Download Chapter 5 Electric Fields in Material Space

Chapter 5
Electric Fields in Material Space
Islamic University of Gaza
Electrical Engineering Department
Dr. Talal Skaik
2012
1
Introduction
• In chapter 4, Electrostatic fields in free space were considered.
• This chapter covers electric fields in materials.
• Materials are generally classified as conductors and nonconductors.
• Non-conducting materials are unusually referred as insulators or
dielectrics.
• Materials may be classified in terms of their conductivity σ , in
mhos per meter (  / m ), or siemens per meter (s/m).
2
Properties of Materials
• A material with high conductivity (σ>>1) is referred to as metals.
• A material with small conductivity (σ<<1) is referred to as
insulator (or dielectric).
• A material with conductivity lies between those of metals and
insulators are called semiconductors.
• Table B.1 in Appendix B (Values of conductivity of common
materials).
 Copper and aluminium are metals.
 Silicon and germanium are semiconductors.
 Glass and rubber are insulators.
3
Properties of Materials
 Some conductors exhibit infinite conductivity at temperature
near absolute zero (T=0 k), and they are called superconductors.
 Dielectric materials have few electrons available for conduction
of current.
 Metals have abundance of free electrons.
4
5.3 Convection and Conduction Currents
• Electric current is generally caused by motion of electric charges.
• The current in (Amperes) through a given area is the electric
charge passing through the area per unit time.
dQ
I=
dt
• In a current of one ampere, charge is being transferred at a rate
of one coulomb per second.
5
Convection and Conduction Currents
• If current ΔI flows through a planar surface ΔS, the current
∆I
density is: J =
∆S
• If current is normal to the surface:
∆I = J ∆S
• If the current is not normal to the surface:
∆I = J ⋅ ∆S
• The total current flowing through a surface S is:
=
I
∫ J ⋅ ∆S
S
6
Convection Current
 Doesn’t involve conductors and consequently doesn’t satisfy
ohms law.
 It occurs when current flows through an insulating medium such
as liquid, vacuum.
 Ex) beam of electrons in vacuum tube is a convection current.
7
Convection
Current
• If there is a flow of charge of density ρV at velocity u=uyay the
current is:
∆Q ρV ∆V ρV ∆S ∆y
∆I=
=
= ρV ∆S u y
=
∆t
∆t
∆t
∆I
=
J y = ρV u y
∆S
In General: J=ρV u


Convection Current Density J=ρV u
8
Conduction Current
 Requires a conductor.
 A conductor has a large number of free electrons that provide
conduction current due to an impressed electric field.
 When electric field is applied, the force on an electron with


charge –e is : F = − eE
 In free space, the electron would accelerate.
 In material, the electron suffers continual collisions.
 The electron will move with different velocities between
collisions.
9
Conduction Current
 The average velocity is called the drift velocity u.
eτ 

u= − E
m
 τ : average time between collisions.
 m: mass of electron.
 Drift velocity is directly proportional to the applied field.
 If there are n electrons per unit volume, ρV =-ne
2 



