part 1:electrostatics

PART 1

ELECTROSTATICS

2

ELECTROSTATICS

2.1 COULUMB’S LAW

2.2 ELECTRIC FIELD INTENSITY

2.3 LINE, SURFACE & VOLUME CHARGES

2.4 ELECTRIC FLUX DENSITY

2.5 GAUSS’S LAW

2.6 ELECTRIC POTENTIAL

2.7 BOUNDARY CONDITIONS

2.8 CAPACITANCE

3

INTRODUCTION

Electromagnetics is the study of the effect of charges at rest and charges in motion.

Some special cases of electromagnetics:

Electrostatics: charges at rest

Magnetostatics: charges in steady motion (DC)

Electromagnetic waves: waves excited by charges in time-varying motion

4

Maxwell’sequations

Fundamental laws of classical electromagnetics

Special cases

Electro-statics

Magneto-statics

Electro-magnetic

waves

Kirchoff’s Laws

Statics: 0t

d

Geometric Optics

TransmissionLine

TheoryCircuitTheory

Input from other

disciplines

INTRODUCTION (Cont’d)

• Electrical phenomena caused by friction

are part of our everyday lives, and can be

understood in terms of electrical charge.

• The effects of electrical charge can be

observed in the attraction/repulsion of

various objects when “charged.”

5


• Charge comes in two varieties called “positive” and “negative.”

• Objects carrying a net positive charge attract those carrying a net negative charge and repel those carrying a net positive charge.

• Objects carrying a net negative charge attract those carrying a net positive charge and repel those carrying a net negative charge.

• On an atomic scale, electrons are negatively charged and nuclei are positively charged.

6


• Electric charge is inherently quantized such that the charge on any object is an integer multiple of the smallest unit of charge which is the magnitude of the electron charge e = 1.602 10-19 C.

• On the macroscopic level, we can assume that charge is “continuous.”

7


8

COULUMB’S LAW

In the late 18th century, Colonel Charles

Augustus Coulomb invented a sensitive

torsion balance that he used to

experimentally determine the force

exerted in one charge by another.

9

COULUMB’S LAW (Cont’d)

He found that the force is proportional to the

product of two charges, inversely proportional

to the square of the distance between the

charges and acts in a line containing the two

charges.

221

RQQkF

10

The proportional constant, k is:

mF

mF

36101085.8

912

0

Where the free space permittivity with a value given by:

041 rk ,


11

Charge Q1 exerts a vector force F12 in Newton's (N) on charge Q2,

122120

2112

4a

RF

QQ


12


If more than two charges, use the principle ofsuperposition to determine the force on aparticular charge.

If there are N charges, Q1,Q2...QN locatedrespectively at point with position vectorsr1,r2...rN the resultant force F on a charge Qlocated at point r is the vector sum of the forcesexerted on Q by each charges Q1,Q2...QN

13


N

N

N

NQQ

QQQQ

rrrr

rr...

rrrr

rrrrrr

rrF

20

2

22

20

2

1

12

10

1

4

44

N

k k

kkQQ1

304 rr

rrF

Or generally,

14

322320

2332

122120

2112

3212

4

4

,

aR

F

aR

F

FFF

QQ

QQTOTAL

For example,


2112 FF

15

EXAMPLE 1

Suppose 10nC charge Q1 located at (0.0, 0.0, 4.0m)and a 10nC charge Q2 located at (0.0, 4.0m, 0.0). Findthe force acting on Q2 from Q1.

16

To employ Coulomb’s Law, first find vector 12R

zy

zy

aa

aa rrR

44

441212

Magnitude of 12R

2432

44 2212

R

SOLUTION TO EXAMPLE 1

17

And zyzy aa

aaRRa

21

21

2444

12

1212

Then

nN

QQ

zy

zy

aa

aa

aR

F

0.40.4

21

21

2410854.84

10101010

4

212

99

122120

2112

SOLUTION TO EXAMPLE 1 (Cont’d)

18

2.2 ELECTRIC FIELD INTENSITY

If Q1 is fixed to be at origin, a second charge

Q2 will have force acting on Q1 and can be

calculated using Coulomb’s Law. We also

could calculate the force vector that would

act on Q2 at every point in space to generate

a field of such predicted force values.

