electromagnetics theory (see 2523) electrostatic field

ELECTROMAGNETICS THEORY(SEE 2523)

ELECTROSTATIC FIELD

INTRODUCTION

An electrostatic field is produced by a

static charge distribution.

A magnetostatic field is produced by the moving charges or a constant current

flow (DC current)

AC current produced the electromagnetic field.

Charges are measured in Coulombs (C). The smallest unit of electric charge is found on the

negatively charged electron and positively charged proton. The electron mass, qe= -1.602x10-19 (C).

So, 1 C would represent about 6x1018 electrons.

In real world, electric charges

can be found

at a point

along a line

on a surface

in a volume or

combination the above distributions

2.1: ELECTRIC CHARGES

2.1.1 A POINT CHARGE

Fig. 2.1: A point charge

The concept of a point charge is used when the dimensions of an electric charge distribution are very small compared to the distance to neighboring electric charges.

2.1.2 A LINE CHARGE

A line charge denotes as the electric charge distribution along a thin line.

Fig 2.2: A line charge

)/(lim0

mCQ

A line charge density :

Total charge along the line :

dQ

l

Q

2.1.3 A SURFACE CHARGE

A surface charge is defined as charges distribution on a thin sheet.

Fig. 2.3: A surface charge

Q

S)2(C/msQ

ss

lim

0

A surface charge density :

Total charge on the surface :

dss s

Q

2.1.4 A VOLUME CHARGE

A volume charge means electric charges in a volume. This volume charge can be viewed as a cloud of charged particles.

Fig. 2.4: A volume charge

Q

V

A volume charge density :

Total charge in the volume :

)3(C/mvQ

vv

lim

0

dvv v

Q

Expressed mathematically,

21QQF 2

12

1R

F 212

21R

QkQF and or

2.2 COULOMB’S LAW

Coulomb’s law states that the force F between two point

charges Q1 and Q2 is : Directly proportional to the

magnitude of the both products of the charges and

inversely proportional to the square of the distance R

between them.

Fig. 2.5: Coulomb’s law

12ˆ

212

212 R

aR

QkQF

ok 4

1

(N)12

ˆ212

421

2 Ra

Ro

QQF

where

o is known as the permittivity of free space

A unit vector, then the force :

The force, on Q1 due to Q2 :

12

12

12ˆ

R

R

Ra

1F

(N)123

124

212

RRo

QQF

(N)2213

214

211

FRRo

QQF

Things need to considered in Coulomb’s law :

>> The force expressed by Coulomb’s law is a mutual force, for

each of the two charges experiences a force of the same magnitude,

although of opposite direction.

>> Coulomb’s law is linear, if the charge multiply by n, the force is

also multiplied by the same factor, n.

>> If there are more than two point charges, the principle of

superposition is used. The force on a charge in the presence of several

other charges is the sum of the forces on that charges due to each of the

other charges acting alone.

2.3 ELECTRIC FIELD INTENSITY DUE TO POINT CHARGE

Consider one charge fixed in position, Q1 and test charge,

Qt that exists everywhere around Q1 , shown in Fig. 2.6.

From the definition of Coulomb’s law, there will exist force

on Qt cause by Q1.

Fig. 2.6

The electric field intensity E is the force per unit charge when placed in the electric field.

V/m)(N/C,1

ˆ21

41

tRa

tRo

Q

tQtFE

tF

For n point charges, the electric field intensity at a point in

space is equal to the vector sum of the electric field intensities

due to each charge acting alone. (Following superposition

theorem). Shown in Fig. 2.7.

Fig. 2.7

The total field at point P :

kRa

n

k kRo

kQ

nRanRo

nQRa

Ro

QRa

Ro

QRa

Ro

Q

n

k kEnEEEEE

ˆ1 24

ˆ24

....3

ˆ23

43

2ˆ

22

42

1ˆ

21

41

1...

321

E

We have :

Thus, the electric field for a line charge :

dQ

Ra

R

dE ˆ

20

4

2.4 ELECTRIC FIELD INTENSITY DUE TO THE LINE CHARGE

Fig. 2.8

Ex:4. Find the electric field intensity about the finite line

charge of uniform distribution along the z axis as shown

in Fig. 2.8.

