EKT241 - Electromagnetic Theory Chapter 3 - Electrostatics

EKT241 - Electromagnetic TheoryEKT241 - Electromagnetic TheoryChapter 3 - ElectrostaticsChapter 3 - Electrostatics

Chapter ObjectivesChapter Objectives

Application of Maxwell’s equations Using of Coulomb’s law Using of Gauss’s law Conductivities of metals Boundary conditions that leads to tangential

components of D, E, and J Capacitance of a two-conductor body Electrostatic energy density stored in a

dielectric medium

Chapter OutlineChapter Outline

Maxwell’s EquationsCharge and Current DistributionsCoulomb’s LawGauss’s LawElectric Scalar PotentialElectrical Properties of MaterialsConductorsDielectricsElectric Boundary ConditionsCapacitanceElectrostatic Potential EnergyImage Method

3-1)3-2)3-3)3-4)3-5)3-6)3-7)3-8)

3-9)3-

10)3-11)3-12)

Maxwell EquationsMaxwell Equations• Maxwell Equations consist of four differential equations, • (1) the law of electro-magnetic induction (Faraday), • (2) Ampere-Maxwell's law,• (3) Gauss's law of electric field, and

• (4) Gauss's law of magnetic field.

(1) rot E = -∂B/∂t (2) rot H = (∂D/∂t) + i

where D is electric flux B is magnetic flux density and i is electric current

density which will be explained later. The latter two equations are expressed

in terms of an operation called "divergence" of D and of B.

(3) div D = ρ (4) div B = 0

where ρ is electric charge density

t

t

DHH

0B

BE

D v

3-1 3-1 Maxwell’s EquationsMaxwell’s Equations

• Maxwell’s equations:

where E, D = electric field quantities ε = electrical permittivity of the material

µ = magnetic permeability of the material ρv = electric charge density per unit volume

J = current density per unit area

Relationship:

D = εE

B = µH

Those equations describe the interrelationship between electric field, magnetic field, electric charge, and electric current.

3-1 3-1 Maxwell’s EquationsMaxwell’s Equations

• In static case, ∂/∂t = 0.• Maxwell’s equations is reduced toElectrostatics

Magnetostatics

0

E

vD

JH

B

0

Maxwell’s EquationsMaxwell’s Equations

Gauss' Law for ElectricityGauss' Law for Electricity

• The electric flux out of any closed surface is proportional to the total charge enclosed within the surface.

• The integral form of Gauss' Law finds application in calculating electric fields around charged objects.

• In applying Gauss' law to the electric field of a point charge, one can show that it is consistent with Coulomb's law.

• While the area integral of the electric field gives a measure of the net charge enclosed, the divergence of the electric field gives a measure of the density of sources.

Gauss' Law for MagnetismGauss' Law for Magnetism

• The net magnetic flux out of any closed surface is zero. This amounts to a statement about the sources of magnetic field. For a magnetic dipole, any closed surface the magnetic flux directed inward toward the south pole will equal the flux outward from the north pole. The net flux will always be zero for dipole sources. If there were a magnetic monopole source, this would give a non-zero area integral.

Faraday's Law of InductionFaraday's Law of Induction• The line integral of

the electric field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.

• This line integral is equal to the generated voltage or emf in the loop, so Faraday's law is the basis for electric generators. It also forms the basis for inductors and transformers.

Ampere's LawAmpere's Law

• In the case of static electric field, the line integral of the magnetic field around a closed loop is proportional to the electric current flowing through the loop. This is useful for the calculation of magnetic field for simple geometries.

3-2 Charge and Current Distributions3-2 Charge and Current Distributions

• Charge may be distributed over a volume, a surface or a line.