19

It becomes convenient to define electric field intensity E1 or force per unit charge as:

2

121 Q

FE

This field from charge Q1 fixed at origin results

from the force vector F12 for any arbitrarily

chosen value of Q2

ELECTRIC FIELD INTENSITY (Cont’d)

20

Coulomb’s law can be rewritten as

to find the electric field intensity at any point in space resulting from a fixed charge Q.

RQ a

RE 2

04


21

Let a point charge Q1 = 25nC be located

at P1 (4,-2,7). If ε = ε0, find electric field

intensity at P2 (1,2,3).

EXAMPLE 2

22

By using the electric field intensity,

RQ a

RE 2

04

This field will be:

12120

9

41025 aR

E


23

zyx aaarrR 4431212

4112 R

Where,

and

aaa

RR

aR

E

zyx

QQ

4434110854.84

1025

44

2312

9

123120

122120


24


If there are N charges, Q1,Q2...QN locatedrespectively at point with position vectorsr1,r2...rN the electric field intensity at point r is:

N

N

N

NQQrrrr

rrrrrr

rrE

2

01

12

10

1

4..

4

N

k k

kkQ1

304

1rr

rrE

25

FIELD LINES

The behavior of the fields can be visualized usingfield lines:

Field vectors plotted within a regular grid in 2D space surrounding a point charge.

26

Some of these fieldvectors can easily bejoined by field lines thatemanate from thepositive point charge.

The direction of the arrow indicates the direction of electric fields

The magnitude is given by density of the lines

FIELD LINES (Cont’d)

27

The field lines terminated

at a negative point charge

The field lines for a pair

of opposite charges

FIELD LINES (Cont’d)

28

2.3 LINE,SURFACE & VOLUME CHARGES

Electric fields due to continuous charge distributions:

29

To determine the charge for each distributions:

Line charge:

LL

L

dlQ

dldQ

Surface charge:

SS

S

dSQ

dSdQ

Volume charge:

VV

V

dVQ

dVdQ

LINE,SURFACE & VOLUME CHARGES (Cont’d)

30

Infinite Length of Line Charge:

To derive the electric field intensity at any

point in space resulting from an infinite

length line of charge placed conveniently

along the z-axis

LINE CHARGE

31

Place an amount of charge in coulombs along the z axis.

The linear charge density is coulombs of charge per meter length,

Choose an arbitrary point P where we want to find the electric field intensity.

mC

L

zP ,,

LINE CHARGE (Cont’d)

32


The electric field intensity is:

zzEEE aaaE

But, the field is only vary with the radial distance from the line.

There is no segment of charge dQ anywhere on the z-axis that will give us . So,E

zzEE aaE

33


Consider a dQ segment a distance z above radial axis, which will add the field components for the second charge element dQ.

The components cancel each other (by symmetry) , and the adds, will give:

zE

E

aE E

34


Recall for point charge,

RQ a

RE 2

04

For continuous charge distribution, thesummation of vector field for each chargesbecomes an integral,

RdQ a

RE 2

04

35


The differential charge,

dzdldQ

L

L

The vector from source to test point P,

z

R

zR

aa aR

36


Which has magnitude, and a

unit vector,

22 z R

22 z

z zR

aaa

So, the equation for integral of continuous charge distribution becomes:

2222204 z

z

z

dz zL

aaE

37


Since there is no component,

a

aE

23220

2322

0

4

4

z

dz

z

dz

L

L

za

38

Hence, the electric field intensity at any point ρaway from an infinite length is:

aE

02L

For any finite length, use the limits on the integral.


39

EXAMPLE 3

Use Coulomb’s Law to

find electric field

intensity at (0,0,h) for

the ring of charge, of

charge density,

centered at the origin in

the x-y plane.

