Electrostatics

Chapter 3

• The electromagnetic constitutive parameters of a material

medium are its electrical permittivity 𝜀,magnetic

permeability 𝜇 and conductivity 𝜍.

** A material is said to be homogeneous if 𝜀, 𝜇 and 𝜍

do not vary from point to point.

** A material is said to be Isotropic if 𝜀, 𝜇 and 𝜍 do not

vary with direction.

3-6 Electric Properties of Materials

• The conductivity of material is a measure of how

easily electron can travel through the material

under the influence of an external electric field.

• Materials are classified as:

conductors(metals).

dielectrics(insulators).

semiconductors.

based on the magnitudes of their conductivity 𝜎.

• Upon applying an external electric field, the

electrons in the conductor migrate from one atom

to the next along a direction opposite that of the

external field. The average velocity of the

electrons are called "electron drift velocity ue "

which gives rise to a “conductor current”.

A perfect dielectric : 𝜍 = 0

A perfect conductor : 𝜍 = ∞

Note :

The unit of conductivity 𝝈 is :

S/m (Siemens per meter)

OR (𝟏 𝛀.𝒎)

(𝑺 = 𝑨/𝑽 (ampere per volt) or 𝟏

𝛀 )

Typical metals : 𝝈 = 𝟏𝟎𝟔 𝐭𝐨 𝟏𝟎𝟕 𝑺

𝒎

Typical dielectrics : 𝝈 = 𝟏𝟎−𝟏𝟎 𝐭𝐨 𝟏𝟎−𝟏𝟕 𝑺

𝒎

Semiconductors : 𝝈 is in between metals and

dielectrics such as Germanium(Ge) has 𝝈 = 𝟐. 𝟐 𝑺/𝒎

• The conductivity of a material depends on

temperature and the presence of impurities.

• At very low temperatures in the region of

absolute zero, some conductors become

“superconductors“.

3-7 Conductors

• The drift velocity of electrons in a conducting

material is related to the externally applied electric

field 𝐸 through :

𝑢𝑒 = −𝜇𝑒𝐸 (𝑚 𝑠 )

Electric field Electron

mobility

with units

of 𝒎𝟐 𝑽. 𝒔

Drift

velocity

𝑢ℎ = 𝜇ℎ𝐸 ( 𝑚 𝑠 )

hole

mobility

• In a semiconductor, current flow is due to the

movements of both electrons and holes and since the

holes are positive charge carriers, the “hole drift

velocity” 𝑢ℎ is in the same direction of 𝐸 ,

• The current density 𝐽 in a medium containing a volume

density 𝜌𝑣 moving with velocity 𝑢 is :

𝐽 = 𝜌𝑣𝑢

• The current density consists of components from both

electrons ( 𝐽 𝑒 ) and holes ( 𝐽 ℎ ).

• Thus the total conduction current density 𝐽 is :

𝐽 = 𝐽 𝑒 + 𝐽 ℎ = 𝜌𝑣𝑒𝑢𝑒 + 𝜌𝑣ℎ𝑢ℎ (𝐴 𝑚2 )

• Using 𝑢 = 𝜇𝐸 :

𝐽 = (−𝜌𝑣𝑒𝜇𝑒 + 𝜌𝑣ℎ𝜇ℎ)𝐸

,Where 𝜌𝑣𝑒 = −𝑁𝑒𝑒 and 𝜌𝑣ℎ = 𝑁ℎ𝑒

𝑒 = 1.6 × 10−19 𝐶

# of electrons

per unit

volume

# of holes per

unit volume

• The conductivity of the material, 𝜍, is defined as :

𝜍 = 𝑁𝑒𝜇𝑒 + 𝑁ℎ𝜇ℎ 𝑒 (𝑆 𝑚 ) (semiconductor)

= −𝜌𝑣𝑒𝜇𝑒 + 𝜌𝑣ℎ𝜇ℎ

• For a good conductor usually :

𝑁ℎ𝜇ℎ ≪ 𝑁𝑒𝜇𝑒

𝜍 = −𝜌𝑣𝑒𝜇𝑒 = 𝑁𝑒𝜇𝑒𝑒 (𝑆 𝑚 ) (conductor)

• In either case :

𝐽 = 𝜍𝐸 ( 𝐴 𝑚2) (ohm′s law − point form)

𝑨

𝒎𝟐 =𝟏

𝜴𝒎×𝑽

𝒎

→ 𝜴 =𝑽

𝑨 (𝒐𝒉𝒎′𝒔 𝒍𝒂𝒘)

Perfect dielectric : 𝐽 = 0

(𝜍 = 0,regardless of 𝐸)

Perfect conductor : 𝐸 = 0

(𝜍 = ∞,regardless of 𝐽 )

• A perfect conductor is an “equipotential medium” .it means that

the electric potential is the same at every point in the conductor.

