Download Gauss`s Law

Reading: Chapter 28
E
Q
Q

k
e 2
4πεo r 2
r

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
 E  EA cos 
 E  EA
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  EA  EA cos  0
or

A
E

 E  EA   EA cos  0
6
Electric Flux
E
1
 E  1  2

area  A1
A1
area  A2

A2
90o  
2  EA2  EA2 cos(90o   )  EA2 sin 
2
1  EA1  EA1 sin 
 E  EA1 sin   EA2 sin 
E
if

then

A1
flux is positive
E 
A1
 E  EA1 sin 
2  0
A2
 E   EA1 cos   EA2 cos 
flux is negative
A2
A2 and E are
orthogonal
7
Electric Flux
• In the more general case, look
at a small flat area element
E  Ei Ai cos θi  Ei  Ai
• In general, this becomes
E  lim
Ai 0
 E  A
i
i


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 Ai point in different
directions
 At each point, they are
perpendicular to the surface
 By convention, they point
outward
E  lim
Ai 0
 E  A
i
i


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 A3 , A4 , A5 , and A6
Then
3  EA3  0
4  EA4  0
5  EA5  0
6  EA6  0
1
E
 E  1  2
1  EA1  EA1 cos(90   )   EA1 sin 
o
4
E
90o  
A1

A3
1
3
2
4
2  EA2  EA2
A4
 E  E ( A2  A1 sin  )
but
A2  A1 sin 
A2
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.
q
E  ke 2
r
Then
Spherical
surface
   Ei dAi E  dAi 
i
i
q
 EA0  E 4 r  4 r ke 2 
r
q
Gauss’s Law
 4 ke q 
2
2
0
 does not depend on r
ONLY BECAUSE
1
E 2
12 r
Electric Flux: Closed Surface
E and A
have opposite directions at every point.
|q|
E  ke 2
r
Spherical
surface
Then
   Ei dAi   E  dAi 
i
i
|q|
  EA0   E 4 r  4 r ke 2 
r
q Gauss’s Law
 4 ke | q |
2
2

0
 does not depend on r
ONLY BECAUSE
1
E 13 2
r
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

qin
E   E  dA 
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 
qin
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  q2  q3  q4
0
q4
q6
q7
16
Gauss’s Law
 Gauss’s law states
E 
 E  dA 
qin
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 
qin
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  EA0   EA0
1   2  EA0
A1
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
qin
   E  dA 
εo
q5

q1  q2  q3  q4
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
Gaussian Surface – Sphere
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
   Ei dAi E  dAi EA0  E 4 r 2
i
Then
- definition of the Flux
i
q

 4 r E
2
q
E  ke 2
r
22
qin
   E  dA 
εo
Gauss’s Law: Applications
• 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
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
qin Q
 E   E  dA   EdA  4πr E 

εo εo
2
E
Q
Q

k
e 2
4πεo r 2
r
The electric field is the same as
for the point charge Q
24
Gauss’s Law: Applications
Spherically Symmetric Charge Distribution
Q
Q
E
 ke 2
2
4πεo r
r
The electric field is the same as
for the point charge Q !!!!!
Q
a

For r > a
Q

For r > a
25
Gauss’s Law: Applications
Spherically Symmetric Charge Distribution
SYMMETRY:
A
E
E - direction - along the radius
E - depends only on radius, r
• Select a sphere as the
gaussian surface, r < a
qin 
Q
4 3
a
3
4 3
r3
r  Q 3  Q
3
a
qin
 E   E  dA   EdA  4πr E 
εo
2
qin
Qr 3 1
Q
E
 ke 3 2  ke 3 r
2
4πεo r
a r
a
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
r1
E 2
the same
solid angle
E1
r2
q1  A1
q2  A2
A1  r12 
A2  r22 
q1
A1
 r12 
E1  ke 2  ke 2  ke
 ke
r1
r1
r12
q2
A2
 r22 
E2  ke 2  ke 2  ke
 ke
2
r2
r2
r2
E1  E2
E1  E2  0
A2
Only because in Coulomb law
1
E 2
r
29
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:
qin
E   E  dA   EdA  E  2πr  
εo
qin  λ
λ
E  2πr  
ε
o
The end view
λ
λ
E
 2ke
2πεr
r
o
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
A3
h
A5
h
A6
dA
A1
E1
E1  E2  E
A1  A2  A
The flux through this surface is 0
  E1 A1  E2 A2  EA  EA  2EA
A


0
0
qin
A
2 EA 
0

E
2 0
does not depend on h
33
Gauss’s Law: Applications

E
2 0
E  2 ke

r
34
Gauss’s Law: Applications
4 3
Q  πa ρ
3
4 3
πa ρ 4
Q
E  ke 3 r  ke 3 3 r  πk e ρr
a
a
3
4
E  πk e ρr
3
35
Example
Find electric field inside the hole
R
r

a
36
4
E  πk e ρr1
3
R
4
E   πk e ρr2
3
R
r


a

r1
r

r2  
4
E  πk e ρr
3
R

r1 r2
a

4
4
4
4
E  πk e ρr1  πk e ρr2  πk e ρr
( 1  r2 )  πk e ρa  const
37
3
3
3
3
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
Q
E1  ke 3 r
a
Q
E 2  k e 2 r
8r
E1  E2  0
Q
Q
ke 3 r  ke 2
a
8r
3
a
r3 
8
a
r 
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
qin
   E  dA 
εo
Then
qin  0
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


EA

and
E

• Gauss’s law:
E
ε
o
ε
o
44
Conductor in Electrostatic Equilibrium
E
σ 
σ
ε
o
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



E 0



r1


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



E 0



r1


The total charge
inside this Gaussian
surface is q, so the
electric field is
q 0







q
E  ke 2
r

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



E 0



r1


q
E  ke 2
r

q 0






E  ke
q
r2

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



E 0



r1


Qq
E  ke 2
r

q 0






E  ke
q
r2

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