ne
τ
 The conduction current density is: J=ρ u =
E =σ E
V
m
 
J=σ E
10
5.4 Conductors
• A conductor has abundance of charge that is free to move.
• A perfect conductor (σ=∞) can not contain an electrostatic field
within it.
Isolated Conductor
11
Conductors
E 0 and −
=
∇V 0, which implies V=0. Thus,
• Inside conductor=
a conductor is an equipotential body.
• According to gauss law,
ε ∫ E ⋅ dS= ∫ ρV dV
S
V
• If E=0, then the charge density ρV = 0.
Inside an isolated perfect conductor, E=0, ρ V =0, Vab =0
(Vab is potential difference between two points a and b in the
conductor)
12
Conductors
• When a potential difference is applied at the ends of the
conductor, E≠0 inside the conductor. ( conductor not isolated,
wired to a source).
• An electric field must exist inside the conductor to sustain the
flow of current.
• The direction of the electric field E produced is the same as the
direction of the flow of positive charges or current I. This
direction is opposite to the direction of the flow of electrons.
• The opposition to the flow is called Resistance.
13
• To derive Resistance:
V
The magnitude of electric field is given by E =
l
I
Assuming conductor has uniform cross section of area S, J =
S
σV
I
= σ E=
S
l
l
R=
V
l
R= =
σS
I σS
ρC l
l
1
R=
=
⇒ Resistivity
, ρC=
S
σS
σ
If the cross section of the conductor is not uniform:
V
R= =
I
The power P (in watts):
− ∫ E ⋅ dl
∫ σ E ⋅ dS
P=
∫ E ⋅ J dv or
v
P=
I 2R
14
Example 5.1
1
3
, calculate the current passing through
2
cos
a
sin
a
A/m
θ
+
θ
(
)
r
φ
3
r
If J
• (a) a hemispherical shell of radius 20 cm, 0<Ѳ<π/2, 0<φ<2π
• (b) a spherical shell of radius 10 cm.
Solution:
2
=
I =⋅
J
dS
dS
r
,
sin θ dφ dθ ar
∫
π /2 2π
(a ) I
π /2
1
2
2
=
θ
θ
φ
θ
r
d
d
(
2
cos
)
sin
2π ∫ cos θ sin θ dθ
3
∫
∫
0.2
=
r
r
r =θ 0
θ 0=
φ 0
=
r =0.2
Let u=sinθ , du = cos θ dθ
=
I
π /2
4π sin θ
= 10
=
π 31.4 A
0.2 2 0
2
( b) 0 < θ < π ,
π
4π sin θ
= 0
0.1 2 0
2
I
r=
0.1
Alternatively for this case:
I= 
∫ J ⋅ dS =
S
∫ ∇ ⋅ J dv =
v
0 , since ∇ ⋅ J = 0
15
Example 5.3
• A wire of diameter 1 mm and conductivity 5 x 107 S/m has 1029 free
electrons per cubic meter when an electric field of 10 mV/m is applied.
Determine
• (a) the charge density of free electrons.
• (b) the current density
• (c) the current in the wire
• (d) the drift velocity of the electrons (take e=-1.6 x 10-19 C)
• Solution
−1.6 × 1010 C / m3
ne (1029 )( −1.6 × 10−19 ) =
( a ) ρV ==
σE =
( b) J =
(5 × 107 )(10 × 10-3 ) =
500 kA / m 2
 d  5π
( c ) I =JS =(5 × 10 )(π )   = × 10-6 × 105 =0.393 A
4
2
J
5 × 105
−5
=
×
J ρV u,=
u =
m/s
( d )sin ce=
3.125
10
10
ρV 1.6 × 10
2
5
16
Example 5.4
• A lead (σ=5 x 107 S/m ) bar of square cross section has a hole
bored along its length of 4 m so that its cross section becomes
that of figure 5.5. find the resistance between the square ends.
• Solution
l
R=
σS
1
where S =d 2 − π r 2 =32 − π  
2
2
π

=  9 −  cm 2
4

4
Hence R=
5 × 106 ( 9 − π / 4 ) × 10−4
= 974 µΩ
17
5.5 Polarization in Dielectrics
• In dielectric materials, charges are not able to move about freely,
they are bound by finite forces. (displacement will take place
when external force is applied).
• Atoms or molecules are electrically neutral since positive and
negative charges have equal amounts.
• When Electric field is applied, positive charges move in the
direction of E, and negative charges move in the opposite
direction.
• The molecules are deformed from their original shape, and each
molecule gets some dipole moment.
+Q (nucleus)
-Q (electron)
18
Polarization in Dielectrics
• The dipole moment is
 
