symmetry. Gauss’s Law is always true, but typically only useful in these symmetric situations.

### 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 0:00
• Objectives 0:07
• Electric Flux 1:16
• Amount of Electric Field Penetrating a Surface
• Symbol
• Point Charge Inside a Hollow Sphere 4:31
• Place a Point Charge Inside a Hollow Sphere of Radius R
• Determine the Flux Through the Sphere
• Gauss's Law
• Total Flux
• Gauss's Law 9:10
• Example 1 9:53
• Example 2 17:28
• Example 3 22:37
• Example 4 25:40
• Example 5 30:49
• Example 6 45:06

### Transcription: Gauss's Law

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about Gauss’s law.0003

In our last lesson, we spent quite a bit of time doing derivations of electric fields due to continuous point charge distributions.0008

Gauss’s law is one of my favorite lessons in Physics.0016

After we have gone through that really tough lesson and you survived it, we are going to show you a much simpler way to deal with a bunch of those situations.0019

Our objectives, number 1, understand the relationship between electric field and electric flux.0028

We will start by talking about what electric flux is.0033

Secondly, calculate the flux of an electric field to a variety of surfaces.0037

State and apply the relationship between flux and lines of force.0041

State Gauss’s law in integral form and apply it qualitatively.0045

Apply Gauss’s law to determine the electric field for planar, spherical, and cylindrical asymmetric charge distributions,0049

that is going to be a big portion of the lesson.0056

Almost without a doubt that is going to be involved in at least one of the free response questions on your AP exam.0058

Finally, apply Gauss’s law and determine the charge density or total charge on the surface in terms of the electric field near that surface.0066

Let us start off by talking about what is electric flux.0074

Electric flux is the amount of electric field penetrating a surface, we will get it the symbol Φ.0079

If we have some random surface, whatever surface happens to be,0086

what we will do is we will define a small piece of that surface, and from that, coming out of it perpendicular to it, we will define a vector da.0090

Where the direction of da, the area is perpendicular to that surface.0105

Assume we have some amount of electric field coming out of that surface as well, coming out through that area.0108

The angle between these we will call Θ.0117

Then this little bit of electric flux coming through that red portion, coming through our little bit of da,0122

we will define as dΦ the flux = the electric field dotted with that area vector da or E da cos Θ for dealing with the magnitude0128

which implies then that the total flux is going to be the integral about little bit of electric flux0147

which is going to be the integral / the entire area of all those little bits of E da.0156

Almost think of this like airflow or rain on a cookie sheet.0165

You have a cookie sheet held flat and it is raining on it, the amount of rain that is hitting that cookie sheet would be equivalent to the flux.0169

Flux goes through the surface that does not work with rain and cookie sheet, it made up the idea.0177

If you start tilting it in different directions, different amounts of rain hit the cookie sheet or go through it, went through a great perhaps, something like that.0181

What happens if you have a close surface instead of just this open surface?0193

Let us draw a close surface, something like that.0198

You will have to try and picture that in 3D because my drawing skills are poor at best.0203

In this case, we could define some little piece over here as da.0211

You could define say something over here as da.0220

Once again, the total flux is going to be the integral over the close surface, a shorthand notation integral over a closed surface of E⋅ da.0229

You are integrating over the entire thing to enclose it.0243

By convention, normal to close surfaces point from the inside to the outside.0246

Total flux through a close surface is positive.0254

If there is more flux from inside to outside, it is positive if there is more going out than going in.0257

If more is going than coming out, we would call that a negative flux.0262

This is just a convention as far as science go.0266

Let us take a look now at a point charge inside a hollow sphere to help us get an idea of how the Gauss’s law thing really works.0270

We will place a point charge inside a hollow sphere of radius R, let us draw ourselves a nice happy little sphere of some radius R, in its center we are going to put a point charge.0279

Let us make it +q.0295

It is pretty easy to see that electric field lines are going to be symmetric due out, assuming I can draw a little bit better sphere.0298

