Lesson 7 Gauss’s Law and Electric Fields

Lesson 7Gauss’s Law and Electric

Fields

Class 18Today, we will:• learn the definition of a Gaussian surface• learn how to count the net number of field lines passing into a Gaussian surface• learn Gauss’s Law of Electricity• learn about volume, surface, and linear charge density• learn Gauss’s Law of Magnetism• show by Gauss’s law and symmetry that the electric field inside a hollow sphere is zero

Section 1Visualizing Gauss’s Law

Gaussian Surface•A Gaussian surface is

–any closed surface–surface that encloses a volume

•Gaussian surfaces include:–balloons–boxes–tin cans

•Gaussian surfaces do not include:–sheets of paper–loops

Counting Field Lines•To count field lines passing through Gaussian surfaces:

–Count +1 for every line that passes out of the surface.–Count ─1 for every line that comes into the surface.

+1

─1

Electric Field LinesWe have a +2 charge and a ─2 charge.

Electric Field LinesWhat is the net number of field lines passing through the Gaussian surface?


+8



+8



─8



─8



0



0

Electric Field LinesFrom the field lines coming out of this box, what can you tell about what’s inside?

Electric Field LinesThe net charge inside must be +1 (if we draw 4 lines per unit of charge).

Gauss’s Law of Electricity

The net number of electric field lines passing through a Gaussian surface is proportional to the charge enclosed within the Gaussian surface.

Section 2Charge Density

Charge DensityVolume: ρ =

Surface: σ =

Linear: λ =

Charge

Volume

Charge

Area

Charge

Length

Charge DensityIn general, charge density can vary with position. In this case, we can more carefully define density in terms of the charge in a very small volume at each point in space. The density then looks like a derivative:

dVdq

Vqr

rV

0

lim

You need to understand what we mean by this equation, but we won’t usually need to think of density as a derivative.

Section 3Gauss’s Law of Magnetism

Gauss’s Law and Magnetic Field Lines

If magnetic field lines came out from point sources like electric field lines, then we would have a law that said:

The net number of magnetic field lines passing through a Gaussian surface is proportional to the magnetic charge inside.

N


But we have never found a magnetic monopole.

- The thread model suggests that there is no reason we should expect to find a magnetic monopole as the magnetic field as we know it is only the result of moving electrical charges.

- The field line model suggests that there’s no reason we shouldn’t find a magnetic monopole as the electric and magnetic fields are both equally fundamental.


What characteristic would a magnetic monopole field have?


What characteristic would a magnetic monopole field have?

monopoletesttest BvqF


All known magnetic fields have field lines that form closed loops.

So what can we conclude about the number of lines passing through a Gaussian surface?

Gauss’s Law of MagnetismThe net number of magnetic field lines passing through any Gaussian surface is zero.

Section 4Gauss’s Law and Spherical

Symmetry

Spherically Symmetric Charge Distribution

The charge density, ρ, can vary with r only.

Below, we assume that the charge density is greatest near the center of a sphere.


Outside the distribution, the field lines will go radially outward and will be uniformly distributed.


The field is the same as if all the charge were located at the center of the sphere!

Inside a Hollow SphereNow consider a hollow sphere of inside radius r with a spherically symmetric charge distribution.

Inside a Hollow SphereThere will be electric field lines outside the sphere and within the charged region. The field lines will point radially outward because of symmetry. But what about inside?

Inside a Hollow SphereDraw a Gaussian surface inside the sphere. What is the net number of electric field lines that pass through the Gaussian surface?

Inside a Hollow SphereThe total number of electric field lines from the hollow sphere that pass through the Gaussian surface inside the sphere is zero because there is no charge inside.

How can we get zero net field lines?1. We could have some lines come in and go out again…

… but this violates symmetry!

How can we get zero net field lines?2. We could have some radial lines come in and other radial lines go out…

… but this violates symmetry, too!

3. Or we could just have no electric field at all inside the hollow sphere.

How can we get zero net field lines?

3. Or we could just have no electric field at all inside the hollow sphere.

How can we get zero net field lines?

This is the only way it can be done!

Conclusion: the static electric field inside a hollow charged sphere with a spherically symmetric charge distribution must be zero.

