PH 222-3A Spring 2010

Gauss’ Law

Lectures 3-4

Chapter 23

(Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)

1

Chapter 23

Gauss’ Law

In this chapter we will introduce the following new concepts:

The flux (symbol Φ ) of the electric field

Symmetry

Gauss’ law

We will then apply Gauss’ law and determine the electric field

generated by:

An infinite, uniformly charged insulating plane

An infinite, uniformly charged insulating rod

A uniformly charged spherical shell

A uniform spherical charge distribution

We will also apply Gauss’ law to determine the electric field inside and

outside charged conductors. 2

n̂

Consider an airstream of velocity that is aimed at a loop

of area . The velocity vector is at angle with respect to the

ˆloop normal . The product cos is know

n as

v

A v

n vA

Flux of a Vector.

the . In this

example the flux is equal to the volume flow rate through the loop (thus the

name flux).

depends on . It is maximum and equal to for 0 ( perpendicular

to the loop

vA v

flux

Note 1 :

plane). It is minimum and equal to zero for 90 ( parallel

to the loop plane).

cos . The vector is parallel to the loop normal and h as

magnitude equal to .

v

vA v A A

A

Note 2 :

3

n̂

n̂n̂

Consider the closed surface shown in the figure.

In the vicinity of the surface assume that we have

a known electric field . The flux of the

electric field thro h

ug

E

Flux of the Electric Field.

the surface is defined as follows:

1. Divide the surface into small "elements" of area .

2. For each element calculate the term cos .

3. Form the sum .

4. Take the limit of the sum a

A

E A EA

E A

2Flux SI unit: N m / C

s the area 0.

The limit of the sum becomes the integral:

The circle on the integral sign indicates that

the surface is closed. When we apply Gauss'

la

A

E dA

Note 1 :

w the surface is known as "Gaussian."

is proportional to the net number of

electric field lines that pass through the surface.

Note 2 :

E dA 4

n̂

n̂n̂

0 enc

0 enc

The flux of through any closed surface net charge enclosed by the surface.

Gauss' law can be formulated as follows:

In equ ation form: Equivalently:

E q

q

Gauss' Law

0 encε E dA q

0 encq

0 encε E dA q

Gauss' law holds for closed surface.

Usually one particular surface makes the problem of

determining the electric field very simple.

When calculating the net charge inside a c

Note 1 : any

Note 2 : losed

surface we take into account the algebraic sign of each charge.

When applying Gauss' law for a closed surface

we ignore the charges outside the surface no matter how

large they are

.

Note 3 :

Examp

1 0 1 2 0 2

3 0 3 4 0 4

1 2 3 4

Surface : , Surface :

Surface : 0, Surface : 0

We refer to , , , as "Gaussian surfaces."

S q S q

S S q q

S S S S

le :

Note :5

n̂

dA

Gauss' law and Coulomb's law are different ways

of describing the relation between electric charge

and electric field in static cases. One can derive

Coulomb's law from Gauss

Gauss' Law and Coulomb's Law

' law and vice versa.

Here we will derive Coulomb's law from Gauss' law.

Consider a point charge . We will use Gauss' law

to determine the electric field generated at a point

at a distance from

q

E

P r q. We choose a Gaussian surface

that is a sphere of radius and has its center at .r q

Coulomb's law Gauss' law

2

2

0 enc 0

We divide the Gaussian surface into elements of area . The flux for each element is:

cos 0 Total flux 4

From Gauss' law we have: 44

dA

d EdA EdA EdA E dA E r

qq q r E q E

r

2

0

This is the same answer we got in Chapter 22 using Coulomb's law.

6

eE

v

F

We shall prove that the electric field inside a conductor vanishes.

Consider the conductor shown in the figure to the left. It is an

experimental fact that such an

The Electric Field Inside a Conductor

object contains negatively charged

electrons, which are free to move inside the conductor. Let's

assume for a moment that the electric field is not equal to zero.

In such a case a nonvanishing force is exerted by the

field on each electron. This force would result in a nonzero

velocity , and the moving electrons would constitute an electric

current. We will see in subsequent chapters th

F eE

v

at electric currents

manifest themselves in a variety of ways:

(a) They heat the conductor.

(b) They generate magnetic fields around the conductor.