• Volume charge density ρv is defined as

• Total charge Q volume V is given by

3-2.1 Charge Densities3-2.1 Charge Densities

3

0C/m lim

dv

dq

v

qv

v

C dvQv

v

3-2.1 Charge Densities3-2.1 Charge Densities

• Surface charge density

• Line charge density

C/m lim

C/m lim

0

2

0

dl

dq

l

q

ds

dq

s

q

ll

ss

Example 3.1 Line Charge DistributionExample 3.1 Line Charge Distribution

Calculate the total charge Q contained in a cylindrical tube of charge oriented along the z-axis. The line charge density is , where z is the distance in meters from the bottom end of the tube. The tube length is 10 cm.

The total charge Q is

C 102 21.0

02

1.0

0

1.0

0

zzdzdzQ l

zl 2

SolutionSolution

3-2.2 Current Densities3-2.2 Current Densities

• Current density is defined as

• For surface S, total current flowing through is

• There are 2 types of current:1) Convection current (generated by actual movement

of electrically charged matter; does not obey Ohm’s law)

2) Conduction current (atoms of conducting material do not move; obeys Ohm’s law)

2A/m uJ v

A dsJIS

3-3 Coulomb’s Law3-3 Coulomb’s Law

• Coulomb’s law states that:

where = unit vector from p to q

• For an electric field E at a given point in space,

where D = εE ε = εR ε0

ε0 = 8.85 × 10−12 ≈ (1/36π) × 10−9 (F/m)

(V/m) 4

RE2R

q

R

(N) E'F q

3-3.1 Electric Field due to Multiple Point 3-3.1 Electric Field due to Multiple Point ChargesCharges

• Total electric field E at any point in space is

• In general for case of N point of charges,

3

2

223

1

11

0 RR

R-R

RR

R-R

4

1E qq

V/m 4

1

13

N

i i

ii

RR

RRqE

Example 3.3 Electric Field due to Two Point Example 3.3 Electric Field due to Two Point ChargesCharges

Two point charges with and are located in free space at (1, 3,−1) and (−3, 1,−2), respectively, in a Cartesian coordinate system.

Find (a) the electric field E at (3, 1,−2) and (b) the force on a 8 × 10−5 C charge located at that point. All distances are in meters.

C102 51

q C104 52

q

Solution 3.3 Electric Field due to Two Point Solution 3.3 Electric Field due to Two Point ChargesCharges

The electric field E with ε = ε0 (free space) is given by

The vectors are

a) Hence,

b) We have

V/m 10108

2ˆ4ˆˆ 5

0

zyxE

2ˆˆ3ˆ ,2ˆˆ3ˆ ,ˆ3ˆˆ 21 zyxRzyxRzyxR

3

2

223

1

11

0 RR

R-R

RR

R-R

4

1E qq

N 1027

4ˆ8ˆ2ˆ10

108

2ˆ4ˆˆ108 10

0

5

0

53

zyxzyxEqF

3-3.2 Electric Field due to Multiple Point 3-3.2 Electric Field due to Multiple Point ChargesCharges

• Total electric field due to 3 types of distribution:

on)distributi (line '

'ˆ4

1

on)distributi (surface '

'ˆ4

1

on)distributi (volume '

'ˆ4

1

'2

'

'2

'

'2

l

l

l

S

s

S

v

v

v

R

dlRdEE

R

dsRdEE

R

dvRdEE

Example 3.5 Electric Field of a Circular Disk of Example 3.5 Electric Field of a Circular Disk of ChargeCharge

Find the electric field at a point P(0, 0, h) in free spaceat a height h on the z-axis due to a circular disk of charge in the x–y plane with uniform charge density ρs as shown, and then evaluate E for the infinite-sheet case by letting a→∞.

solution 3.5 Electric Field of a Circular Disk of solution 3.5 Electric Field of a Circular Disk of ChargeCharge

A ring of radius r and width dr has an area ds = 2πrdr

The charge is

The field due to the ring is

The total field at P is

For an infinite sheet of charge with a =∞,

rdrdsdq ss 2

rdrhr

hzdE s

2

4ˆ

2/3220

22

002/322

0

12

ˆ2

ˆha

hz

hr

rdrhzE s

as

charge ofsheet infinite 2

ˆ0

szE

3-4 Gauss’s Law3-4 Gauss’s Law

• Gauss’s law states that D through a surface is proportional to enclosed Q.

