chapter 28. gauss’s law the nearly spherical shape chapter

Chapter 28. Gauss’s Law

The nearly spherical shape

of the girl’s head determines

the electric field that causes

her hair to stream outward.

Using Guass’s law, we can

deduce electric fields,

particularly those with a

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

particularly those with a

high degree of symmetry,

simply from the shape of the

charge distribution.

Chapter Goal: To

understand and apply

Gauss’s law.1

Topics:

• Symmetry

• The Concept of Flux

• Calculating Electric Flux

Chapter 28. Gauss’s Law


• Gauss’s Law

• Using Gauss’s Law

• Conductors in Electrostatic Equilibrium

2

Chapter 28. Reading Quizzes


Chapter 28. Reading Quizzes

3

The amount of electric field

passing through a surface is

called

A. Electric flux.


A. Electric flux.

B. Gauss’s Law.

C. Electricity.

D. Charge surface density.

E. None of the above.

4

The amount of electric field

passing through a surface is

called

A. Electric flux.


A. Electric flux.

B. Gauss’s Law.

C. Electricity.

D. Charge surface density.

E. None of the above.

5

Gauss’s law is useful for calculating

electric fields that are

A. symmetric.

B. uniform.


B. uniform.

C. due to point charges.

D. due to continuous charges.

6

Gauss’s law is useful for calculating

electric fields that are

A. symmetric.

B. uniform.


B. uniform.

C. due to point charges.

D. due to continuous charges.

7

Gauss’s law applies to

A. lines.

B. flat surfaces.


B. flat surfaces.

C. spheres only.

D. closed surfaces.

8

Gauss’s law applies to

A. lines.

B. flat surfaces.


B. flat surfaces.

C. spheres only.

D. closed surfaces.

9

The electric field inside a

conductor in electrostatic

equilibrium is


A. uniform.

B. zero.

C. radial.

D. symmetric.

10

The electric field inside a

conductor in electrostatic

equilibrium is


A. uniform.

B. zero.

C. radial.

D. symmetric.

11

Chapter 28. Basic Content and Examples


Chapter 28. Basic Content and Examples

12

Symmetry and Electric Field

We say that a charge distribution is symmetric if there are a group of geometric transformations that do not cause any physical change.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.13

The Concept of Flux


The Concept of Flux


An electric field passing through a

surface


16

The electric Flux of a constant electric

field


17

The electric Flux of a Nonuniform

Electric Field

i

i

ii

i

e AE )(rr

∑∑ ⋅=Φ=Φ δδ


∫ ⋅=Φsurface

e AdErr

The Flux Through a Curved Surface


Electric Flux

Definition:

• Electric flux is the product of the

magnitude of the electric field and the

surface area, A, perpendicular to the field

• ΦE = EA

• The field lines may make some angle θ

with the perpendicular to the surface


20

with the perpendicular to the surface

• Then ΦE = EA cos θ Errrr

EEAΦ =Φ =Φ =Φ =

Errrr

cosE

EA θθθθΦ =Φ =Φ =Φ =

normal

θθθθ

θθθθ

Electric Flux

Definition:

• Electric flux is the scalar product of electric

field and the vector

• Errrr

θθθθ

θθθθ

Arrrr

EAΦ =Φ =Φ =Φ =rrrrrrrr

Arrrr


21

cos 0E

EA EA θθθθΦ = = >Φ = = >Φ = = >Φ = = >rrrrrrrr

orErrrr

cos 0E

EA EA θθθθΦ = = − <Φ = = − <Φ = = − <Φ = = − <rrrrrrrr

θθθθ

θθθθ

Arrrr

Electric Flux

• In the more general case, look at a

small flat area element

• In general, this becomes

cosE i i i i iE A θ E A∆Φ = ∆ = ⋅ ∆r r


22

• The surface integral means the

integral must be evaluated over the

surface in question

• The units of electric flux will be

N.m2/C

0surface

limi

E i iA

E A E dA∆ →

Φ = ⋅ ∆ = ⋅∑ ∫r

r r r r

Electric Flux: Closed Surface

• A positive point charge, q, is

located at the center of a sphere of

radius r

• The magnitude of the electric field

everywhere on the surface of the

sphere is

Spherical

surface


23

sphere is

E = keq / r2

• Electric field is perpendicular to the

surface at every point, so

has the same direction as

at every point.

