chapter 5 electric fields in material spacesite.iugaza.edu.ps/tskaik/files/emi_chapter5.pdf ·...

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 non-

conductors.

• Non-conducting materials are unusually referred as insulators or

dielectrics.

• Materials may be classified in terms of their conductivity σ , in

mhos per meter ( ), or siemens per meter (s/m).

2

/ m

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

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

Properties of Materials

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.

• In a current of one ampere, charge is being transferred at a rate

of one coulomb per second.

5

dQIdt

=

• If current ΔI flows through a planar surface ΔS, the current

density is:

• If current is normal to the surface:

• If the current is not normal to the surface:

• The total current flowing through a surface S is:

6

Convection and Conduction Currents

IJS

∆=∆

I J S∆ = ∆

J SI∆ = ⋅∆

J SS

I = ⋅∆∫

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

• If there is a flow of charge of density ρV at velocity u=uyay the current is:

8

Convection Current

=

In General: J= u

V VV y

y V y

V

Q V S yI S ut t tIJ uS

ρ ρ ρ

ρ

ρ

∆ ∆ ∆ ∆∆ = = = ∆

∆ ∆ ∆∆

= =∆

Convection Current Density J= uVρ

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 :

In free space, the electron would accelerate.

In material, the electron suffers continual collisions.

The electron will move with different velocities between

collisions.

9

F Ee= −

The average velocity is called the drift velocity u.

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

The conduction current density is:

10

Conduction Current

u Eemτ

= −

2

J= u E = EVne

mτρ σ=

J= Eσ

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

• Inside conductor , which implies V=0. Thus,

a conductor is an equipotential body.

• According to gauss law,

• If E=0, then the charge density ρV = 0.

(Vab is potential difference between two points a and b in the

conductor)

12

Conductors

E 0 0and V= −∇ =

E S= VS V

d dVε ρ⋅∫ ∫

VInside an isolated perfect conductor, E=0, =0, =0abVρ

• 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

Conductors

• To derive Resistance: The magnitude of electric field is given by

Assuming conductor has uniform cross section of area S,

If the cross section of the conductor is not uniform:

The power P (in watts):

14

2E J or v

P dv P I R= ⋅ =∫

E=

1, ResistivityCC

I VS l

V lRI S

llRS S

σσ

σρ ρ

σ σ

=

= =

= = = ⇒

VEl

=IJS

=

E

E S

dlVRI dσ

− ⋅= =

⋅∫∫

lRSσ

=

If , calculate the current passing through

• (a) a hemispherical shell of radius 20 cm, 0<Ѳ<π/2, 0<φ<2π

• (b) a spherical shell of radius 10 cm.

Solution:

15

2

/2 2 /22

3 0.20.20 0 0

/22

0

2

0

, sin

1 2( ) ( 2cos ) sin 2 cos sin

Let u=sin , cos

4 sin 10 31.4 0.2 2

( ) 0 , 0.1

4 sin 00.1 2

r

rr

I J dS dS r d d a

a I r d d dr r

du d

I A

b r

I

π π π

θ φ θ

π

π

θ φ θ

θ θ φ θ π θ θ θ

θ θ θ

π θ π

θ π

π θ

=== = =

= ⋅ =

= =

=

= = =

< < =

= =

∫

∫ ∫ ∫

( ) 33

1 2cos a sin a A/mrJr φθ θ= +

Example 5.1

Alternatively for this case:

I= 0 , since 0S v

J dS J dv J⋅ = ∇ ⋅ = ∇ ⋅ =∫ ∫

• 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

16

29 19 10 3

7 -3 2

25 -6 5

55

10

( ) (10 )( 1.6 10 ) 1.6 10 C / m

( ) (5 10 )(10 10 ) 500 /

5( ) (5 10 )( ) 10 10 0.3932 4

5 10( )sin , 3.125 10 /1.6 10

V

VV

a neb J E kA m

dc I JS A

Jd ce J u u m s

ρ

σ

ππ

ρρ

−

−

= = − × = − ×

= = × × =

= = × = × × =

×= = = = ×

×

Example 5.3

• 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

17

( )

22 2 2

2

6 4

1 32

94

4 R=5 10 9 / 4 10

974

lRS

where S d r

cm

Hence

σ

π π

π

πµ

−

=

= − = −

= −

× − ×

= Ω

Example 5.4

• 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.

