Section 4 6 TRƯỜNG ĐIỆN TỪ

Slide Presentations for ECE 329,
Introduction to Electromagnetic Fields,
to supplement “Elements of Engineering
Electromagnetics, Sixth Edition”
by

Nannapaneni Narayana Rao

Edward C. Jordan Professor of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, Urbana, Illinois, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India


4.6
Boundary Conditions


4.6-3

Why boundary conditions?
Medium
1
Inc.
wave

Ref.
wave

Medium
2

Trans.
wave


4.6-4

Maxwell’s equations in integral form must be satisfied
regardless of where the contours, surfaces, and volumes are.
Example:

C3

C1

Medium 1

C2

Medium 2


4.6-5

Example of derivation of boundary conditions
d
C E  d l  dt S B  d S
Medium 1
an

Lim
ad  0
bc  0



abcda

as

a

b

d

c

E  d l 

Lim
ad  0
bc  0

Medium 2

d
area B  d S

dt abcd

4.6-6

Eab ab   Ecd cd  0
Eab  Edc 0

aab  E1  E2  0

as × an  E1  E2  0

as  an × E1  E2  0
or,

an × E1  E2  0

Et1  Et 2 0


4.6-7

Summary of boundary conditions

an × E1  E2  0

or

Et1  Et 2 0

an × H1  H2  J S or

Ht1  H t 2  J S

an  D1  D2  S or

Dn1  Dn 2 S

an  B1  B2  0

or

Bn1  Bn 2 0


4.6-8

Perfect Conductor Surface
(No time-varying fields inside a perfect conductor. Also
no static electric field; may be a static magnetic field.)
Assuming both E and H to be zero inside, on the surface,

an × E = 0

or

Et 0

an × H = J S or

Ht  J S

an  D   S

or

Dn S

an  B 0

or

Bn 0


4.6-9

an

E
+

-

E

 
an
JS

H

 

H

JS


4.6-10

Dielectric-Dielectric Interface
S 0, J S 0

an × E1  E2  0

or

Et1 Et 2

an × H1  H2  0

or

Ht1 Et 2

an  D1  D2  0

or

Dn1 Dn 2

an  B1  B2  0

or

Bn1 Bn 2


4.6-11

an

Medium 1, e0

Dn1

En1

Dn2

En2

Bn1

Hn1

Bn2

Hn2

Et1
Et2
Medium 2, 3e0

an
Medium 1, m0

Ht1
Ht2

Medium 2, 2m0


4.6-12

Example:
D4.11 At a point on a perfect conductor surface,





(a) D D0 ax  2a y  2az and pointing away from
the surface. Find S . D0 is positive.

D D0 ax  2a y  2az 
an  
D
D0 ax  2a y  2az
D

D
S an  D =
D=
D
D

2

 D D0 ax  2a y  2az 3D0


4.6-13





(b) D D0 0.6 ax  0.8 a y and pointing toward the
surface. D0 is positive.

D
an 
D

D
D
S an  D = 
D = 
D
D

2

 D  D0 0.6 ax  0.8 a y
 D0


4.6-14

Example:

E1 E0 az for r  a.

(a) At 0, 0, a ,

an az

E1 is entirely normal.
 D2 D1 2  0 E1

D2
E2  2E1 2 E0 az
0

z

r>a
e 2 =e 0

(0, 0, a)

a a 

0,
,


2
2



(0, a, 0)
re 1 = 2e 0

y


4.6-15

(b) At 0, a, 0 ,

an a y

E1 is entirely tangential

E2 E1 E0 az
a a 


,
(c) At  0,
,
2 2

1
an  a y  az 
2
an × E2  E1  0 
 Solve.
an  D2  D1  0 