No such effects have ever been observed, thus the original

assumption that there exists a nonzero electric field inside

the conductor. We conclude that :

The electrostatic electric field inside a conductor is equal to zero.E 7

Consider the conductor shown in the figure that has

a total charge . In this section we will ask the question:

Where is this charge located? To answer the question

we wil

q

A Charged Isolated Conductor

l apply Gauss' law to the Gaussian surface shown

in the figure, which is located just below the conductor

surface. Inside the conductor the electric field 0.

Thus 0 ( ).

From Gaus

E

E A

eq. 1

enc

0

enc

s's law we have: ( ).

If we compare eq. 1 with eq. 2 we get = 0 .

q

q

eq. 2

Thus no charge exists inside the conductor. Yet we know that the conductor

has a nonzero charge . Where is this charge located? There is only one place

for it to be: On the surface of the conductor.

q

electrostatic charges can exist inside a conductor.

charges reside on the conductor surface.

No

All 8

Consider the conductor shown in the figure that has

a total charge . This conductor differs from the one

shown on the previous page in one aspect: It has a

q

An Isolated Charged Conductor with a Cavity

cavity. We ask the question: Can charges reside

on the walls of the cavity?

As before, Gauss's law provides the answer.

We will apply Gauss' law to the Gaussian surface shown

in the figure, which is

enc

0

located just below the conductor

surface. Inside the conductor the electric field 0.

Thus 0 ( ).

From Gauss's law we have: ( ).

If we compare eq. 1 with eq. 2 we ge

E

E A

q

eq. 1

eq. 2

enct = 0.q

Conclusion :

There is no charge on the cavity walls. All the excess charge

remains on the outer surface of the conductor.

q

9

1n̂3n̂

2n̂

S1S2S3

The electric field inside a conductor is zero. This is

not the case for the electric field outside. The

electric field vector is perpendicular to the E

The Electric Field Outside a Charged Conductor

conductor

surface. If it were not, then would have a component

parallel to the conductor surface.

E

E

Since charges are free to move in the conductor, would cause the free

electrons to move, which is a contradiction to the assumption that we

have stationary charges. We will apply Gauss' law using

E

1 2 3 1 2 3

1

the cylindrical closed

surface shown in the figure. The surface is further divided into three sections

, , and as shown in the figure. The net flux .

cos 0

S S S

EA EA

2

enc3

0

enc enc

0 0

cos90 0

0 (because the electric field inside the conductor is zero).

1 . The ratio is known as surface charge density .

EA

qEA

qE

AE

q

A

0

E

10

Symmetry. We say that an object is symmetric under a particular

mathematical operation (e.g., rotation, translation, …) if to an observer

the object looks the same before and after the operation.

Note: Symmetry is a primitive notion and as such is very powerful.

Featureless

sphere

Rotation axis

ObserverRotational symmetryExample of Spherical

Symmetry

Consider a featureless beach ball

that can be rotated about a

vertical axis that passes through

its center. The observer closes his

eyes and we rotate the sphere.

When the observer opens his

eyes, he cannot tell whether the

sphere has been rotated or not.

We conclude that the sphere has

rotational symmetry about the

rotation axis. 11

Featureless

cylinder

Rotation axis

Observer

Rotational symmetry A Second Example of Rotational

Symmetry

Consider a featureless cylinder that can

rotate about its central axis as shown in

the figure. The observer closes his eyes

and we rotate the cylinder. When he

opens his eyes, he cannot tell whether

the cylinder has been rotated or not. We

conclude that the cylinder has rotational

symmetry about the rotation axis.

12

Observer

Magic carpet

Infinite featureless

plane

Translational

symmetryExample of Translational

Symmetry:

Consider an infinite featureless

plane. An observer takes a trip

on a magic carpet that flies

above the plane. The observer

closes his eyes and we move the

carpet around. When he opens

his eyes the observer cannot tell

whether he has moved or not.

We conclude that the plane has

translational symmetry.

13

Recipe for Applying Gauss’ Law

1. Make a sketch of the charge distribution.

2. Identify the symmetry of the distribution and its effect on

the electric field.

3. Gauss’ law is true for any closed surface S. Choose one

that makes the calculation of the flux as easy as possible.