P=Qd
Where d is the distance vector from –Q to +Q of the dipole.
• If there are N dipoles, the total dipole moment due to the
electric field is:
N
Q1d1 +Q 2d 2 +  +Q N d N = ∑ Q k d k
k =1
• As a measure of intensity of polarization, define Polarization P (in
coulombs per meter squared) as the dipole moment per unit
volume of the dielectric:
N
P = lim
∆v → 0
∑Q d
k =1
k
k
∆v
19
Polarization in Dielectrics
Two groups of dielectrics:
• Nonpolar: nonpolar dielectric molecules do not posses dipoles
until the application of electric field.
• Examples: hydrogen, oxygen, nitrogen,
• Polar: molecules have built-in permanent dipoles that are
randomly oriented. When external E is applied, dipole moments
are aligned parallel with E.
Polarization of a polar molecule:
(a) permanent dipole (E = 0),
(b) alignment of permanent
dipole (E ≠ 0).
• Examples: water, sulfer dioxide
20
Polarization in Dielectrics
• When polarization occurs, two charge densities are formed:
(1) An equivalent surface charge density ρ ps is formed over the
surface of the dielectric.
ρ ps= P ⋅ a n
where an is unit normal to the surface.
(2) An equivalent volume charge density ρ pv is formed throughout
the dielectric.
ρ pv = −∇ ⋅ P
Notes:
ρ ps and ρ pv are called bound (or polarization) surface and
volume charge densities, respectively, as a distinct from free
surface and volume charge densities ρ s and ρ v .
• Bound charges are those that are not free to move within the
dielectric material; they are caused by the displacement that
occurs on a molecular scale during polarization.
21
Polarization in Dielectrics
• If the entire dielectric were electrically neutral prior to
application of E and if we have not added free charge, the
dielectric will remain electrically neutral. Thus the total charge of
the dielectric material remains zero.
Qbs : total bound surface charge
Qbv : total bound volume charge
Qbs = ∫ ρ ps dS = 
∫ P.dS
Qbv =−Qbs = ∫ ρ pv dv =− ∫ ∇.Pdv
v
ρ ps= P ⋅ a n
ρ pv = −∇ ⋅ P
v
total charge= 
∫ ρ psdS + ∫ ρ pv dv = Qbs − Qbs = 0
S
v
22
Polarization in Dielectrics
• If the dielectric region contains free charge with volume charge
density of ρV , the total volume charge density is:
ρ=
ρv + ρ pv
t
=
∇⋅E
1
ε0
(ρ
v
+=
ρ pv )
1
ε0
( ρv − ∇ ⋅ P )
∇ ⋅ ( ε 0 E+P ) = ρv = ∇ ⋅ D
⇒ D=ε 0 E + P
• The application of E to the dielectric material causes the flux
density to be greater than it would be in free space.
• Polarization is proportional to the applied electric field:
P=χ eε 0 E
• Where χ e is the susceptibility of the material.
Susceptibility of a material describes its response to an applied field.
23
5.6 Dielectric Constant and Strength
D=ε 0 E + P = ε 0 E + χ eε 0 E
D=ε 0 (1 + χ e ) E=ε 0ε r E
D=ε E
, ε =ε 0ε r
ε
, ε r =(1 + χ e ) =
ε0
• ε is called the permittivity of the dielectric.
• εo is the permittivity of free space. ε0=10-9/36π
• εr is the relative permittivity.
F/m.
The dielectric constant (or relative permittivity) ε r is the ratio of
the permittivity of the dielectric to that of free space.
• Table B.2 in appendix B, values of dielectric constants of some
materials.
24
• No dielectric is ideal. When the electric field in a dielectric is
sufficiently large, it begins to pull electrons completely out of the
molecules, and dielectric becomes conducting.
• Dielectric breakdown occurs when a dielectric becomes
conducting.
• The dielectric strength is the maximum electric field that a
dielectric can tolerate or withstand without electrical breakdown.
25
5.7 Linear, Isotropic, and homogeneous
Dielectrics
• A material is linear if D varies linearly with E.
• Materials for which ε (or σ) does not vary in the region being
considered (the same at all points , i.e. independent of (x,y,z) are
said to be homogeneous.
• Materials are inhomogeneous (or monohomogeneous) when ε is
dependent on the space coordinates.
• Materials for which D and E are in the same direction are said to
be isotropic.
• For anisotropic (or nonisotropic) materials, D, E, and P are not
parallel. ε or χe has nine components.
26
Linear, Isotropic, and homogeneous Dielectrics
 Dx  ε xx
 D  = ε
 y   yx
 Dz   ε zx
ε xy ε xz   E x 
 
