Application of Gauss’s Law: Differential Volume Element

President University Erwin Sitompul EEM 4/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 4

Engineering Electromagnetics

http://zitompul.wordpress.com


Application of Gauss’s Law: Differential Volume ElementChapter 3 Electric Flux Density, Gauss’s Law, and DIvergence

We are now going to apply the methods of Gauss’s law to a slightly different type of problem: a surface without symmetry.

We have to choose such a very small closed surface that D is almost constant over the surface, and the small change in D may be adequately represented by using the first two terms of the Taylor’s-series expansion for D.

The result will become more nearly correct as the volume enclosed by the gaussian surface decreases.


0x

0( )f x

0x x x

0( ) ( )f x x f x

0( ) ( )f x f x x

20 0 00

( ) ( ) ( )( ) ( ) ( ) ( )

1! 2! !

nnf x f x f x

f x f x x x xn

A point near x0

Only the linear terms are used for the linearization

Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence

Taylor’s Series Expansion


Application of Gauss’s Law: Differential Volume ElementChapter 3 Electric Flux Density, Gauss’s Law, and DIvergence

Consider any point P, located by a rectangular coordinate system.

The value of D at the point P may be expressed in rectangular components:

0 0 0 0x x y y z zD D D D a a a

We now choose as our closed surface, the small rectangular box, centered at P, having sides of lengths Δx, Δy, and Δz, and apply Gauss’s law:

Sd Q D S

front back left right top bottomSd D S



Application of Gauss’s Law: Differential Volume ElementWe will now consider the front surface

in detail.The surface element is very small, thus

D is essentially constant over this surface (a portion of the entire closed surface):

front frontfront D S

front xy z D a

,frontxD y z

The front face is at a distance of Δx/2 from P, and therefore:

,front 0 rate of change of with 2x x x

xD D D x

0 2x

x

DxD

x



We have now, for front surface:

Application of Gauss’s Law: Differential Volume Element

0front

2x

x

DxD y z

x

In the same way, the integral over the back surface can be found as:

back backback D S

back ( )xy z D a

,backxD y z

,back 0 2x

x x

DxD D

x

0back

2x

x

DxD y z

x


If we combine the two integrals over the front and back surface, we have:



front back xD x y z

x

right left yD y x z

y

top bottom zD z x y

z

Repeating the same process to the remaining surfaces, we find:

These results may be collected to yield:

S

yx zDD D

d x y zx y z

D S

S

yx zDD D

d Q vx y z

D S



Charge enclosed in volume yx zDD D

v vx y z

Application of Gauss’s Law: Differential Volume ElementThe previous equation is an approximation, which becomes

better as Δv becomes smaller.For the moment, we have applied Gauss’s law to the closed

surface surrounding the volume element Δv, with the result:



Application of Gauss’s Law: Differential Volume ElementExample

Let D = y2z3 ax + 2xyz3 ay + 3xy2z2 az pC/m2 in free space. (a) Find the total electric flux passing through the surface x = 3, 0 ≤ y ≤ 2, 0 ≤ z ≤ 1 in a direction away from the origin. (b) Find|E| at P(3,2,1). (c) Find the total charge contained in an incremental sphere having a radius of 2 μm centered at P(3,2,1).

ψ SSd D S(a)

1 2 2 3 3 2 2

0 0 3

2 3 x y z xz y x

y z xyz xy z dydz

a a a a

1 2 2 3

0 0y z dydz

2 11 13 4

3 40 0y z

2

3pC




(b) 2 3 3 2 22 3x y zy z xyz xy z D = a a a2 3 3 2 2(2) (1) 2(3)(2)(1) 3(3)(2) (1)P x y z D = a a a

24 12 36 C mx y z p = a a a

2 2 2(4) (12) (36)P PD D =238.158 C mp

0

PP

D

E

2

12

38.158 C m

8.854 10

p

4.31V m




(c) yx zDD D

Q vx y z

yx zP

P

DD DQ v

x y z

43 2 3 6 3 33

321

0 2 6 pC m (2 10 ) mxyz

xz xy z

43 2 6 3

30 2(3)(1) 6(3)(2) (1) (2 10 ) Cp

272.61 10 C


We shall now obtain an exact relationship, by allowing the volume element Δv to shrink to zero.