4. Use Gauss’ law to determine the electric field vector:

enc

0

q

14

S1

1n̂

2n̂

S2

S3

3n̂

Consider the long rod shown in the figure. It is uniformly

charged with linear charge density . Using symmetry

arguments we can show that the

Electric Field Generated by a Long, Uniformly Charged Rod

electric field vector points

radially outward and has the same magnitude for points at

the same distance from the rod. We use a Gaussian surface

that has the same symmetry. It is a cylinder of rad

r

S ius

and height whose axis coincides with the charged rod.

r

h

1 2

3 1 2 3 1 3

We divide S into three sections: Top flat section , middle curved section ,

and bottom flat section . The net flux through is . Fluxes and

vanish because the electric field i

S S

S S

en2

0

0 0

c

0

s at right angles with the normal to the surface:

2 cos0 2 2 . From Gauss's law we have: .

If we compare these two equations we get: .2

2

q hrhE rhE rhE

hr EE

rh

02E

r

15

1n̂

2n̂

3n̂ S1

S2S3

We assume that the sheet has a positive charge of

surface density . From symmetry, the electric field

vector is peE

Electric Field Generated by a Thin, Infinite,

Nonconducting Uniformly Charged Sheet

rpendicular to the sheet and has a

constant magnitude. Furthermore, it points away from

the sheet. We choose a cylindrical Gaussian surface

with the caps of area on either side of the sheet as

shown

S

A

1 2

3

in the figure. We divide into three sections:

is the cap to the right of the sheet, is the curved

surface of the cylinder, and is the cap to the left of the

sheet. The net flux through

S

S S

S

S 1 2 3

1 3 2

enc

0 0

0 0

is .

cos0 . 0 ( = 90 )

2 . From Gauss's law we have:

2 .2

EA EA

q AEA

AEA E

02E

16

S

S'

A

A'

1 1

The electric field generated by two parallel conducting infinite planes is charged with

surface densities and - . In figs. and we show the two plates isolated so that

one does not influence the

a b

charge distribution of the other. The charge spreads out

equally on both faces of each sheet. When the two plates are moved close to each

other as shown in fig. , then the charges on one plate attrc act those on the other. As

a result the charges move on the inner faces of each plate. To find the field between

the plates we apply Gauss' law for the cylindrical surface , which has caps of area

iE

S

enc 10

0 0

enc 1

1

0

0

0

0

.

2The net flux To find the field outside

the plates we apply Gauss' law for the cylindrical surface ', which has caps of area .

The net

2= .

flux

i i

A

q AE A E

S A

E

qE A

1

0

0

0 0. E

1

0 0

2iE

0 0E

17

1 Consider a Gaussian surface

that is a sphere with radius and whose

center coincides with that of the ch

S

r R

The Electric Field Generated by a Spherical Shell

of Charge and Radius

Inside the shell :

q R

2 enc

0

2

arged shell.

The electric field flux 4 0.

Thus

Consider a Gaussian surface

that is a sphere with radius and whose

center coincides with that of the ch r

0.

a ge

i

i

qr E

S

r R

E

Outside the shell :

2 enc0

0 0

0 2

0

d shell.

The electric field flux 4 .

Thus

Outside the shell the electric field is the

same as if all the charge of the shell were

concentrated at the shell center

4

.

.q

Er

q qr E

Note :

1n̂iE

2n̂0E

0iE

0 2

04

qE

r

18

S2

S1

iE2n̂

oE1n̂

1 Consider a Gaussian surface

that is a sphere with radius and whose

center coincides with that

S

r R

Electric Field Generated by a Uniformly Charged Sphere

of Radius and Charge

Outside the sphere :

R q

0 2

2

0 enc 0

2

0

0

of the charged shell.

The electric field flux 4 / /

Thus

Consider a Gaussian surface

that is a sphere with radius and whose

center coincides w h

.

it

4

r E q q

S

qE

r

r R

Inside the sphere :

3

0

2 enc

0

3 32

enc 3 33 3 0

that of the charged shell.

The electric field flux 4 . Since the charge is

uniform 44 4

3 3

s .hu 4

T i

i

enci

qr E

q q r r qq q

q

r ER

R

ER

Rr

r

19

0 2

04

qE

r

3

04i

qE r

R

R

3

04

q

R

r

E

O

S2

S1

iE2n̂

0E1n̂

Electric Field Generated by a Uniformly Charged Sphere

of Radius and Charge

Summary :

R q

20