Transcript Слайд 1 - Georgia State University

Reading: Chapter 28
E 
Q
4π ε o r
2
 ke
Q
r
2
For r > a

Gauss’s Law
1
Chapter 28
Gauss’s Law
2
Electric Flux
Definition:
• Electric flux is the product of the
magnitude of the electric field and the
surface area, A, perpendicular to the
field
• ΦE = EA
• The field lines may make some angle θ
with the perpendicular to the surface
• Then ΦE = EA cos θ
E
normal


 E  EA
E
 E  E A cos 
3
Electric Flux: Surface as a Vector
Vector, corresponding to a Flat Surface of Area A, is determined
by the following rules:
 the vector is orthogonal to the surface
 the magnitude of the vector is equal to the area A
The first rule
A
90
o
 the vector is orthogonal to the surface
normal
does not determine the direction of A .
There are still two possibilities:
Area = A
A
or
90
o
90
o
A
You can choose any of them
4
Electric Flux: Surface as a Vector
If we consider more complicated surface then the directions of
vectors should be adjusted, so the direction of vector is a smooth
function of the surface point
correct
or
A2
A1
wrong
or
A3
5
Electric Flux
Definition:
• Electric flux is the scalar product of
electric field and the vector A
•   EA
A

E

 E  E A  E A co s   0
or

E
A

 E  E A   E A co s   0
6
Electric Flux
1
 E  1   2
area  A1
E

A1
area  A2

A2
90  
o
 2  E A 2  E A 2 cos(90   )  E A 2 sin 
o
2
 1  E A1  E A1 sin 
flux is positive
 E  E A1 sin   E A2 sin 
E
if

A1

E

 E  E A1 sin 
2  0
A2
then
A1
 E   E A1 cos   E A2 cos 
flux is negative
A2
A 2 and E are
orthogonal
7
Electric Flux
• In the more general case, look
at a small flat area element
  E  E i  Ai c o s θ i  E i   Ai
• In general, this becomes
 E  lim
 Ai  0
E
i
  Ai 

E  dA
surface
• The surface integral means the
integral must be evaluated over
the surface in question
• The units of electric flux will be
N.m2/C2
8
Electric Flux: Closed Surface
The vectors  A i point in different
directions
 At each point, they are
perpendicular to the surface
 By convention, they point
outward
 E  lim
 Ai  0
E
i
  Ai 

E  dA
surface
9
Electric Flux: Closed Surface
2
Closed surface
 E  1   2   3  4  5  6
6
3
5
E is orthogonal to A 3 , A 4 , A 5 , and A 6
Then
 3  E A3  0
 4  E A4  0
 5  E A5  0
 6  E A6  0
1
E
 E  1   2
4
E
A3
90  
o
1
 1  E A1  E A1 co s(9 0   )   E A1 sin 
o
A1
3
2

 2  E A2  E A2
A2
4
A4
 E  E ( A2  A1 sin  )
but
A2  A1 sin 
A1

A2
Then
E  0
10
(no charges inside closed surface)
Electric Flux: Closed Surface
• A positive point charge, q, is
located at the center of a sphere
of radius r
• The magnitude of the electric
field everywhere on the surface
of the sphere is
Spherical
surface
E = keq / r2
• Electric field is perpendicular to
the surface at every point, so
E has the same direction as
A at every point.
11
Electric Flux: Closed Surface
E has the same direction as A at every point.
E  ke
q
r
2
Then
 

Spherical
surface
E i dA i  E  dA i 
i
i
 E A0  E 4 r  4 r k e
2
 4 k e q 
q
0
2
q
r
2

Gauss’s Law
 does not depend on
r
ONLY BECAUSE
E 
1
12 r
2
Electric Flux: Closed Surface
E and A have opposite directions at every point.
|q |
E  ke
r
2
Spherical
surface
Then
 
 E dA
i
i
  E  dA i 
i
i
  E A0   E 4 r   4 r k e
2
  4  k e | q |
q
0
2
|q |
r
2


Gauss’s Law
 does not depend on
r
ONLY BECAUSE
E 13
1
r
2
Gauss’s Law
 The net flux through any closed surface
surrounding a point charge, q, is given by q/εo
and is independent of the shape of that
surface
 The net electric flux through a closed surface
that surrounds no charge is zero
Ai
q
E
Ai
 
q
0
E
q
 
q
0
14
Gauss’s Law
 Gauss’s law states
E 

 E  dA 
q in
εo
qin is the net charge inside the surface
 E is the total electric field and may have contributions
from charges both inside and outside of the surface
Ai
q
E
Ai
 
q
0
E
q
 
q
0
15
Gauss’s Law
 Gauss’s law states
E 

E  dA 
q in
εo
 qin is the net charge inside the surface
 E is the total electric field and may have contributions from
charges both inside and outside of the surface
q5
q2
 
q1
q3
q1  q 2  q 3  q 4
0
q4
q6
q7
16
Gauss’s Law
 Gauss’s law states
E 

E  dA 
q in
εo
 qin is the net charge inside the surface
 E is the total electric field and may have contributions
from charges both inside and outside of the surface
q5
q2
q1
  0
q3
q4
q6
q7
17
Gauss’s Law
 Gauss’s law states
E 

