E.1

Static Electric Fields

Coulomb’s Law

Gauss’s Law

Electric Potential

Electrical Properties of Materials– Conductors– Dielectrics

Capacitance

24ˆ

Rq

πεRΕ =

vρ=⋅∇ D

V−∇=E

EJ σ=ED ε=

E.2

Introduction

Example: Electric field due to a chargeFrom the field distribution, we can guess that the divergence ofthe electric field is non-zero and the curl of the electric field is zero.

vρ=⋅∇ D

0=×∇ E

E.3

Coulomb’s Law

Coulomb’s law1. An isolated charge q induces an electric field E at every

point in space and is given by

ε is the electrical permittivity of the medium

mVR

q /4

ˆ2πε

RΕ =

E.4

Coulomb’s Law

2. In the presence of an electric field E at a give point in space, the force acting on a test charge q’ is

For a material with electrical permittivity ε, we have

Nq ΕF '=

EED orεεε ==

mFo /10854.8 12−×=ε

E.5

Electric Field due to Charges

The expression for the electric field due to a single charge canbe extended to find the field due to multiple point charges as well as continuous charge distributions.

The electric field at position vector R caused by charges q1, q2,…, qN located at points with position vectors R1, R2,…, RN is given by

Note: is the unit vector from qi to the observation point.

Demonstrations: D4.1, D4.2, M4.1 Q1-2, M4.2 Q1, M4.3 Q1-2

mVqN

i i

i

i

i /)(4

11

2∑= −

−

−=

RRRR

RRΕ

πε

ii RRRR −− )(

E.6

Electric Field due to Charges

Electric field due to a volume charge distribution

∫∫ ==⇒

==

vv

v

v

Rdvd

mVRdv

Rdqd

2/

22

''ˆ

41

/'4'ˆ

'4'ˆ

ρπε

περ

πε

'RΕΕ

'RRΕ

E.7

Electric Field due to Charges

The electric field due to charges on a surface is

The electric field due to charges on a line is

∫=' 2'

'ˆ4

1S

s

Rdsρ

πε'RΕ

∫=' 2'

'ˆ4

1l

l

Rdlρ

πε'RΕ

E.8

Gauss’s Law

Gauss’s Law states that the outward flux of D through a surface is proportional to the enclosed charged Q.

∫ =⋅

=⋅∇

S

v

QdSD

D ρ Differential form

Integral form

E.9

Example: point charge

Construct a spherical surface with centre at q and radius R. Electric field is the same everywhere on the surface. Applying integral form of Gauss’s law gives

RDE

sRsD

ˆ4

/

4

ˆ

2

2

Rq

qDR

dsD

dDd

R

s R

s Rs

πεε

π

==∴

==

=

⋅=⋅

∫∫∫

E.10

Example: infinite long line of charge

Since the line of charge is infinite in extent and is along the z-axis, D must be in the radial r-direction and must not depend on or z.

rDE

rrsD

ˆ2

/

2

ˆˆ0

2

0

r

hrhD

dzrdDd

l

lr

h

z rs

περε

ρπ

φπ

φ

==∴

==

⋅=⋅ ∫ ∫∫ = =

E.11

Example: Spherical charge distribution

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

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

=

Applying Gauss’s Law to the Gaussian surface,

rDE

srsD

ˆ3

/

4

ˆ2

o

vo

r

s Rs

rQDr

dDd

ερε

π

==∴

==

⋅=⋅ ∫∫

E.12

Example: Spherical charge distribution

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

vaQ ρπ 3

34

=

Applying Gauss’s law,

rDE

srsD

ˆ3

/

4

ˆ

2

3

2

ra

QDr

dDd

o

vo

r

s Rs

ερε

π

==∴

==

⋅=⋅ ∫∫

Demonstrations: D4.9, D4.10, M4.6 Q2-3

E.13

Electric Potential

∫ =

=×∇

Cd 0

0

lE.

E

According to vector identity, for any scalar V

0)( ≡∇×∇ V

The electrostatic field is irrotational or conservative, i.e.

So we can define a scalar electric potential V

∫∞ ⋅−=⇒

−∇=P

dV

V

lE

EThe integral is independent of the path taken

E.14

Electric Potential

Electric potential due to point charges:

Electric potential due to volume distribution

Electric potential due to surface distribution

Electric potential due to line distribution

∑= −

=N

i i

iqV14

1RRπε

∫='

''4

1v

v dvR

V ρπε

∫='

''4

1S

s dsR

V ρπε

∫='

''4

1l

l dlR

V ρπε

Demonstrations: M4.1 Q.3 M4.2 Q.2, M4.6 Q.1

E.15

Electrical Properties of Materials

The electromagnetic constitutive parameters of a material medium are – Electrical permittivity ε– Electrical permeability µ– Conductivity σ

Throughout this subject, we assume the materials are– Homogenous

• Constitutive parameters do not vary from point to point– Isotropic

• Constitutive parameters are independent of direction

E.16

Conductors

A conductor has a large number of loosely attached electrons in the outermost shells of the atoms.

Upon applying an external electric field, the electrons migrate from one atom to the next along a direction opposite that of theexternal field.

Their movement gives rise to a conduction current

– Point form of Ohm’s law

EJ σ=

E.17

Conductors

Conductivity of some common materials

Perfect dielectric: σ=0 and then J = 0 regardless of E

Perfect conductor:σ=∞ and then E = 0 regardless of J

E.18

Dielectrics

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.

E.19

Dielectrics

A dielectric medium polarized by an external electric field

E.20

Dielectrics

The polarization (or bound) charges can also contribute to the electric field E as the free charges.

It can be shown that D is related to E and P according to the following equation:

PED += oεIn most materials, P is in the same direction and proportional to E 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 +=

E.21

Dielectric Strength

If the applied electric field exceeds the dielectric strength of the material, it will free the electronics completely from the molecules in the form of a conduction current. – Spark may occur– Dielectric breakdown

E.22

Capacitance

Capacitance between two conductors is defined as:

)(FVQC =

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

E.23

Example 4-11: Parallel-Plate Capacitor

Because of the applied voltage difference, charge +Qaccumulates uniformly on the top plate and –Q accumulates uniformly on the lower plate.In the dielectric medium between the plates, the charges induces a uniform electric field (fringing field is ignored).

E.24

Example 4-11: Parallel-Plate Capacitor

Applying Gauss’s law, the electric field can be determined:

AQzzEz

s

εερ

/ˆ/ˆ

ˆ

−=−=−=E

( )

dA

dAQQ

EdQ

d

QVQC

d

εε

=

==

⋅−=

=

∫

/

0lE

E.25

Example: 4-12: Coaxial Line

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

rlQ

QrlDd rs

πεε

π

2ˆ/

2

rDE

sD

−==⇒

−==⋅∫

E.26

Example: 4-12: Coaxial Line

( )abl

Q

drrl

Q

dV

b

a

b

a

/ln2

ˆ2

ˆ

πε

πε

=

⋅

−−=

⋅−=

∫

∫rr

lE

( )abl

VQC

/ln2πε

=

=∴

E.27

Exercise