Download Chapter 5 Electric Field in Material Space

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Fall 2016
ECE 305 Electromagnetic Theory
Chapter 5 Electric Fields In Material Space
Qiliang Li
Dept. of Electrical and Computer Engineering,
George Mason University, Fairfax, VA
1
§ 5.1 Introduction
Free Space / Vacuum: ε0
conductors
Real material space
nonconductors
Nonconductors – insulators or dielectrics
Susceptibility, permittivity, linearity,
isotropy, homogeneity, dielectric
strength, and relaxation time
2
§ 5.2 Properties of Materials
• Conductivity, σ
unit: Siemens per meter ( S/m )
metals: σ >> 1
Insulators: σ << 1
in between  Semiconductors
As T  0K, some conductors have σ ∞
Called: superconductors
What is “Semiconductor”?
3
Examples
MOSFET
conductor
Semiconductor
Optical fiber
4
§ 5.3 Convection and Conduction
Current
• The current (in amperes) through a given
area is the electric charge passing
through the area per unit time. 1 A = 1
C/s
𝑑𝑄
𝐼=
𝑑𝑡
Current density 𝐽 =
∆𝐼
∆𝑆
or 𝐼 =
𝑱 ∙ 𝑑𝑺
5
• Different between convection current
and conduction current
• Convection current does not involve
conductors, therefore, does not satisfy
Ohm’s law. It occurs when current flows
through an insulating medium such as
liquid, rarefied gas or a vacuum
6
ΔS
ρV (charge density)
z
y
Δy
x
∆𝑄 𝜌𝑉 ∆𝑉 𝜌𝑉 ∆𝑆∆𝑦
∆𝐼 =
=
=
= 𝜌𝑉 ∆𝑆 𝑢𝑦
∆𝑡
∆𝑡
∆𝑡
∆𝐼
𝐽𝑦 =
= 𝜌𝑉 𝑢𝑦
∆𝑆
7
• In general: 𝑱 = 𝜌𝑉 𝒖
So, this equation is the general form of
both convection and conduction current
density
For conduction current, 𝜌𝑉 can be
expressed by electron density – see next
page
8
• Conduction current requires a conductor
𝑭 = −𝑒𝑬
From momentum Ft = mu:
𝑚𝒖
= −𝑒𝑬
𝜏
τ: the average time interval between
collisions – electron travel in a real materials
space
e
9
• Therefore, 𝒖 =
𝑒𝜏
− 𝑬
𝑚
If there are n electrons per unit volume
𝜌𝑉 = −𝑛𝑒
So
𝑛𝑒 2 𝜏
𝑱 = 𝜌𝑉 𝒖 =
𝑬
𝑚
or
𝑱 = 𝜎𝑬
10
§ 5.4 Conductors
• A perfect conductor (𝜎 = ∞) cannot contain
E-field within it.
Ee
Ei
Ee
ρV=0
E=0
Ee is applied external field, Ei is internal field
--isolated conductor: V= constant
11
• For a conductor whose ends are
maintained at potential difference V