Errrr

Arrrr


has the same direction as at every point.

Spherical

surface

Errrr

Arrrr

2e

qE k

r====

Then

i i i

i i

E dA E dAΦ = = =Φ = = =Φ = = =Φ = = =∑ ∑∑ ∑∑ ∑∑ ∑rrrrrrrr


24

2 2

0 2

0

4 4

4

i i

e

e

qEA E r r k

r

qk q

π ππ ππ ππ π

ππππεεεε

= = = == = = == = = == = = =

= == == == =

∑ ∑∑ ∑∑ ∑∑ ∑

does not depend on rΦΦΦΦ

Gauss’s Law

BECAUSE2

1E

r∝∝∝∝


and have opposite directions at every point.

Spherical

surface

Errrr

Arrrr

2

| |e

qE k

r====

Then

E dA E dAΦ = = − =Φ = = − =Φ = = − =Φ = = − =∑ ∑∑ ∑∑ ∑∑ ∑rrrrrrrr


25

2 2

0 2

0

| |4 4

4 | |

i i i

i i

e

e

E dA E dA

qEA E r r k

r

qk q

π ππ ππ ππ π

ππππεεεε

Φ = = − =Φ = = − =Φ = = − =Φ = = − =

= − = − = − == − = − = − == − = − = − == − = − = − =

= − == − == − == − =

∑ ∑∑ ∑∑ ∑∑ ∑

does not depend on rΦΦΦΦ

−−−−

Gauss’s Law

ONLY BECAUSE2

1E

r∝∝∝∝

EXAMPLE 28.1 The electric flux inside a

parallel-plate capacitor

QUESTION:


26




27




28




29

Gauss’s Law

Gauss’s law states

qin is the net charge inside the surface

E is the total electric field and may have contributions

inE

o

qE dA

εΦ = ⋅ =∫

r r


30


from charges both inside and outside of the surface

q q

0

q

εεεεΦ =Φ =Φ =Φ =

0

q

εεεεΦ =Φ =Φ =Φ =

iArrrr

Errrr

Errrr

iArrrr





What is the electric flux here?

inE

o

qE dA

εΦ = ⋅ =∫

r r

q


31

1q

0Φ =Φ =Φ =Φ =

2q

3q

4q

5q

7q6

q

Gauss’s Law:





inE

o

qE dA

εΦ = ⋅ =∫

r r


32


q−−−−

0Φ =Φ =Φ =Φ =2q

q

4q

5q

7q6

q

Gauss’s Law: Applications

Chapter 28


33


Although Gauss’s law can, in theory, be solved to find

E for any charge configuration, in practice it is limited to

symmetric situations

To use Gauss’s law, you want to choose a Gaussian

surface over which the surface integral can be simplified

and the electric field determined

Take advantage of symmetry

Remember, the gaussian surface is a surface you

in

o

qE dA

εΦ = ⋅ =∫

r r


34

Remember, the gaussian surface is a surface you

choose, it does not have to coincide with a real surface

1q

1 2 3 4

0

q q q q

εεεε

+ + ++ + ++ + ++ + +Φ =Φ =Φ =Φ =

2q

3q

4q

5q

6q

SYMMETRY:

Gauss’s Law: Point Charge

++++q

Errrr

Errrr

- direction - along the radius

Errrr

Gaussian Surface – Sphere

Only in this case the magnitude of


35

Errrr

- depends only on radius, r

2e

qE k

r====

0

q

εεεεΦ =Φ =Φ =Φ = - Gauss’s Law

2

04

i i i

i i

E dA E dA EA E rππππΦ = = = =Φ = = = =Φ = = = =Φ = = = =∑ ∑∑ ∑∑ ∑∑ ∑rrrrrrrr

- definition of the Flux

Then 2

0

4q

r Eππππεεεε

====

Only in this case the magnitude of electric field is constant on the

Gaussian surface and the flux can be

easily evaluated

Gauss’s Law: Applications in

o

qE dA

εΦ = ⋅ =∫

r r

• Try to choose a surface that satisfies one or more of these conditions:

– The value of the electric field can be argued from symmetry to be

constant over the surface

– The dot product of E.dA can be expressed as a simple algebraic product

EdA because E and dA are parallel

– The dot product is 0 because E and dA are perpendicular


36

– The field can be argued to be zero over the surface

correct Gaussian surface

++++

wrong Gaussian surface

Spherically Symmetric Charge Distribution


The total charge is Q

• Select a sphere as the gaussian

SYMMETRY:

Errrr


Errrr


Errrr

A∆∆∆∆rrrr


37


surface

• For r >a

2 in

2 2

4

4

E

o o

e

o

q QE dA EdA πr E

ε ε

Q QE k

πε r r

Φ = ⋅ = = = =

= =

∫ ∫r r

The electric field is the same as for the point charge Q



2 24e

o

Q QE k

πε r r= = The electric field is the same as

for the point charge Q !!!!!

Q

Qa

≡≡≡≡For r > a


38

≡≡≡≡

≡≡≡≡For r > a



SYMMETRY:

Errrr


Errrr


Errrr

A∆∆∆∆rrrr



39


surface, r < a

33

33

4

4 3

3

in

Q rq r Q Q

aa

ππππ

ππππ

= = <= = <= = <= = <

2 in

3

in

2 3 2 3

4

1

4

E

o

e e

o

qE dA EdA πr E

ε

q Qr QE k k r

πε r a r a

Φ = ⋅ = = =

= = =

∫ ∫r r



• Inside the sphere, E varies

linearly with r

E→ 0 as r→ 0

• The field outside the sphere is


40

• The field outside the sphere is

equivalent to that of a point

charge located at the center of

the sphere

Field due to a thin spherical shell


• Use spheres as the gaussian surfaces

• When r > a, the charge inside the surface is Q and

E = keQ / r2

• When r < a, the charge inside the surface is 0 and E = 0


41

Field due to a thin spherical shell


• When r < a, the charge inside the surface is 0 and E = 0

1r

1A∆∆∆∆

1 1q Aσσσσ∆ = ∆∆ = ∆∆ = ∆∆ = ∆

2 2q Aσσσσ∆ = ∆∆ = ∆∆ = ∆∆ = ∆

2

1 1A r∆ = ∆Ω∆ = ∆Ω∆ = ∆Ω∆ = ∆Ω 2

2 2A r∆ = ∆Ω∆ = ∆Ω∆ = ∆Ω∆ = ∆Ω

E∆∆∆∆rrrr

2q A rσ σσ σσ σσ σ∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω


422

A∆∆∆∆

2r

the same

solid angle1

E∆∆∆∆rrrr

2E∆∆∆∆rrrr

2

1 1 1

1 2 2 2

1 1 1

e e e e

q A rE k k k k

r r r

σ σσ σσ σσ σσσσσ

∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω

2

2 2 2

2 2 2 2

2 2 2

e e e e

q A rE k k k k

r r r

σ σσ σσ σσ σσσσσ

∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ ∆ ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω∆ = = = = ∆Ω

1 2E E∆ = ∆∆ = ∆∆ = ∆∆ = ∆

1 20E E∆ + ∆ =∆ + ∆ =∆ + ∆ =∆ + ∆ =

r rr rr rr r

2

1E

r∝∝∝∝Only because in Coulomb law

Field from a line of charge


• Select a cylindrical Gaussian surface

– The cylinder has a radius of r and a

length of ℓ

• Symmetry:

E is constant in magnitude (depends only


43

E is constant in magnitude (depends only

on radius r) and perpendicular to the

surface at every point on the curved part

of the surface

The end view

Field from a line of charge


dArrrr The flux through this surface is 0

The flux through this surface:


44The end view

( ) in2E

o

qE dA EdA E πr

εΦ = ⋅ = = =∫ ∫

r rl

( )2

22

o

e

o

λE πr

ε

λ λE k

πε r r

=

= =

ll

inq λ= l

Field due to a plane of charge


• Symmetry:

E must be perpendicular to the plane and

must have the same magnitude at all

points equidistant from the plane

• Choose a small cylinder whose axis is

dArrrr


45

• Choose a small cylinder whose axis is

perpendicular to the plane for the

gaussian surface

The flux through this surface is 0

Field due to a plane of charge


dArrrr5

Arrrr

6Arrrr

3Arrrr

4Arrrr

2Arrrr

h

h

2Errrr


46

The flux through this surface is 0

63

A

1Arrrr

1Errrr

1 1 2 22E A E A EA EA EAΦ = + = + =Φ = + = + =Φ = + = + =Φ = + = + =

r rr rr rr rr rr rr rr r

1 2E E E= == == == =

1 2A A A= == == == =

0 0

inq Aσσσσ

ε εε εε εε εΦ = =Φ = =Φ = =Φ = =

0

2A

EAσσσσ

εεεε====

02

Eσσσσ

εεεε==== does not depend on h



47

02

Eσσσσ

εεεε====

2e

E kr

λλλλ====

Conductors in Electric Field

Chapter 28


48

Electric Charges: Conductors and Isolators

Electrical conductors are materials in which

some of the electrons are free electrons

These electrons can move relatively freely

through the material

Examples of good conductors include copper,

aluminum and silver


49

Electrical insulators are materials in which all

of the electrons are bound to atoms

These electrons can not move relatively freely

through the material

Examples of good insulators include glass, rubber

and wood

Semiconductors are somewhere between

insulators and conductors

Electrostatic Equilibrium

Definition:

when there is no net motion of charge

within a conductor, the conductor is said to

be in electrostatic equilibrium

Because the electrons can move freely through the


50

Because the electrons can move freely through the

material

no motion means that there are no electric forces

no electric forces means that the electric field

inside the conductor is 0

If electric field inside the conductor is not 0, then

there is an electric force and, from the second

Newton’s law, there is a motion of free electrons.

0E ≠≠≠≠rrrr

F qE====r rr rr rr r

F qE====r rr rr rr r

Conductor in Electrostatic Equilibrium

• The electric field is zero everywhere inside the conductor

• Before the external field is applied, free

electrons are distributed throughout the

conductor


51

• When the external field is applied, the

electrons redistribute until the magnitude

of the internal field equals the magnitude

of the external field

• There is a net field of zero inside the

conductor


• If an isolated conductor carries a charge, the charge resides on its surface

Electric filed is 0,

so the net flux through

Gaussian surface is 0


52

in

o

qE dA

εΦ = ⋅ =∫

r r

Then in 0q =


• The electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/εo

• Choose a cylinder as the gaussian surface

• The field must be perpendicular to the surface

– If there were a parallel component to E,

charges would experience a force and


53

charges would experience a force and

accelerate along the surface and it would not be

in equilibrium

• The net flux through the gaussian surface is

through only the flat face outside the conductor

– The field here is perpendicular to the surface

• Gauss’s law:

E

o o

σA σEA and E

ε εΦ = = =


o

σE

ε=

0E =

0σ >

0σ <


54

0σ >

0σ <

0σ <

Chapter 28. Summary Slides


Chapter 28. Summary Slides

55

General Principles


56

General Principles


57

Important Concepts


58

Important Concepts


59

Important Concepts


60

Applications


61

Chapter 28. Questions


Chapter 28. Questions

62

This box contains


A. a net positive charge.

B. a net negative charge.

C. a negative charge.

D. a positive charge.

E. no net charge.

63

This box contains


A. a net positive charge.

B. a net negative charge.

C. a negative charge.

D. a positive charge.

E. no net charge.

64


The total electric flux through this box isA. 6 Nm2/C.

B. 4 Nm2/C.

C. 2 Nm2/C.

D. 1 Nm2/C.

E. 0 Nm2/C.

65Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

A. 6 Nm2/C.

B. 4 Nm2/C.

C. 2 Nm2/C.

D. 1 Nm2/C.

E. 0 Nm2/C.

The total electric flux through this box is

66

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