L

40


By inspection, the ring

charges delivers only

and contribution

to the field.

component will be

cancelled by symmetry.

zdEEd

Ed

41

RdQ a

RE 2

04

Each term need to be determined:


The differential charge,

addldQ

L

L

42

The vector from source to test point,

z

R

haR

aa aR


unit vector,

22 ha R

22 ha

ha zR

aaa


43


The integral of continuous charge distribution becomes:

2222204 ha

ha

ha

ad zL aaE

zL h

ha

ad aE2

322

04

44


Rearranging,

2

02322

04z

L dha

ah aE

Easily solved,

zL

ha

ah aE2

32202

45

EXAMPLE 4

An infinite length line of charge

exists at x = 2m and z = 4m. Find the

electric field intensity at the origin.

mnC

L 0.4

46


Sketch in three dimensions and the cross section:

47


The vector from line charge to the origin:

zx aaaR 42

Which has magnitude, and a unit

vector,

20R

zxR aaaa204

202

48

Inserting into the infinite line charge equation:

mV

zx

zx

L

aa

aa

aE

4.142.7

2042

2010854.82104

2

12

90


49

SURFACE CHARGE

Infinite Sheet of Surface Charge:

To derive the electric field intensity at point

P at a height h above a charge sheet of

infinite area (x-y plane).

The charge distribution, is inS 2mC

50

SURFACE CHARGE (Cont’d)

51


Consider a differential charge,

dddSdQ

S

S

The vector from surface charge to the origin:

zhaaR

52


unit vector,

22 h

h zR

aaa

22 h R

Where, for continuous charge distribution:

RdQ a

RE 2

04


53

The equation becomes:

2222204 ha

h

h

dd zS aaE

zS h

h

dd aE2

322

04

Since only z components exists,


54


zS

zS

zS

zS

hh

dhh

h

ddh

aE

a

a

aE

0

0

2122

0

0

2322

0

2

0 0 23

220

2

2

24

4

55


A general expression for the field from a sheet

charge is:

NS aE02

Where is the unit vector normal from the

sheet to the test point.Na

56

EXAMPLE 5

An infinite extent sheet of charge

exists at the plane y = -2m. Find the electric

field intensity at point P (0, 2m, 1m).

210mnC

S

57


Sketch the figure:

58


The unit vector directed away from the sheet and toward the point P is ya

mV

y

y

NS

a

a

aE

565

10854.821010

2

12

90

59

VOLUME CHARGE

A volume charge is distributed over a

volume and is characterized by its volume

charge density, inV 3mC

The total charge in a volume containing a

charge distribution, is found by

integrating over the volume:V

V

V dVQ

60

EXAMPLE 6

Find the total charge

over the volume with

volume charge density,

3105

5m

Ce zV

61


V

V dVQ The total charge, with volume:

dzdddV Thus,

C

dzdde

dVQ

z

z

VV

14

01.0

0

2

0

04.0

02.0

10

10854.7

55

62

VOLUME CHARGE (Cont’d)

To find the electric field intensity resulting from a volume charge, we use:

RV

RdVdQ a

Ra

RE 2

02

0 44

Since the vector R and will vary over the volume, this triple integral can be difficult. It can be much simpler to determine E using Gauss’s Law.

V

63

2.4 ELECTRIC FLUX DENSITY

Consider an amount of charge

+Q is applied to a metallic

sphere of radius a.

Enclosed this charged sphere

using a pair of connecting

hemispheres with bigger radius.

64

ELECTRIC FLUX DENSITY (Cont’d)

The outer shell is grounded. Remove the ground

then we could find that –Q of charge has

accumulated on the outer sphere, meaning the

+Q charge of the inner sphere has induced the –Qcharge on the outer sphere.

65

Electric flux, extends from the positivecharge and casts about for a negative charge. Itbegins at the +Q charge and terminates at the–Q charge.

psi

The electric flux density, D in is:2mC

RaR

D 204

where ED


66

This is the relation between D and E, where is the material permittivity. The advantage of using electric flux density rather than using electric field intensity is that the number of flux lines emanating from one set of charge and terminating on the other, independent from the media.

We can find the total flux over a surface as:

dS D


67

We could also find the electric flux density, Dfor:

Infinite line of charge:

Where

aE

02L So,

aD2

L


68

Where

So, NS aD

2

NS aE02

Infinite sheet of charge:

Volume charge distribution:

RV dV a

RE 2

04 So, R

V dV aR

D 24


69

EXAMPLE 7

Find the amount of electric flux through the surface at z = 0 with

and

mymx 3050 ,

243 mCxxy zx aaD

70


The differential surface vector is

zdxdyd aS

We could have chosen but the

positive differential surface vector is pointing in

the same direction as the flux, which give us a

positive answer.

zdxdyd aS

71


Therefore,

C

xdxdy

dxdyxxy

dS

x y

zzx

aaa

D

150

4

435

0

3

0

Why?!