𝑬 ≠ 0 inside the conductor
𝑉
𝐸=
𝑙
𝐼
𝑉
𝐽 = = 𝜎𝐸 = 𝜎
𝑆
𝑙
𝐼
𝑆
=
𝑉
𝜎
𝑙
𝑅 =
𝑉
𝐼
=
𝑙
𝜎𝑆
or 𝑅 =
𝜌𝑐 𝑙
𝑆
Where ρc = 1/σ is the resistivity
12
• For a conductor with nonuniform cross
section
𝑉
𝑅= =
𝐼
𝑉
𝑆
𝑬 ∙ 𝑑𝒍
𝜎𝑬 ∙ 𝑑𝑺
Power P (in watts): rate of change of energy
W (in Joules)
𝑃=
𝜌𝑉 𝑑𝑉 𝑬 ∙ 𝒖 =
𝑉
𝑬 ∙ 𝜌𝑉 𝒖𝑑𝑉 =
𝑬 ∙ 𝑱𝑑𝑉
𝑉
13
Power density wP (in W/m3)
𝑑𝑃
𝑤𝑃 =
= 𝑬 ∙ 𝑱 = 𝜎|𝑬|2
𝑑𝑉
For a conductor with uniform cross-section
𝑃=
𝑬 ∙ 𝑑𝒍
𝐿
𝑱 ∙ 𝑑𝑺 = 𝑉𝐼
𝑆
Or
𝑃 = 𝐼 2 𝑅 --- the common
form of Joule’s law
14
Example 5.1: If 𝑱 =
1
𝑟3
2𝑐𝑜𝑠𝜃𝒂𝒓 + 𝑠𝑖𝑛𝜃𝒂𝜽 𝐴/𝑚2
Calculate the current passing through
(a) A hemispherical shell of radius = 20 cm, 0 <
𝜋
𝜃 < ; 0 < 𝜙 < 2𝜋
2
𝐼=
Solve:
𝑆
𝑱 ∙ 𝑑𝑺
where 𝑑𝑆 = 𝑟 2 𝑠𝑖𝑛𝜃𝑑𝜙𝑑𝜃𝒂𝒓
So
𝐼=
4𝜋 𝑠𝑖𝑛2 𝜃
(θ
0.2 2
𝜋
2
0
2𝜋 1
2𝑐𝑜𝑠𝜃
0 𝑟3
𝑟 2 𝑠𝑖𝑛𝜃𝑑𝜙𝑑𝜃|𝑟=0.2 =
from 0 to π/2)=31.4 (A)
15
Continue example 5.1
(b) A spherical shell of radius 10 cm
4𝜋 𝑠𝑖𝑛2 𝜃
𝐼=
(θ from 0 to π)=0
0.1 2
Example 5.4: A lead (𝜎 = 5 × 106 𝑆/𝑚) bar of
square cross section has a circular hole. L=4 m
Find the resistance.
Solve: 𝑅 = 𝑙/𝜎𝑆
b/c: 𝑆 = 𝑑 2 − 𝜋𝑟 2
1 cm
So, 𝑅 =
4
5×106 ×
𝜋
9−
4
×10−4
3 cm
16
§ 5.5 Polarization in Dielectrics
Dipole moment p= 𝑄𝒅
Polarization in dielectrics 𝑷 = lim
𝑁
𝑘=1 𝑄𝑘 𝒅𝒌
∆𝑉→0
∆𝑉
This is dipole moment per unit volume
So, potential 𝑑𝑉 =
𝑷∙𝒂𝒓 𝑑𝑉 ′
4𝜋𝜖0 𝑅 2
where P dV’ is the dipole moment (dV’ is diff vol.)
b/c: 𝛻
So
′
1
𝑅
=
𝒂𝑹
𝑅2
𝑷∙𝒂𝑹
𝑅2
= 𝜵′ ∙
𝑷
𝑅
−
𝜵′ ∙𝑷
𝑅
17
• Therefore, 𝑉 =
𝑷∙𝒂 ′
𝑹
𝑆 ′ 4𝜋𝜖0 𝑅
𝑑𝑆 ′ +
−𝛻 ′ ∙𝑷
′
𝑑𝑉
𝑉 ′ 4𝜋𝜖0 𝑅
Compared to Eq. (4.68) and (4.69)
The total positive bound charges on surface S:
𝑄𝑏 =
𝑷 ∙ 𝑑𝑺
The charge that remains inside S is
−𝑄𝑏 = −
𝛻 ∙ 𝑷𝑑𝑉
𝑉
Total volume charge density ρt:
𝜌𝑡 = 𝜌𝑉 + 𝜌𝑝𝑉 = 𝛻 ∙ 𝜖0 𝐸
18
Therefore, 𝜌𝑉 = 𝛻 ∙ 𝜖0 𝑬 − 𝜌𝑝𝑉 = 𝛻 ∙ (𝜖0 𝑬 + 𝑷)
From this equation, we have:𝐃 = (𝜖0 𝑬 + 𝑷)
For some dielectrics, 𝑷 = 𝑋𝑒 𝜖0 𝑬
where Xe is electric susceptibility
19
§ 5.6 Dielectric Constant and Strength
𝑫 = 𝜖0 1 + 𝑋𝑒 𝑬 = 𝜖0 𝜖𝑟 𝑬
Or
𝑫 = 𝜖0 𝜖𝑟 𝑬 = 𝜖𝑬
where 𝜖 = 𝜖0 𝜖𝑟 , the permittivity
𝜖𝑟 is the dielectric constant: 𝜖𝑟 = 1 + 𝑋𝑒 =
𝜖
𝜖0
Dielectric strength is the maximum electric field
that a dielectric can tolerate or withstand
without electric breakdown.
