Chapter 7. Poisson’s and Laplace’s equationsocw.sogang.ac.kr/rfile/2013/course13-Electromagnet...Poisson’s and Laplace’s equations. EMLAB 27.1 Derivation of Poisson’s & Laplace’s

EMLAB

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Chapter 7. Poisson’s and Laplace’s equations

EMLAB

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7.1 Derivation of Poisson’s & Laplace’s equations

EDD ,

)equationsPoisson'(2

V

VE

)()( VE

VV 2

for homogeneous medium

2

2

2

2

2

2

ˆˆˆˆˆˆ

ˆˆˆ

z

V

y

V

x

V

z

V

y

V

x

V

zyx

z

V

y

V

x

VV

zyxzyx

zyx

Laplace operator has different forms

for different coordinate systems.

EMLAB

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Laplace’s equations

The differential equation for source-free

region becomes a Laplace equation.

)equations'Lapace(0V2

2

2

2

2

2

2 11

z

VVVV

2

2

2

2

2

22

z

V

y

V

x

VV

2

2

222

2

2

2

sin

1sin

sin

11

V

r

V

rr

Vr

rrV

(rectangular coordinate)

(cylindrical coordinate)

(spherical coordinate)

EMLAB

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7.2 Uniqueness theorem Solution of a differential equation

The solution that satisfies the differential equation and its

boundary condition is unique regardless of the solution

procedure.

conditionboundary

),( yxVb

)y,x(Vinterior)(

)()boundary(,0

)()boundary(,0

21

22

2

11

2

VV

fVV

fVV

r

r

022

2

VV

e ddW

EE

boundary:S To prove the uniqueness theorem, we assume that there exist

two distinct solutions that satisfy the same boundary condition.

Then, the difference of the those two solutions will have zero

value at the boundary and will have non-zero value in the

interior region.

0)interior(

0)boundary(,02

21

VV

0(interior)

This situation is contradictory to the original assumption that

there exist two distinct solution that satisfy the same boundary

condition.

0constant(interior)0(interior)

)0(02

)(2

)(22

22

Sondd

dd

SV

VV

a

EMLAB

5

Laplace’s equations

0r

Vr

r

1

r

1V

r

1)r(V)3(

0V11

V

a/bln

/bln)(V)2(

0exsinexsinV

exsin)y,x(V)1(

2

2

2

2

yy2

y

• The potentials V on the left side satisfy the same Laplace equation.

• Although they satisfy the same differential equation, the solutions are

different. The reason is that the boundary conditions are different.

• This example explains us that the exact solution to the differential

equation is obtained only when we are given the differential equation as

well as the boundary condition.

0zyx 2

2

2

2

2

2

V0 V0

V1

V0

V1

V1

)1( )2( )3(

ba

1

10

EMLAB

6

Example : Laplace eqs.

Sd

z

x

0V

V0

Unlike the procedures in the previous chapters, the potential V is

first obtained solving Laplace equation. Then, using the potential, E,

D, , Q, C are obtained.

