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Tiêu đề Boundary Conditions
Tác giả Nannapaneni Narayana Rao, Edward C. Jordan
Người hướng dẫn Distinguished Amrita Professor of Engineering
Trường học University of Illinois at Urbana-Champaign
Chuyên ngành Electrical and Computer Engineering
Thể loại Bài giảng
Thành phố Urbana
Định dạng
Số trang 15
Dung lượng 319 KB

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4 6 Boundary Conditions Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edw[.]

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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

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Boundary Conditions

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Why boundary conditions?

Medium

Inc.

wave Ref.

wave

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regardless of where the contours, surfaces, and volumes are Example:

C2

C1

C3

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Example of derivation of boundary conditions

d

dt



ElBS

abcda

d

dt

Medium 1

Medium 2

a

c d

b

n

a

s

a

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1 2 0

t t

EE

or,

   

0 0 0 0

ab

E ab E cd

 1 2  0

a × E E

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Summary of boundary conditions

a × E E

a × H H J

aD D

 1 2  0 or 1 2 0

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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,

or 0

a ×E = 0

or

n S HtJS

a × H = J

or

n S D n S

a D

0 or 0

a B

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  E

E

 

n

a

S

n

a

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Dielectric-Dielectric Interface

0, 0

a × E E

a × H H

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Medium 1, e0

Medium 2, 3e0

E t1

E t2

E n1

E n2

D n1

D n2

n

a

Medium 1, m0

Medium 2, 2m0

H t1

B n1

B n2

H t2

H n1

H n2 n

a

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

(a) and pointing away from

the surface Find DD0  x  2 y S 2 0z is positive 

0

0

2 2

2 2

n

D D

 

 

 

D a

2

 

D

D

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and pointing toward the

surface D0 is positive

0 0.6 x 0.8 y

D

(b)

a

D

2

0 0

D D



D

D

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1 is entirely normal.

E

2 1 2 e0 1

D   D E

(a)

nz

At 0, 0, , a

2

0

e

y

z

(0, 0, a)

r < a

(0, a, 0)

r > a

0, ,

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(c)

1 is entirely tangential

E

ny

2  1 E0 z

1 2

0

Solve 0

n n

  

  

a × E E

At 0, , ,

2 2

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