The Electric Field inside a Hollow Sphere

0E

Class 19Today, we will:• learn how to use Gauss’s law and symmetry to find the electric field inside a spherical charge distribution• show that all the static charge on a conductor must reside on its outside surface• learn why cars are safe in lightning but cows aren’t


Electric field lines do not start or end outside charge distributions, but that can start or end inside charge distributions.


What is the electric field inside a spherically symmetric charge distribution?


Inside the distribution, it is difficult to draw field lines, as some field lines die out as we move inward. – We need to draw many, many field lines to keep the distribution uniform as we move inward.


But we do know that if we drew enough lines, the distribution would be radial and uniform in every direction, even inside the sphere.


Let’s draw a spherical Gaussian surface at radius r.

r


Now we split the sphere into two parts – the part outside the Gaussian surface and the part inside the Gaussian surface.

r r


The total electric field at r will be the sum of the electric fields from the two parts of the sphere.

r r


Since the electric field at r from the hollow sphere is zero, the total electric field at r is that of the “core,” the part of the sphere within the Gaussian surface.

r r


Outside the core, the electric field is the same as that of a point charge that has the same charge as the total charge inside the Gaussian surface.

r


r

204

1)(r

qrE enc

Inside a spherically symmetric charge distribution, the static electric field is:

Example: Uniform DistributionA uniformly charged sphere of radius R has a total charge Q. What is the electric field at r < R ?

Example: Uniform Distribution

204

1)(r

qrE enc

A uniformly charged sphere of radius R has a total charge Q. What is the electric field at r < R ?

rSince the charge density is uniform:

VV

Qq encenc

Example: Uniform Distribution

30

334

334

20

20

20

41

14

1

14

14

1)(

RQr

Rr

Qr

VVQ

r

rqrE

enc

enc

Section 5Gauss’s Law and

Conductors

Gauss’s Law and ConductorsTake an arbitrarily shaped conductor with charges on the outside.

+

++

+

+

++

+

Gauss’s Law and ConductorsThe static electric field inside the conductor must be zero. – Draw a Gaussian surface inside the conductor.

+

++

+

+

++

+

Gauss’s Law and ConductorsNo field lines go through the Gaussian surface because E=0. Hence, the total enclosed charge must be zero.

+

++

+

+

++

+

Gauss’s Law and ConductorsThe same must be true of all Gaussian surfaces inside the conductor.

+

++

+

+

++

+

Surface Charge and ConductorsWhat if there are no charges on the outside and the net charge of the conductor is zero?

-- The volume charge density inside the conductor must be zero and the surface charge density on the conductor must also be zero.

Surface Charge and ConductorsWhat if there are no charges on the outside and there is net charge on the surface of a conductor?

+

+

+

++

+

+

+++

+

+

++

+

+

Surface Charge and ConductorsThe charge distributes itself so the field inside is zero and the surface is at the same electric potential everywhere.

+ ++

+

+

+ ++++

+

+

+

+

+

+

Example: Surface Charge on a Spherical Conductor

A spherical conductor of radius R has a voltage V. What is the total charge? What is surface charge density?

Example: Surface Charge on a Spherical Conductor

A spherical conductor of radius R has a voltage V. What is the total charge? What is surface charge density?

RV

RVR

AQ

VRQRQRV

RrrQrV

02

0

0

0

0

44

44

1)(

,4

1)(

On the outside, the potential is that of a point charge.

On the surface, the voltage is V(R).

Take Two Conducting Sphereswith the Same Voltage

The smaller sphere has a larger charge density.

+

+

+

+

+

+

+

+

++

+

+

++

Now Connect the Two SpheresThe charge density is greater near the “pointy” end.

The electric field is also greater near the “pointy” end.

++

+

+

++

+

+

++

+

+

+

+

Edges on ConductorsCharge moves to sharp points on conductors.

Electric field is large near sharp points.

Smooth, gently curved surfaces are the best for holding static charge.

Lightning rods are pointed.

A Hollow ConductorWhat if there’s a hole in the conductor?

+

++

+

+

++

+

A Hollow ConductorDraw a Gaussian surface around the hole.

+

++

+

+

++

+

A Hollow ConductorThere is no net charge inside the Gaussian surface.