We could then say that the little bit of electric flux through some little piece of this da0309

then we define that da here is going to be equal to the electric field dotted with da.0319

which as we said previously is just E da cos Θ.0331

If it is a point charge inside a hollow sphere, anywhere in the electric field goes through that sphere,0338

it is going to be perpendicular uniformly around the entire sphere.0343

That means that Θ, everywhere the angle between those two was going to be 0.0347

cos of Θ is going to be = 1.0354

Therefore, the differential of that total flux, that little bit of flux is going to be just E da.0358

Our total flux then it is going to be the integral of all those little bits of flux or the integral over this close surface around the entire sphere of E da.0367

Since it is a point charge inside the center of a sphere the electric field is going to be the same everywhere as you go around the entire area.0381

For the purposes of this problem, E is a constant so that E comes out of the integral sign, E integral over the close surface of da.0389

The integral over the close surface of that differential of area, as you go around the entire surface,0399

it is just going to give you that entire area so that becomes Ea.0404

And as you know the area of a sphere = 4π R².0410

We can write then that the total flux is going to be = E × 4π R².0421

Even further though, we know the electric field.0435

We just spent a whole lesson on it.0439

The electric field due to that point charge = Q/ 4 π E₀ R² in the direction of r ̂ coming out from that point charge.0442

We could then write the flux as being = we have got our 4π R² × our electric field Q/ 4π E₀ R² and all of that then must be =,0458

Let us see, 4π 4π R² R² = Q/ E₀, simplifies very nicely.0486

Which implies then that the total flux = the integral/ the close surface of E⋅ da which was equal to the total charge enclosed by our sphere ÷ E₀.0499

There is a poor mans derivation of what we are going to call Gauss’s law,0518

the first of Maxwell's four equations which form the backbone of this entire E and M course.0525

Gauss’s law, you got to know it front words, backwards, upside down, and underwater.0531

The total flux is the integral over the close surface of E ⋅ da which is the total charge enclosed by your surface ÷ E₀.0536

Looking at this in a little bit more detail, what is this law good for?0549

It is extremely useful for finding the electric field due to charge distributions in cases0554

where you have symmetry and you need specifically 1 of 3 types of symmetry.0559

You need spherical symmetry, planar symmetry, or cylindrical symmetry, for this to be a very useful law.0563

Gauss’s law always holds, it is always true.0570

However, it is really only useful for the most part in these cases.0573

What it states, the total flux is the integral/ the close surface of E ⋅ da which is the total charge enclosed by that surface ÷ Ε 0, that constant.0578

Let us do some examples.0591

Find the electric field inside and outside a thin hollow shell of uniformly distributed charged Q.0594

Let us see if we can draw this first.0602

Find the electric field inside and outside a thin hollow shell of uniformly distributed charged Q.0605

There is our shell of charge with some amount Q.0615

What we are going to do is we are going to choose a Gaussian surface and vision this imaginary close surface,0621

in this case it is going to be a sphere to give us the most possible symmetry.0630

First we are going to do that inside the shell then we are going to do it outside the shell,0633

whenever we want to know the electric field strength.0638

By symmetry, the electric field at all points on this Gaussian surface, on this Gaussian sphere must be the same and must point radially in or out.0640

Let us look inside this sphere.0654

We will define a Gaussian sphere here in green, this mental sphere that we have made up to help us solve the problem and0656

we will say that at some radius that we will call Ri which is inside the radius of our shell of charge there in blue with radius R.0663

Here is our Gaussian surface in green and we are looking for the electric field inside.0677

Where Ri is less than R.0684

Let us even spell that out Ri is less than R for this analysis.0686

Once we do that, we can say that the integral over the close surface of E ⋅ da = Q enclose by Gaussian surface ÷ E₀.0693

Gauss’s law we start off by writing it, then we also know that the area over here of our Gaussian surface0705

is just going to be 4π Ri² because we picked a sphere inside the shell of charge.0717