5.5 Polarization in Dielectrics

18

+Q (nucleus) -Q (electron)

• The dipole moment is

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:

• 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:

19

P=Qd

1 1 2 21

Q d +Q d + +Q d Q dN

N N k kk=

= ∑

10

Q dP lim

N

k kk

v v=

∆ →=

∆

∑

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.

• Examples: water, sulfer dioxide 20


Polarization of a polar molecule: (a) permanent dipole (E = 0), (b) alignment of permanent dipole (E ≠ 0).

• When polarization occurs, two charge densities are formed:

(1) An equivalent surface charge density is formed over the surface of the dielectric.

where an is unit normal to the surface.

(2) An equivalent volume charge density is formed throughout the dielectric.

Notes:

and are called bound (or polarization) surface and volume charge densities, respectively, as a distinct from free surface and volume charge densities .

• 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

psρ

nP apsρ = ⋅

pvρ

Ppvρ = −∇⋅

psρ pvρ

and s vρ ρ


• 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.

22

: total bound surface charge: total bound volume charge

= P. S

= .P

total charge= 0

bs

bv

bs ps

bv bs pvv v

ps pv bs bsS v

QQ

Q dS d

Q Q dv dv

dS dv Q Q

ρ

ρ

ρ ρ

=

= − = − ∇

+ = − =

∫ ∫∫ ∫

∫ ∫


nP apsρ = ⋅

Ppvρ = −∇⋅

• If the dielectric region contains free charge with volume charge density of ρV , the total volume charge density is:

• 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:

• Where is the susceptibility of the material.

Susceptibility of a material describes its response to an applied field.

23

( ) ( )

( )0 0

0

1 1E P

E+P D

t v pv

v pv v

v

ρ ρ ρ

ρ ρ ρε ε

ε ρ

= +

∇⋅ = + = −∇⋅

∇ ⋅ = = ∇ ⋅ 0 D= E + Pε⇒

e 0P= Eχ εeχ


5.6 Dielectric Constant and Strength

• ε is called the permittivity of the dielectric.

• εo is the permittivity of free space. ε0=10-9/36π F/m.

• εr is the relative permittivity.

• Table B.2 in appendix B, values of dielectric constants of some materials.

24

( )0 0 e 0

0 e 0

D= E + P E + ED= 1 E= Er

ε ε χ εε χ ε ε

=

+

( )0 e0

D= E , , 1r rεε ε ε ε ε χε

= = + =

The dielectric constant (or relative permittivity) is the ratio of the permittivity of the dielectric to that of free space.

rε

• 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

• 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

x xx xy xz x

y yx yy yz y

z zx zy zz z

D ED ED E

ε ε εε ε εε ε ε

=

Linear, Isotropic, and homogeneous Dielectrics

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:

28

/2 /2

/2 /22 3

/2 /2

for each of the six faces , there is a surface charge .for the face located at x=L/2:

P a2

6totalbound surface charge is : = 6 32

bound volum

ps

ps x x L x L

L L

s ps psS L L

aLax

aLQ dS dydz L aL

ρ

ρ

ρ ρ

= =

− −

= ⋅ = =

= = =∫ ∫ ∫

3

3 3

e charge density is P ( ) 3

total bound volume charge is = 3 3

Hence total charge is 3 3 0

pv

v pv

t s v

a a a a

Q dv a dv aL

Q Q Q aL aL

ρ

ρ

= −∇ ⋅ = − + + = −

= − = −

= + = − =

∫ ∫

Example 5.5

• 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

29

( )

( )

