Intro.1

Reference

Vector Analysis (Self-study)Chapter 2

Static Electric Fields (3 Weeks)Chapter 3.1 OverviewChapter 3.3 Coulomb’s LawChapter 3.4 Gauss’s Law and ApplicationsChapter 3.5 Electric PotentialChapter 3.6 Material Media in Static Electric FieldChapter 3.7 Electric Flux Density and Dielectric ConstantChapter 3.8 Boundary conditions for Electrostatic FieldsChapter 3.9 Capacitance and CapacitorsChapter 3.11 Poisson’s and Laplace’s Equations

Intro.2

III. VECTOR ANALYSIS

3.1 Vector algebra

• addition

• associative law A+(B+C) = (A+B)+C

• commutative law A+B = B+A

• multiplication by scalar

• distributive law a( B+C) = aB + aC

aB B

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• Scalar (or Dot) Product:

A.B = B.A

A.(B+C) = A.B + A.C

ABBA θcos=A.B

• Vector (or Cross) Product: )sin( ABBA θnaBA ≡×

AxB

A an

θAB B

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CB)AC)(BAABABBA××≠××

×≠×−=×(

Note for cross products,

Cartesian Co-ordinate System (unit vectors ax ay az)

zzyyxx BABABA ++=A.B

zyx

zyx

BBBAAA

zyx aaaBA =×

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Differential displacement vector dl

dl = dx ax + dy ay + dz az dl

Example: To integrate F along the path y=x2 from a (1,1) to b (2,4), where

( )( )yxyx aaaaF.dl dydxy

xyb

a

b

a

++=

=

∫∫

2

∫ ∫ =+=2

1

4

1 315dyydx

yx aaF += y

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Differential surface vector dS:

+ve in the ‘outward’ normal direction to the

surface element for an enclosed volume

Example:

Find F.dS for the surface ABCD for the given F in the following unit cube.

dS

A B

az ax F

C D ay

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∫ ∫ ∫ +−=

−=

−+=

1

0

1

0

)2(

2)2(

dydzzyd

dydzdzy

SF.

aS

aaF

x

yx

23

22

1

0

1

0

2−=

+−= ∫ dyzyz

Note the minus sign in dS because positive direction is outward from the enclosed volume

for surface ABCD

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Cylindrical coordinate system

zzxyyx

=

=

+=

−1

22

tanφ

ρ z ϕ x ρ y

φρ

ρφ

φρ

aaa

aaa

aaa

=×

=×

=×

z

z

z

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=

++=

++=

−

−

xy

zyxz

zyxr

1

222

1

222

tan

cos

φ

θ

Spherical coordinate system

z θ r ϕ x y

r

r

r

aaa

aaa

aaa

=×

=×

=×

φθ

θφ

φθ

Intro.10

3.2 Scalar and Vector Calculus

Integrals of scalars and vectors over volumes, surfaces and lines often used in electromagnetics. Examples:

∫V

vdvρ

∫S

dSD.

∫C

dlE.

ρv is the charge density (per unit volume)

D is the electric flux density (or electric displacement)

E is the electric field intensity

Total charge enclosed within volume V

Total electric flux passing through the surface S

Potential difference between two points on the line

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Gradient of scalar field f

zyx aaagradzf

yf

xfff

∂∂

+∂∂

+∂∂

=∇=

Gradf is a vector in the direction of maximum increase of the field f.

namaxl

ff∂∂

=∇

an is unit vector in the direction of maximum increase of f

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Divergence of a vector field A:

v

ddiv S

v ∆≡ ∫

→∆

SA.A

0lim

If we consider A as a flux density (per unit surface area), the closed surface integral represents the net flux leaving the volume ∆v

In rectangular coordinates,

zA

yA

xAdiv zyx

∂∂

+∂

∂+

∂∂

=∇= .AA

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Divergence Theorem:

If A is a vector, then for a volume V surrounded by a closed surface S,

∫ ∫=∇V S

ddv SA..A

The above integral represents the net flex leaving the closed surface S if A is the flux density

V

S

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Curl of a vector field: Curl of a vector A is a measure of the tendency of A to “push or pull” around a closed path that encircles a point. Component of curlA in direction ai

i

C

Sii S

dcurl i

i ∆≡×∇≡

∫→∆

lA.AA

0lim)()(

∆Si Ci ai

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In rectangular coordinates

zyx AAAzyx ∂

∂∂∂

∂∂

=×∇

zyx aaa

A

max

0lim

∆=×∇ ∫

→ S

dC

∆S

lA.aA n

Since the maximum value of any component of a vector is equal to the magnitude of the vector,

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Stokes’s theorem: For an open surface S bounded by a contour C,

∫ ∫=×∇S C

dd lA.S.A )( C S

The line integrals from adjacent cells cancel leaving the only the contribution along the contour C which bounds the surface S.

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IV. ELECTROSTATIC FIELDS (Time-Invariant)

The electric field intensity E is defined as the force on an unit positive charge at a point.

D is the electric flux density (or electric displacement density). The direction of D is that of the electric flux at the point, and its magnitude is the no.of flux lines passing through an unit normalsurface area.

4.1 Gauss’s Law :

∫ =

=∇

S

v

QdSD.

.D ρ Point form

Integral form

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ρv is the space volume charge density (coulomb per unit volume) at a point. Q is the total charge enclosed within a closed surface S. Unit of D is C/m2.

Relation between D and E:

EED εεε == ro

εo is permittivity of free space (vacuum), εr is the relative permittivity (or dielectric constant) of the medium material, and ε is the permittivity of the medium material. (Unit of εo andε is Farad/metre)

In free space,

ED o

r

εε

== 1

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• Gauss’s law can determine the electric field pattern due to anydistribution of charges.