02

2

2

2

2

2

2

22

zzyx

d

S

dV

SV

V

QC

d

SVSdaQ

d

V

d

V

d

V

d

V

zd

Vz

s

S

s

s

s

0

0

0

0

0

0

0

0

)6(

)5(

ˆˆbottom)(

ˆˆtop)()4(

ˆ)3(

ˆ)2(

)()1(

DzDn

DzDn

zED

zE

If the plates are wide enough to ignore the variation of

electric field along x and y directions

0,0

yx

constant),( BABAzAz

zd

Vz

d

VA

BVBAdd

00

0

)(

0)0(,)(

Using the boundary conditions on the two plates,

EMLAB

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Example 2

)/ln(

)/ln()(

)/ln(

ln,

)/ln(

ln)(,0ln)(

,conditionsboundary theUsing

constants),(ln,0

0,0

symmetricaxially anddirectionz in the infinite be toassumed iscylinder The

0111

000

0

2

2

2

2

2

2

2

2

2

2

ab

bVV

ba

bVB

ba

VA

VBaAaVBbAbV

BABAVAVV

z

VV

V

z

VVVV

)/ln(

2)6(

2)/ln(

1)5(

)/ln(

1ˆˆ)(

)/ln(

1ˆˆ)()4(

ˆ)/ln(

1)3(

ˆ)/ln(

1)2(

)/ln(

)/ln()()1(

0

0

0

0

0

0

ab

L

V

QC

aLaba

VdaQ

abb

Vb

aba

Va

ab

V

ab

VV

ab

bVV

S

s

s

s

DρDn

DρDn

ρED

ρE

x

y

r

ab

V00V

EMLAB

8

2tanln

2tanln

)(0,

2tanln

02

,2

tanln)(

,conditionsboundary theUsing

2tanln

2tan

2sec

2

1

2cos

2cos

2sin2

2cos

2sin2

sin

) ,(sin

,sin0sin

0,0

symmetricaxially are anddirection radial thealonginfinity toextend surfaces The

0sinsin

1

sin

1sin

sin

11

00

0

2

2

2

2

2

2

2

222

2

2

2

VVBV

A

BVVBAV

A

d

AdAdAdA

BABdA

VAVV

V

r

V

V

r

V

r

V

rr

Vr

rrV

상수는

0V

V0

2cotln

r2

V

QC)6(

dr

2cotln

V2

sinr

drdsinr

2cotln

VdaQ)5(

2cotlnsinr

Vˆˆ)a()4(

ˆ

2tanlnsinr

V)3(

ˆ

2tanlnsinr

VˆV

r

1V)2(

2tanln

2tanln

V)(V)1(

1

r

0

0

r

0

2

0

0

S

s

0s

0

0

0

11

DθDn

θED

θθE

Example 3

EMLAB

Example : Poisson eq. Diode (simple model)

V

P

NNx

Px

P-type N-type

VV 2

V

dx

Vd

2

2

1Cxdx

dV V

02

2

dx

Vd02 C

dx

dV03 CV

1Cxdx

dVE P

x

)(,

0)(

1

1

pP

xpP

pP

px

xxExC

Cxdx

dVxE

)( pP xx

dx

dV

2

2)(2

)( CxxxV pP

0)( 2 CxV p

2)(2

)( pP xxxV

xE

PN

NxPx

)( pN xx

dx

dV

2

2)(2

)( CxxxV NN

2

)(2

)(2

)0(

22

2

2

2

2

PPNN

PP

NN

xxC

xCxxV

2)(

2)(

222 PPNN

NN xx

xxxV

V

NxPx

P-type region

N-type region

9

EMLAB

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Method of separation of variables

)()()(),,(

02

2

2

2

2

2

zZyYxXzyx

zyx

0)(1

011

0111

0111

,by Divided

0

22

2

222

2

2

2

2

22

2

22

2

2

2

2

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

Zdz

Zd

dz

Zd

Z

Ydy

Yd

dz

Zd

Zdy

Yd

Y

Xdx

Xd

dx

Xd

Xdz

Zd

Zdy

Yd

Y

dz

Zd

Zdy

Yd

Ydx

Xd

X

XYZ

dz

ZdXY

dy

YdXZ

dx

XdYZ

zyx

s.conditioinboundary thefrom fixed becan ,,,,,

coshsinh)(

cossin)(

cossin)(

0)(

0

0

0111

2222

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

FEDCBA

zFzEzZ

xDxCyY

xBxAxX

Zdz

Zd

Ydy

Yd

Xdx

Xd

dz

Zd

Zdy

Yd

Ydx

Xd

X

•The original equation on the left is split into three

equations containing single variable. The

equations are related to one another through

variables and .

• , can be fixed using boundary conditions.

