3. ELECTROSTATICSRuzelita Ngadiran

Chapter 4 Overview

Maxwell’s equations

Where;

E = electric field intensityD = electric flux densityρv = electric charge density

per unit volumeH = magnetic field intensityB = magnetic flux densityt

t

DJH

0B

BE

D v

Maxwell’s equations:3

Maxwell’s Equations

God said:

And there was light!

Maxwell’s equations

Relationship:

D = ε E

B = µ H

ε = electrical permittivity of the material

µ = magnetic permeability of the material

Maxwell’s equations:

t

t

DJH

0B

BE

D v

5

Maxwell’s equations

For static case, ∂/∂t = 0. Maxwell’s equations is reduced to:

Electrostatics Magnetostatics

0

E

D vJH

B

0

6

Charge and current distributions

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

Electric field due to continuous charge distributions:

7

Charge and current distributions

3

0C/m lim

dv

dq

v

qv

v

C VdQv

v

Volume charge density, ρv is defined as:

Total charge Q contained in a volume V is:

8

Charge and current distributions

Surface charge density

2

0C/m lim

ds

dq

s

qs

s

• Total charge Q on a surface:

C SdQS

S

9

Charge and current distributions

Line charge density

C/m lim0 dl

dq

l

ql

l

• Total charge Q along a line

C ldQl

l

10

Charge Distributions

Volume charge density:

Total Charge in a Volume

Surface and Line Charge Densities

Example 1

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.

zl 2

12

Solution to Example 1

The total charge Q is:

C 01.021.0

02

1.0

0

1.0

0

zzdzdzQ l

13

Example 2

Find the total charge

over the volume with

volume charge

density:

3105

5m

Ce zV

14

Solution to Example 2

The total charge Q:

C

dzdde

dVQ

z

z

V

V

10854.7

5

14

01.0

0

2

0

04.0

02.0

105

15

Current Density

For a surface with any orientation:

J is called the current density

Convection vs. Conduction

Coulomb’s Law

Electric field at point P due to single charge

Electric force on a test charge placed at P

Electric flux density D

For acting on a charge

For a material with electrical permittivity, ε:

D = ε E

where:ε = εR ε0

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

For most material and under most condition, ε is constant, independent of the magnitude and direction of E

19

At point P, the electric field E1 due to q1 alone:

At point P, the electric field E1 due to q2 alone:

(V/m)

4

-3

1

111

RR

RRE

q

(V/m)

4

-3

2

222

RR

RRE

q

E-field due to multipoint charges

20

Electric Field Due to 2 Charges

Example 3

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)(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

22

Solution to Example 3

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

The vectors are:

2ˆˆ3ˆ ,2ˆˆ3ˆ ,ˆ3ˆˆ 21 zyxRzyxRzyxR

32

223

1

11

21

--

4

1

RR

RR

RR

RR

EEE

qq

23

Solution to Example 3

a) Hence,

b) We have

V/m 10108

2ˆ4ˆˆ 5

0

zyx

E

N 1027

4ˆ8ˆ2ˆ10

108

2ˆ4ˆˆ108 10

0

5

0

53

zyxzyxEF q

24

Electric Field Due to Charge Distributions

Field due to:

Cont.

Example 4

Find the electric field at a point P(0, 0, h) in free space at 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. Then evaluate E for the infinite-sheet case by letting a→∞.

28

Solution to Example 4

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

The charge is:

The field due to the ring is:

dqrdrds ss 2

rdr

hr

hdE s

2

4ˆ

2/3220

z

29

Solution to Example 4 The total electric field at P is

With plus sign corresponds to h>0, minus sign corresponds to h<0.

For an infinite sheet of charge with a =∞,

22

002/322

0

12

ˆ2

ˆha

h

hr

rdrh sa

s

zzE

charge ofsheet infinite 2

ˆ0

szE

30

Gauss’s Law

Application of the divergence theorem gives:

Example 5

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

32

Solution to Example 5

Construct a cylindrical Gaussian surface.The integral is:

Equating both equations, and re-arrange, we get:

(1) .... 2

ˆˆ0

2

0

rhDQ

dzrdDQ

r

h

z

r

rr

(2) .... But hρQ l

rD l

r 2

33

Solution to Example 5

Then, use , we get:

Note: unit vector is inserted for E due to the fact that E is a vector in direction.

charge of line infinite 2

ˆˆ000 r

D lr

rr

DE

space freefor 0ED

r̂r̂

34

Applying Gauss’s Law

Construct an imaginary Gaussian cylinder of radius r and height h:

Electric Scalar Potential

Minimum force needed to move charge against E field:

Electric Scalar Potential

Electric Potential Due to Charges

In electric circuits, we usually select a convenient node that we call ground and assign it zero reference voltage. In free space and material media, we choose infinity as reference with V = 0. Hence, at a point P

For a point charge, V at range R is:

For continuous charge distributions:

Relating E to V

Cont.

