lecture 9: electrostatics

Lecture 9: Electrostatics

Electric Flux Density, Gauss Law, Applications of Gauss Law, Chapter 4: pages 126-137

@Copyright Dr. Mohamed Bakr, EE 2FH3, 2014

Electric Flux Density D (C/m2)


the electric field is material-dependent and this makes it not

suitable for calculating flux which should be material-

independent

define D=E

0D Ein vacuum

the principle of superposition applies to D as well

2

1

4

v R

v

dvR

aD

2

1

4r

Q

r E a

2

1

4r

Q

r D a

Electric Flux Density (Cont’d)


multiple point charges

2 21 1

1 ( ) 1 ( )( ) ( )

4 4| | | |

N Nn n n n

n n

n nn n

Q Q

r rE r a D r a

r r r r

line charge

dz ldQ dz

( , ,0)P

R

z

xy0

L

1R

2R

1

2

z

1P

2P

2 1

2 1

cos cos4

sin sin4

l

lz

D

D

infinite sheet of uniform charge:

2

sn

D a

The Electric Flux (C)


general expression for differential

flux through a differential surface

the total flux through a surface

is given by

flux through a closed surface

d d D s

S

d D s

d D s

flux lines show the direction and

density of the flux

ds

na

high density

low density

D

Electric Flux (Cont’d)


electric flux through a closed surface is a measure of the

electric sources in the enclosed volume.

0

source (positive flux)

sink (negative flux) no source (zero net flux)

0

0

0

S

d D s

Gauss’ Law

the electric flux through a closed surface is equal to the electric

charge enclosed by that surface

no charge enclosed means no flux

using Divergence theorem, we have


v

S v

d Q dv D s

v

S v v

d dv dv D s D

v D

integral form

differential form

Cases of Gauss’ Law


net electric flux =5.0nC net electric flux =0

Applications of Gauss Law


Gauss’ law makes solutions to problems with planar, cylindrical

or spherical symmetry easy

procedure: choose an integration surface so that

D is everywhere either normal or tangential to surface

normal: ; tangential: 0d Dds d D s D s

when normal to surface, D is also constant on surface

S S

d D ds D S D s

Case Study: A Point Charge


field is in the ar direction

and depends only on r

select Gaussian surface as a

sphere centred at the charge

24r r

S S

d ds QrD D D s

24r

QD

r 24

r

QE

r

Case Study: An Infinite Line Charge

because of symmetry, field is in the a

Direction and depends only on

choose Gaussian surface as a cylinder


2 L

S S

d ds l lD D D s

2

LD

2

LE

due to symmetry, result is

obtained in a simple way!

Case Study: because of symmetry, field is

normal to plane and changes

only along z

choose Gaussian surface as

shown (side integrals cancel)


2 2 sz z

S top bottom S

d d d ds A AD D D s D s D s

2

szD

2

szE

Case Study: A Uniformly Charged Sphere


field is in the ar direction

and depends only on r

select Gaussian surface as

a sphere centred at the

charge

2 344 ,

3r r v

S S

d ds r arrD D D s

,3

r v

rr aD

A Uniformly Charged Sphere (Cont’d)


R

rD

3

va

0

21/ R

a

3

( )

2 3

3

2 2

4

3

4( ) 4

3

( )34

v v

S r v

r v

vr

d Q dv a

D r r Q a

Q aD r

r r

D s

outside the sphere

outside the sphere, the field is identical to a point charge at

the centre

lecture 9: electrostatics

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