Prof. D. WiltonECE Dept.

Notes 13

ECE 2317 ECE 2317 Applied Electricity and MagnetismApplied Electricity and Magnetism

Notes prepared by the EM group,

University of Houston.

Divergence -- Physical ConceptDivergence -- Physical Concept

r < a:

2 30

20

44

3

C/m3

encl

S

r v

vr

D n dS Q

r D r

rD

x

y

z

v = v0

a

r

Start by considering a sphere of uniform volume charge density

The electric field is calculated using Gauss's law:

r > a: 2 30

320

2

44

3

C/m3

r v

vr

r D a

aD

r

Divergence -- Physical Concept (cont.)Divergence -- Physical Concept (cont.)

x

y

z

v = v0

ar

2ˆ 4 r

S

D n dS r D

30

4

3 va (r > a)

(r < a)

Flux through a spherical surface:

30

4

3 vr

20 C/m3v

r

rD

320

2C/m

3v

r

aD

r

Divergence -- Physical Concept (cont.)Divergence -- Physical Concept (cont.)

(r < a)

(r > a)

More flux lines are added as the radius increases (as long as we stay inside the charge region).

0S

D n dS

Observation:

V

S

Divergence -- Physical Concept (cont.)Divergence -- Physical Concept (cont.)

0S

D n dS

The net flux out of a small volume V inside the charge region is not zero.

Divergence is a mathematical way of describing this.

0 0

1ˆdiv lim lim

V VS

D D n dSV V

V

Definition of divergence:

Gauss’s Law -- Differential FormGauss’s Law -- Differential Form

Note: the limit exists independent of the shape of the volume (proven later).

0

1div lim

VS

D D n dSV

V

v (r)

encl v

S

D n dS Q r V

0

div lim encl

V

v

QD r

Vr

Apply divergence definition to small volume inside a region of charge

Gauss’s Law -- Differential FormGauss’s Law -- Differential Form

Gauss’s Law -- Differential Form (cont.)Gauss’s Law -- Differential Form (cont.)

div vD r r

The electric Gauss law: This is one of Maxwell’s equations.

0 0

0

avg

0

div lim lim

1lim

lim

encl

V V

vVV

vV

v

QD r

V V

dVV

r

Alternatively,

ExampleExample

Choose V to be small sphere of radius r:

x

y

z

v = v0

a

r

0

1div lim

VS

D D n dSV

34

3V r

3

2 200

44 4

3 3v

r v

S

r rD n dS D r r

30

0 00 03

41div lim lim4 3

3

vv v

V V

rD

r

Verify that the differential form of Gauss’s law gives the correct result at the origin for the example of a sphere of uniform volume charge density.V

Calculation of DivergenceCalculation of Divergence

,0,02

,0,02

0, ,02

0, ,02

0,0,2

0,0,2

x

S

x

y

y

z

z

xD n dS D y z

xD y z

yD x z

yD x z

zD x y

zD x y

Assume point of interest is at the origin for simplicity.

The integrals over the 6 faces are approximated by “sampling” the integrand at the centers of the faces.

0

1div lim

VS

D D n dSx y z

x

y

z

(0,0,0)

x

y

z

Calculation of Divergence (cont.)Calculation of Divergence (cont.)

1

,0,0 ,0,02 2

0, ,0 0, ,02 2

0,0, 0,0,2 2

S

x x

y y

z z

D n dSx y z

x xD D

xy y

D D

y

z zD D

z

,0,02

,0,02

0, ,02

0, ,02

0,0,2

0,0,2

x

S

x

y

y

z

z

xD n dS D y z

xD y z

yD x z

yD x z

zD x y

zD x y

0

1div lim

VS

D D n dSx y z

Calculation of Divergence (cont.)Calculation of Divergence (cont.)

0

,0,0 ,0,02 2

div lim

0, ,0 0, ,02 2

0,0, 0,0,2 2

x x

V

y y

z z

x xD D

Dx

y yD D

y

z zD D

z

Hence div yx zDD D

Dx y z

For arbitrary origin, just add x,y,z to coordinate quantities in parentheses!

