Dr. Ahmed Said Eltrass

Electrical Engineering Department

Alexandria University, Alexandria, Egypt

Fall 2015

Part-I: Electro-Static

Lecture 5

Office hours: Sunday (10:00 to 12:00 a.m )

4th floor, Electrical Engineering Building

Application of Gauss’s Law: Differential Element Volume

• Apply Gauss’s law to a problem that does not possess any

symmetry

• Without symmetry, a Gaussian surface cannot be chosen such

that the normal component of D is constant or zero

everywhere on the surface

• Choose a very small closed surface (Δν ) where D is almost

constant over the surface

• In this procedure, we will not obtain a solution for D

• Instead we will obtain some valuable information about D

Chapter 3 (Continued)

Electric Flux Density, Gauss’s

Law, and Divergence

Gauss’ Law in the Point Form (Differential Form)

zyx

DΔxDdsD

x

DΔxDD

xDΔx

DD

PΔx/

aDD

zyDdsD

azyaDSDdsD

xx

xx

xx

x

xx

2

2

with of change of rate2

hence and , from 2at is faceFront

constantely approximat is D small, very iselement surface The

0

0

0

front

front x,

front x,

front x,front x,

front x,

front

front x,front

front

front

front back

front back

back

back x,

back x,

back

back x,back

back

back

22

integrals two theseCombine

2

2

hence and , from 2at is faceBack

)(

surfaceback over the integral thenowConsider

00

0

0

zyΔxx

DdsDdsD

zyx

DΔxDzy

x

DΔxDdsDdsD

zyx

DΔxDdsD

x

DΔxDD

PΔx/

zyDdsD

azyaDSDdsD

x

xx

xx

xx

xx

xx

top bottom

right left

procedure same theFollowing

zyΔxz

DdsDdsD

zyΔxy

DdsDdsD

z

y

zyΔxz

D

y

D

x

DQS zyx

S

dD

parts three theCombining

The expression is an approximation which becomes better as

Δν becomes smaller

Δvz

D

y

D

x

DQS zyx

S

dD

2

39-

C/m 2cossin

origin. at the located 10 of volumelincrementaan in

enclosed charge totalfor the valueeapproximatan Find -7

zy

x

x

x azayeayeD

m

Example:

Allowing the volume element Δν to shrink to zero, an exact

relationship can be derived

The divergence of the vector flux density D is the outflow of flux

from a small closed surface per unit volume as the volume shrinks

to zero.

Maxwell’s First Equation vD

div

vΔv

S

Δv

zyx

Δv

Q

Δv

S

z

D

y

D

x

D

limlim

00

dD

vS

Δv Δv

S

div

dD

DD of Divergence lim0

Gauss’ Law in

the Point Form

Δvz

D

y

D

x

DQS zyx

S

dD

• The physical meaning of this form is defining the divergence of electric

flux density vector as per unit volume flux emerging from an incremental

volume (point) due to the charge (source) distribution in space.

• The divergence refers to a flux that diverges (or converges)

as determined by the charge distribution.

• In the absence of ρv, the divergence of the electric flux is zero, which

indicates that the flux does not diverge (or converge).

• The presence of a charge distribution at the location (ρv ≠0) will cause the

flux to diverge (positive ρv) or converge (negative ρv causes negative

divergence).

• A positive divergence for any vector quantity indicates a source

of that vector quantity at that point. Similarly, a negative divergence

indicates a sink.

Maxwell’s First Equation

vD

div

Maxwell’s First Equation

Will be given in exams

vD

div

• In different coordinate systems

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.

vS

dvS DdD

Proof

vS

v

v

v

S

v

v

S

dvS

dvS

dvQ

QS

DdD

D :equationfirst sMaxwell’ From-

dD

where

dD :law Gauss from Starting

enclosed

enclosed

The total flux crossing the closed surface is equal to the integral of

the divergence of the flux density throughout the closed volume

22 2

3 and 0 and 2, and 01, and 0

planes by the formed ipedparallelepr rectangula Over the

field for the theoremdivergence theof sidesboth Evaluate -8

C/maxaxyD

zyx

D

yx

Example: