Chapter 23

Finding the Electric Field - II

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)

n̂

n̂

Consider an airstream of velocity which is aimed at a loop

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

ˆloop normal . The product cos known a t

s h

v

v

n vA

Flux of a vector.

e . 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 pl

vA v

flux

Note 1 :

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

to the loop plane).

cos The vector is parallel to the loop normal and has

magnitute equal

to .

v

vA v A A

A

Note 2 :

(23-2)

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 as t

A

E A EA

E A

2Flux SI unit:

he area 0

The limit of the sum becomes the integral:

The circle on the intergral sign indicates that

the closed surface is closed. When we appl

N

G

m /

y aus

Cd

A

E A

Note 1 :

s'

law the surface is known as "Gaussian"

is proportinal to the net number of

electric field lines that pass through the surface

Note 2 :

E dA

(23-3)

n̂

n̂n̂

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

Gauss' law can be formulated as follows:

In equation form: Equivalently:

o enc

o enc

E q

q ε

Gauss' Law

o encE dA q

o encq

o 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 :

Exampl

1 1 2 2

3 3 4 4

1 2 3 4

Surface S : , Surface S :

Surface S : 0 , Surface S : 0

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

o o

o o

q q

q q

e :

Note :

(23-4)

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

' laws 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

P at a distance from

q

E

r

. We choose a Gaussian surface

which is a sphere of radius and has its center at .

q

r q

Coulomb's law Gauss' law

2

2o o

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

cos0 Total flux 4

From Gauss' law we have: 44enc

d EdA EdA EdA E dA E r

qq q r E q E

r

2o

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

(23-5)

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)

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 will

q

A charged isolated conductor :

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 Gauss's l

E

E A

eqs.1

aw we have: ( )

If we compare eqs.1 with eqs.2 we get: = 0

enc

o

enc

q

q

eqs.2

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

has a non-zero 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 (23-7)

Consider the conductor shown in the figure that has

a total charge . This conductor differs from the one

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

q

An isolated charged conductor with a cavity :

cavity. We ask the question whether charges can 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 located just below the conductor

surface. Inside the conductor the electric field 0

Thus 0 ( )

From Gauss's law we have: ( )

If we compare eqs.1 with eqs.2 we

enc

o

E

E A

q

eqs.1

eqs.2

get: = 0encq

Conclusion :

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

remains on the outer surface of the conductor

q

(23-8)

1̂n3n̂

2n̂

S1S2S3

The electric field inside a conductor is zero. This is

not the case for the electric field outside. The

elecric 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 th

E

1 2 3 1 2 3

1

e cylindrical closed

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

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

cos0 .

S S S

EA EA

2

3

cos90 0

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

1 The ratio is known as surface charge density

enc

o

enc enc

oo

EA

qEA

qE

AE

q

A

o

E

(23-9)

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)

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. Then, 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

(23-11)

observer

magic carpet

infinite featureless plane

Translational symmetry

Example 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

(23-12)

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)

S1

1̂n

2n̂S2

S3

3n̂

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

charged with linera charge density . Using symmetry

arguments we can show that th

Electric field generated by a long, uniformly charged rod

e electric field vector points

radially outwards and has the same magnitude for points at

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

S that has the same symmetry. It is a cylinder of r

r

adius

and height whose axis coincides with the charged rod.

r

h

1 2

3 1 2 3 1 2

We divide S into three sections: Top flat section S . Middle curved section S .

Bottom flat section S . The net flux through S is . Fluxes and

vanish because the electric field is

3

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: 22

o

enc

o o

o

q hrhE rhE rhE

hrhE E

r

(23-14)2 o

Er

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 :

1̂niE

2n̂oE

0iE

24oo

qE

r

(23-17)

S2

S1

iE

2n̂

oE

1̂n

1 Consider a Gaussian surface S

which is a sphere with radius and whose

center coincides with that

r R

Electric field generated by a uniformly charged sphere

of radius R and charge q

Outside the sphere :

2

2

2

of the charge shell.

The electric field flux 4 / /

Thus

Consider a Gaussian surface S

which is a sphere with radius and whose

center coincides w

4

ith

o enc o

oo

or E q

r R

qE

r

q

Inside the sphere :

2

3 3 32

3 3

3

3

that of the charge shell.

The electric field flux 4

4 /4

4 /

T4

hus

enci

o

enc io

io

qr E

r R R q

qE

q q q rr

R

R

r

Er

(23-18)

24oo

qE

r

34io

qE r

R

R

34 o

q

R

r

E

O

S2

S1

iE

2n̂

oE

1̂nElectric field generated by a uniformly charged sphere

of radius R and charge q. Summary

(23-19)