DivergenceChapter 3 Electric Flux Density, Gauss’s Law, and DIvergence

Syx zdDD D Q

x y z v v

D S

0 0lim limSyx z

v v

dDD D Q

x y z v v

D S

The last term is the volume charge density ρv, so that:

0lim Syx z

vv

dDD D

x y z v

D S



Let us no consider one information that can be obtained from the last equation:

0lim Syx z

v

dDD D

x y z v

D S

This equation is valid not only for electric flux density D, but also to any vector field A to find the surface integral for a small closed surface.

0lim Syx z

v

dAA A

x y z v

A S



This operation received a descriptive name, divergence. The divergence of A is defined as:

0Divergence of div lim S

v

d

v

A S

A A

“The divergence of the vector flux density A is the outflow of flux from a small closed surface per unit volume as the volume shrinks to zero.”

A positive divergence of a vector quantity indicates a source of that vector quantity at that point.

Similarly, a negative divergence indicates a sink.



div yx zDD D

x y z

D

1 1div ( ) z

D DD

z

D

22

1 1 1div ( ) (sin )

sin sinr

Dr D D

r r r r

D

Rectangular

Cylindrical

Spherical



ExampleIf D = e–xsiny ax – e–x cosy ay + 2z az, find div D at the origin and P(1,2,3)

div yx zDD D

x y z

D sin sin 2x xe y e y 2

Regardles of location the divergence of D equals 2 C/m3.


Maxwell’s First Equation (Electrostatics)Chapter 3 Electric Flux Density, Gauss’s Law, and DIvergence

We may now rewrite the expressions developed until now:

div yx zDD D

x y z

D

0div lim S

v

d

v

D S

D

div vD Maxwell’s First EquationPoint Form of Gauss’s Law

This first of Maxwell’s four equations applies to electrostatics and steady magnetic field.

Physically it states that the electric flux per unit volume leaving a vanishingly small volume unit is exactly equal to the volume charge density there.



The Vector Operator and The Divergence TheoremDivergence is an operation on a vector yielding a scalar, just

like the dot product.We define the del operator as a vector operator:

x y zx y z

a a a

( )x y z x x y y z zD D Dx y z

D a a a a a a

yx zDD D

x y z

D

Then, treating the del operator as an ordinary vector, we can write:

div yx zDD D

x y z

D = D =


The Vector Operator and The Divergence TheoremChapter 3 Electric Flux Density, Gauss’s Law, and DIvergence

1 1( ) z

D DD

z

D

22

1 1 1( ) (sin )

sin sinr

Dr D D

r r r r

D

Cylindrical

Spherical

The operator does not have a specific form in other coordinate systems than rectangular coordinate system.

Nevertheless,



We shall now give name to a theorem that we actually have obtained, the Divergence Theorem:

vol volSvd Q dv dv D S D

volSd dv D S D

The first and last terms constitute the divergence theorem:

“The integral of the normal component of any vector field over a closed surface is equal to the integral of the divergence of this vector field throughout the volume enclosed by the closed surface.”



ExampleEvaluate both sides of the divergence theorem for the field D = 2xy ax + x2 ay C/m2 and the rectangular parallelepiped fomed by the planes x = 0 and 1, y = 0 and 2, and z = 0 and 3.

volSd dv D S D

3 2 3 2

0 10 0 0 0

3 1 3 1

0 20 0 0 0

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

SS x x x x

y y y y

d dydz dydz

dxdz dxdz

D S D a D a

D a D a

0 2( ) ( )y y y yD D 0( ) 0,x xD

3 2

10 0( )

SS x xd D dydz D S

3 2

0 02ydydz 12 C

But

Divergence Theorem



2(2 ) ( )xy xx y

D =

3 2 1

vol 0 0 0(2 )

z y xdv y dxdydz

D

2y

21 32

0 00x y z

The Vector Operator and The Divergence Theorem

12 C

vol12 C

Sd dv D S D


Homework 4D3.6. D3.7. D3.9. All homework problems from Hayt and Buck, 7th Edition.

Deadline: 8 February 2011, at 07:30.


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Application of Gauss’s Law: Differential Volume Element