We have as much symmetry as possible.0726

Then we can state that the left hand side E is constant everywhere.0729

The integral of da is just going to be the area, left hand side becomes ea = Q enclosed/ E₀.0735

We did not have to do any real integration when you are using Gauss’s law, vast majority of the time if you have to do any real integration over here,0743

you probably did not pick the greatest surface.0750

Ea = Q enclosed/ E₀ which implies then that E × our area 4π Ri² = the charge enclosed ÷ E₀.0753

Or we can say then that E = charge enclosed / 4π E₀ Ri².0769

What makes this one especially easy though is the charge enclose by our Gaussian sphere is 0,0784

all the charges outside in this blue shell of charge there is nothing inside the Gaussian sphere.0789

Q enclosed = 0 therefore, the electric field = 0 inside that hollow shell of charge.0794

Gauss’s law defined the electric field inside where the 0 everywhere.0807

Let us take a look at the second area.0812

We are going to look when we are looking outside that hollow shell of charge.0814

If you are looking at the green one, let us take a look at this purple one.0819

We will set a Gaussian surface, another sphere, that hollow shell I should say, outside our shell of charge.0823

We will call this one Ro, as radius Ro.0834

We are looking where Ro is greater than R or outside that hollow shell of charge.0841

We will start off by rating Gauss’s law again, integral / the close surface of E ⋅ da = charge enclosed ÷ E₀.0850

The same thing again and the same derivation as well, E is a constant everywhere because we chose our Gaussian surface very carefully.0864

The integral over the close surface of da is just going to be a, so the Ea = Q enclosed/ E₀.0873

Where this area is the area of your Gaussian sphere, the area of that sphere is going to be 4π Ro² = Q enclosed/ E₀0884

which implies then that the electric field is going to be our Q enclosed/ 4π E₀ R 0².0906

The total charge enclosed + Q that is going to be Q ÷ 4π E₀ R 0².0922

For any R that is an Ro, that is outside this hollow shell of charge.0934

This is the same as if all the charge was placed in a point in the center of the sphere.0942

You could treat it like it is a point charge once you are outside, it does not matter.0947

What shape it is, as long as you can use Gauss’s law in it, this could be any amount of charge, any size diameter shell,0951

as long as you are outside you can treat it all as if all of that charge was condensed in a single point inside the center of the shell.0958

See how powerful this is? how much slicker this is?0967

You can just see the beauty and all the symmetry of the math as you do these sorts of problems.0972

We will go through a couple more examples to help clarify it but let us make a graph quickly of the electric field vs. radius just to help finish this one off.0976

There is our radius R, there is our electric field, and what we have is inside that value of R, let us make this R,0992

I will write it in blue even so it corresponds to our shell of charge here.1003

There is the radius of our shell of charge, inside that the electric field is 0 and outside that is Q/ 4π E₀ R 0².1009

What we have is a discontinuity where we got something like this, where that is proportional to α sign or fish sign, whatever you want to call it.1020

It is proportional to 1/ R² as long as you are outside R.1030

You can use Gauss’s to help you find these so quickly and so easily.1040

Let us do another example.1046

Find the electric field due to an infinite plane of uniform charge density Σ.1049

We have got an infinite plane, let us draw that here.1055

We are applying a charge that goes on and on and on and we have given it sum surface charge density Σ which is total charge ÷ area.1063

Since it is infinite, we cannot know the total charge or areas.1072

We will find a ratio of those and define the Σ.1075

Choose our Gaussian surface, you got to be smart about this and having done these sorts of things1079

before I know that a good way to do this is to pick a cylinder.1085

I'm going to pick a cylinder, I think almost of a soda can, centered that the plane.1089

There is the top and it also extends to the same amount underneath.1107

There is our Gaussian surface, a cylinder and we will set it so that it is some distance D above and D below that infinite plane.1112

By symmetry, the electric field at all points on the cylinder must point perpendicular to the plane through the caps of the cylinder.1127