ε yy ε yz  E y
 
ε zy ε zz   E z 
• A dielectric material is linear if D= εE and ε does not change with
the applied E field.
• A dielectric material is homogeneous if D= εE and ε does not
change from point to point.
• A dielectric material is isotropic if D= εE and ε does not change
with direction.
27
Example 5.5
A dielectric cube of side L and center at the origin has polarization
P=ar where a is a constant and r=xax+yay+zaz. Find all bound
charge densities and show that the total bound charge vanishes.
Solution:
for each of the six faces , there is a surface charge ρ ps .
for the face located at x=L/2:
aL
ax x L /2 =
ρ ps =
=
P ⋅ a x x L=
=
/2
2
L /2 L /2
6aL 2
dydz =
L 3aL3
ρ ps dS 6 ∫ ∫ ρ ps=
totalbound surface charge is : Qs = ∫ =
2
S
− L /2 − L /2
bound volume charge density is ρ pv = −∇ ⋅ P = −( a + a + a ) = −3a
total bound volume charge is Qv = ∫ ρ pv dv =
−3a ∫ dv =
−3aL3
Hence total charge is Qt = Qs + Qv = 3aL3 − 3aL3 = 0
28
Example 5.6
• The electric field intensity in polystyrene (εr =2.55) filling the space
between the plates of parallel-plate capacitor is 10kV/m. The
distance between the plates is 1.5 mm. Calculate:
• (a) D (b) P (c) the surface charge density of free charge on the plates.
• (d) the surface density of polarization charge.
• (e) the potential difference between the plates.
• Solution
10−9
(a) D=ε 0ε r E=
225.4 nC/m 2
× (2.55) ×104 =
36π
10−9
(b) P=χ eε 0 E = ( ε r − 1) ε 0 E= (1.55) ×
×104= 137 nC/m 2
36π
(c) ρ s =D ⋅ a n = Dn =225.4 nC/m 2
(d) ρ ps =P ⋅ a n =Pn =137 nC/m 2
(
)
(e)V =Ed =
104 1.5 ×10−3 =
15 V
29
Example 5.7
A dielectric sphere (εr =5.7) of radius 10 cm has a point charge of 2pC
placed at its centre. Calculate:
(a) The surface density of polarization charge on the surface of the sphere.
(b) The force exerted by the charge on a -4pC point charge placed on the
sphere.
• Solution
(a) Assuming point charge is located at the origin,
E=
Q
4πε 0ε r r
2
ar
χ eQ
P=χ eε 0 E =
a
2 r
4πε r r
ρ ps =P ⋅ a r
(b) F
(ε − 1) Q =
= r
4πε r r 2
Q1Q2
a
=
2 r
4πε 0ε r r
(4.7)2 ×10−12
2
13.12
pC/m
=
4π (5.7)100 ×10−4
(−4)(2) ×10−24
a r = −1.263 a r pN
−9
10
4π ×
(5.7)100 ×10−4
36π
30
Example 5.8
Find the force with which the plates of a parallel-plate capacitor
attract each other. Also, determine the pressure on the surface of
the plate due to the field.
Solution
ρs
E=
a n where a n is a unit normal to the plate
2ε
and ρ s is the surface charge density.
ρs
ρ s2 S
F=QE ρ=
an
an
=
s S.
2ε
2ε 0ε r
or
ρ s2 S Q 2
=
F =
2ε
2ε S
ρ s2
The pressure is force per area =
2ε 0ε r
31
5.8 Continuity Equation and Relaxation Time
From principle of charge conservation, the time rate of decrease of charge
within a given volume must be equal to the net outward current flow
through the surface of the volume.
I OUT
dQin
=
−
∫S J ⋅ dS =
dt
where Qin is the total charge enclosed by the closed surface.
∂ρ
dQ
d
− in =
− ∫ ρv dv =
− ∫ v dv
dt
dt V
∂t
V
Using divergence theorem,
∫ J ⋅ dS = ∫ ∇ ⋅ J dv
S
→ ∫ ∇ ⋅ J dv = − ∫
V
V
∂ρv
dv
∂t
V
∂ρv
∇⋅J = −
∂t
∂ρv
⇒ ∇⋅J = −
∂t
This equation is called continuity of current equation or continuity equation.
32
Continuity Equation and Relaxation Time
Consider introducing a charge at some interior point of a given
material (conductor or dielectric).
From Ohm's law J=σ E
ρv
From Gauss's law ∇ ⋅ D=ρv and hence ∇ ⋅ E=
ε
substituting in the continuity equation,
∂ρ
∇⋅J = − v
∂t
∂ρv
∇ ⋅ σ E= −
∂t
∂ρv σ
∂ρv
σρv
0
+ ρv =
= −
∂t ε
∂t
ε
∂ρv σ
+ ρv =
0
∂t ε
33
Continuity Equation and Relaxation Time
∂ρv σ
0
+ ρv =
∂t ε
∂ρv
σ
Separating the variables:
=
− ∂t
ρv
ε
σt
and integrating both sides: ln ρv =
− + ln ρv 0
ε
where ln ρv 0 is a constant of integration.
ε
− t /T
ρv ρ=
where Tr
v0 e
σ
r
Tr is the time constant in seconds.
ρv = ρv 0 e
− t / Tr
34
Continuity Equation and Relaxation Time
• When we introduce a volume charge density at an interior point in a
material, it decays resulting in a charge movement from the interior point
at which it was introduced to the surface of the material.
• The time constant Tr of this decay is called relaxation time or
rearrangement time.
• Relaxation time is the time it takes for a charge placed in the interior of
a material to drop to e-1 or 36.8 percent of its initial value.
• It is very short for good conductors and very long for good dielectrics.
• For a good conductor the relaxation time is so short that most of the
charge will vanish from the interior point and appear at the surface within
a short time.
• For a good dielectric the relaxation time is very long that the introduced
charge remains wherever placed for times up to days.
35
5.9 Boundary conditions
• When the field exists in a medium consisting of two different
media, the conditions the field must satisfy are called boundary
conditions.
• For the electrostatic field the following boundary conditions are
important:
• Dielectric – dielectric interface.
• Conductor – dielectric.
• Conductor – free space.
We will use Maxwell Equations:
∫ E ⋅ dl =0, ∫ D ⋅ dS=Q
enc
36
Dielectric-dielectric Boundary
conditions
• Consider the boundary between two dielectrics with permittivities
37
Dielectric-dielectric Boundary conditions
The fields in the two media can be expressed as:
E1 =
E1t + E1n
E2 =
E 2t + E 2 n
Apply the equation
0 to the path abcda in the figure
∫ E ⋅ dl =
L
∆h
∆h
∆h
∆h
− E2n
− E 2t ∆w + E 2 n
+ E1n
= 0
E ⋅ dl = E1t ∆w − E1n