20
§ 5.7 is not required
§ 5.8 Continuity equation and
relaxation time
𝑑𝑄𝑖𝑛
𝑱 ∙ 𝑑𝑺 = −
=−
𝑑𝑡
𝐼𝑜𝑢𝑡 =

𝜵∙𝑱=
𝜕𝜌𝑉
−
𝜕𝑡
b/c: 𝐽 = 𝜎𝐸 , and 𝛻 ∙ 𝐸 =
So, 𝛻 ∙ 𝜎𝐸 =
𝜎𝜌𝑉
=
𝜖
−𝑡/𝑇𝑟
So, 𝜌𝑉 = 𝜌𝑉0 𝑒
𝑉
𝑑𝜌𝑉
𝑑𝑉
𝑑𝑡
𝜕𝜌𝑉
−
𝜕𝑡
𝜌𝑉
𝜖

𝜕𝜌𝑉
𝜌𝑉
=
𝜎
− 𝜕𝑡
𝜖
, where 𝑇𝑟 = 𝜖/𝜎
21
§ 5.9 Boundary Conditions
• Use Maxwell’s Equations
𝑬 ∙ 𝑑𝒍 = 0
𝑫 ∙ 𝑑𝑺 = 𝑄𝑒𝑛𝑐
A. Dielectric-Dielectric
E1n
Boundary Conditions
Consider L: abcd
𝑆
𝑬 ∙ 𝑑𝒍 = 0
E1t
E2
E2n
ε1
E1
E2t
Δh
ε2
d
a
ΔW
b
c
22
𝑬 ∙ 𝑑𝒍 = 0
∆ℎ
∆ℎ
∆ℎ
= 𝐸1𝑡 ∆𝑤 − 𝐸1𝑛
− 𝐸2𝑛
− 𝐸2𝑡 ∆𝑤 + 𝐸2𝑛
2
2
2
∆ℎ
+ 𝐸1𝑛
2
𝐷1𝑡
𝐷2𝑡
𝐸1𝑡 = 𝐸2𝑡 
=
𝜖1
𝜖2
Use cylindrical Gaussian surface
∆𝑄 = 𝜌𝑆 ∆𝑆 = 𝐷1𝑛 ∆𝑆 − 𝐷2𝑛 ∆𝑆
 𝐷1𝑛 − 𝐷2𝑛 = 𝜌𝑆
If 𝜌𝑆 = 0 (no free surface charge), 𝐷1𝑛 = 𝐷2𝑛
23
therefore, 𝜖1 𝐸1𝑛 = 𝜖2 𝐸2𝑛
From 𝐸1𝑡 = 𝐸2𝑡 and 𝜖1 𝐸1𝑛 = 𝜖2 𝐸2𝑛
𝑡𝑎𝑛𝜃1

𝑡𝑎𝑛𝜃2
=
𝜖𝑟1
𝜖𝑟2
θ1
B. Conductor-Dielectric
θ2
Boundary Conditions
① No electric field may exist within a conductor
𝜌𝑉 = 0 ,
𝐸=0
②b/c E = −𝛻𝑉 = 0  there can be no potential
difference between any two point in a conductor
24
③ 𝐷𝑡 = 𝜖0 𝜖𝑟 𝐸𝑡 = 0, 𝐷𝑛 = 𝜖0 𝜖𝑟 𝐸𝑛 = 𝜌𝑆
E-field must be external to the conductor
C. Conductor-Free space boundary conditions
𝐷𝑡 = 𝜖0 𝐸𝑡 = 0, 𝐷𝑛 = 𝜖0 𝐸𝑛 = 𝜌𝑆
Example 5.9 and 5.10 will discussed in the class.
25
Related documents