72

EXAMPLE 8

Determine D at (4,0,3) if there is a point

charge at (4,0,0) and a line

charge along the y axis.

mC5

mCm 3

73


How to visualize ?!

74

Let total flux,

LQTOTAL DDDWhere DQ is flux densities due to point charge

and DL is flux densities due to line charge.

Thus,

R

RQ

Q

Q

aR

aR

ED

2

20

00

4

4


75


Where,

za

R3

3,0,00,0,43,0,4

Which has magnitude, and a unit vector,

zz

R aaa 3

33R

76

So,

2

3

2

138.0

94105

4

mCm

Q

z

z

RQ

a

a

aR

D


77

And aD

2LL

Where, 5

340,0,03,0,40,0,03,0,4 zx aaa

So,

218.024.0

534

523

mCmzx

zxL

aa

aaD


78

Therefore, total flux:

242240

18.024.0318.0

mC

zx

zxz

LQTOTAL

aa

aaa DDD


79

2.5 GAUSS’S LAW

If a charge is enclosed, the net flux passing through the enclosing surface must be equal to the charge enclosed, Qenc.

Gauss’s Law states that:

The net electric flux through any closed surface isequal to the total charge enclosed by that surface

encQd SD

80

It can be rearranged so that we have relation

between the Gauss’s Law and the electric flux.

VVenc

enc

dVQ

Qd

SD

V

VS

dVdQ SD

GAUSS’S LAW (Cont’d)

81


Gauss’s Law is useful in finding the fields for

problems that have high degree of symmetry.

• Determine variables influence D and what

components D present

• Select an enclosing surface, called Gaussian

Surface, whose differential surface is directed

outward from the enclosed volume and is

everywhere (either tangent or normal to D)

82

GAUSS’S LAW APPLICATION (Cont’d)

Use Gauss’s Law to determine electricfield intensity for each cases below:

Point Charge

Infinite length of Line Charge

Infinite extent Sheet of Charge

83

POINT CHARGE

• Point Charge:

It has spherical coordinate

symmetry, where the field

is everywhere directed

radially away from the

origin. Thus,

rrD aD

84

For a gaussian surface, we could find the differential surface vector is:

rddrd aS sin2So,

ddrD

ddrDd

r

rrr

sin

sin2

2

aaSD

POINT CHARGE (Cont’d)

85


Since the gaussian surface has a fixed radius,

Dr will be constant and can be taken from

integration to yield

r

r

Dr

ddrDd

2

0

2

0

2

4

sin

SD

86


By using Gauss’s Law, where:

QrD

Qd

r

enc

24

SD

So, which leads to expected result:24 r

QDr

rrQ aE 2

04 See page 16!

87

INFINITE LENGTH LINE OF CHARGE

• Infinite length line of charge:

Find D and then E at any point zP ,,

A Gaussian surface containing the point P is placed around a section of an infinite length line of charge density L

occupying the z-axis.

88


An element of charge dQ along the line will give Dρ and Dz. But second element of dQ will result in cancellation of Dz. Thus,

aD DThe flux through the closed surface is:

side

bottomtop

d

ddd

SD

SDSDSD

89

Where,

aS , aS

, aS

dzddddd

ddd

sidezbottom

ztop

Then, we know that Dρ is constant on the side of gaussian surface

DhdzdD

dzdDdh

z

side

22

0 0

aaSD


90

The charge enclosed by the gaussian surface:

hdzQ L

h

Lenc 0

encL QhDhd 2SD

We know that,

So,

2LD Thus, as

expected: aE

02L


91

INFINITE EXTENT SHEET OF CHARGE

• Infinite extent sheet of charge:

Determine the field everywhere resulting from an infinite extent sheet of charge ρS

placed on the x-y plane at z = 0.

Locate a point at which we want to find the field along the z axis at height h.

92

Gaussian surface must contain this point and surround some portion of the charged sheet.

A rectangular box is employed as the Gaussian surface surrounding a section of sheet charge with sides 2x, 2y and 2z

SHEET OF CHARGE (Cont’d)

93

Only a DZ component will be present, and the

charge enclosed is simply:

xy

dydxdSQ

S

y

y

x

xSS

4

No flux through the side of the box, so find

the flux through the top and bottom surface


94

z

bottomzzz

topzzz

bottomtop

Dxy

dxdyD

dxdyD

ddd

42

aa

aa

SDSDSD


Notice that the answer is independent of the height of the box.