Ra

Ro

dEd ˆ

24

2-'2

ˆ-'-ˆ

zzrR

zzzrrR

Solution:

∫ 2/32)-'(24

'ˆ)-'(-∫ 2/3

2)-'(24

'ˆ

∫ ˆ)-'(-ˆ2/3

2)-'(24

'

b

azzro

dzzzzb

azzro

dzrr

b

azzzrr

zzro

dzE

Using 2/1222

∫ 2/322

,2/1

22

1-∫ 2/3

22

xcc

x

xc

dx

xcxc

xdx

Hence :

2)-(22cos,

2)-(21cos

2)-(2)-(

2sin,

2)-(2)-(

1sin

zbr

r

azr

rzbr

zb

azr

az

mVrz

rr

oE /

1cos-

2cosˆ

1sin

2sinˆ

4

Hence :

2/1)2)-(2(1-

2/1)2)-(2(1ˆ

2/1)2)-(2()-(-

2/1)2)-(2()-(ˆ

4

2/1)2)-'(2(ˆ

2/1))2-'(2(2)-'(ˆ

4

zarzbrz

zarza

zbrzb

rr

o

b

azzrz

zzrrzzrr

oE

If the line is infinite :

(V/m)rr

E ˆ0

2

1

2 2/= =

mVrz

rr

oE /

1cos-

2cosˆ

1sin

2sinˆ

4

We obtain :

We have :

Thus, the electric field for a surface charge

dsS S

Q ∫

Ra

s R

dsSE ˆ∫ 20

4

2.5 ELECTRIC FIELD INTENSITY DUE TO THE SURFACE CHARGE

)(C/m2

s

Fig. 2.9

Ex:6. Find the electric field intensity along z axis on a disc with

a radius a(m) which is located on xy plane with uniform charge

distribution as shown in Fig. 2.9.

s R

aR

dssE ˆ2

04

22ˆˆ

rzR

rrzzR

Using cylindrical coordinate :

Solution:

szr

rrdrrdss

zr

zzdrrds

srrzz

zr

drrdsE

2/322

04

ˆ2/3

220

4

ˆ

ˆˆ2/3

220

4

)(

The component is cancel because the charge distribution is

symmetry.

Only component exists.

r

z

r

2/1

221

2ˆ1

2/122

12

ˆ

za

zoszz

zao

zsz

a

zro

zszza

zr

rdrdo

zsE

0

2/122

1)2(4

ˆˆ0 2/3

22

2

04

02

ˆ szE For an infinite sheet of charge :

02

ˆ snE In general, it can be

summarized to be :

The electric field is normal to the sheet and independent of

the distance between the sheet and the point.

In the parallel plate capacitor, the electric field existing

between the two plates having equal and opposite charge is

given by

The electric field is zero for both above and below plates.

nsnsnsE ˆ0

)ˆ-(0

2-

ˆ0

2

We have

Thus, the electric field intensity for a volume charge is

dvv vQ ∫

Ra

v R

dvvE ˆ∫ 20

4

2.6 ELECTRIC FIELD INTENSITY DUE TO THE

VOLUME CHARGE

)3(C/mv

Fig. 2.10

Ex:8. Find the electric field intensity for a sphere with a radius

a(m) and with uniform charge density, shown in

Fig. 2.10.

The electric field at P due to elementary volume charge is

where

Ed

Ra

Ro

ddrdrv

Ra

Ro

dvvEd

ˆ24

)sin2(

ˆ24

rzaR

ˆsinˆcosˆ

Solution:

Due to the symmetry of the charge distribution, the Ex and Ey

becomes zero.

Only Ez exists. Thus,

Using Cosine law,

cosˆ dEzEzE

cos2-222cos2-222

zRRzr

zrrzR

It is convenient to evaluate the integral in terms of R and r, so

Differentiating (keeping z and r fixed) then

zRrRz

zrRrz

22-22

cos

22-22

cos

dcos

zrRdRd sin

0 )( )-( rzRrz where and

Ra

Ro

ddrdrvEd ˆ24

)sin2(

Substituting all the equations into electric field equation yields

Ra

Ro

ddrdrvEd ˆ24

)sin2(

vazo

drra

zo

v

draa

RrzRr

zo

v

a

r

rz

rzRdRdr

Rrzr

zo

v

da

r

rz

rzR RzRrRzdrzr

RdRov

zE

334

2412

04

24

00

)2-2(-24

0 22-21

28

)2(

2

0 0 21

22-22

4

cosˆ dEzEzE

Because and then electric field at

P(0,0,z) is given by

Due to the symmetry of the charge distribution, the electric field at any point can be written as

vQv

3

34 av

zzo

QE ˆ24

),,( rP

rro

QE ˆ24

va

zozE

3

34

241

Which is identical to the electric field at the same point due

to a point charge Q located at the origin.