• Differential and integral form of Gauss’s law is defined as

law sGauss' vρD

law sGauss' Q sD S

d

Example 3.6 Electric Field of a Line of Example 3.6 Electric Field of a Line of ChargeCharge

Use Gauss’s law to obtain an expression for E in freespace due to an infinitely long line of charge with uniform charge density ρl along the z-axis.

Construct a cylindrical Gaussian surface and the integral is

which gives the result

SolutionSolution

hrπhDhdzrdrDr l

h

z

rlr

0

2

0

2 or ˆˆ

charge of line infinite 2

ˆˆ000 r

rD

rD

E lr

3-5 Electric Scalar Potential3-5 Electric Scalar Potential

• Electric potential energy is required to move unit charge between the 2 points.

• Potential difference along any path is obtained by

3-5.1 Electric Potential as a Function of 3-5.1 Electric Potential as a Function of Electric FieldElectric Field

2

1

1221

P

P

dlEVVV

3-5.1 Electric Potential as a Function of Electric 3-5.1 Electric Potential as a Function of Electric FieldField

• Kirchhoff’s voltage law states that net voltage drop around a closed loop is zero.

• Line integral E around closed contour C is

• The electric potential V at any point is given by

ticsElectrosta 0C

dlE

(V) 2

1

P

P

dlEV

3-5.2 Electric Potential due to Point 3-5.2 Electric Potential due to Point ChargesCharges

• For N discrete point charges, electric potential is

• For a continuous charge distribution, we have

3-5.3 Electric Potential due to Continuous 3-5.3 Electric Potential due to Continuous DistributionsDistributions

on)distributi (line ''4

1)R(

on)distributi (surface ''4

1)R(

on)distributi (volume ''4

1)R(

'

'

'

l

l

S

s

v

v

dlR

V

dsR

V

dvR

V

V 4

1

1

N

i i

i

RR

qRV

3-5.4 Electric Field as a Function 3-5.4 Electric Field as a Function of Electric of Electric PotentialPotential• To find E for any charge distribution easily,

• Poisson’s and Laplace’s equations are use to find V where boundaries are known.

3-5.5 Poisson’s Equation3-5.5 Poisson’s Equation

VE

equation sPoisson' 2

VV

equation sLaplace' 02 V

3-6 Electrical Properties of Materials3-6 Electrical Properties of Materials

• Different materials have different conductivity.

3-7 Conductors3-7 Conductors

• Conductivity of the material, σ, is defined as

where µe = electron mobility (m/s)

µh = hole mobility (m/s)

Ne = number of free electrons

Nh = number of free holes

e = 1.6 × 10−19 C (absolute charge)

torsemiconduc S/m e μNμN

μρμ- ρσ

hhee

hvheve

conductor S/m eμNμ ρ eeeve

3-7 Conductors3-7 Conductors

• Point form of Ohm’s law states that

• Properties for perfect dielectric and conductor:

law sOhm' A/m 2EJ

Jofregardless,0 :with conductor Perfect

Eofregardless,0 :0 with dielectricPerfect

E

J

Example 3.8 Conduction Current in a Copper Example 3.8 Conduction Current in a Copper WireWire

A 2-mm-diameter copper wire with conductivity of5.8 × 107 S/m and electron mobility of 0.0032 (m2/V·s)is subjected to an electric field of 20 (mV/m).

Find (a) the volume charge density of free electrons, (b) the current density, (c) the current flowing in the wire, (d) the electron drift velocity, and (e) the volume density of free electrons.

