Download Chapter 23

CHAPTER-23
Gauss’ Law
1
CHAPTER-23 Gauss’ Law
Topics to be covered
The flux (symbol Φ ) of the electric field
Gauss’ law
Application of Gauss’ law to 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
Application of Gauss’ law to determine the electric
field inside and outside charged conductors.
2
Ch 23-1 Gauss’ Law
Gaussian surface:
 a hypothetical
(imaginary) surface
enclosing a charge
distribution.
Gauss’ law: relates the
electric field at points
on a closed Gaussian
surface to the net
charge enclosed by that
surface
3
Ch 23-2 Flux
Flux :
Volume flow rate
(Volume flux)  of air
through a small square loop
with area A
  = v cos A =v.A
* where v is air velocity,
* A is area vector , with
magnitude equal to loop area
and direction is normal to
plane of loop
* and  is angle between v and
A
4
Ch 23-3 Flux of an Electric Field
 Electric Flux through a Gaussian
surface immersed in a nonuniform electric field
 Gaussian surface is divided into
element of area A with
corresponding electric field E at
that location then electric flux

  =  E. A
 If A becomes smaller then
  = surf E.dA
 Net fux through the cylindrical
surface
= -EdA+0+EdA=0
5
Ch 23-3 Flux of an Electric Field
 Electric Flux through a
Gaussian surface
immersed in a nonuniform electric field
 Gaussian surface is
divided into element of
area A with
corresponding electric
field E at that location
then electric flux 
 =  E. A
 If A becomes smaller
then
 = surf E.dA
 Net fux through the
cylindrical surface
= -EdA+0+EdA=0
6
Ch 23-4 Gauss’ Law
Gauss Law: relates the net
flux  of an electric field
through a closed Gaussian
surface and the net charge
qenc that is enclosed by the
surface
0  = qenc but  = surf E.dA
0 surf E.dA = qenc
 = surf E.dA=qenc/0
7
Ch 23-4 Gauss’
Law
 Point charges enclosed in the
surface
 Surface S1: lines of E field
paralell to area vector dA;
E.dA is positive; qenc is
positive
 Surface S2: lines of E field
antiparalell to area vector
dA; E.dA is negative; qenc is
negative
 Surface S3: lines of E field
antiparalell to area vector dA
in upper half and paralell in
lower half; E.dA is zero; qenc
is zero
 Surface S4: Similar to S3
and net charge in the surface
is Zero
8
Ch 23-5 Gauss’ Law and Coulomb Law
 Gauss Law: surf E.dA= qenc/ 0
If E is constant at the surface
then
 surf E.dA= Esurf dA = EA=qenc/ 0
 For a sphere A= 4R2
 Then E= qenc/ A0 = qenc /4 0 R2
 E= qenc /4 0 R2=k qenc/R2
dA
n̂
9
Ch 23-6 A Charged
Isolated Conductor
 A Charged Conductor
If an excess charge is placed on
an isolated conductor, that
amount of charge will move
entirely to the surface of the
conductor.
 An Isolated Conductor with a
Cavity
 There is no net charge on the
cavity wall of a charged
conductor
10
Induced Charges
A charged spherical
shell with charge 100e enclosing
another point
charge -50e
Charge on shell
inner surface: +50e
Charge on shell
outer surface: 150e
11
Ch 23-6 A Charged Isolated Conductor
 Electric Field outside the Surface
of a Conductor
 Imagine a tiny Gaussian cylindrical
surface embedded in the conductor
with one end cap inside the
conductor. E-field field through
this section is zero.
 Evaluate  =surf E.dA= qenc/ 0 on
the surface
 For the cylinder the net flux is only
through the end cap lying outside
the conductor, where E A . Then
 = EA=qenc/0 =A/0
 E=/0
where  is suface charge density

( charge per unit area)
12
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.
Rotational symmetry
Featureless
sphere
Rotation axis
(23-10)
Observer
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. 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
13
axis.
Rotational symmetry
Featureless
cylinder
Rotation axis
Observer
(23-11)
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.
14
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
(23-12)
symmetry.
Translational
symmetry
Observer
Magic carpet
Infinite featureless
plane
15
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:

qenc
0
(23-13)
16
Ch 23-7 Applying Gauus’ Law:
Cylindrical Symmetry








Electric field E at a distance r from
the axis of a infinitely long plastic rod
with uniform positive linear charge
density .
Imagine a tiny Gaussian cylindrical
surface coaxial with the rod with two
end caps of the cylindrical surface.
Flux of E-field through the end caps
is zero where E  A. Net flux of Efield through the circumference where
E A.
= surf E.dA= E surf dA= qenc/0
For circumference surf dA= 2rh
Then EA= E 2rh= qenc/0
E=(qenc/2rh 0)= 1/2r0*(qenc /h)
E=(1/20)  /r =2k  /r; = qenc/h
S1
n̂1
S2
n̂2
S3
n̂3
17
Ch 23-8 Applying Gauss’ Law:
Planar Symmetry
 Non conducting Sheet
 A thin, infinite, non conducting sheet
with a uniform positive charge density
 on one side only.
 E at a distance r in front of the
sheet. Choose a Gaussian cylinder
with end caps passing through the
surface
 E field  to cylinder circumference
net flux zero through it
 E field  to end caps. Net flux
through caps
 = surf E.dA= qenc/0=A/0
 surf E.dA=EA+EA= 2EA=A/0
 E=/20 (Sheet of charge )
n̂2
S2
S3
n̂3
n̂1
S1
18
Ch 23-8 Applying Gauss’ Law: Planar
Symmetry
Ei 
2 1
0
 Two conducting plates with charge density 1
 All charges on the two faces of the plates
 For two oppositely charged plates placed near each other, E
field outer side of the plates is zero while inner side the Efield= 2 1/0
19
Ch 23-8 Applying Gauss’ Law: Planar
Symmetry
 Two non-conducting
plates with charge
density +
 All charges on the one face
of the plates
 For two oppositely charged
plates placed near each
other, E field outer side of
the plates is EL or ER with
EL= ER= E(+)- E(+) and
between the plates the field
 EB = E(+)+ E(-)
20
Ch 23-9 Applying Gauss’ Law:
Spherical Symmetry
Shell Theorem:
 A shell of uniform charge
attract or repel a charged
particle that is outside the shell
as if all the shell charge were
concentrated at the center of
the shell.
Eo=kq/r2 (r>R)
 If a charged particle is located
inside a shell of uniform charge,
there is no electrostatic force
on the particle from the shell.
Ei=0 (r<R)
n̂2 E0
n̂1
Ei
21
Ch 23-9 Applying Gauss’ Law:
Spherical Symmetry
 Electric field outside a
uniform sphere of charge
•
E=kq/r2
( r>R)
 Electric field inside a
uniform sphere of charge
•
E=kqr/R3
( r>R)
22
Suggested problems Chapter 23
23