UNIVERSITY OF HORMUUD

FIELD THEORY

CHAPTER FOUR:

ELECTROSTATICSPart 3

Engr Burhan Omar Sheikh Ahmed

M.Eng (Electrical-Electronics and Telecommunications)

Universiti Teknologi Malaysia (UTM

ELECTROSTATICS

CHAPTER FOUR

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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

2.9 POISSON'S AND LAPLACE'S EQUATIONS

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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.

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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

–Q charge on the outer sphere.

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Faraday’s experiment found that the total charge on the outer sphere was equal in magnitude to the original charge placed on the inner sphere.

He concluded that there is some sort of displacement from the inner sphere to the outer which was independent of the medium, and we refer to this flux as displacement, displacement flux or simply electric flux.

Faraday showed, that there is direct proportionality between the electric flux and charge.

ELECTRIC FLUX DENSITY (Cont’d)

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From previous Electric Field equations we have shown that the electric field intensity is dependent on the medium in which the charge is placed.

Anew vector field D independent of the medium is defined by

The electric flux density, D in is:2mC

ED

ELECTRIC FLUX DENSITY (Cont’d)

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Electric flux, extends from the positive

charge and casts about for a negative charge.

It begins at the +Q charge and terminates at

the –Q charge.

psi

ELECTRIC FLUX DENSITY (Cont’d)

We can find the total flux over a surface as:

dS D

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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.

ELECTRIC FLUX DENSITY (Cont’d)

ED

10

We could also find the electric flux density, D

for:

Infinite line of charge:

Where

aE02L So,

aD

2L

ELECTRIC FLUX DENSITY (Cont’d)

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Where

So, NS aD

2

N

S aE02

Infinite sheet of charge:

Volume charge distribution:

RV dV

aR

E2

04

So, RV dV

aR

D2

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ELECTRIC FLUX DENSITY (Cont’d)

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EXAMPLE 1

Find the amount of electric flux through

the surface at z = 0 with

and

mymx 3050 ,

243 mCxxy zx aaD

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SOLUTION TO EXAMPLE 1

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

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SOLUTION TO EXAMPLE 1 (Cont’d)

Therefore,

C

xdxdy

dxdyxxy

dS

x y

zzx

aaa

D

150

4

43

5

0

3

0

Why?!

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EXAMPLE 2

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

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

charge along the y axis.

mC 5

mCm 3

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SOLUTION TO EXAMPLE 2

How to visualize ?!

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Let total flux, LQTOTAL DDD

Where DQ is flux densities due to point charge

and DL is flux densities due to line charge.

Thus, 0 0 20

2

4

4

Q R

R

QD E a

R

Q a

R

SOLUTION TO EXAMPLE 2 (Cont’d)

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SOLUTION TO EXAMPLE 2 (Cont’d)

Where,

za

R

3

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

Which has magnitude, and a unit vector,

zz

R aa

a 3

3

3R

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So,

2

3

2

138.0

94

105

4

mCm

Q

z

z

RQ

a

a

aR

D

SOLUTION TO EXAMPLE 2 (Cont’d)

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And then0 2

LLD E a

Where, 5

34

0,0,03,0,4

0,0,03,0,4 zx aaa

So,

218.024.0

5

34

52

3

mCmzx

zxL

aa

aaD

SOLUTION TO EXAMPLE 2 (Cont’d)

02LE a

::::::::::::::

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Therefore, total flux:

242240

18.024.0318.0

mC

zx

zxz

LQTOTAL

aa

aaa

DDD

SOLUTION TO EXAMPLE 2 (Cont’d)