∫
2
2
2
2
abcda
0
E1t ∆w − E 2t ∆w =
→ E1t =
E 2t
E1t = E 2t
•Tangential components of E are equal at the boundary.
•Et undergoes no change on the boundary and it is continuous
across the boundary.
38
Dielectric-dielectric Boundary conditions
E1t = E 2t
→
D1t
ε1
=
D2t
ε2
→ ε 2 D1t = ε1 D2t
• The tangential component of D under goes some change across the
boundary.
• So D is said to be discontinuous across the boundary.
• The boundary conditions for the normal components are obtained
by applying Gauss’s law on a small pill box shaped volume as in the
next figure.
Q
∫ D ⋅ dS =
S
D1n ∆S − D2 n ∆S =∆Q =ρ S ∆S
ρS
D1n − D2 n =
Assuming ∆h → 0
D1n − D2 n =
ρS
39
Dielectric-dielectric Boundary conditions
40
Dielectric-dielectric Boundary conditions
D1n − D2 n =
ρS
If no free charge exists at the boundary, ρS =0
D1n = D2 n
D1n = D2 n
•
Normal components of D are equal at the boundary.
•
Dn undergoes no change on the boundary and it is continuous
across the boundary.
D1n = D2 n
→
ε1 E1n = ε 2 E2 n
→
E1n ε 2
=
E2 n ε1
• The normal component of E undergoes some change
across the boundary.
• So En is said to be discontinuous across the boundary.
41
Dielectric-dielectric Boundary conditions
If the field on one side is known, we can find the field on the other side.
θ1 : angle between E1 and normal.
θ2 : angle between E2 and normal.
42
Dielectric-dielectric Boundary conditions
This is called law of refraction
43
Conductor-dielectric Boundary conditions
Apply the equation
∫ E ⋅ dl =0 to the path abcda in the figure
∆h
∆h
∆h
∆h
E ⋅ dl =Et ∆w − En
− 0⋅
− 0 ⋅ ∆w + 0 ⋅
+ En
=
0