95

Then we have:

2

442

Sz

Sz

D

xyDxy

Qd

SD

zS aD

2

or

And electric field intensity, as expected:

NS aE02


96


Related to Gauss’s Law, where net flux isevaluated exiting a closed surface, is theconcept of divergence.

Expression for divergence by applying Gauss’sLaw might be too lengthy to derive, but it can bedescribed as:

V D

97

The expression is also called the point form of Gauss’s Law, since it occurs at some particular point in space. For instance,

Plunger stationary – no net movement of molecules

Plunger moves up – net movement where air molecules diverging air is expanding

Plunger pushes in – net flux is negative and molecules diverging air is compressing


98

EXAMPLE 9

Suppose:

Find the flux through the surface of a cylinder

with and by evaluating the

left side and the right side of the divergence

theorem.

aD 2

hz 0 a

99


Remember the divergence theorem?

V

dVdS DD

We can first evaluate the left side of the divergence theorem by considering:

side

bottomtop

d

ddd

SD

SDSDSD

100

A sketch of this cylinder is shown withdifferential vectors.

The integrals over the top and bottom surfaces are each zero, since:

0 zaa


101

Thus,

3

2

0 0

2

2 ha

dzd

ddh

z

side

a

SDSD


102

For evaluation of the right side of the divergence

theorem, first find the divergence in cylindrical

coordinate:

31

1

3

D D


103

Performing a volume integration on this divergence,

3

2

0 0 0

2

2

3

3

ha

dzdd

dzdd

dV

a h

z

V

D

This is the same!


104

2.6 ELECTRIC POTENTIAL

To develop the concept of electric potential and show its relationship to electric field intensity.

In moving the object from point a to b, the work can be expressed by:

b

adW LF

dL is differential length vector along some portion of the path between a and b

105

ELECTRIC POTENTIAL (Cont’d)

The work done by the field in moving the charge from a to b is

b

afieldE dQW LE

If an external force moves the charge against the field, the work done is negative:

b

adQW LE

106

We can defined the electric potential difference, Vba

as the work done by an external source to move a charge from point a to point b as:

b

aba d

QWV LE

Where,abba VVV


107


Consider the potential difference between twopoints in space resulting from the field of apoint charge located at origin, where theelectric field intensity is radially directed, thenmove from point a to b to have:

b

arr

b

aba dr

rQdV aaLE 2

04

108

Thus,

ab

br

arba

VVab

Q

rQV

114

4

0

0

The absolute potential at some finite radius from a point charge fixed at the origin:

rQV

04


109

If the collection of charges becomes a continuous distribution, we could find:

rdQV

04

Where,

rdVV

rdSV

rdLV

V

S

L

0

0

0

4

4

4

Line charge

Surface charge

Volume charge


110

N

NQ

QQV

rr...

rrrr

0

20

2

10

1

4

44

N

k k

kQV104

1rr

Or generally,

The principle of superposition, where applied to electric field also applies to potential difference.


111


Based on figure, if a closed path is chosen, the integral will return zero potential:

Three different paths to calculate work moving from the origin to point P against an electric field.

0 LE d

112

EXAMPLE 10

Two point charges -4 μC and 5 μC are

located at (2,1-,3) and (0,4,-2)

respectively. Find the potential at

(1,0,1).

113


Let and CQ 41 CQ 52

So,

20

2

10

144 rrrr

QQV

Where,

262,4,12,4,01,0,1

62,1,13,1,21,0,1

2

1

rr

rr

114


264105

64104

441,0,1

0

6

0

620

2

10

1

rrrrQQV

Therefore,

kVV 872.51,0,1

115

The electrostatic potentialcontours from a point chargeform equipotential surfacessurrounding the point charge.The surfaces are alwaysorthogonal to the field lines.The electric field can bedetermined by finding the max.rate and direction of spatialchange of the potential field.


116

Therefore,

VEThe negative sign indicates that the field is pointing in the direction of decreasing potential.

