Chapter 23Chapter 23Gauss’s LawGauss’s Law

Summer 2008Summer 2008

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.

(23-1)

Last Time: Definition – Sort of –

Electric Field Lines

Field Lines Electric Field

Last time we showed that

Ignore the Dashed Line … Remember last time .. the big plane?

00

00

0

0

E=0 0 E=0

We will use this a lot!

NEW RULES Imagine a region of space where the ELECTRIC

FIELD LINES HAVE BEEN DRAWN. The electric field at a point in this region is

TANGENT to the Electric Field lines that have been drawn.

If you construct a small rectangle normal to the field lines, the Electric Field is proportional to the number of field lines that cross the small area.

The DENSITY of the lines. We won’t use this much

n̂

n̂

So far …

The electric field exiting a closed surface seems to be related to the charge inside.

But … what does “exiting a closed surface mean”?

How do we really talk about “the electric field exiting” a surface? How do we define such a concept? CAN we define such a concept?

Mr. Gauss answered the question with..

The “normal component” of the ELECTRIC FIELD

E

nEn

nEnE

nE

)cos(

)cos(

E

E

n

n

DEFINITION FLUX

E

nEn

)cos(

)(

AE

nEAE

AFlux n

“Element” of Flux of a vector E leaving a surface

dAdd

also

dd NORMAL

nEAE

AEAE

For a CLOSED surface:n is a unit OUTWARD pointing vector.

Definition of TOTAL FLUX through a surface

dA

is surface aLEAVING Field

Electric theofFlux Total

out

surfaced

nE

Visualizing Flux

ndAEflux

n is the OUTWARD pointing unit normal.

Remember

ndAEflux

)cos(nEnE

n E

A

Component of Eperpendicular tosurface.

This is the fluxpassing throughthe surface andn is the OUTWARDpointing unit normalvector!

ExampleCube in a UNIFORM Electric Field

L

E

E is parallel to four of the surfaces of the cube so the flux is zero across thesebecause E is perpendicular to A and the dot product is zero.

Flux is EL2

Total Flux leaving the cube is zero

Flux is -EL2

Note sign

area

Simple Example

0

22

0

20

20

20

44

1

4

1

4

1

4

1

qr

r

q

Ar

qdA

r

q

dAr

qdA

Sphere

nE

r

q

Gauss’ Law

n is the OUTWARD pointing unit normal.

0

0

enclosedn

enclosed

qdAE

qndAE

q is the total charge ENCLOSEDby the Gaussian Surface.

Flux is total EXITING theSurface.

Line of Charge

L

Q

L

Q

length

charge

Materials

Conductors Electrons are free to move. In equilibrium, all charges are a rest. If they are at rest, they aren’t moving! If they aren’t moving, there is no net force on them. If there is no net force on them, the electric field must be

zero.

THE ELECTRIC FIELD INSIDE A CONDUCTOR IS ZERO!

eE

v

F

We shall prove the 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 o

The electric field inside a conductor

bject contains negatively charged

electrons which are free to move inside the conductor. Lets

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

In such a case an non-vanishing force F

is exerted by the

field on each electron. This force would result in a non-zero

velocity and the moving electrons would constitute an electric

current. We will see in subserquent chapters tha

eE

v

t electric currents

manifest themselves in a variety of ways:

a) they heat the conductor

b) they generate magnetic fields around the concuctor

No such effects have ever been observed. Thus the original

assumption that there exists a non-zero electric field inside

the conductor. Thus we conclude that :

The electrostatic electric field inside a conductor is equal to zeroE

(23-6)

Isolated Conductor

Electric Field is ZERO inthe interior of a conductor.

Gauss’ law on surface shownAlso says that the enclosedCharge must be ZERO.

All charge on a Conductor must reside onThe SURFACE.

Charged Conductors

E=0

E

---

-

-

Charge Must reside onthe SURFACE

0

0

E

or

AEA

Very SMALL Gaussian Surface

Charged Isolated Conductor

The ELECTRIC FIELD is normal to the surface outside of the conductor.

The field is given by:

Inside of the isolated conductor, the Electric field is ZERO.

If the electric field had a component parallel to the surface, there would be a current flow!!!

0

E

Isolated (Charged) Conductor with a HOLE in it.

0

0Q

dAEn

Because E=0 everywhereinside the conductor.

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

remains on the outer surface of the conductor

q

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 symmetry

Example 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. Then, 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.

(23-10)

Line of Charge

From SYMMETRY E isRadial and Outward

rE

hrhE

qdAEn

0

0

0

2

2

Infinite Sheet of Charge

cylinderE

h

0

0

2

E

AEAEA

We got this sameresult from thatugly integration!

1̂n

2n̂

3n̂ S1

S2S3

We assume that the sheet has a positive charge of

surface density . From symmetry, the electric field

vector is E

Electric field generated by a thin, infinite,

non - conducting uniformly charged sheet.

perpenducular to the sheet and has a

constant magnitude. Furthermore it points away from

the sheet. We choose a cylindrical Gaussian surface S

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

show

A

1 2

3

n in the figure. We devide S into three sections:

S is the cap to the right of the sheet. S is the curved

surface of the cylinder. S is the cap to the left of the

sheet. The net flux through S 1 2 3

1 2 3

is

cos0 . 0 ( =90 )

2 . From Gauss's law we have:

2 2

enc

o o

o o

EA EA

q AEA

AE EA

2 o

E

(23-15)

S

S'

A

A'

1 1

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

surface densities and - . In fig.a and b we shown the two plates isolated so that

one does not influence the cha

rge distribution of the other. The charge speads out

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

other as shown in fig.c, then the charges one one plate attract 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 S which has caps of area .

The

iE

A

1

1 1

12 net flux To find the field outside

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

The net flux 0

2

enci o

o oi

enco

o

o

oo

q AE A E

A

q

E

E EA

0

12i

o

E

0oE

(23-16)

1 Consider a Gaussian surface S

which is a sphere with radius and whose

center coincides with that of the

r R

The electric field generated by a spherical shell

of charge q and radius R.

Inside the shell :

2

2

charge shell.

The electric field flux 4 0

Thus

Consider a Gaussian surface S

which is a sphere with radius and whose

center coincides with that of the char

0

g

enci

o

i

qr E

r R

E

Outside the shell :

2

2

e 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

co

4

ncentrated at the shell center.

enco

o o

oo

q

r

qr E

qE

Note :

1̂n iE

2n̂ oE

0iE

24oo

qE

r

(23-17)

Insulators

In an insulator all of the charge is bound. None of the charge can move. We can therefore have charge anywhere in

the volume and it can’t “flow” anywhere so it stays there.

You can therefore have a charge density inside an insulator.

You can also have an ELECTRIC FIELD in an insulator as well.

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

o

q

(23-13)

A uniformly charged cylinder.

R

r

RE

hRqrhE

Rr

rE

hrrhE

Rr

0

2

0

2

0

0

0

2

2

)()2(

2

)()2(