∫
2
2
2
2
abcda
→ Et ∆w
= 0
→ E=
0
t
→
D=
0
t
=
Dt ε=
0
0ε r Et
→
ε 0ε r E=
D=
0
t
t
44
Conductor-dielectric Boundary conditions
To find boundary conditions for normal components, apply gauss law:
∫ D ⋅ dS = Q
Dn ∆S − 0 ⋅ ∆S = ∆Q
S
∆Q ρ S ∆S
=
Dn =
= ρS
∆S
∆S
Assuming ∆h → 0
=
Dn ε=
ρS
0ε r En
45
Conductor-dielectric Boundary conditions
Notes
• No electric field exists inside a conductor.
• Since E = −∇V , there is no potential difference between any
two points in the conductor.
• The electric field must be external to the conductor and must be
normal to its surface.
46
Conductor-Free space Boundary conditions
Special case of Conductor-Dielectric conditions.
Replacing ε r = 1:
ε=
ρS
0ε r En
=
ρS
Dn ε=
0 En
Dn
=
Dn ε=
ρS
0 En
=
Dt ε=
0
0 Et
Dt ε=
As in the earlier case =
0
0 Et
47
Example 5.9
• Two extensive homogeneous isotropic dielectrics meet on plane
z=0. for z>0, εr1 =4 and for z<0, εr2 =3. a uniform Electric field
E1=5ax-2ay+3az kV/m exists for z ≥0. Find:
•
•
•
•
(a) E2 for z≤0.
(b) The angles E1 and E2 make with the interface.
(c) the energy densities (in J/m3) in both dielectrics.
(d) the energy with a cube of side 2 m centred at (3,4,-5).
48
Example 5.9 - solution
E1 =E1n +E1t
E1n =
E1.a n =
E1.a z =→
3 E1n =
3a z
E1t
E=
5a x -2a y
1 -E1n
E=
E=
5a x -2a y
2t
1t
ε 0ε r 2 E 2 n =
ε 0ε r1E1n
D 2n =→
D1n
E2n
=
ε r1
4
4
=
E1n =
E1n =
3a z 4a z
εr2
3
3
E 2 =E=
5a x -2a y + 4a z
2t +E 2 n
kV/m
−−−−−−−−−−−−−−−−−−−−−−−
(b) α1 =90-θ1
, α 2 =90-θ 2
tan α1=
E1n
tan α 2=
E 2n
E1t
E 2t
=
3
=
25 + 4
3
→ α1= 29.1 , θ1= 60.9
29
=
tan θ1 ε r1
4


→ α 2= 36.6 , θ 2= 53.4 . (Note:
=
is satisfied)
49
tan θ 2 ε r 2
29
Example 5.9 - solution
(c) The energy densities in (J/m 3 ) in both dielectrics:
w E1 = 12 ε1 E1 =
2
1
2
w E 2 = ε 2 E2 =
1
2
2
(
)
( 3 ×μJ/m) × (25 + 4 + 16) × 10 =
−9
6
4μJ/m
(25
4
9)
10
× 10
×
+
+
×
= 672
36π
10−9
36π
1
2
6
3
597
3
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
(d) The energy with a cube of side 2 m centred at ( 3, 4, −5) .
{2 ≤ x ≤ 4,
3 ≤ y ≤ 5, -6 ≤ z ≤ −4}, z=-5 (Region 2)
W
=
E = ∫ w E 2 dv
4
5
−4
w )dzdydx
∫ ∫ ∫ (=
E2
w=
4.776 mJ
E 2 (2)(2)(2)
x = 2 y = 3 z = −6
50
Example 5.10
• Region y<0 consists of a perfect conductor while region y>0 is a
dielectric medium (εr1 =2). If there is a surface charge 2 nC/m2 on the
conductor, determine E and D at:
• (a) A(3,-2,2)
• (b) B(-4,1,5)
• Solution
(a) Point A(3,-2,2) is in the conductor
since y=-2<0 at A. Hence: E=0=D
------------------------(b) B(-4,1,5) is in the dielectric medium
since y=1>0 at B.
D n =ρ s = 2 nC/m 2
D 2×10-9
E=
=π10a-9
a y =36
ε 0 ε r 36π (2)
y
=113.1 a y V/m
51