94 2

0

94 2

e 0 0

2n

2n

4 3

10(a) D= E= (2.55) 10 225.4 nC/m36

10(b) P= E = 1 E (1.55) 10 137 nC/m36

(c) D a 225.4 nC/m

(d) P a 137 nC/m

(e) 10 1.5 10 15 V

r

r

s n

ps n

D

P

V Ed

ε επ

χ ε ε επ

ρ

ρ

−

−

−

× × =

− = × × =

= ⋅ = =

= ⋅ = =

= = × =

Example 5.6

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

30

( )

20

ee 0 2

122

2 4

241 2

920

(a) Assuming point charge is located at the origin,

E a4

P= E a4

1 (4.7)2 10P a 13.12 pC/m4 4 (5.7)100 10

( 4)(2) 10(b) F a104 4 (5.7)100 1036

rr

rr

rps r

r

rr

Qr

Qr

Qr

Q Qr

πε εχχ επεε

ρπε π

πε ε ππ

−

−

−

−−

=

=

− ×= ⋅ = = =

×

− ×= =

× × 4a 1.263 a pNr r= −

Example 5.7

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

31

2

0

2 2

2

0

E a where a is a unit normal to the plate 2

and is the surface charge density.

F=QE . a a2 2

2 2

The pressure is force per area = 2

sn n

s

s ss n n

r

s

s

r

SS

orS QF

S

ρερ

ρ ρρε ε ε

ρε ε

ρε ε

=

= =

= =

Example 5.8

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.

where Qin is the total charge enclosed by the closed surface.

This equation is called continuity of current equation or continuity equation.

32

5.8 Continuity Equation and Relaxation Time

J S inOUT

S

dQI ddt

= ⋅ = −∫

Using divergence theorem, J S J

J

J

in vv

V V

S V

v

V V

v

dQ d dv dvdt dt t

d dv

dv dvt

t

ρρ

ρ

ρ

∂− = − = −

∂

⋅ = ∇ ⋅

∂→ ∇⋅ = −

∂

∂⇒∇⋅ = −

∂

∫ ∫

∫ ∫

∫ ∫

J v

tρ∂

∇ ⋅ = −∂

Consider introducing a charge at some interior point of a given material (conductor or dielectric).

33

Continuity Equation and Relaxation Time

From Ohm's law J= E

From Gauss's law D= and hence E=

substituting in the continuity equation,

J

E=

0

vv

v

v

v v

vv

t

t

t

t

σρρε

ρ

ρσ

σρ ρερ σ ρ

ε

∇ ⋅ ∇ ⋅

∂∇ ⋅ = −

∂∂

∇ ⋅ −∂

∂= −

∂∂

+ =∂

0vvt

ρ σ ρε

∂+ =

∂

34

0

0

/0

0

Separating the variables:

and integrating both sides: ln ln

ln is a constant of integration.

where

i

r

vv

v

v

v v

v

t Tv v r

r

t

t

t

where

e T

T

ρ σ ρε

ρ σρ ε

σρ ρε

ρερ ρσ

−

∂+ =

∂∂

= − ∂

= − +

= =

s the time constant in seconds.


/0 rt T

v v eρ ρ −=

• 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:

36

encE =0, D S=Q dl d⋅ ⋅∫ ∫

Dielectric-dielectric Boundary conditions

• Consider the boundary between two dielectrics with permittivities

37


38

•Tangential components of E are equal at the boundary. •Et undergoes no change on the boundary and it is continuous across the boundary.

1 1 1 2 2 2

1 1 2 2 2 1

The fields in the two media can be expressed as:E E E E E E

Apply the equation E l 0 to the path abcda in the figure

E l E E E E E E 02 2 2 2

E

t n t n

L

t n n t n nabcda

d

h h h hd w w

= + = +

⋅ =

∆ ∆ ∆ ∆⋅ = ∆ − − − ∆ + + =

∫

∫

1 2

1 2

E 0E E

t t

t t

w w∆ − ∆ =→ = 1 2E Et t=

• 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.