On the sides, it should be pretty easy to see by symmetry everything is going to cancel out,1134

We are only going to end up having to worry about these caps our cylinder but let us go through that step by step.1138

Let us start by writing Gauss’s law, the integral / the close surface of E ⋅ da = our enclosed charge ÷ E₀,1146

which implies then since we know Σ is Q/ a and Q = Σ a.1161

We can write then the left hand side, this total flux I'm going to break up into pieces.1170

I'm going to take a look at the flux through the top, the flux through the sides, and the flux through the bottom cap.1176

That is the flux through the top + the electric flux through the bottom + total electric flux through the sides to give me my total1183

which is this left hand side and that must be equal to our charge enclosed is Σ × a, where that is going to be a ÷ Ε 0.1201

Of course, by symmetry we can look and see that the flux to the sides of this are all going to cancel out so the total flux through the sides is going to be 0.1220

We can say then that the flux through the top + the electric flux through the bottom must = Σ a/ E₀.1229

The flux through the top and the flux to the bottom must be the same because they are symmetric.1248

This one was pointing down and this one was pointing up.1256

We are worried about magnitudes for the time being.1257

The flux through the top = the flux through the bottom which by the way has to be the electric field at that point × the area so that = Ea.1260

Flux through the top Ea + flux in the bottom Ea gives us 2 Ea = Σ a / E₀ or solving for the electric field then, the electric field just = Σ / 2 E₀.1275

There is no dependents whatsoever on D and we have seen this before when we talk about infinite planes and infinite lines, it does not matter.1295

Since, it is infinite the only thing you are worried about is that surface charge density.1303

How much simpler is that in the derivation that we did with that disk that we grew bigger and bigger1308

and bigger in the last lesson to figure out the electric field due to an infinite plane.1313

It is much slicker derivation right here using Gauss’s law.1318

Let us do a graph of the electric field vs. distance again just so that we are consistent.1322

There is D, there is our electric field, and since there is no dependence on D, our graphs looks like a nice straight line at Σ/ 2 E₀.1335

Very good, let us see if we can extend this a little further and talk about parallel plates.1353

Find the electric field outside between 2 opposite recharged parallel planes or plates.1360

Let us draw our plates first, we will draw a top plate, we will draw a bottom plate.1365

We will say that this one has charged + charge density + Σ.1381

This one must have - Σ as its charge density.1385

What we are going to do is we are going to look at the electric field due to the different plates in different regions.1390

If I look at the electric field due to just the top one, the + Σ over here on the left,1396

the electric field due to the top plate must be going away since that is a positive charge.1402

On this side of it must be going away and over here, we are still looking at the electric field just due to this one so it must be going away.1408

Everywhere the magnitude of that is Σ / 2 E₀.1417

Let us take a look at the electric field due to just the bottom plate in these regions.1424

As a negative charged density it is going to have the field lines going toward it.1430

Up in this region we have Σ/ 2 E₀ toward it, the same thing here still toward the bottom plate and1436

we are looking at just the electric field due to the bottom plate for now.1446

And down here we are going toward that negatively charged plate.1449

When we put these together, if we want to find the total or the net electric field up above the 2 plates,1453

we have Σ / 2 E₀ up and Σ/ 2 E₀ down which gives us an electric field = 0 there.1460

In between the 2 plates, we have Σ / 2 E₀ down due to the top plate.1468

Σ / 2 E₀ down due to the bottom plate so our total electric field here is going to be Σ/ E₀ headed down.1473

At the very bottom, we have Σ/ 2 E₀ down, Σ / 2 E₀ up, so the electric field is 0.1488

What we really find after all of this is, is that between the parallel plates above the plates no electric field,1496

below the plates no electric field and between the plates Σ/ E₀ going from the positive to negative.1504

It is important to note that this is not accurate when you get near the ends of real plates.1512

If it is an infinite plane of charge, absolutely works great.1517

When you get to the edges you actually get some sort of fringe in effects, like this.1520

