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Elektromagnetika TeknikCHAPTER 3 ELECTRIC FLUX AND DIVERGENCE

Bab 3Elemen Volume Diferensial, Divergensi dan

Persamaan Pertama Maxwell

Engineering Electromagnetics

Aplikasi Hukum Gauss: Elemen Volume Diferensial

Chapter 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. We intend eventually to allow this volume to approach zero.

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


Taylor’s Series Expansion


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:

S

d Q D S

front back left right top bottomS

d D S



We 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 2

x x x

xD D D x

02

xx

DxD

x



We have now, for front surface:

0front

2

xx

DxD y z

x

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

back backback

D Sback ( )xy z D a

,backxD y z

,back 02

xx x

DxD D

x

0back

2

xx

DxD y z

x


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


front back xD

x y zx

right left

yDy x z

y

top bottom zD

z x yz

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

dan


Charge enclosed in volume yx z

DD Dv v

x y z


The previous equation is an approximation, which becomes better as Δv becomes smaller, and in the following section the volume Δv will be let to approach zero.

For the moment, we have applied Gauss’s law to the closed surface surrounding the volume element Δv.

The result is the approximation stating that:



Let D = y2z3 ax + 2xyz3 ay + 3xy2z2 az nC/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 an approximate value for the total charge contained in an incremental sphere having a radius of 2 mm centered at P(3,2,1).

ψ SS

d D S(a)

1 2

2 3 3 2 2

0 03

2 3 x y z xz y

x

y z xyz xy z dydz

a a a a

1 22 3

0 0y z dydz

2 11 13 4

3 40 0y z

2

3nC

Contoh soal


(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 nC mx y z = a a a

2 2 2(4) (12) (36)P PD D =

238.158nC m

0

P

P

D

E2

12

38.158 nC m

8.854 10

4.31 kV m

Solusi


(c) yx zDD D

Q vx y z

yx zP

P

DD DQ v

x y z

43 2 3 3 3 33

321

0 2 6 nC m (2 10 ) mxyz

xz xy z

43 2 3 3

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

152.61 10 C

Solusi

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


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

Divergence

Divergence


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 SA 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.

Divergence


div yx z

DD D

x y z

D

1 1div ( ) z

D DD

z

D

2

2

1 1 1div ( ) (sin )

sin sinr

Dr D D

r r r r

D

Persegi

Silinder

Bola

Divergence


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

div yx z

DD D

x y z

D sin sin 2x xe y e y 2

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

Contoh Soal

Maxwell’s First Equation (Electrostatics)


We may now rewrite the expressions developed until now:

div yx z

DD D

x y z

D

0div lim S

v

d

v

D SD

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.

elektromagnetika teknikpersamaan pertama maxwell engineering electromagnetics. aplikasi hukum gauss:...

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