E  dA 
q in
εo
 qin is the net charge inside the surface
 E is the total electric field and may have contributions
from charges both inside and outside of the surface
q5
q2
  0
q
q
q4
q6
q7
18
Gauss’s Law: Problem
What is the flux through surface 1
1   2  0
 2  E A0   E A0
A1
 1    2  E A0
1
E
A2
A0
2
A3
19
Chapter 28
Gauss’s Law: Applications
20
Gauss’s Law: Applications
 Although Gauss’s law can, in theory, be solved to find
E for any charge configuration, in practice it is limited to
symmetric situations
 To use Gauss’s law, you want to choose a Gaussian
surface over which the surface integral can be simplified
and the electric field determined
 Take advantage of symmetry
 Remember, the gaussian surface is a surface you
choose, it does not have to coincide with a real surface
 
 E  dA 
q in
εo
q5
 
q1  q 2  q 3  q 4
0
q6
q2
q1
q3
q4
21
Gauss’s Law: Point Charge
E

q
SYMMETRY:
E - direction - along the radius
E
 
 
- depends only on radius, r
q
0

Only in this case the magnitude of
electric field is constant on the
Gaussian surface and the flux can be
easily evaluated
- Gauss’s Law
E i dA i  E  dA i  E A 0  E 4  r
i
Then
Gaussian Surface – Sphere
2
- definition of the Flux
i
q

 4 r E
2
E  ke
q
r
2
22
Gauss’s Law: Applications
 
 E  dA 
• Try to choose a surface that satisfies one or more of these conditions:
– The value of the electric field can be argued from symmetry to be
constant over the surface
– The dot product of E.dA can be expressed as a simple algebraic
product EdA because E and dA are parallel
– The dot product is 0 because E and dA are perpendicular
– The field can be argued to be zero over the surface
correct Gaussian surface
wrong Gaussian surface

23
q in
εo
Gauss’s Law: Applications
Spherically Symmetric Charge Distribution
The total charge is Q
SYMMETRY:
E
A
E - direction - along the radius
E
- depends only on radius, r
• Select a sphere as the
gaussian surface
• For r >a
E 
E 
 E  d A   E d A  4π r
Q
4π ε o r
2
 ke
Q
r
2
2
E 
q in
εo

Q
εo
The electric field is the same as
for the point charge Q
24
Gauss’s Law: Applications
Spherically Symmetric Charge Distribution
E 
Q
4π ε o r
2
 ke
Q
r
The electric field is the same as
for the point charge Q !!!!!
2
Q
a
For r > a

Q
For r > a

25
Gauss’s Law: Applications
Spherically Symmetric Charge Distribution
E
SYMMETRY:
A
E - direction - along the radius
E
- depends only on radius, r
• Select a sphere as the
gaussian surface, r < a
q in 
Q
4
3
a
4
3
r  Q
3
3
r
3
a
3
Q
E 
 E  d A   E d A  4π r
E 
q in
4π ε o r
2
 ke
Qr
a
3
3
1
r
2
 ke
2
E 
Q
a
3
q in
εo
r
26
Gauss’s Law: Applications
Spherically Symmetric Charge Distribution
• Inside the sphere, E varies
linearly with r
E → 0 as r → 0
• The field outside the sphere
is equivalent to that of a point
charge located at the center
of the sphere
27
Gauss’s Law: Applications
Field due to a thin spherical shell
• Use spheres as the gaussian surfaces
• When r > a, the charge inside the surface is Q and
E = keQ / r2
• When r < a, the charge inside the surface is 0 and E = 0
28
Gauss’s Law: Applications
Field due to a thin spherical shell
• When r < a, the charge inside the surface is 0 and E = 0
 A1
 q 2    A2
 A1  r1  
 A 2  r2  
2
r1
E2
the same
solid angle
 q 1    A1
E1  ke
 q1
2
1
r
 ke
2
  A1
2
1
r
 r1  
2
 ke
2
1
r
 k e  
E1
E 2  ke
r2
 q2
2
2
r
 ke
  A2
2
2
r
 r2  
2
 ke
2
2
r
 k e  
E1  E 2
E1  E 2  0
 A2
Only because in Coulomb law
1
E  29
2
r
Gauss’s Law: Applications
Field from a line of charge
• Select a cylindrical Gaussian surface
– The cylinder has a radius of r and
a length of ℓ
• Symmetry:
E is constant in magnitude (depends
only on radius r) and perpendicular
to the surface at every point on the
curved part of the surface
The end view
30
Gauss’s Law: Applications
Field from a line of charge
dA
The flux through this surface is 0
The flux through this surface:
E 
 E  d A   E d A  E  2π r  
q in
εo
q in  λ
E  2πr
E 
The end view