(cont.)

Poisson’s & Laplace’s Equations

In the absence of charges:

Conductivity

Conductivity – characterizes the ease with which charges can move freely in a material.

Perfect dielectric, σ = 0. Charges do not move inside the material

Perfect conductor, σ = ∞. Charges move freely throughout the material

43

Conductivity Drift velocity of electrons, in a conducting

material is in the opposite direction to the externally applied electric field E:

Hole drift velocity, is in the same direction as the applied electric field E:

where: µe = electron mobility (m2/V.s)

µh = hole mobility (m2/V.s)

(m/s) Eu ee

(m/s) Eu hh

eu

hu

44

Conductivity

Conductivity of a material, σ, is defined as:

where ρve = volume charge density of free electrons

ρvh = volume charge density of free holes

Ne = number of free electrons per unit volume

Nh = number of free holes per unit volume e = absolute charge = 1.6 × 10−19 (C)

torsemiconduc S/m e μNμN

μρμ- ρσ

hhee

hvheve

conductor S/m eμNμ ρ eeeve

45

Conductivity

Conductivities of different materials:

46

Conductivity

ve = volume charge density of electronshe = volume charge density of holese = electron mobilityh = hole mobilityNe = number of electrons per unit volumeNh = number of holes per unit volume

Conduction Current

Conduction current density:

Note how wide the range is, over 24 orders of magnitude

Resistance

For any conductor:

Longitudinal Resistor

G’=0 if the insulating material is air or a perfect dielectric with zero conductivity.

Joule’s Law

The power dissipated in a volume containing electric field E and current density J is:

For a resistor, Joule’s law reduces to:

For a coaxial cable:

Piezoresistivity

The Greek word piezein means to press

R0 = resistance when F = 0F = applied forceA0 = cross-section when F = 0 = piezoresistive coefficient of material

Piezoresistors

Wheatstone Bridge

Wheatstone bridge is a high sensitivity circuit for measuring small changes in resistance

Dielectric Materials

Polarization Field

P = electric flux density induced by E

Electric Breakdown

Electric Breakdown

Boundary Conditions

Summary of Boundary Conditions

Remember E = 0 in a good conductor

Conductors

Net electric field inside a conductor is zero

Field Lines at Conductor Boundary

At conductor boundary, E field direction is always perpendicular to conductor surface

Capacitance

Capacitance

For any two-conductor configuration:

For any resistor:

Application of Gauss’s law gives:

Q is total charge on inside of outer cylinder, and –Q is on outside surface of inner cylinder

Tech Brief 8: Supercapacitors

For a traditional parallel-plate capacitor, what is the maximum attainable energy density?

= permittivity of insulation materialV = applied voltage = density of insulation materiald = separation between plates

Mica has one of the highest dielectric strengths ~2 x 10**8 V/m. If we select a voltage rating of 1 V and a breakdown voltage of 2 V (50% safety), this will require that d be no smaller than 10 nm. For mica, = 60 and = 3 x 10**3 kg/m3 .

Hence:

W = 90 J/kg = 2.5 x10**‒2 Wh/kg.

By comparison, a lithium-ion battery has W = 1.5 x 10**2 Wh/kg, almost 4 orders of magnitude greater

Energy density is given by:

A supercapacitor is a “hybrid” battery/capacitor

Users of Supercapacitors

Energy Comparison

Electrostatic Potential Energy

Electrostatic potential energy density (Joules/volume)

Total electrostatic energy stored in a volume

Energy stored in a capacitor

Image Method

Image method simplifies calculation for E and V due to charges near conducting planes.

1.For each charge Q, add an image charge –Q2.Remove conducting plane3.Calculate field due to all charges