prof. sang-yong jungcontents.kocw.net/kocw/document/2014/sungkyunkwan/... · 2016-09-09 · 5.4...

L o g o www.emlab.kr

Prof. Sang-Yong Jung

( 23432B 031-299-4952 www.emlab.kr [email protected] )

Engineering Electromagnetics (Week 9)

L o g o

Ch5. Current, Conductors, and Dielectrics 2

5.1_ Current and Current Density

5.2_ Continuity of Current

5.3_ Metallic Conductors

5.4_ Conductor Properties and Boundary Conditions

5.5_ The Method of Images

5.6_ Semiconductors

5.7_ The Nature of Dielectric Materials

5.8_ Boundary Conditions for Perfect Dielectric Materials

전기기기연구실 (www.emlab.kr) Electric Machines LAB (www.emlab.kr)

5.1 Current and Current Density 3

□ Current : I=𝑑𝑄

𝑑𝑡

□ Current density( J) : yields current in Amps when it is integrated over a cross-

sectional area.

𝐼 = J ∙ 𝑑𝐒

𝑆

□ Consider a charge Q, occupying volume v, moving in the positive x direction at

velocity vx

∆𝑄 = 𝜌𝑣 ∆𝑣 = 𝜌𝑣 ∆𝑆 ∆𝐿

𝑄 = 𝜌𝑣 ∆𝑆 ∆𝑥

J =𝜌𝑣𝐯

∆𝐼 =∆𝑄

∆𝑡= 𝜌𝑣 ∆𝑆

∆𝑥

∆𝑡


5.2 Continuity of Current 4

□ Charge Qi is escaping from a volume through closed surface S, to form current density

J. The total current is:

I = = =

the minus sign is needed to produce positive outward flux,

while the interior charge is decreasing with time.

- Apply the divergence theorem:

I = = = =

𝛻 ∙ J ∆𝑣 = −𝜕𝜌𝑣

𝜕𝑡 ∆𝑣

−𝑑

𝑑𝑡 𝜌𝑣𝑑𝑣

𝑣𝑜𝑙

−dQidt

J

𝑠

∙ 𝑑𝑺

J

𝑠

∙ 𝑑𝑺 𝛻 ∙ J 𝑑𝑣

𝑣𝑜𝑙

−𝜕𝜌𝑣𝜕𝑡

𝑣𝑜𝑙

𝑑𝑣

𝛻 ∙ J = −𝜕𝜌𝑣

𝜕𝑡

−𝑑

𝑑𝑡 𝜌𝑣 𝑑𝑣

𝑣𝑜𝑙


5

a) Conductors exhibit no energy gap between valence and conduction bands.

So electrons move freely.

b) Insulators show large energy gaps, requiring large amounts of energy to lift electrons

into the conduction band. When this occurs, the dielectric breaks down.

c) Semiconductors have a relatively small energy gap, so modest amounts of energy

(applied through heat, light, or an electric field) may lift electrons from valence to

conduction bands.

5.3 Metallic Conductors


6

□ Free electrons move under the influence of an electric field. The applied force on an

electron of charge Q = -e will be

𝑭 = −𝑒𝑬

□ When forced, the electron accelerates to an equilibrium velocity, known as

the drift velocity:

𝐯𝑑 = −𝜇𝑒𝐄 ( e : the electron mobility )

□ The drift velocity is used to find the current density through:

J= −𝜌𝑒𝜇𝑒𝑬= 𝜎𝑬

(Conductivity : 𝜎= −𝜌𝑒𝜇𝑒 )



7

□ Consider the cylindrical conductor, with voltage V applied across the ends.

Current flows down the length, and is assumed to be uniformly distributed over the

cross-section, S.

Using Ohm’s Law : V=IR

The resistance of the cylinder: R=𝐿

𝜎𝑆


− 𝐄 ∙ 𝑑𝐋 = −𝐄 ∙ 𝑑𝐋 = −𝐄 ∙ 𝐋𝑏𝑎

𝑎

𝑏

= E ∙ 𝐋𝑎𝑏

𝑎

𝑏

→ V=EL 𝑉𝑎𝑏=

𝐼 = − J ∙ 𝑑𝐒 = J𝑆

𝑠

→ J = 𝐼

𝑆 = 𝜎𝐸 = 𝜎

𝑉

𝐿 → 𝑉 =

𝐿

𝜎𝑆I

𝑅 =𝑉𝑎𝑏𝐼=− 𝐄 ∙ 𝑑𝐋𝑎

𝑏

𝜎𝐄 ∙ 𝑑𝐒

𝑠


5.4 Conductor Properties and Boundary Condition 8

□ Consider a conductor ( excess charge),

1. Charge can exist only on the surface as a surface charge density.

(s is not in the interior.)

2. Electric field cannot exist in the interior, nor can it possess a tangential

component at the surface.

→ The surface of a conductor is an equipotential.


9 5.4 Conductor Properties and Boundary

□ Over the rectangular integration path, use 𝐄 ∙ 𝑑𝐋 = 0

𝐸𝑡=0

+𝑏

𝑎 +𝑐

𝑏 +𝑑

𝑐 = 0𝑎

𝑑

𝐸𝑡∆𝑤 − 𝐸𝑁,𝑎𝑡 𝑏 1

2∆ℎ + 𝐸𝑁,𝑎𝑡 𝑎

1

2∆ℎ=0 (h → zero)


10

5.4 Conductor Properties and Boundary

- Gauss’ Law is applied to the cylindrical surface shown below:

𝐃 ∙ 𝑑𝐒 =

𝑡𝑜𝑝+

𝑏𝑜𝑡𝑡𝑜𝑚+

𝑠𝑖𝑑𝑒𝑠

𝑠= Q

𝐷𝑁∆𝑆 = 𝑄 = 𝜌𝑠∆𝑆 or 𝐷𝑁= 𝜌𝑠

D𝑡 = E𝑡 = 0

D𝑁 = 𝜖0E𝑁 = 𝜌𝑠


11 5.4 Conductor Properties and Boundary

□ Summary

- The static electric field intensity inside a conductor → 0

the surface of a conductor → normal direction

- The conductor surface is an equipotential surface.

At the surface : (Tangential E = 0)

(Normal D = the surface charge density)

𝑬 × 𝒏|𝑠 = 0

𝐃 ∙ 𝒏|𝑠 = 𝜌𝑠


12 5.5 The Method of Images

□ In the electric dipole, a surface along the plane of symmetry is an equipotential (V = 0)

- The boundary conditions and charges

(surface along the plane of symmetry) = (grounded conducting plane)

→ identical in the upper half spaces , but not in the lower half


13

□ Example

In this case, we are to find the surface charge density on the conducting plane at the point

(2,5,0). A 30-nC line charge lies parallel to the y axis at z = 3.

5.5 The Method of Images


14

- Referring to the figure,

Add the two field :


𝐸+ =𝜌𝐿2𝜋𝜀0𝑅+

𝐚𝑅+ = 30 × 10−9

2𝜋𝜖0 13 2𝐚𝑥 − 3𝐚𝑧

13

𝐸− =30 × 10−9

2𝜋𝜖0 13 2𝐚𝑥 + 3𝐚𝑧

13

𝐸 =−180 × 10−9𝐚𝑧2𝜋𝜖0(13)

= −249𝐚𝑧 𝑉/𝑚

𝑅+ = 2𝐚𝑥 − 3𝐚𝑧 𝑅− = 2𝐚𝑥 + 3𝐚𝑧


15

- The electric field at point P:

𝐸 = −249𝐚𝑧 𝑉/𝑚

D=𝜖0𝐸 = −2.20𝐚𝑧 𝑛𝐶/𝑚2

- To find the charge density, use 𝐃 ∙ 𝐧𝑠 = 𝜌𝑠 ( 𝐧𝑠= 𝐚𝑧 )

𝐃 ∙ 𝐧𝑠 = −2.20𝐚𝑧 ∙ 𝐚𝑧= −2.20 𝑛𝐶/𝑚2



16 5.6 Semiconductors

□ Two types of current carriers (an intrinsic semiconductor material )

1) electrons 2) holes.

- Many semiconductor properties may be described by treating the hole.

( = a positive charge of e, a mobility, μh, and an effective mass )

- Both carriers move in an electric field, they move in opposite directions;

→ each contributes a component of the total current (same direction)

- The conductivity : a function of both hole and electron concentrations and mobilities,

𝜎 = −𝜌𝑒𝜇𝑒 + 𝜌ℎ𝜇ℎ


17 5.6 Semiconductors

- Differences between the metallic conductors and the intrinsic semiconductors

Temperature ↑ → The conductivity intrinsic semiconductor ↑

metallic conductor ↓

- Intrinsic semiconductors satisfy the point form of Ohm’s law

→ The conductivity = constant (current density and the direction of the current density)

- By adding very small amounts of impurities.

→ The number of charge carriers and the conductivity - ↑↑

- Doping : Donor (additional electrons) : n-type semiconductors,

Acceptors (extra holes ) : p-type materials.


18 5.7 The Nature of Dielectric Materials

□ Electric Dipole and Dipole Moment

- In dielectric, charges are held in position (bound), and ideally there are no free charges

that can move and form a current. Atoms and molecules may be polar (having

separated positive and negative charges), or may be polarized by the application of an

electric field.

□ A polarized atom or molecule possesses a dipole moment (p)

→ p = Q (the charge magnitude) × d (the positive and negative charge separation)

Dipole moment is a vector that points from the negative to the positive charge.



□ Model of a Dielectric

A dielectric can be modeled as an ensemble of bound charges in free space, associated

with the atoms and molecules that make up the material.

Some of these may have intrinsic dipole moments, others not. In some materials (such

as liquids), dipole moments are in random directions.



□ The number of dipoles is expressed as a density, n dipoles per unit volume.

The Polarization Field of the medium :

[dipole moment/volume] or [C/m2]

𝑃 = lim∆𝑣→0

1

∆𝑣 𝐩𝑖

𝑛∆𝑣

𝑖=1



□ An electric field may increase the charge separation in each dipole, and possibly

re-orient dipoles so that there is some aggregate alignment.

- The effect is to increase P.

𝑃 = lim∆𝑣→0

1

∆𝑣 𝐩𝑖

𝑛∆𝑣

𝑖=1

= nP



□ Consider an electric field applied at an angle to a surface normal as shown.

The resulting separation of bound charges (or re-orientation) leads to positive bound

charge crossing upward through surface of area S, while negative bound charge crosses

downward through the surface.

Dipole centers that lie within the range (1/2) d cos above or below the surface will

transfer charge across the surface.


23

- The total bound charge that crosses the surface is given by:

∆𝑄𝑏 = 𝑛𝑄𝑑 cos 𝜃 ∆𝐒 = 𝑛𝑄𝐝 ∙ ∆𝐒 = 𝐏 ∙ ∆𝐒

P

5.7 The Nature of Dielectric Materials


24

□ A charge within the closed surface consisting of bound charges, qb , and free charges, q.

The total charge will be the sum of all bound and free charges. Using the Gauss’ Law in

terms of the total charge, QT as:

- (𝑄𝑏 : bound charge, Q : free charge)

( , )

𝑄𝑇 = 𝑄𝑏 + 𝑄

𝑄𝑏 = − 𝐏

𝑆

∙ 𝑑𝐒 𝑄𝑇 = 𝜖0𝐄 ∙ 𝑑𝐒

𝑆

Q = 𝑄𝑇 − 𝑄𝑏 = (𝜖0𝐄 + 𝐏) ∙ 𝑑𝐒

𝑆

𝐷 = 𝜖0𝐸 + 𝑃



25

□ A stronger electric field results in a larger polarization in the medium.

In a linear medium, the relation between P and E is linear :

(𝜒𝑒: the electric susceptibility of the medium)

(the dielectric constant, or relative permittivity : 𝜖𝑟 = 𝜒𝑒 + 1)

- Leading to the overall permittivity of the medium:


𝑃 = 𝜖0𝜒𝑒𝐸

𝐷 = 𝜖0𝐸 + 𝜒𝑒𝜖0𝐸 = 𝜒𝑒 + 1 𝜖0𝐸

𝐷 = 𝜖0𝐸 = 𝜖𝐸 where 𝜖 = 𝜖𝑟𝜖0


26

□ In an isotropic medium, the dielectric constant is invariant with direction of the applied

electric field.

This is not the case in an anisotropic medium (usually a crystal) in which the dielectric

constant will vary as the electric field is rotated in certain directions. In this case, the

electric flux density vector components must be evaluated separately through the

dielectric tensor. The relation can be expressed in the form:

𝐃𝑥 = 𝜖𝑥𝑥𝐄𝑥 + 𝜖𝑥𝑦𝐄𝑦 + 𝜖𝑥𝑧𝐄𝑧

𝐃𝑦 = 𝜖𝑦𝑥𝐄𝑥 + 𝜖𝑦𝑦𝐄𝑦 + 𝜖𝑦𝑧𝐄𝑧

𝐃𝑧 = 𝜖𝑧𝑥𝐄𝑥 + 𝜖𝑧𝑦𝐄𝑦 + 𝜖𝑧𝑧𝐄𝑧



27

□ Taking the previous results and using the divergence theorem

(The point form expressions)

1. Bound Charge :

2. Total charge :

3. Free charge :


𝑄𝑏 = 𝜌𝑏𝑑𝑣 = 𝐏 ∙ 𝑑𝐒 → 𝛻 ∙ 𝐏 = −𝜌𝑏

𝑆

𝑣

𝑄𝑇 = 𝜌𝑇𝑑𝑣 = 𝜖0𝐄 ∙ 𝑑𝐒 → 𝛻 ∙ 𝜖0𝐄 = 𝜌𝑇

𝑆

𝑣

𝑄 = 𝜌𝑣𝑑𝑣 = 𝐃 ∙ 𝑑𝐒 → 𝛻 ∙ 𝐃 = 𝜌𝑣

𝑆

𝑣


28 5.8 Boundary Conditions for Perfect Dielectric Materials

□ Boundary Condition for Tangential Electric Field

- The E is conservative :

𝐸 ∙ 𝑑𝐿 = 0

𝐸𝑡𝑎𝑛1∆𝑤 − 𝐸𝑡𝑎𝑛2∆𝑤 = 0

𝐸𝑡𝑎𝑛1−𝐸𝑡𝑎𝑛2 = 0

𝐸𝑡𝑎𝑛1 = 𝐸𝑡𝑎𝑛2


29

- Apply Gauss’ Law :

If the charge density is zero :

𝐷𝑁1 = 𝐷𝑁2

5.8 Boundary Conditions for Perfect Dielectric Materials

𝑄 = 𝐷 ∙ 𝑑𝑆

𝑆

𝐷𝑁1∆𝑆 − 𝐷𝑁2∆𝑆 = ∆𝑄 = 𝜌𝑠∆𝑆

𝐷𝑁1 − 𝐷𝑁2 = 𝜌𝑠


30

- Find the relation between the angles 1 and 2, assuming no charge density on the

surface.

The normal components of D will be continuous across the boundary


𝐷𝑁1 = 𝐷1 cos 𝜃1 =𝐷2 cos 𝜃2 =𝐷𝑁2


31

- Then, with tangential E continuous across the boundary

taking the ratio of the two underlined equations :


𝐷𝑡𝑎𝑛1𝐷𝑡𝑎𝑛2=𝐷1 sin 𝜃1𝐷2 sin 𝜃2

=𝜖1𝜖2 or 𝜖2𝐷1 sin 𝜃1 = 𝜖1𝐷2 sin 𝜃2

tan 𝜃1tan 𝜃2

=𝜖1𝜖2

prof. sang-yong jungcontents.kocw.net/kocw/document/2014/sungkyunkwan/... · 2016-09-09 · 5.4...

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