• Since 𝐸 = 0 everywhere in the perfect conductor, the voltage

difference 𝑉21 = 0

Example

The conductor shown in the figure is applied to an

electric field of (20 mV m ) . Find :

(a) Volume charge density 𝜌𝑣of free electrons.

(b) Current density J.

(c) The current flowing in the wire.

(d) The electron drift velocity.

(e) Volume density of free electrons 𝑁𝑒.

solution

3-7.1 Resistance

*** What are 𝐽 and 𝐸 directions?!

𝐼 , 𝐽 , and 𝐸 have the same direction from high potential to low potential as shown in the figure.

• Using the point form of Ohm’s law, we can derive an expression

for the resistance R of a conductor of length 𝑙 and uniform

cross section A .

𝑉 = 𝑉1 − 𝑉2

= − 𝐸. 𝑑𝑙 = − 𝑥 𝐸𝑥 . 𝑥 𝑑𝑙 = −𝐸𝑥 𝑥1 − 𝑥2 = 𝐸𝑥𝑙𝑥1

𝑥2

𝑥1

𝑥2

𝑙 = 𝑥2 − 𝑥1

𝐸 = 𝒙 𝐸𝑥

𝑉 = 𝐸𝑥 𝑥2 − 𝑥1 = 𝐸𝑥 𝑙

• The current flowing through the cross section A at 𝑥2 is

𝐼 = 𝐽 . 𝑑𝑠 𝐴

= 𝜍𝐸. 𝑑𝑠 𝐴

= 𝜍𝐸𝑥𝐴

∴ R =𝑉

𝐼=

𝐸𝑥𝑙

𝜍𝐸𝑥𝐴=

𝑙

𝜍𝐴 (Ω)

• For any arbitrary shape, the resistance R can be

expressed as :

𝑅 = 𝑉

𝐼 =

− 𝐸 .𝑑𝑙 𝑙

𝐽 .𝑑𝑠 𝑠

= − 𝐸 .𝑑𝑙 𝑙

𝜎𝐸. 𝑑𝑠 𝑠

** The reciprocal of R (1

𝑅 ) is called the conductance G

and the unit of G is (Ω−1) or siemens (S) .

** For linear resistor:

𝐺 =1

𝑅=

𝜎𝐴

𝑙 (S)

Example 1

solution

Example 2 Conductance of Coaxial Cable

Obtain an expression for G′, the conductance per unit length of

the insulation layer .

• Since the current is radial, the area through which the

current flows is 𝐴 = 2𝜋𝑟𝑙. Hence,

𝐽 = 𝒓 𝐼

𝐴= 𝒓

𝐼

2𝜋𝑟𝑙

𝐽 = 𝜍𝐸 → 𝐸 =𝐽

𝜎= 𝒓

𝐼

2𝜋𝜎𝑟𝑙

• Current 𝐼 flows from higher potential to lower potential .

𝑉𝑎𝑏 = − 𝐸. 𝑑𝑙 = − 𝐼

2𝜋𝜎𝑙

𝑟 .𝑟 𝑑𝑟

𝑟=

𝐼

2𝜋𝜎𝑙

𝑑𝑟

𝑟=

𝐼

2𝜋𝜎𝑙ln(

𝑏

𝑎

𝑏

𝑎

𝑎

𝑏

𝑎

𝑏)

solution

• The conductance per unit length is then:

𝐺′ =𝐺

𝑙=

1

𝑅𝑙=

𝐼

𝑉𝑎𝑏𝑙=

𝐼

𝐼2𝜋𝜍𝑙

ln(𝑏𝑎) 𝑙

=2𝜋𝜍

ln(𝑏𝑎) (𝑆 𝑚 )

3-7.2 Joule's Law

Let’s now consider the power dissipated in a conducting medium

in the presence of an electrostatic field 𝐸. The electric force acting

on charge 𝑞𝑒 and 𝑞ℎ are:

𝐹 𝑒 = 𝑞𝑒𝐸 = 𝜌𝑣𝑒∆𝑣𝐸 , ∆𝑣 is the element of volume.

𝐹 ℎ = 𝑞ℎ𝐸 = 𝜌𝑣ℎ∆𝑣𝐸 The energy(work) expended by electric field in moving 𝑞𝑒 by

distance ∆𝑙𝑒 is :

∆𝑤 = 𝐹𝑒. ∆𝑙𝑒 + 𝐹ℎ. ∆𝑙ℎ

• The power measured in watts(W) is:

∆𝑃 =∆𝑤

∆𝑡= 𝐹 𝑒 .