By applying to the potential field:

rr rQ

rQ

rV aaE 2

00 44


117

IMPORTANT!!Three ways to calculate E:

If sufficient symmetry, employ Gauss’s Law.

Use the Coulomb’s Law approach.

Use the gradient equation.

118

EXAMPLE 11

Consider a disk of charge ρS, find the

potential at point (0,0,h) on the z-axis and

then find E at that point.

119


Find that,

dddSdQ

S

S

and 22 hr

With rdQV

04then,

a

SrddV

0

2

004

120

Let and leads to

integral then,

22 hu

How to calculate the integral?

ddu 2

duu 21

hah

hV

S

aS

22

0

0

22

0

2

2


121

To find E, need to know how V is changing with position. In this case E varies along the z-axis, so simply replace h with z in the answer for V, then proceed with the gradient equation.

zS

zS

z

az

z

az

zzVV

aa

aE

2202201

212

21

2


122

BOUNDARY CONDITIONS

So far we have considered the existence of electric

field in a region consisting of two different media,

the condition that the field must satisfy at the

interfacing separating the media called “boundary

condition”. Thus, we could see how the fields

behave at the boundary between a pair of

dielectrics or between a dielectric and a conductor.

123

First boundary condition can be determined by performing a line integral of E around a closed rectangular path,

BOUNDARY CONDITIONS

124

Fields are shown in each medium along with normal and tangential components. For static fields,

0 LE d

Integrate in the loop clockwise starting from a,

0 a

d

d

c

c

b

b

adddd LELELELE

BOUNDARY CONDITIONS

125

Evaluate each segment,

221

2

02

0

21

10

1

hEE

dLEdLEd

wEdLEd

NN

h

NNNh

NNN

c

b

T

w

TTT

b

a

aaaaLE

aaLE

BOUNDARY CONDITIONS

126

221

2

01

0

22

2

0

2

hEE

dLEdLEd

wEdLEd

NN

h

NNNh

NNN

a

d

Tw

TTT

d

c

aaaaLE

aaLE

Summing for each segment, then we have the first boundary condition:

21 TT EE

BOUNDARY CONDITIONS

127

Second boundary condition can be determined by applying Gauss’s Law over a small pillbox shaped Gaussian surface,

BOUNDARY CONDITIONS

128

encQd SD

BOUNDARY CONDITIONS

The Gauss’s Law,

Where,

sidebottomtop

dddd SDSDSDSD

The pillbox is short enough, so the flux passes through the side is negligible.

129

So, only top and bottom where:

SDdSDd

SDdSDd

NNNNbottom

NNNNtop

22

11

aaSD

aaSD

Which sums to:

encNN QSDD 21

BOUNDARY CONDITIONS

130

And the right side of Gauss’s Law,

SdSQ SSenc

Thus, it leads to the second boundary condition:

SNN DD 21

This is when the normal direction from medium 2 to medium 1.

BOUNDARY CONDITIONS

131

If the normal direction is from medium 1 to medium 2,

SNN DD 12

BOUNDARY CONDITIONS

Generally,

S 2121 DDa

132

For a boundary conditions between a dielectric and a good conductor,

0TEBecause in a good conductor, E = 0. And since the electric flux density is zero inside the conductor,

SN D

BOUNDARY CONDITIONS

133

EXAMPLE 12

Consider that the field E1 is known as:

Find the field E2 in the other dielectrics.

mVzyx aaaE 5431

134


135

BOUNDARY CONDITIONS

We can employ Poisson’s and Laplace’sequations to help find the potential functionwhen conditions at the boundaries are specified.

From divergence theorem expression,

V DBy considering ,ED

V E

136

From the gradient expression,

VEWhich gives us Poisson’s equation,

VV 2

In charge free medium in which , it becomes Laplace’s equation:

0V

02 V

BOUNDARY CONDITIONS

137

EXAMPLE 13

Determine the electric potential in the dielectric region between a pair of concentric spheres that have a potential difference Vab.