+

++

+

+

++

+

A Hollow ConductorIs there surface charge on the surface of the hole?

+

++

+

+

++

+

+

+

A Hollow ConductorThere is no field surrounding the charge to hold the charges fixed, so the charges migrate and cancel each other out.

+

++

+

+

++

+

Charge on a ConductorStatic charge moves to the outside surface of a conductor.

+

++

+

+

++

+

Lightning and CarsWhy is a car a safe place to be when lightning strikes?

Note: Any car will do – it doesn’t need to be a Cord….

Lightning and CarsIs it the insulating tires?

Lightning and Cars

If lightning can travel 1000 ft through the air to get to your car, it can go another few inches to go from your car to the ground!

Is it the insulating tires?

Lightning and CarsA car is essentially a hollow conductor.

Charge goes to the outside.

The electric field inside is small.

Lightning and CarsA car is essentially a hollow conductor.

Charge goes to the outside.

The electric field inside is small.

How should a cow stand to avoid injury when lightning strikes nearby?

Physicist’s Cow

I

d

Cow

Earth

Physicist’s Cow

I

d

Cow

Earth

RVP

2

When d is bigger, the resistance along the ground between the cow’s feet is bigger, the voltage across the cow is bigger, and the current flowing through the cow is bigger.

How should a cow stand to avoid injury when lightning strikes nearby?

So the cow should keep her feet close together!

Class 20Today, we will:• learn how integrate over linear, surface, and volume charge densities to find the total charge on an object• learn that flux is the mathematical quantity that tells us how many field lines pass through a surface

Section 6Integration

Gauss’s Law of ElectricityThe net number of electric field lines passing through a Gaussian surface is proportional to the enclosed charge.

But, how do we find the enclosed charge?

Charge and Density is valid when?Vq

Charge and Density when ρ is uniform.If ρ is not uniform over the whole volume, we find some small volume dV where it is uniform. Then:

If we add up all the little bits of dq, we get the entire charge, q.

Vq

dVdq

dVdqq

Integration

The best way to review integration is to work through some practical integration problems.

Integration

The best way to review integration is to work through some practical integration problems.

Our goal is to turn two- and three- dimensional integrals into one-dimensional integrals.

Fundamental Rule of Integration

Identify the spatial variables on which the integrand depends.You must slice the volume (length or surface) into slices on which these variables are constant.

Fundamental Rule of Integration

When integrating densities to find the total charge, the density must be a constant on the slice or we cannot write dVdq

Fundamental Rule of IntegrationExamples

Square in x-y plane

Cylinder

Sphere r

zr

yx


dAdqx ,Consider a very thin slice.

Is constant on this slice?


dAdqx ,Consider a very thin slice.

Is constant on this slice?


Square in x-y plane

Cylinder

Sphere r

zr

yx


Square in x-y plane

Cylinder

Sphere r

zr

yx


Square in x-y plane

Cylinder

Sphere r

zr

yx


Square in x-y plane

Cylinder

Sphere r

zr

yx


Square in x-y plane

Cylinder

Sphere r

zr

yx

Rules for Areas and Volumes of SlicesMemorize These!!!

Square in x-y plane

Disk

Cylinder

Sphere drrdV

dzrdV

drLrdV

drrdA

dxLdA

2

2

4

2

2


Square in x-y plane

Disk

Cylinder

Sphere drrdV

dzrdV

drLrdV

drrdA

dxLdA

2

2

4

2

2


Square in x-y plane

Disk

Cylinder

Sphere drrdV

dzrdV

drLrdV

drrdA

dxLdA

2

2

4

2

2


Square in x-y plane

Disk

Cylinder

Sphere drrdV

dzrdV

drLrdV

drrdA

dxLdA

2

2

4

2

2


Square in x-y plane

Disk

Cylinder

Sphere drrdV

dzrdV

drLrdV

drrdA

dxLdA

2

2

4

2

2

Rules for Areas and Volumes of Slices

Memorize These!!!Square in x-y plane

Disk

Cylinder

Sphere drrdV

dzrdV

drLrdV

drrdA

dxLdA

2

2

4

2

2

Let’s Do Some Integrals

Charge on a CylinderA cylinder of length L and radius R has a charge

density where is a constant and z is the distance from one end of the cylinder. Find the

total charge on the cylinder.