• If the boundary condition becomes complex, three dimensional Laplace

equation is very difficult to solve. This is because the Laplace equation is a

partial differential equation that contains three variables x, y and z.

• If the boundary is parallel to coordinate axes, the solution to Laplace

equation can be represented as a multiplication of three functions that contain

only one variable.

• This method is called as a method of separation of variables.

EMLAB

11

Example : Separation of variables

0VV0

V0

V0 a

b

)y,x(V

)y(Y)x(X)y,x(

0yx 2

2

2

2

If the boundary extends to infinity in

the z-direction, the derivative with

respect to z becomes zero.

0

1n

n

1n

n

2

2

2

2

2

2

2

2

2

2

Vb

ynsin

b

ansinhA)y,a(V

b

ynsin

b

xnsinhA)y,x(V

),3,2,1n(b

nnb

0bsinxsinhA)b,x(V

0D

0D)xcoshBxsinhA()0,x(V

0B

0)ycosDysinC(B)y,0(V

)ycosDysinC(

)xcoshBxsinhA()y,x(V

ycosDysinCY

xcoshBxsinhAX

0Ydy

Yd

0Xdx

Xd

0dy

Yd

Y

1

dx

Xd

X

1

1n

0

0

m

0m

m0

b

0

0

b

0

0m

0

1n

n

b

y)1n2(sin

b

a)1n2(sinh)1n2(

b

x)1n2(sinh

V4)y,x(V

m;

b

amsinhm

V4

b

amsinh

])1(1[

m

V2A

])1(1[m

bV

b

ymcos

m

bVdy

b

ymsinV

2

b

b

amsinhA

b~0 b

ymsin

Vb

ynsin

b

ansinhA)y,a(V

홀수은

적분하면까지곱하고를양변에

nmfor2

b

nmfor0

b

y)mn(sin

mn

1

b

y)mn(sin

mn

1

2

b

dyb

y)mn(cos

b

y)mn(cos

2

1dy

b

ymsin

b

ynsin

b

0

b

0

b

0

EMLAB

12

Result-Matlab code

% potential_demo.m

% 변수 분리 방법에 의해 푼 전압을 그림.

clear,clf;

hold on;

imax = 30;

jmax = 30;

a=2.5; b = 1;

for i=1:imax+1

for j=1:jmax+1

x(i,j) = (i-1)*a/imax;

y(i,j) = (j-1)*b/jmax;

z(i,j) = Vseries(x(i,j),y(i,j));

end

end

colormap jet;

surf(x,y,z);

shading interp;

%caxis([0,100]);

colorbar('vert');

view([0,90]);

%Vseries.m

% 변수 분리 방법에 의한 전압 계산

function f=Vseries(x,y)

V=100;

f=0;

a=2.5;

b=1;

for i=1:10

n=2*i-1;

f=f+sinh(n*pi*x/b)/sinh(n*pi*a/b)*sin(n*pi*y/b)*(1/n);

end

f=f*4*V/pi;

6.12 문제

A

d

z

x

0V

V0

d10

x

0V

V0

(a) 전지가 연결되어 있으므로 전압은 그대로.

(b) 전기장을 적분한 것이 전압인데,

전압은 그대로인 채 거리만

늘어났으므로 전기장의 세기는 1/10 배로 됨.

(c) D는 전기장에 비례하므로 1/10

배.

(d) 전하 Q는 D에 비례하므로 마찬가지로 1/10.

(e) _s도 Q에 비례하므로 1/10.

(f) 전기 에너지는

2

2

1CVWE

13

EMLAB

문제 7.6

z

0

d

]/[ 3

0 mC

(a) Determine the potential field between plates.

21

201

0

0

2

22

2, CzCzCz

dz

d

dz

d

02

)(

0)0(

21

20

2

CdCddz

Cz

)(2

)(

2,0

2)(

0)0(

0

011

20

2

zdzz

dCdCddz

Cz

(b) Determine the electric field E between plates.