39

1 21 2 2 1 1 2

1 2

E E t tt t t t

D D D Dε εε ε

= → = → =

1 2

1 2

D S

Assuming 0S

n n S

n n S

d Q

D S D S Q S hD D

ρρ

⋅ =

∆ − ∆ = ∆ = ∆ ∆ →− =

∫

1 2n n SD D ρ− =


40

• Normal components of D are equal at the boundary. • Dn undergoes no change on the boundary and it is continuous across the boundary.


41

• The normal component of E undergoes some change across the boundary. • So En is said to be discontinuous across the boundary.

1 2

S

1 2

If no free charge exists at the boundary, =0n n S

n n

D D

D D

ρρ

− =

=1 2n nD D=

1 21 2 1 1 2 2

2 1

nn n n n

n

ED D E EE

εε εε

= → = → =


42

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.


43

This is called law of refraction

Conductor-dielectric Boundary conditions

44

0

Apply the equation E =0 to the path abcda in the figure

E = 0 0 0 02 2 2 2

0 0 0 0

t n nabcda

t t t t r t

dl

h h h hdl E w E w E

E w E D D Eε ε

⋅

∆ ∆ ∆ ∆⋅ ∆ − − ⋅ − ⋅∆ + ⋅ + =

→ ∆ = → = → = → = =

∫

∫

0 0t r tD Eε ε= =

To find boundary conditions for normal components, apply gauss law:

D S 0 Assuming 0

nS

Sn S

d Q D S S Q h

Q SDS S

ρ ρ

⋅ = ∆ − ⋅∆ = ∆ ∆ →

∆ ∆= = =∆ ∆

∫


45 0n r n SD Eε ε ρ= =


46

Notes • No electric field exists inside a conductor.

• Since , 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.

E V= −∇

r

0

0

0

Replacing 1:

As in the earlier case 0

n r n S

n n S

t t

D ED E

D E

εε ε ρε ρ

ε

== == =

= = 47

Conductor-Free space Boundary conditions Special case of Conductor-Dielectric conditions.

0

0 0n n S

t t

D ED E

ε ρε

= == =

• 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

1 1n 1t

1n 1 n 1 1n

1t 1 1n x y

1t x y

2n 1n 0 2 2 0 1 1

11 1

2

2 2t 2 x

2t

2

y

1 1 2 2

E

E

E =E +EE E .a E .a 3 E 3aE E -E 5a -2a

E 5a -2a

D D E E4 4E E 3a 4a3 3

E =E +E 5a -2a 4a kV/m

(b) =90- , =90-

ta

z z

r n r n

rn n z z

r

n z

n

ε ε ε εεε

α θ α θ

= = = → =

= =

= =

= → =

= = = =

= +

− − − − − − − − − − − − − − − − − − − − − − −

1n1 1 1

1t

2n 1 12 2 2

2t 2 2

E 3 3n 29.1 , 60.9E 25 4 29E tan4tan 36.6 , 53.4 . (Note: is satisfied) E tan29

r

r

α α θ

θ εα α θθ ε

= = = → = =+

= = → = = =

49

Example 5.9 - solution

50

( )( )

9

9

3

2 6 3101 11 1 12 2 36

2 6 3101 12 2 22 2 36

(c) The energy densities in (J/m ) in both dielectrics:

w = E 4 (25 4 9) 10 672 μJ/m

w = E 3 (25 4 16) 10 597 μJ/m

(d) The energy with a

E

E

π

π

ε

ε

−

−

= × × + + × =

= × × + + × =

− − − − − − − − − − − − − − − − − − − − − − − − − − − − − − −

( )

( )4 5 4

E 2 2 22 3 6

cube of side 2 m centred at 3,4, 5 .

2 4, 3 5, -6 4 , z=-5 (Region 2)

W = w w w (2)(2)(2) 4.776 mJE E Ex y z

x y z

dv dzdydx−

= = =−

−

≤ ≤ ≤ ≤ ≤ ≤ −

= = =∫ ∫ ∫ ∫

Example 5.9 - solution

• 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

51

-9

2n

-9

y100 r 36π

(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 mediumsince y=1>0 at B.D = 2 nC/m

D 2×10E= = a =36π aε ε (2)

sρ =

y y=113.1 a V/m

Example 5.10

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