Calculation of Divergence (cont.)Calculation of Divergence (cont.)

div yx zDD D

Dx y z

0

1div lim

VS

D D n dSV

The divergence of a vector is its “flux per unit volume”

““del operator”del operator”

x y zx y z

: sin cos

: sin cos

sin sin cos

ˆ ˆsin sin cos

d dx x

dx dxd d

x x x x xdx dx

d dx x x x x x x

dx dx

d dx y x z x z x

dx dx

(1)

(2)

(3)

Examples of derivative operators:

scalar

vector

The “del” operator is a vector differential operator

scalar -> vector

vector->scalar

vector->vector

scalar -> scalar

ExampleExample

, , sin 3V x y z x x y y z xy

Find V

, , sin 3V x y z x y z x x y y z xyx y z

, , sin 3V x y z x y xyx y z

, , cos 3 0 3 cosV x y z x x

““del operator” (cont.)del operator” (cont.)Now consider:

x y z

yx z

D x y z xD yD zDx y z

DD Dx x y y z z

x y z

yx zDD D

Dx y z

divD D

Hence

sovD

Gauss's law :

Note: No unit vectors appear!

Summary of Divergence FormulasSummary of Divergence FormulasRectangular:

Cylindrical:

Spherical:

yx zDD D

Dx y z

1 1 zD D

D Dz

22

1 1 1sin

sin sinr

DD r D D

r r r r

Note the dot the del

operator is important; any

symbol following it tells

us how to use and read it:

"gradient"

"divergence"

"curl"

after

The divergence of a vector is its “flux per unit volume”

ExampleExample

x

y

z

v = v0

a

r < a :

0vD Note: This agrees with the electric Gauss law.

22

2 02

202

1

1

3

1

r

v

v

D r Dr r

rr

r r

rr

r

Evaluate the divergence of the electric flux vector inside and outside a sphere of uniform volume charge density, and verify that the answer is what is expected from the electric Gauss law.

0

3v r

D r

Example (cont.)Example (cont.)

x

y

z

v = v0

a

r > a :3

023

v aD r

r

0D

Note: This agrees with the electric Gauss law.

32 0

2 2

30

2

1

3

10

3

v

v

aD r

r r r

a

r r

Maxwell’s EquationsMaxwell’s Equations

0v

BE

tD

H Jt

D

B

Faraday’s law

Ampere’s law

electric Gauss law

magnetic Gauss law

Divergence TheoremDivergence Theorem

V

S

V S

A dV A n dS

The volume integral of “flux per unit volume” equals the total flux!

( , , )A A x y z an arbitrary

vector function of position

In words, for a vector :A

Divergence Theorem (cont.)Divergence Theorem (cont.)

Proof:

V

0

1

limn

N

rVnV

A dV A V

rn is the center of cube n

From the definition of divergence:

0

1lim

1

n

n

n

r VS

S

A A n dSV

A n dSV

Hence:

0 0

1 1

lim limn

n

N N

rV Vn nV S

AdV V A A n dS

Divergence Theorem (cont.)Divergence Theorem (cont.)

Hence: the surface integral cancels on all INTERIOR faces.

0

1

limn

N

VnV S

AdV A n dS

2n

1n

1 2

A n

Consider two adjacent cubes:

is opposite on the two faces

V

Divergence Theorem (cont.)Divergence Theorem (cont.)

01

0outsidefaces

lim

lim

n

n

N

VnV S

VS

A dr A n dS

A n dS

V S

AdV A n dS

Hence: 0

outsidefaces

limn

VV S S

A dV A n dS A n dS

Therefore: (proof complete)

V

Divergence Theorem (cont.)Divergence Theorem (cont.)

n

n

The vol. integral of the “flux per unit volume” is the “flux”

ExampleExample

x

y

z

1 [m]

2 [m]3 [m]

3A x x

Verify the divergence theorem using the box.

3 3 2 3 0 2

18S

A n dS x x x x

3 3

yx zAA A

Ax y z

xx

3 3 3 1 2 3 18V V

A dV dV V

Given:

From the divergence theorem:

Validity of Divergence DefinitionValidity of Divergence Definition

0

1div lim

SV

D D dSV

n

V

S

0

0

1div lim

1lim

r

VV

rV

DV

D

dV

DV

D

V

n̂

Is this limit independent of the shape of the volume?

r

Hence, the limit is the same regardless of the shape of the limiting volume.

This is valid for any volume, so let V V (a small volume inside the original volume)

Gauss’s Law Gauss’s Law (Conversion between forms)(Conversion between forms)

V

enclQ

S

encl

S

D n dS Q

V 0

v

v

V V

v

D dV dV

D V V

Divergence theorem:

Hence:

v

V S

encl v

V

D dV D n dS

Q dV

vD

Gauss’s Law Gauss’s Law (Summary of two forms)(Summary of two forms)

encl

S

D n dS Q

vD

Divergence theorem

Integral (volume) form of Gauss’s law

Differential (point) form of Gauss’s law

Divergencedefinition