∆𝑙 𝑒

∆𝑡+ 𝐹 ℎ .

∆𝑙 ℎ

∆𝑡

=𝐹 𝑒 . 𝑢𝑒 + 𝐹 ℎ. 𝑢ℎ

= (𝜌𝑣𝑒𝐸. 𝑢𝑒 + 𝜌𝑣ℎ𝐸𝑢ℎ)∆𝑣

=𝐸 . 𝐽 ∆𝑣

𝑃 = 𝐸 . 𝐽 𝑑𝑣 𝑣

(𝑊) (Joule’s law)

• Since 𝑗 = 𝜍𝐸 → 𝑃 = 𝜍 𝐸 2𝑑𝑣𝑣

𝑣 = 𝑙𝐴 → separating the integral:

𝑃 = 𝜍 𝐸 2𝑑𝑣

𝑣 = 𝜍𝐸𝑥𝑑𝑠𝐴 𝐸𝑥𝑑𝑙 𝑙

→ 𝑃 = 𝜍𝐸𝑥𝐴 𝐸𝑥𝑙 = 𝐼𝑉 (𝑊)

• Using 𝑉 = 𝐼𝑅 → 𝑷 = 𝑹𝑰𝟐

• In dielectric (insulator),an externally applied electric

field 𝐸𝑒𝑥𝑡 cannot cause mass migration of charges since

they are not able to move freely, but it can “polarize” the

atoms or molecules in the material by distorting the

center of the cloud and the location of the nucleus.

3-8 Dielectrics

• The “induced” electric field ,called a polarization field,

is weaker than and opposite in direction to 𝐸𝑒𝑥𝑡 .

• Each dipole exhibits a dipole moment. The

materials described here are called a nonpolar

materials. Nonpolar molecules become

polarized only when an external electric field

is applied ,and when the field is terminated, the

molecules return to their original unpolarized

state.

• Since 𝐷 and 𝐸 are related by 𝜀0 in free space, the presence

of microscopic dipoles in dielectric material alters that

relationship in that material to :

𝐃 = 𝛆𝟎𝐄 + 𝐏

,where P is called the “electric polarization field”, accounts

for the polarization properties of material.

• The polarization field is produced by the electric field 𝐸

and depends on the material properties.

• In linear, isotropic and homogeneous media, the polarization field is

directly proportional to 𝐸 and is expressed as :

𝑷 = 𝜺𝟎𝒙𝒆𝑬 ,where 𝑥𝑒 is called electric susceptibility of the material.

*** Note 𝑥𝑒 is a dimensionless quantity.

𝑫 = 𝜺𝟎𝑬 + 𝜺𝟎𝒙𝒆𝑬

= 𝜺𝟎 𝟏 + 𝒙𝒆 𝑬

𝐷 = 𝜺𝑬

• Which defines the permittivity 𝜀 of the material

as

𝜀 = 𝜀0𝜀𝑟 = 𝜀0(1 + 𝑥𝑒)

** For air 𝜀𝑟≅ 1.0006 at see level.

• The polarized atom or molecule may be

represented by an electric dipole consisting of

charge +𝑞 at the center of the nucleus and

charge −𝑞 at the center of electric cloud. Each

such dipole sets up a small electric ,pointing

from the positively charged nucleus to the

center of the equally but negatively charged

electron cloud.

Dielectric Breakdown

• If a dielectric material is placed in a very

strong electric field (exceeds a certain critical

value ,known as the dielectric strength Eds

,electrons can be torn from their corresponding

nuclei causing large currents to flow and

damaging the material. This phenomenon is

called dielectric breakdown.

*** dielectric strength Eds is the highest magnitude of E that the material can

sustain without breakdown.

Dielectric Breakdown (Cont’d)

• The dielectric strength of a material may vary by several orders of magnitude depending on various factors including the exact composition of the material as well as other factors such as temperature and humidity. Some typical values of dielectric strength for some common insulators are:

• Usually dielectric breakdown does not permanently

damage gaseous or liquid dielectrics, but does ruin solid

dielectrics.

Dielectric Breakdown (Cont’d)

• Capacitors typically carry a maximum voltage rating.

Keeping the terminal voltage below this value insures

that the field within the capacitor never exceeds Eds

for the dielectric.

• If 𝑉 is sufficiently large so that 𝐸 exceeds the

dielectric strength of air, ionization occurs and

discharge (lightning) follows.