You got to have very big plates and either neglect the edges or have infinite plates for this to make sense.1525

But it is a pretty good approximation for most plates especially in the region not near the very edges of those plates.1530

Let us take a look at another example.1538

Find the electric field strength in the distance R from an infinitely long uniformly charged wire of linear charge density Λ.1542

We have done wires before, if I recall we got quite a bit of math and be quite a bit simpler here using Gauss’s law.1550

Let us start by drawing our wire and we are going to give it some linear charged density Λ.1558

We have got to pick a Gaussian surface that is going to give us the most possible symmetry.1573

In this case for a wire, what I'm going to choose is a cylinder that is centered on that wire.1580

Let us choose a cylinder, something like that for our Gaussian surface.1587

What is nice there is we got a cap on this end, a cap on this end.1598

Just by observation we can see by symmetry that those are going to completely cancel out and1601

all we are going to have to worry about is the pieces going through the sides of our cylinder,1607

through the edges of our can not the top or the bottom.1612

We will start off by writing Gauss’s law, the integral over the close surface of E ⋅ Da = the total enclosed charge, the charge enclosed by our Gaussian surface ÷ Ε 0.1615

This is the total flux but we are going to break that flux up into a piece through the left cap, a piece through the right cap,1632

and our piece through the sides I’m just going to write that as the cylinder portion for our flux must = Q enclosed / E₀.1641

As we just mentioned, by symmetry and we can state that the flux through the left + the flux through the right must = 0.1655

All we have to worry about is the flux through the sides of our cylinder.1671

When I'm calling cylinder portion = the total enclosed charge through our cylinder ÷ E₀.1675

To find the flux due our cylinder, first thing we have to realize is because it has the wire in the middle,1683

the electric field is going to be the same anywhere on the edges of that cylinder, anywhere around it.1689

That nice piece of symmetry again as we chose our Gaussian surface very wisely.1694

We need to know the area of that, however.1700

If we have the edges of the can to find the area of it, we really need to do is think about cutting it and spreading it out to.1703

We got a cylinder, we got to spread it out somehow.1710

What would the dimensions of that B once you spread it out?1713

The way we are going to do that is we will say that the flux through the cylinder is going to be,1718

if the radius of the top of the cap is 2 π R and we cut it,1726

that is going to be one dimension of our rectangle we get when we spread out that canned peas.1730

That will be 2π R and the other dimension is going to be some L that we have not defined yet, the length of our can.1736

We have got that and we have to multiply that all of course by our electric field to get the total flux.1747

There is our Ea which is the flux through that.1753

We can then write that we have our 2 π RL × our electric field = our enclosed charge ÷ E₀.1757

We will go back to our linear charge density Λ which we know Λ is total charge ÷ length.1771

Then our Q enclosed, the amount of charge enclosed by our cylinder is just going to be the linear charge density × this length L.1781

Then we have on the left hand side 2 π RL × our electric field = Q enclosed on the right Λ L ÷ E₀.1793

Or solving for the electric field E = Λ/ 2 π E₀ R.1808

Same as the answer we reached using that point charge summation in the previous lesson1827

that is so much simpler, easier, slicker, quicker, and much more intuitive too.1833

Gauss’s law is a very powerful tool.1840

Let us take a look at some of old AP questions to hone our skills here and we will start off by looking at the 2008 E and M exam free response number 1.1845

And you can look that up on www.google.com or college web sites, it is available for you to download and printout.1858

Here is a link to it currently, I would highly recommend taking a minute, printing it out, and taking as couple seconds1863

and seeing if you can solve it on your own before you check the answers.1869

If you get stuck that is perfectly fine then come back to the video and play along with it.1873

If you got the whole thing, just check your answers as you go through and see what you got right and what might need more work.1877

Use this is a practice opportunity.1882

In this question, we are starting off with a metal sphere that have radius A containing a charge + Q1886

and it is surrounded by an uncharged concentric metallic shell and a radius B and outer radius C.1894