)2(2

ˆ 0 dzz

E

14

EMLAB

문제 7.10

z

0

d

]/[ 3

0 mC

b

(a) Determine the potential field between plates.

Region I

21

201

0

0

2

22

2, CzCzCz

dz

d

dz

d

Region II

Region II

211

2

22

,

00

CzCCdz

d

dz

d

1010

21 CCbDD zz

bCC

0

01

0

1

2121

2021

2)()( CbCCbCbbVbV

0,00)()0( 21221 CdCCdVV

)()(2

)(

)()(

)(2

2)(

0

2

02

01

bzbdb

zdbzV

bzb

z

bdb

bdbbzzV

R

R

R

15

EMLAB

문제 7.25

yb

VyaV 0),(

V0

V0

V0 a

b

)y,x(V

)y(Y)x(X)y,x(

0yx 2

2

2

2

z축으로 무한히 뻗어 있는 구조라고 가정하면 z에 대한 미분은 0이

된다.

yb

V

b

yn

b

anAyaV

b

yn

b

xnAyxV

nb

nnb

bxAbxV

D

DxBxAxV

B

yDyCByV

yDyC

xBxAyxV

yDyCY

xBxAX

Ydy

Yd

Xdx

Xd

dy

Yd

Ydx

Xd

X

n

n

n

n

0

1

1

2

2

2

2

2

2

2

2

2

2

sinsinh),(

sinsinh),(

),3,2,1(

0sinsinh),(

0

0)coshsinh()0,(

0

0)cossin(),0(

)cossin(

)coshsinh(),(

cossin

coshsinh

0

0

011

1

1

0

1

0

10

0

2

0

0

0

0

1

sin

sinh

sinh)1(2

),(

;

sinh

)1(2

)1(sincos

sin2

sinh

~0 sin

sinsinh),(

n

n

m

m

m

b

b

m

n

n

b

yn

b

ann

b

xnV

yxV

m

b

amm

VA

m

bV

b

ym

m

b

b

ym

m

yb

b

V

dyb

ymy

b

Vb

b

amA

bb

ym

yb

V

b

yn

b

anAyaV

홀수은

적분하면까지곱하고를양변에

nmfor2

b

nmfor0

b

y)mn(sin

mn

1

b

y)mn(sin

mn

1

2

b

dyb

y)mn(cos

b

y)mn(cos

2

1dy

b

ymsin

b

ynsin

b

0

b

0

b

0

16

EMLAB

반지름이 0.2mm 인 평행한 원통 도체 2개로

이루어진 전선의 중심간 거리가 2mm일 때 다음을 구하라. 도체를 둘러 싼 매질의 비유전율은 3이고, 전기 전도도는 1.5 mS/m이다.

(1) 단위 길이 당 capacitance.

(2) 단위 길이 당 저항.

(3) 단위 길이 당 누설 전류.

V1003r

2mm 0.2mm

V0

mAR

VI

RRC

pFC

bh

L

bh

L

V

L

V

QC

bhb

bhhV

b

bhheK

b

bhhK

K

K

h

b

K

Kab

K

Kah

K

Kay

K

Kax

Keyax

yax

yax

yax

r

rV

yaxryaxr

r

r

rr

r

rr

r

rV

r

r

r

L

LL

r

L

r

L

V

V

r

L

r

L

r

L

r

L

r

L

L

r

L

r

205)3(

487)2(

4.36)1(

)/(cosh)/(cosh

2

)/(coshln2

,1

2

1

2,

1

1

1

2

1

1

)(

)(

)(

)(log

4log

2

)(,)(

ln2

ln2

ln2

.2

0

1

0

1

0

1

0

22

0

222

1

22

1

1

1

1

1

1

1

2

1

12

2

1

1

1

4

22

22

22

22

01

2

0

22

2

22

1

1

2

202

0

201

0

10

0

0

문제 6.31 17

EMLAB