Chapter 23 Finding the Electric Field - II In this chapter we will introduce the following new...
-
Upload
alexis-moore -
Category
Documents
-
view
229 -
download
0
Transcript of Chapter 23 Finding the Electric Field - II In this chapter we will introduce the following new...
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)