How do you slice the cylinder?

What is the volume of each slice?

4z

Charge on a Cylinder

5

52

0

42

24

LRq

dzzRdqq

dzRzdVdqL

Charge on a SphereA sphere of radius R has a charge density where is a constant. Find the total charge on the sphere.

How do you slice the sphere?

What is the volume of each slice?

r

Charge on a Sphere

44

0

3

2

44

4

4

RRq

drrdqq

drrrdVdqR

Section 7Gauss’s Law and Flux

Field Lines and Electric Field

EAk

N

ANkE

1

This is valid when1) .A is the area of a section of a perpendicular

surface.2)The electric field is constant on A.

Field Lines and Electric Field

EAk

N

ANkE

1

This is valid when1) A is the area of a section of a perpendicular

surface.2)The electric field is constant on A.-- But E is a constant on A only in a few cases of

high symmetry: spheres, cylinders, and planes.

Electric Flux

enc

enc

enc

qEA

qEAk

N

qN

1

Gauss’s Law states that:

EA is called the electric flux. We write it as or just .

E

Electric Flux

enc

enc

enc

qEA

qEAk

N

qN

1

Gauss’s Law states that:

EA is called the electric flux. We write it as or just .Flux is a mathematical expression for number of field lines passing through a surface!

E

Electric Flux and a Point Charge

22

0

44

1 rrq

EA

Lets calculate the electric flux from a point charge

passing through a sphere of radius r.

Electric Flux and a Point Charge

0

22

0

1

44

1

qrrq

qEA

Gauss’s law says this is proportional to the charge enclosed in the sphere!

Electric Flux and Gauss’s Law

encq0

1

This means that we can write Gauss’s Law of Electricity as

A Few Facts about FluxFor our purposes, we will (almost) always calculate flux through a section of perpendicular surface where the field is constant. So we will evaluate flux simply as:

EA

A Few Facts about FluxBut we do need to find a more general expression for flux so you’ll know what it really means…

An Area VectorWe wish to define a vector area. To do this

1)we need a flat surface.2)the direction is perpendicular to the plane of

the area. (Don’t worry about the fact there are two choices of

direction that are both perpendicular to the area – up and down in the figure below.)

3) the magnitude of vector is the area.A

A Few Facts about FluxFirst, Let’s consider the flux passing through a frame oriented perpendicular to the field.

A

EA0

A Few Facts about FluxIf we tip the frame by an angle θ, the angle between the field and the normal to the frame, there are fewer field lines passing through the frame.

A

A

EA0 cosEA

A Few Facts about FluxOr, using the vector area of the loop, we may write:

A

AAE

EA0 cosEA

A Few Facts about Flux

only holds when the frame is flat and the field is uniform.What if the surface (frame) isn’t flat, or the electric field isn’t uniform?

AE

Area Vectors on a Gaussian Surface1)We must take a small region of the surface dA

that is essentially flat.2) We choose a unit vector perpendicular to the

plane of dA going in an outward direction.

Ad

A Few Facts about Flux

The flux through this small region is:

AdEd

A Few Facts about FluxTo find the total flux, we simply add up all the contributions from every little piece of the surface.

AdEd

Recall that the normal to each small area is taken to be in the outward direction.

A Few Facts about FluxThus, the most general equation for flux through a surface is:

AdE

If we take the flux through a Gaussian surface, we usually write the integral sign with a circle through it to emphasize the fact that the integral is over a closed surface:

AdE

Class 21Today, we will:• learn how to use Gauss’s law to find the electric fields in cases of high symmetry

• insdide and outside spheres• inside and outside cylinders• outside planes

Section 7Gauss’s Laws in Integral

Form

Gauss’s Law of ElectricityIntegral Form

The number of electric field lines passing through a Gaussian surface is proportional to the charge enclosed by the surface.

dVAdESS

0

1

encq0

1

We can make this simple expression look much more impressive by replacing the flux and enclosed charge with integrals:

Gauss’s Law of MagnetismIntegral Form

The number of magnetic field lines passing through a Gaussian surface is zero

0S AdB

0 B

With the integral for magnetic flux, this is:

Gauss’s Law of ElectricityTee-Shirt Form

dVAdESS

0

1

0qAdE

This can be written in many different ways. A popular form seen on many tee-shirts is:

Gauss’s Law of ElectricityTee-Shirt Form

dVAdESS

0

1

0qAdE

This can be written in many different ways. A popular form seen on many tee-shirts is:

This is a good form of Gauss’s law to use if you want to impress someone with how smart you are.