It wants us to first determine the induced charge on the inner surface of the metallic shell.1900

The way I will start to do that is looking at party A here, we have some solid positively charged + Q piece there1907

and outside of that we have another conducting shell.1918

+Q in the middle and out here we are ask to find what is the charge on the inner surface of this shell.1931

Over here at this radius that they call B, that is B.1938

They are defining that as A.1943

The way I would do that to begin with is I would first choose Gaussian surface that is inside this shell and apply Gauss’s law.1949

I'm going to make a shell here that is actually contained, runs through the middle of that conducting shell.1958

There is my Gaussian surface, that sphere there in purple.1968

The integral over the close surface of E ⋅ Da as the enclosed charged ÷ E 0.1972

To do this next step, you probably need to have seen this before either in a previous Physics class.1985

You have to know that the electric field inside a conductor is 0.1991

When you know that the electric field inside a conductor is 0 this becomes very simple.1996

We will get that to that in a later lesson.2003

If you seen at a previous algebra based physics course that would be a great start here.2005

Once you know that then you can say the left hand side, if you know the electric field is 0, the left hand side 0 down with something must be 0 = Q enclosed ÷ E₀.2012

Therefore, Q enclosed by our Gaussian surface must be 0.2025

Q enclosed includes + Q from our inner sphere + let us call it QB, whatever we have on the inner surface of this shell must be = 0.2032

If Q + QB = 0 that implies then to me that QB must be = -Q. Our answer is –Q.2052

If you have + Q here on the outside of this sphere, you must have a -Q charge on the inside there using Gauss’s law.2063

The trick is knowing where to put your Gaussian surface.2076

Let us take a look at the second part there, find the induced charge on the outer surface of the metallic shell.2080

To do that, I'm going to choose a Gaussian surface now that is outside that shell.2087

I’m going to use that red Gaussian surface for part 2.2093

Integral now over the close surface of E ⋅ da which is Q enclose/ E₀,2100

always starting by writing Gauss’s law must be = the total enclosed charge must be + Q.2107

We know that because it says this is neutral in the hole and that is your positive + Q.2115

Once we do that, we can realize that Q enclosed = we have got our QA which we called + Q over there.2122

We have our QB and we now have our QC which is what we will call the charge on the outside of the sphere.2139

This implies then that our + Q, our total enclosed charge Q enclosed + Q must be = we know QA is + Q.2153

We know QB is -Q + whatever QC is.2167

The only way that will work, if we have + Q total, + Q and -Q is going to be 0,2174

That implies then that QC must be = +Q.2179

The charge on the outside here must be + Q which only makes sense by the law of conservation of charge.2186

If we said we had -Q on this inner surface and the whole thing is neutral we must have + Q on that outer surface.2192

There is how we would do part A.2200

Let us take a look and slide over to doing part B now.2203

It asks us to determine expressions for the magnitude of the electric field as a function of radial distance for different regions.2210

For the first region, we will call region I, we have R is less than A.2217

We are inside that inner sphere.2224

In order to do these problems, I'm first going to define now a volume charged density where Ρ = Q/V = Q/.2228

I'm not sure I'm going to need them because we got conductors 4π A³.2242

I will define it but I'm not sure we are going to need it now that I think about it.2247

Let us take a look here for are less than A, Gauss’s law integral / the close surface E ⋅Da = the total charge enclosed ÷ E₀.2250

The left hand side this becomes E × the area, the right hand side is our charge enclosed which is 0.2264

Therefore, we can state that the electric field is 0 inside that metal sphere,2274

which of course we knew because they already said just a few moments ago that the electric field inside the conductor must be 0.2280

We proved it.2286

Let us move on to part 2.2289

We are looking in that region between the solid sphere and that hollow conducting shell, or that solid conducting shell I should say.2293

