Chapter 6 Dielectrics and Capacitance

President University Erwin Sitompul EEM 8/1

Lecture 8

Engineering Electromagnetics

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com

2 0 1 3


Chapter 6Dielectrics and Capacitance

Engineering Electromagnetics


The Nature of Dielectric MaterialsA dielectric material in an electric field can be viewed as a free-

space arrangement of microscopic electric dipoles, a pair of positive and negative charges whose centers do not quite coincide.

These charges are not free charges, not contributing to the conduction process. They are called bound charges, can only shift positions slightly in response to external fields.

All dielectric materials have the ability to store electric energy. This storage takes place by means of a shift (displacement) in the relative positions of the bound charges against the normal molecular and atomic forces.

Chapter 6 Dielectrics and Capacitance


The Nature of Dielectric MaterialsThe mechanism of this charge displacement differs in various

dielectric materials.Polar molecules have a permanent displacement existing

between the centers of “gravity” of the positive and negative charges, each pair of charges acts as a dipole.

Dipoles are normally oriented randomly, and the action of the external field is to align these molecules in the same direction.


Nonpolar molecules does not have dipole arrangement until after a field is applied.

The negative and positive charges shift in opposite directions against their mutual attraction and produce a dipole which is aligned with the electric field.


The Nature of Dielectric MaterialsEither type of dipole may be described by its dipole moment p:


Qp d

total1

n v

ii

p p

If there are n dipoles per unit volume, then there are nΔv dipoles in a volume Δv. The total dipole moment is:

We now define the polarization P as the dipole moment per unit volume:

01

1lim

n v

iv

iv

P p

The immediate goal is to show that the bound-volume charge density acts like the free-volume charge density in producing an external field ► We shall obtain a result similar to Gauss’s law.

n nQ p d


The Nature of Dielectric MaterialsTake a dielectric containing nonpolar molecules. No molecules

has p, and P = 0 throughout the material.Somewhere in the interior of the dielectric we select an

incremental surface element ΔS, and apply an electric field E.The electric field produces a moment p = Qd in each molecule,

such that p and d make an angle θ with ΔS.


Due to E, any positive charges initially lying below the surface ΔS and within ½dcosθ must have crossed ΔS going upward.

Any negative charges initially lying above the surface ΔS and within ½dcosθ must have crossed ΔS going downward.



The Nature of Dielectric MaterialsFor n molecules/m3, the net total charge (positive and negative)

which crosses the elemental surface in upward direction is:

cosbQ nQd S

bQ P S

b SQ d P S

bQ nQ d S

The notation Qb means the bound charge. In terms of the polarization, we have:

If we interpret ΔS as an element of a closed surface, then the direction of ΔS is outward.

The net increase in the bound charge within the closed surface is:


The Nature of Dielectric MaterialsSeeing some similarity to Gauss’s law, we may now generalize

the definition of electric flux density so that it applies to media other than free space.

We write Gauss’s law in terms of ε0E and QT, the total enclosed charge (bound charge plus free charge):


0T SQ d E ST bQ Q Q

T bQ Q Q 0( )S

d E P SCombining the last three equations:

We may now define D in more general terms:

0 D E P• There is an added term to D when a

material is polarized


The Nature of Dielectric MaterialsFor equations with volume charge densities, we now have:


b bvQ dv

vvQ dvT TvQ dv

With the help of the divergence theorem, we may transform the equations into equivalent divergence relationships:

b P

0 T Ev D


The Nature of Dielectric MaterialsChapter 6 Dielectrics and Capacitance

To utilize the new concepts, it is necessary to know the relationship between E and P.

This relationship will be a function of the type of material. We will limit the discussion to isotropic materials for which E and P are linearly related.

In an isotropic material, the vectors E and P are always parallel, regardless of the orientation of the field.

The linear relationship between P and E can be described as:0e P E

0 0e D E E

1r e

0 r D E E

0 r

χe : electric susceptibility,

a measure of how easily

a dielectric polarizes in response to an electric field

0( 1)e E

We now define:

εr : relative permittivity


The Nature of Dielectric Materials In summary, we now have a relationship between D and E

which depends on the dielectric material present:


D E

0 r

v D

Sd Q D S



ExampleWe locate a slab of Teflon in the region 0 ≤ x ≤ a, and assume free space where x < 0 and x > a. Outside the Teflon there is a uniform field Eout = E0ax V/m. Find the values for D, E, and P everywhere.

in 0 in2.1D E

in 0 in1.1P E

,teflon 2.1r 2.1 1 1.1e

out 0 xEE a

out 0 0 xED a

out 0P • No dielectric materials outside 0 ≤ x ≤ a

• No relations yet established over the boundary• This will be discussed in the next section


Boundary Conditions for Perfect Dielectric MaterialsConsider the interface between two dielectrics having

permittivities ε1 and ε2, as shown below.


We first examine the tangential components around the small closed path on the left, with Δw<< and Δh<<< :

0d E Ltan1 tan 2 0E w E w

tan1 tan 2E E


Boundary Conditions for Perfect Dielectric MaterialsThe tangential electric flux density is discontinuous,


tan1 tan 2tan1 tan 2

1 2

D DE E

tan1 1

tan 2 2

D

D

1 2N N SD S D S Q S

1 2N N SD D

The boundary conditions on the normal components are found by applying Gauss’s law to the small cylinder shown at the right of the previous figure (net tangential flux is zero).

• ρS cannot be a bound surface charge density because the polarization already counted in by using dielectric constant different from unity

• ρS cannot be a free surface charge density, for no free charge available in the perfect dielectrics we are considering

• ρS exists only in special cases where it is deliberately placed there


Boundary Conditions for Perfect Dielectric MaterialsExcept for this special case, we may assume ρS is zero on the

interface:


1 2N ND D

1 1 2 2N NE E

The normal component of electric flux density is continuous. It follows that:


Boundary Conditions for Perfect Dielectric MaterialsCombining the normal and the tangential

components of D,

1 1 1 2 2 2cos cosN ND D D D

tan1 1 1 1

tan 2 2 2 2

sin

sin

D D

D D


2 1 1 1 2 2sin sinD D

1 1

2 2

tan

tan

After one division,1 2 1 2


Boundary Conditions for Perfect Dielectric MaterialsThe direction of E on each side of

the boundary is identical with the direction of D, because D = εE.


tan1 tan 2E E1 1 2 2N NE E

1 2 1 2

2E

1E


Boundary Conditions for Perfect Dielectric MaterialsThe relationship between D1 and D2 may be derived as:


2

2 222 1 1 1

1

cos sinD D

2

2 212 1 1 1

2

sin cosE E

The relationship between E1 and E2 may be derived as:


Boundary Conditions for Perfect Dielectric MaterialsChapter 6 Dielectrics and Capacitance

out 0 xEE a

out 0 0 xED a

out 0P

ExampleComplete the previous example by finding the fields within the Teflon.

in out 0 0 xE D D a

in 0 in1.1P E

inin

r 0 D

E 0 0

r 0

xE

a

00.476 xE a

0 00

r 0

1.1 xE

a

0 00.524 xE a

• E only has normal component



ExampleA spherical conducting shell has an excess charge of +10 C. A point charge of –15 C is located at the center of the shell. Use Gauss’s law to calculate the charge on the inner and outer surface of the shell.

–15 C, point charge at the center

+15 C, on inner surface of the shell, counteract the point charge so that no field exists in the conductor

Since total charges in the shell is +10 C, the charges on outer surface must be –5 C

• Inside a conductor, E = 0, static equilibrium.

• No field means no flux, whereas means no enclosed charge for any imaginary surface in the conductor.



–15 C

+15 C

–5 C

20

1 15

4E

r

20

1 5

4E

r

0E

E

r

• What will be the difference if the shell is made of insulator?

► Excess charge given to the shell will not be evenly distributed, never move from initial location of charging.

► The distribution of charges on the inner and outer surface of the shell is not homogenous and cannot be determined

► The field E is not radially homogenous



Example Now, the conducing spherical shell is replaced by an insulating one, with εr = 3. The shell has no excess charge. A point charge of –15 C is still located at the center of the shell. Determine the magnitude of electric field E as function of radius r.

1 2N ND D

1 1 2 2N NE E

• The direction of the field is radially outward only normal component exists

12 1

2N NE E

1

1

3 NE

–15 C

20

1 15

4E

r

20

1 15

4E

r

E

r

20

1 1 15

4rE

r


Boundary Conditions Between a Conductor and a Dielectric

The boundary conditions existing at the interface between a conductor and a dielectric are much simpler than those previously discussed.

First, we know that D and E are both zero inside the conductor.Second, the tangential E and D components must both be zero

to satisfy:


0d E LD E

Finally, the application of Gauss’s law shows once more that both D and E are normal to the conductor surface and that DN = ρS and EN = ρS/ε.

The boundary conditions for conductor–free space are valid also for conductor–dielectric boundary, with ε0 replaced by ε.

0t tD E

N N SD E


We will now spend a moment to examine one phenomena: “Any charge that is introduced internally within a conducting material will arrive at the surface as a surface charge.”

Given Ohm’s law and the continuity equation (free charges only):



J E

v

t

J

We have:v

t

E

v

t

D


If we assume that the medium is homogenous, so that σ and ε are not functions of position, we will have:



v

t

D

Using Maxwell’s first equation, we obtain;

vv t

Making the rough assumption that σ is not a function of ρv, it

leads to an easy solution that at least permits us to compare different conductors.

The solution of the above equation is:( )

0t

v e • ρ0 is the charge density at t = 0• Exponential decay with time constant of ε/σ


Good conductors have low time constant. This means that the charge density within a good conductors will decay rapidly.

We may then safely consider the charge density to be zero within a good conductor.

In reality, no dielectric material is without some few free electrons (the conductivity is thus not completely zero). The charge introduced internally in any of them will eventually reach the surface.



( )0

tv e

t

ρv

ρ0

ρ0/e

ε/σ


Homework 7D6.1. D6.2. D6.3.


All homework problems from Hayt and Buck, 7th Edition.Due: Monday, 10 June 2013.

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Chapter 6 Dielectrics and Capacitance