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)
n̂
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 :
(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 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 :
(23-3) AdE
AdE
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
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 :(23-4)
encqAdE
0
encqAdE
0
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.
(23-5)
SS
dAEEdA
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(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 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(23-7)
AdE
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
(23-8)
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
(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 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. (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. 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
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.
(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
0
q
(23-13)
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 2
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
en3
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
(23-14)
3
2
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 2 3
enc
0 0
0 0
is .
cos 0 . 0 ( = 90 )
2 . From Gauss's law we have:
2 .2
EA EA
q AEA
AEA E
02E
(23-15)
3 2
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 10
0
1
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 flux
2. ii
A
q AE A E
S A
qA
E
E
0 0.0 E
1
0
2iE
0 0E
(23-16)
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
(23-17)
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 :
2 enc
0
3 3 32
enc 3 3 3
3
0
0
that of the charged shell.
The electric field flux 4 .
4 /4
4 /
Thus .4
i
i
i
qr E
r R R qq q q r E
R
qE r
R
r r
(23-18)
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
(23-19)