Here we have R between A and B in that diagram.2305

The integral / the close surface of E ⋅ Da, Q enclosed / E₀.2312

Choosing a Gaussian surface that is in that region which implies that Ea which is going to be E × 4π R²2321

where R is between A and B must equal our total charge enclosed + Q ÷ E₀.2330

Therefore, the electric field was just going to be Q/ 4π E₀ R² in that region.2339

Just like it was a point charge.2349

Moving on to part 3, now we have R between B and C or again we are looking inside that region, we will draw a picture of it to make it clear.2352

There is our solid sphere, here is our conducting sphere, now we want our R somewhere in this region there in red.2367

We can do that easily in a couple of ways.2379

We know already since this is +Q here and –Q on this surface, our total enclosed charge is 0.2382

Therefore, the electric field is going to come out to be 0 or once again we are inside the conductor at this point electric field must be 0, straightforward.2388

Part 4, now what we are going to do is look for R greater than C.2400

We are in some region outside the entire object, the entire setup.2406

In this case, we will start off with Gauss’s law again, integral/ the close surface of E ⋅ Da = D enclosed charge ÷ Ε 0.2413

The left hand side becomes the electric field × the area E × 4π R².2425

Our total charge enclosed is just +Q, therefore the electric field again is Q/ 4π E₀ R².2432

Again, we are treating the entire thing as long as you are outside of that, outside of all those pieces, as if the entire charge was concentrated in a single point.2444

Again, the power of Gauss’s law.2455

It wants us to do a graph of the electric field as a function of distance.2461

It should be pretty straightforward because we just figured out all of the different pieces analytically.2466

Let us draw our graph here.2473

There is our R, there is our electric field, and let us make a couple marks here for A, B, and C, just to help us to limit where things are in the graph.2485

We already know between 0 and A the electric field is 0 and between B and C, the electric field is 0.2508

Those regions where you are inside the conductor electric field is 0 inside the conductor.2516

When we are in between A and B, we are proportional to 1/ R² and when we are outside of C we are again proportional to 1/ R².2521

Fairly straightforward and simple graph, that is part C of the question.2536

And part D, if this is your first Physics course it could be a little bit tricky2542

if you have seen had algebra based Physics before, then you probably had this before and it is a fair question.2549

Either way, it is fair by the time you finish the entire course sequence.2555

Given an electron of mass M and it is caring a charge - E is released from rest at very large distance from the spheres.2559

Find the expression for the speed of the particle at a distance 10 R from the center of the sphere and2565

they really meant to say 10 C from the center of the spheres.2570

What is helpful to know here is that the electrical potential energy due to those point charges is 1/ 4π E₀ × the one charge × the other charge ÷ R.2576

That is the piece that you probably have to know already.2591

I would do this from a conservation of energy perspective.2594

It is a fairly easy way to solve this problem knowing that the kinetic energy + the potential energy is going to remain constant.2597

Because you are starting at very large distance from the spheres we are going to say that is our potential energy level from rest to largest.2608

Since the rest kinetic energy is 0, potential of 0.2616

Therefore, the kinetic energy is going to be ½ ME V² and that is going to be equal to the potential energy Q × the charge E/ 4π E₀ × are distance 10 C.2620

Now it is just an exercise in algebra to find the velocity.2644

V² then is going to be equal to, we got the 2 there, we are going to have Q × E/ 2 π E₀ ME × 10 C.2648

Therefore, V itself is going to be equal to the square root of the QE/ 2 × 10 is going to be 20 π E₀ ME √C².2667

D might be a little tricky if you have not got in the potential energy piece and electrostatics yet.2688

We will get there in a couple of lectures.2693

The rest of it is pretty straightforward application and great practice for Gauss’s law.2696

Let us do one more free response problem here.2703

This one from the 2011 test free response number 1, that was kind of an interesting take on Gauss’s law2706

and whether you really understand how this law works and also a bit of close reading.2714