138


The charge distribution is: 30

mC

rV

This employs Poisson’s equation and the potential is only a function of r,

0

022

02

)(1

r

V

rrrVr

rr

rV

139

Multiply with r2 and integrate to obtain,

ArrrVr

r

0

202

2)(

Dividing both sides with r2 and integrate again,

BrArrV

r

0

02

)(


140

Assume that the Vab consists of a voltage Va on the inner conductor and the outer conductor is grounded. So,

22

0)(

12

)(

0

0

0

0

BbAbVbrV

BaAaVarV

rb

ra


141

ababAabV

BbAbB

aAaVVV

ra

rrbaab

0

0

0

0

0

0

2

22

From (3) you can get:0

02

r

a abba

abVA


(3)

ba

abVbaB a

r

0

02

Thus,

142

Then, the potential between the spheres is

given by:

raVr

rarV a

r

0

02

)(


143

CAPACITANCE

The amount of charge that accumulates as a function of potential difference is called the capacitance.

VQC

The unit is the farad (F) or coulomb per volt.

144

CAPACITANCE (Cont’d)

Two methods for determining capacitance:

Q Method

• Assume a charge +Q on plate a and a charge –Q on plate b.

• Solve for E using the appropriate method (Coulomb’s Law, Gauss’s Law, boundary conditions)

• Solve for the potential difference Vab between the plates (The assumed Q will divide out)

145

V Method

• Assume Vab between the plates.

• Find E , then D using Laplace’s equation.

• Find ρS, and then Q at each plate using conductor dielectric boundary condition (DN = ρS )

• C = Q/Vab (the assumed Vab will divide out)

CAPACITANCE (Cont’d)

146

EXAMPLE 14

Use Q method to find the capacitance for

the parallel plate capacitor as shown.

147


Place charge +Q on the inner surface of the top plate, and –Q charge on the upper surface of the bottom plate, where the charge density,

Use conductor dielectric boundary, to obtain:

dSQ SSQ

S from

zSQ aD from SN D

148

We could find the electric field intensity, E

zrSQ aE0

The potential difference across the plates is:

SQddz

SQ

dV

rz

d

zr

a

bab

00 0

aa

LE


149

Finally, to get the capacitance:

SQd

QVQC

rab

0


dSC r0

150

EXAMPLE 15

Use V method to find the capacitance for a length Lof coaxial line of inner conductor radius a and

outer radius b, filled with dielectric permittivity as

shown.

151


Employ Laplace’s equation to find the potential

field everywhere in the dielectric. Assume that

fringing fields are neglected and the field is only

in function of ρ.

Laplace’s equation becomes:

0

V

152


Integrating twice to obtain:

BAV ln

Apply boundary condition to determine A and B. Let V(b) = 0 and V(a) = Vab to get:

ab

VAbAB abln

,ln So,

abbVV ab

lnln

153

Apply the gradient to obtain E:

aE

E

abV

VV

abln

and also..

aD

abVabr

ln0


154

We can find Q on the inner conductor with:

abVLaL

abaVSQ abrabr

S ln22

ln00


At inner conductor, the flux is directed outward indicating a positive surface charge density,

abaVabr

S ln0

155

Thus, we can find the capacitance:

abL

VQC r

ln2 0


156

EXAMPLE 16

Conducting spherical shells with radius a =10cmand b = 30cm are maintained at a potentialdifference of 100V, such that V(r = a) = 100V andV(r = b) = 0V. Determine:

• The V and E in the region between the shells.

• If εr = 2.5 in the shells, determine the totalcharges induced on the shells and the capacitanceof the capacitor.

157

With:

5.23.01.0

r

mbma

EXAMPLE 16 (Cont’d)

158


Employ Laplace’s equation to find the potential

field and the field is only in function of r.

Laplace’s equation becomes:

01 22

2

rVr

rrV

Multiply by r2,02

rVr

r

159

Integrating once gives

22

rA

rVA

rVr

Integrating again gives

BrAV


160

To obtain the value of constant A and B, use boundary conditions, where:

When,

bAB

BbAVbr

00,

Thus,

rbAV 11


161


When,

ab

VA

abAV

rbAVV

VVar

11

11

11,

0

0

0

0

:

162

Therefore, the potential difference V

ab

rbVV11

11

0


163

So, the electric field intensity E

r

ab

Vr

aE

11

1 02


Apply the gradient to obtain E:

rr rA

rVV aaE 2

164

The total charges Q is:

ddr

abr

V

ddQ

r sin111 2

2

0 02

00

SESD


165

This yields to:

ab

V

dd

ab

VQ

r

r

114

sin11

00

2

0 0

00


166

Thus, substituting the values of a, b and V0

to get the total charges, Q

nC

Q

1723.41.0

13.0

11005.210854.84 12


167

And for the potential difference V = V0.

