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The Electric Flux and the Displacement Charge - Study Guide | EEGR 305, Study notes of Electrical and Electronics Engineering

Material Type: Notes; Professor: Wilkins; Class: Electromagnetic Theo & Appl; Subject: Electrical Engineering; University: Morgan State University; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/07/2009

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Electric Flux Ψdisplacement charge
Flux lines leave +Q and terminate on -Q
equal but opposite charge induced
Ψ= Q coulombs
The amount of flux per unit
area is the flux density
n
aD
ds
dΨ
=
Suppose D is not normal to ds?
s
D
aDn
d
cos D
Ψ
dsdsd
θ
+Q
-Q
anD
ds
θ
Chapter 3
A surface vector
points in the direction
normal to the surface
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff

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Electric Flux Ψ displacement charge

Flux lines leave +Q and terminate on -Q

equal but opposite charge induced

Ψ = Q coulombs

The amount of flux per unit area is the flux density

D (^) ds a n =^ d^ Ψ

Suppose D is not normal to d s?

D s

D an d

D cos = ⋅

d Ψ = ds θ = ⋅ ds

+Q
-Q

a n (^) D

ds

θ

Chapter 3

A surface vector points in the direction normal to the surface

Total flux out of a closed surface is equal to the net charge enclosed within the surface

∫ ∫

vol

v S

S

s ρ dv

s

D d

chargeenclosedQ D d

S

enc

We can use Gauss’s Law to find D as well as using the approach discussed in the previous chapter to find E and henceforth D.

Gauss’ Law

D =ε 0 E

Relationship between D and E :

Point Charge Example

r

s s

s

s s

r

Q

r

Q

D r D

D r d d

Q d D ds

2^ a

2

2

2

0 0

2

D

sin

D S

ππ

∫ ∫

∫ ∫

Gaussian Surface

Q

r

Coaxial Cable Example

Choose gaussian surface of a cylinder for a< ρ <b

ρ

ρ π ρ πρ

ρ φ π ρ

πρ

ρ φ

π

π

s s s s

L s s

s

L s s

a a L D L D

Q ad dz a L

Thetotalch e on the inner conductor

Q D L

Q D dS D d dz

∫ ∫

∫ ∫ ∫

arg :

0

2

0

2

0 0 (^) b a

L

GS for ρ>b

GS for a<ρ<b

Coaxial Cable Example contd.

From before:

QOuter=- 2 π a ρ sLinner 2 π bL ρ s outer= - 2 π aL ρ s inner

If the gaussian surface is chosen for ρ > b:

The total charge =0=Ds 2 πρ L ⇒ Ds = 0

souter s inner b

ρ = −a^ ρ

Differential Volume Element

•Here we assume no symmetry.

•We would like to get an idea of the spatial variation of D.

⋅ =

⋅ = → ( d ñvdv)

v

vdv v v

d v (^) v D S

D S 0

lim 0

lim ∆∆ ∆∆ ∆∆ ∆∆

ρ

The Left Hand Side(LHS) is the divergence of D.

The Right Hand Side (RHS) is ρ v.

The flux flowing out of face 2 is:

y z

x x ∆ ∆

∆ − ) 2

D (^) x (

The net flow is (1) -(2)

z

D

x y z y

D

x y z x

D

x y z

x

D

x y z

x y z

x

The flux flowing out of face 1 is:

y z

x x ∆ ∆

  • ) 2

D (^) x (

Differential Volume Element contd.

Differential Volume Element contd.

v

x y z

v

x y z

z

D

y

D

x

D

v z

D y

D

x

D

ρ

ρ

= ∂

 = 

  

∂ = ∆∆

v

x y z

z

D

y

D

x

D

div = ρ

∴ D =

Divergence in Other Coordinate

Systems

( )

z
D D
div D
D

z

D y

D x

D div x y z ∂

D =

( ) ( ) φ

φ θ

θ θ θ θ

D r

D r

r Dr r r

div sin

1 sin sin

1 2 1 D 2

The Vector Operator Del

x ax^ y ay ∂z a^ z

The vector operator (^) ∇ Del:

In rectangular coordinates

We are already familiar with scalar operations such as the derivative:

... etc.

dx

d

D

dx

d

D 2

2

=^2 =

Start with Gauss’ Law:

∫ ∫

∫ ∫

= ∇ ⋅

= ∇ ⋅

= =

vol

vol

vol

Q

d dv

dv

d dv

D s D

D

D s ρ

The Divergence Theorem

This Theorem allows us to calculate the charge by enclosing it within a surface or integrating over a volume