In this problem, they start off by asking you to use Gauss’s law to prove that the electric field inside a spherical shell2718

with the uniform surface charge of Σ outside surface and no charge anywhere else inside is 0 and describe the Gaussian surface you would use.2728

We only have been doing that for a while now.2738

Let us take a look, here we will call this A.2740

If we use a spherical shell that is inside the charged shell as our Gaussian surface2744

Then we have some charge Σ here and our Gaussian surface inside it.2769

The integral/ the close surface, E ⋅ Da = Q enclosed / E₀.2780

The Q enclosed inside there is 0.2790

Since Q enclosed = 0 we can say that E × the area 4π R² = 0.2793

Therefore, E = 0 nice and quick proof done.2802

For part B, it says the charges are now redistributed so that the surface charge density here is no longer uniform.2809

Is the electric field still 0 everywhere inside the shell?2818

The answer of course is no.2823

When you do that without the symmetry of charges to cancel each other out, there can be an internal electric field.2826

Gauss’s law still holds but it is not the same everywhere.2833

Using Gauss’s law is not overly helpful to you in this point for determining electric field.2836

I would write something like no, without the symmetry of charges to cancel each other out you may have an internal electric field.2841

Something like that should work.2879

They change that and here is where they throw the curveball, something you might not have seen before which makes it a really great question.2884

They give you a small conducting sphere was charged + Q with a center that is in the corner of a cubicle surface.2892

Again, if you have not downloaded and look at the question, make sure you do this.2898

Otherwise, this is not going to make any sense.2902

For which phases of the surface of any of the electric flux through the phase = 0.2904

As you look at the diagram, the only place the electric flux is going to be 0 is when the field lines are running parallel to the surfaces.2910

And of the 6 surfaces on the cube, 3 surfaces are going to be parallel, are going to be touching that points so you have electric field lines parallel to them.2919

No flux through those surfaces and those ones would be ADEH, ABCD, and ABGH.2928

It asks you to explain your reasoning, something like field lines run parallel to the surfaces.2941

Therefore, no flux thorough these surfaces.2963

Moving on, let us take a look at D.2983

At which corners of the surface does the electric field have the least magnitude?2986

That could be a really tricky if you do not read the question very carefully.2992

At which corner do you have the least magnitude of electric field?2997

You have to realize at this point that they are talking about a small conducting sphere whose center is at the corner A.3002

Conducting sphere inside a conductor the electric field is 0 so point A which is inside the conductor has to have 0 electric field.3011

I would state that A is inside the conducting sphere and the electric field is 0 inside the conductor at equilibrium.3022

It is kind of a trick question there if you have not read it very carefully.3051

Let us take a look at part E, find the electric field strength that the positions you have indicated in part D.3057

We just did that and I think they are trying to throw you off again by giving you all these fundamental constants.3065

You do not need them, you already know inside the conductor E = 0.3070

E at point A = 0 done.3075

Finally, last part of the question and last part of this lesson.3079

Given that 1/8 sphere point A is inside the surface, find the electric flux through phase CDEF.3084

Gauss’s law integral / the close surface of E ⋅ Da = Q enclosed / E₀.3092

Our Q enclosed is Q ÷ 8 because only 1/8 of that sphere is inside our cubed.3103

Q enclose is Q/8 so that is Q/ 8 E₀.3113

By symmetry, we have three faces of the cube that have flux running through them.3120

They are all symmetric, they all have the same orientation, the same distance with respect to that point A.3127

Therefore, each one is going to get 1/3 total flux so that is going to be 1/3 of Q/8 E₀ where I would say the flux to each one is going to be Q/24 E₀.3134

Hopefully, that gets you a good start on Gauss’s law, an extremely important part of the course.3156

If you struggle with it, please take your time go back and do some practice problems.3160

You are really going to need this for success in the APC E and M exam.3164

Thank you so much for watching www.educator.com.3169

We will see you again real soon, make it a great day.3172

OR

### Start Learning Now

Our free lessons will get you started (Adobe Flash® required).