So, the capacitance, C

pFVnC

VQC

723.41

1001723.4


PART 1

END

169

PRACTICAL APPLICATION

Laser Printer

Electret Microphone

Electrolytic Capacitors

170

LASER PRINTER

171

• OPC drum : Organic Photoconductive Cartridge –has a special coating will hold electrostatic charge.

• The surface is photoconductive – will discharge if surface hit by light.

i. A portion of drum passes under a negativecharged wire large negative charges induces apositive charge on the drum.

ii. Image to be printed is delivered to this chargedregion by laser and spinning mirror combination.

iii.Wherever the laser light strikes the drum, thephotoconductive material is discharged.

LASER PRINTER (Cont’d)

172

iv. The drum rolls past a toner The toner is blackpowder and positive charge drawn to thoseportions of the charge that have been dischargedby the laser.

v. Paper is fed through same speed as drum itpasses over positively charged wire that gives thepaper a strong negative charge.

vi. The positively charged toner drum is transferred tothe stronger negative charged on the paper thenit passes near a negatively charged wire thatremoves the negative charge from the paper,prevents it from statically clinging to the drum


173

vii.The paper and loose toner powder passed throughheated fuser rollers powder melts into the paperfiber the warm paper exits the printer.

• The drum continues rolling, passing through highintensity light discharges all the photoconductorsto erase the image from the drum, and ready forapplication of positive charge again from coronawire.


174

SUMMARY (1)

•The force exerted on a charge Q1 on charge Q2 in a medium of permittivity ε is given by Coulomb’s Law:

12212

2112

4a

RF

QQ

Where is a vector from charge Q1 to Q2121212 aR R

2

121 Q

FE

•Electric field intensity E1 is related to force F12 by:

175

The Coulomb’s Law can be rewritten as:

RQ a

RE 2

04

For a continuous charge distribution:

RRdQ aE 2

04•For a point charge at origin:

rrQ aE 2

04

SUMMARY (2)

176

•For an infinite length line charge ρL on the z axis

aE

2L

•For an infinite extent sheet of charge ρS

NS aE

2

•Electric flux density related to field intensity by:

ED 0 r

SUMMARY (3)

Where εr is the relative permittivity in a linear, isotropic and homogeneous material.

177

SUMMARY (4)

• Electric flux passing through a surface is given by:

dS D

• Gauss’s Law states that the net electric flux through any closed surface is equal to the total charge enclosed by that surface:

encQd SD

Point form of Gauss’s Law is

V D

178

SUMMARY (5)

• The electric potential difference Vab between a pair of points a and b in an electric field is given by:

ab

b

aba VVdV LE

Where Va and Vb are the electrostatics potentials at aand b respectively.

For a distribution of charge in the vicinity of the origin, where a zero reference voltage is taken at infinite radius:

rdQV4

179

SUMMARY (6)

• E is related to V by the gradient equation:

VEWhich for Cartesian coordinates is:

zyx zV

yV

xVV aaa

• The conditions for the fields at the boundary between a pair of dielectrics is given by:

S 2121 DDa21 TT EE and

180

Where ET1 and ET2 are the electric field components tangential to the boundary, a21 is a unit vector from medium 2 to 1 and ρS is the surface charge at the boundary. If no surface charge is present, the components of D normal to the boundary are equal:

21 NN DD

SUMMARY (7)

At the boundary between a conductor and a dielectric, the conditions are:

0TE and SN D

181

• Poisson’s equation is:

VV 2

Where the Laplacian of V in Cartesian coordinates is given by:

2

2

2

2

2

22

zV

yV

xVV

In a charge free medium, Poisson’s equation reduces to Laplace’s equation

02 V

SUMMARY (8)

182

• Capacitance is a measure of charge storage capability and is given by:

VQC

SUMMARY (9)

For coaxial cable:

For two concentric spheres:

abLC

ln2

abV Lab ln

2

So,

baC

114

baQVab

114

So,

part 1:electrostatics

Education

point r

qn r r n

point charge q1

fixed charge

r r12 r r14

unit charge

qq n r r n

charge qlocated