λ
2πεr
o
λ
εo
 2ke
λ
r
31
Gauss’s Law: Applications
Field due to a plane of charge
• Symmetry:
E must be perpendicular to the plane
and must have the same magnitude
at all points equidistant from the
plane
dA
• Choose a small cylinder whose axis
is perpendicular to the plane for the
gaussian surface
The flux through this surface is 0
32
Gauss’s Law: Applications
Field due to a plane of charge
E2
A2
A4
A5
h
dA
h
A3
A6
A1
E1
E1  E 2  E
A1  A2  A
The flux through this surface is 0
  E 1 A1  E 2 A 2  E A  E A  2 E A
 
q in
0

A
0
2EA 
A
0
E 

2 0
does not depend on h
33
Gauss’s Law: Applications
E 
E  2ke

2 0

r
34
Gauss’s Law: Applications
Q 
4
3
πa ρ
3
4
E  ke
E 
4
3
Q
a
3
r  ke 3
3
πa ρ
a
3
r 
4
3
πk e ρr
πk e ρr
35
Example
Find electric field inside the hole
R
r

a
36
E 
4
3
R
πk e ρr1
E  
4
3
R
r


a
r1

r
r2  

E 
4
3
R

E 
πk e ρr2
4
3
πk e ρr1 
r1
a
4
3
πk e ρr
r2

πk e ρr2 
4
3
πk e ρr
( 1  r2 ) 
4
3
πk e ρa  co n s37t
Example
The sphere has a charge Q and radius a. The point
charge –Q/8 is placed at the center of the sphere.
Find all points where electric field is zero.
E1
r
E2
E1  ke
Q
a
E2  ke
r
3
Q
8r
2
r
E1  E 2  0
ke
r
3

a
Q
a
3
8
3
r  ke
Q
8r
2
r 
a
2
38
Chapter 28
Conductors in Electric Field
39
Electric Charges: Conductors and Isolators
 Electrical conductors are materials in which
some of the electrons are free electrons
 These electrons can move relatively freely
through the material
 Examples of good conductors include copper,
aluminum and silver
 Electrical insulators are materials in which all
of the electrons are bound to atoms
 These electrons can not move relatively freely
through the material
 Examples of good insulators include glass, rubber
and wood
 Semiconductors are somewhere between
insulators and conductors
40
Electrostatic Equilibrium
Definition:
when there is no net motion of charge
within a conductor, the conductor is
said to be in electrostatic equilibrium
Because the electrons can move freely through the
material
 no motion means that there are no electric forces
 no electric forces means that the electric field
inside the conductor is 0
If electric field inside the conductor is not 0, E  0 then
there is an electric force F  qE and, from the second
Newton’s law, there is a motion of free electrons.
F  qE
41
Conductor in Electrostatic Equilibrium
• The electric field is zero everywhere inside the
conductor
• Before the external field is applied,
free electrons are distributed
throughout the conductor
• When the external field is applied, the
electrons redistribute until the
magnitude of the internal field equals
the magnitude of the external field
• There is a net field of zero inside the
conductor
42
Conductor in Electrostatic Equilibrium
• If an isolated conductor carries a charge, the charge
resides on its surface
Electric filed is 0,
so the net flux through
Gaussian surface is 0
 
 E  dA 
Then
q in  0
q in
εo
43
Conductor in Electrostatic Equilibrium
• The electric field just outside a charged conductor is
perpendicular to the surface and has a magnitude of σ/εo
• Choose a cylinder as the gaussian surface
• The field must be perpendicular to the surface
– If there were a parallel component to E,
charges would experience a force and
accelerate along the surface and it would
not be in equilibrium
• The net flux through the gaussian surface is
through only the flat face outside the
conductor
– The field here is perpendicular to the
surface
σA
σ


E
A

a
n
d
E

• Gauss’s law:
E
εo
εo
44
Conductor in Electrostatic Equilibrium
E 
σ 
σ
εo
σ 
E  0
σ 
σ 
σ 
45
Conductor in Electrostatic Equilibrium: Example
Find electric field if the conductor spherical shell has zero charge
r2
r1
q 0
conductor
E  0
46
Conductor in Electrostatic Equilibrium: Example
Find electric field if the conductor spherical shell has zero charge

r2



r1





E  0
The total charge
inside this Gaussian
surface is 0, so the
electric field is 0

q 0







surface charge, total charge is –q<0
surface charge, total charge is q>0
This is because the total charge of the conductor is 0!!!
47
Conductor in Electrostatic Equilibrium: Example
Find electric field if the conductor spherical shell has zero charge

r2



r1





E  0
The total charge
inside this Gaussian
surface is q, so the
electric field is
r
2

q 0

E  ke
q






surface charge, total charge is q>0
This is because the total charge of the conductor is 0!!!
total charge is –q<0
48
Conductor in Electrostatic Equilibrium: Example
Find electric field if the conductor spherical shell has zero charge

r2



q 0






2

r1



r


E  0
E  ke
q
E  ke
q
r
2

surface charge, total charge is q>0
This is because the total charge of the conductor is 0!!!
total charge is –q<0
49
Conductor in Electrostatic Equilibrium: Example
Find electric field if the charge of the conductor spherical shell is Q

r2



q 0






2

r1



r


E  0
E  ke
Q q
E  ke
q
r
2

surface charge, total charge is Q+q>0
This is because the total charge of the conductor is Q!!!
total charge is –q<0
50