• In the simplest case, consider the field at a point P of distance rfrom a point charge of magnitude Q in free space.

P r +Q

Construct a spherical surface with centre at Q and radius r. E is the same everywhere on the surface. Applying integral form of Gauss’s law gives

raE

E

2

2

4

4

rQ

Qr

o

o

πε

επ

=

=

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The force acting a test charge q at point P is then (Coulomb’s law):

raF 241

rqQ

oπε=

Example : Find the electric field due to a spherical charge distribution of radius a with uniform volume charge density ρv in free space.

Gaussian surface of radius r a

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Case (i): For a Gaussian surface of radius r less than the radius a of the charge distribution, the total charge enclosed by the Gaussian surface is:

vrQ ρπ 3

34

=

raE

E

o

o

rr

Qrr

ερ

επ

3)(

)(4 2

=

=

Applying Gauss’s Law to the Gaussian surface,

Case (ii): For a Gaussian surface of radius r greater than a :

ρπ 334 aQ =

Solution:

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4.2 Electric (electrostatic) potential V:

The electrostatic field is conservative, i.e.

∫ =

=×∇

Cd 0

0

lE.

E

Applying Gauss’s law,

raE

E

2

3

2

3)(

)(4

rar

Qrr

o

o

ερ

επ

=

=

According to vector identity, for any scalar V

0)( ≡∇×∇ V

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• So we can define a scalar electric potential V which is easier to work with, V−∇=E

Integrating along a path ∫−=−b

aab dVV lE.

• Note that the integral is independent of the path taken. Physically the potential difference is equal to the work done in moving an unit charge from point A to point B.

• Potential due to N point charges: (self-reading)

∑= −

=N

k

k

o

QV14

1

krrπε

(Hence unit of E is volt/metre)

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4.3 Capacitance

Capacitance between two conductors is defined as:

VQC =

where Q is the charge on the conductor and V is the potential difference between the conductors.

Energy stored in the capacitor = 2

21 CV

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Example: Find the potential difference between two coaxial cyclinders of radius a, b in the following diagram. Assume the surface charge density (per unit area) on the inner conductor is ρs. Hence find the capacitance per unit length of the cylinder.

ρs r b

Construct a Gaussian cylindrical surface of unit length and radius r. The E-field at radius r is:

raE

E

rar

arr

s

s

ερ

επρπ

=

=

)(

2)(2

Solution:

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The potential difference Vab between outer and inner conductor:

∫−=−=a

bbaab drVVV rE ).(

=−= ∫ a

bar

adr sa

b

s lnε

ρε

ρ

Capacitance per unit lengthba

s

VaC ρπ2

=

=

abln

2πε

Intro.27

4.4 Poisson’s and Laplace’s equations:

• used to determine the electrostatic potential based on the charge distribution

• Substituting into Gauss’s law V−∇=Eερ v=⋅∇ E

yields the Poisson’s equation:

ερ v

zV

yV

xV

−=∂∂

+∂∂

+∂∂

2

2

2

2

2

2

ερ vV −=∇ 2

In rectangular co-ordinates,

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• Possion’s and Laplace’s equations can be applied to solve complex electrostatic problems. These equations can be solved either by analytical or numerical methods if we provide the boundary values (of the potential or electric field).

• A special case when space is free of charge (Laplace’s equation)

02 =∇ V

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• Product Solution of Laplace Equation

Assume a particular solution V can be expressed as the product of a function of x alone and a function of y alone:

XYV =

22

2

22

2

1

1

α

α

=−

=

dyYd

Y

dxXd

XBy separating the variables,

The constants A,B,C,D are fixed by the boundary conditions. The general solution is the superimposition of all possible particular solutions.

)sincos)(sinhcosh( yDyCxBxAV αααα ++=

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Example: study particular solutions of the form A = C = 0, BD = V1, so that

yxVV αα sinsinh1=

This solution satisfies the boundary conditions: V = 0 at x = 0, y = 0, y = nπ/α; and V = V1 along the curve

1sinsinh =yx αα

V=V1

V=0

V=0

V=0

x

y

Further Reading:

Series solution (Hayt7.5)

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4.5 Dielectric material

• Dielectrics are insulating materials, and contain bound charges which cannot move freely to generate currents.

• Bound charges can move short distances under an electric field to form electric dipoles. (A dipole is a pair of equal and oppositecharges (+Q,-Q) separated at a small distance r. The dipole moment p is equal to Qrar. )

t E

p +ve -ve

Zero field Under a field

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• From a macroscopic view point, all the minute dipole moments pk add up to result in a net dipole moment per unit volume called the polarization vector P (C/m2):

v

vN

k

v ∆=

∑∆

=

→∆

1

0lim

kpP

where N is the number of dipoles per unit volume.

• The polarization (or bound) charges can also contribute to theelectric field E as the free charges. (Note ρv defined before is free charge density.) It can be shown that D is related to E and P according to the following equation:

volume ∆v pk

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PED += oε

• In most materials, P is in the same direction and proportional to E (P = εoχE), so that

EE εεε == ro

where εo is the permittivity of vacuum, εr is the relative permittivity (or dielectric constant) of the material, ε is the permittivity of the material, and χ is the susceptibility of the material.

EED χεε oo +=

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4.6 Presence of a conductor

Electric current will flow in a conductor when there is an electric field. The current density J (A/m2) is proportional to the electric field intensity E.

where σ is the conductivity of the material. This is another form of Ohm’s Law. The total current I flowing through a surface S is given by:

EJ σ=

∫=S

dI SJ.

In the interior of a perfect conductor, σ is very large so that E can be assumed to be zero.

Intro.35

END