Gauss’s Law of ElectricityPractical Form

dVAdESS

0

1

dVEA 0

1

This is the form of Gauss’s law you will use when you actually work problems.


dVEA 0

1

Now let’s think about what this equation really means!


dVEA 0

1

Electric field onGaussian surface-- Must be the sameeverywhere on thesurface!


dVEA 0

1


Area of the entireGaussian surface – Mustbe a perpendicularsurface (an elementof a field contour)!


dVEA 0

1


Area of the entireGaussian surface – Mustbe a perpendicularsurface (an elementof a field contour)!

Integral ofthe charge density overthe volumeenclosed by theGaussiansurface!

Section 9Using Gauss’s Law to Find

Fields

Problem 1: Spherical Charge Distribution

OutsideBasic Plan:1) Choose a spherical Gaussian

surface of radius r outside the charge distribution.

2)

3) Integrate the charge over the entire charge distribution.

0

24

totalqErEA


Outside

R

total

dVr

E

qErEA

r

02

0

0

2

41

4

R

r


Outside

4

44

1

4

20

0

32

0

0

22

0

Rr

E

drrr

E

drrrr

E

R

R

R

r


InsideBasic Plan:1) Choose a spherical Gaussian

surface of radius r inside the charge distribution.

2)

3) Integrate the charge over the inside of the Gaussian surface only.

0

24

totalqErEA


Inside

r

total

dVr

E

qErEA

r

02

0

0

2

41

4

Rr

0

24

20

0

32

0

0

22

0

44

44

1

rrr

E

drrr

E

drrrr

E

r

r

Rr


Inside

Problem 3: Cylindrical Charge Distribution

OutsideBasic Plan:1) Choose a cylindrical

Gaussian surface of radius r and length L outside the charge distribution.

2)

3) Integrate the charge over the entire charge distribution.

0

2

totalqrLEEA


OutsideBasic Plan:4) Note that there are no field

lines coming out the ends of the cylinder, so there is no flux through the ends!


Outside

R

total

dVrL

E

qrLEEA

r

00

0

7

21

2

Rr

9

22

1

9

0

0

8

0

0

7

0

Rr

E

drrr

E

drrLrrL

E

R

R

R

r


Outside


InsideBasic Plan:1) Choose a cylindrical Gaussian

surface of radius r and length L inside the charge distribution.

2)

3) Integrate the charge over the inside of the Gaussian surface only.

0

2

totalqrLEEA

r

total

dVrL

E

qrLEEA

r

00

0

7

21

2

Rr


Inside

0

89

0

0

8

0

0

7

0

99

22

1

rrr

E

drrr

E

drrLrrL

E

r

r

Rr


Inside

Infinite Sheets of ChargeBasic Plan:1) Choose a box with faces

parallel to the plane as a Gaussian surface. Let A be the area of each face.

2) Find the charge inside the box. No integration is needed.

Problem 5: Infinite Sheet of Charge(Insulator with σ given)

AANote there is flux out both sides of the box!

0

0

2

2

E

AEA

Problem 6: Infinite Sheet of Charge(Conductor with σ on each surface)

AANote there is flux out both sides of the box, and the total charge density is 2σ!

0

0

22

E

AEA

Problem 6: A second way…

A

0inE

0

0

E

AEA

Now there is flux out only one side of the box, but the total charge density inside is just σ!

Problem 7: A Capacitor

There is flux out only one side of the box!

AQE

AAE

00

0

AThe area of the plate is and the area of the box is .

A

AA

A Word to the Wise!

If you can do these seven examples, you can do every Gauss’s law problem I can give you! Know them well!

Documents

Lesson 7 Gauss’s Law and Electric Fields