Solution 3.8 Conduction Current in a Copper Solution 3.8 Conduction Current in a Copper WireWire

a)

b)

c)

d)

e)

3107

C/m 1081.10032.0

108.5

eve

2637 A/m 1016.11020108.5 EJ

A 64.34

1041016.1

66

JAI

m/s 104.610200032.0 53 Eu ee

32919

10

melectrons/ 1013.1106.1

1081.1

eN ve

e

3-7.1 Resistance3-7.1 Resistance

• For resistor of arbitrary shape, resistance R is

• The reciprocal of R is called the conductance G, siemensor S

1

l

A

RG

S

l

S

l

dsE

dlE

dsJ

dlE

I

VR

3-7.2 Joule’s Law3-7.2 Joule’s Law

• Joule’s law states that for a volume v, the total dissipated power is

• Conductor has free electrons.• Dielectric electrons are strongly bounded.• When E exceeds critical value, the material

encounter dielectric breakdown.

3-8 Dielectrics3-8 Dielectrics

law sJoule' W v

JdvEP

3-9 Electric Boundary Conditions3-9 Electric Boundary Conditions

• Tangential component Et of the electric field is defined as

• Boundary condition on the tangential component of Dt is

V/m 21 tt EE

2

2

1

1

tt DD

3-9.1 Dielectric–Conductor Boundary3-9.1 Dielectric–Conductor Boundary

• Boundary condition at conductor surface is

where = normal vector pointing outward

• Electric field lines point away surface when ρs is positive.

n

surfaceconductor at ˆ111 snED

3-9.2 Conductor–Conductor Boundary3-9.2 Conductor–Conductor Boundary

• The normal component of J has to be continuous across the boundary between 2 different media under electrostatic conditions.

• Setting J1n = J2n, we have

ticselectrosta 2

2

1

11 sn ρJ

3-10 Capacitance3-10 Capacitance

• Capacitance is defined as

where V = potential difference (V)

Q = charge (C) C = capacitance (F, farads)

• For a medium with uniform σ and ε,

For C/V V

QC

RC

Example 3.11 Capacitance and Breakdown Voltage of Example 3.11 Capacitance and Breakdown Voltage of Parallel-Plate Parallel-Plate CapacitorCapacitor

Obtain an expression for the capacitance C of a parallel-plate capacitor comprised of two parallel plates each of surface area A and separated by a distance d. The capacitor is filled with a dielectric material with permittivity ε. Also, determine the breakdown voltage if d = 1 cm and the dielectric material is quartz.

Solution 3.11 Capacitance and Breakdown Voltage of Solution 3.11 Capacitance and Breakdown Voltage of Parallel-Plate Parallel-Plate CapacitorCapacitor

The charge density on the upper plate is ρs = Q/A. Hence,

The voltage difference is

Hence, the breakdown voltage is

AQEEzE s / whereˆ

EddzzEzdlEVdd

00

ˆˆ

V 103101030 526 dEV dsbr

3-11 Electrostatic Potential Energy3-11 Electrostatic Potential Energy

• When q travels to a given point, electrostatic potential energy We will be stored.

where we = electrostatic potential energy per unit volume

• For parallel-plate capacitor, the plates are attracted by an electrical force F,

32 J/m 2

1E

v

Ww e

e

N eWF

3-12 Image Method3-12 Image Method

• Image theory states that a charge Q above a grounded perfectly conducting plane is equal to Q and its image –Q with ground plane removed.

Example 3.13 Image Method for Charge Example 3.13 Image Method for Charge above above Conducting PlaneConducting Plane

Use image theory to determine V and E at an arbitrarypoint P(x, y, z) in the region z > 0 due to a charge Q infree space at a distance d above a grounded conductingplane.

From charge Q (0, 0, d) and its image −Q (0, 0,−d) in Cartesian coordinates, we have

SolutionSolution

2/3222

2/3222

032

231

1

0 zyx

zyx

44

1

dzyx

dzyx

dzyx

dzyx

Q

R

QR

R

QRE