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Lesson 11–13Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu

Method of images replaces the original boundary by appropriate image charges inlieu of a formal solution of Poissonʼs or Laplaceʼs equation so that the original problemis greatly simplified.

The basic principle of the method of images is the uniqueness theorem. As long as (1)the solution satisfies Poissonʼs or Laplaceʼs equation and (2) the solution satisfies thegiven boundary condition, the simplest solution should be taken.

3.5 Method of Images

Point charge over grounded plane conductor

where R1 is the distance from ds to the point underconsideration and S is the surface of the entireconducting plane.

dsRzdyx

QzyxV

S

s!++"+

=10

222

04

1

)(4),,(

#

$%$%

By direct solution

)11

(4

),,(0 !+

!=RR

QzyxV

"#

Only valid in the region of y > 0.

By method of images


Example A positive point charge Q is located at distances d1 and d2, respectively,from two grounded perpendicular conducting half-planes, as shown in the figure.Determine the force on Q caused by the charges induced on the planes.




li!! "=

Line Charge And Parallel Conducting Cylinder

Letʼs take a trial solution (an intelligent guess) that

We have found that the E field generated by a line charge is )/( 2 0

mVr

l

r

!"

#aE =

The electric potential at a distance r from a line charge of density can be obtained byintegrating the electric field intensity E

l!

r

rdrr

drEVl

r

r

l

r

r

r

0

00

ln2

1

200

!"

#

!"

#=$=$= %%

Note that the reference point for zero potential, r0, cannot be at infinity. Let us leave r0unspecified for the time being.



l!

The potential at a point on or outside the cylindricalsurface is obtained by adding the contributions ofand . In particular, at a point M on the cylindricalsurface, we have

Equipotential surfaces are specified by

Therefore, triangles OMPi and OPM similar. We have

r

r

r

r

r

rV

il

i

ll

Mln

2ln

2ln

20

0

0

0

0!"

#

!"

#

!"

#=$=

We have simplified the solution by considering the r0 is so large that the distance of the reference point to and is negligible.

i!

l!

i!

Constant=r

ri

OP

OM

OM

OP

PM

MPii ==

orConstant===

d

a

a

d

r

rii

From the above equation we see that if the image line charge , together with , will make the dashed cylindrical surface in the figure equipotential. As the point Mchanges its location on the dashed circle, both ri and r will change; but their ratio remains a constant that equals a/d.

dadi

/2

=

l!

l!"



Two Parallel Conducting Cylinders of The Same Radius

d

aV

lln

20

2

!"

#=

r

rV

il

Mln

20

!"

#=

d

a

a

d

r

rii ==

Since

and

d

aV

lln

20

1

!"

#$=We have and

The capacitance per unit length is)/ln(

0

21 adVVC

l!"#

=$

=

whered

aDdDd

i

2

!=!= from which we obtain )4(2/1 22aDDd !+=

Consequently, )/( )2/(cosh]1)2/()2/ln[(

1

0

2

0mF

aDaDaD

C!

=!+

="#"#



Two Parallel Conducting Cylinders of The Same Radius (cont.)


We now develop a method for solving 3-D problems where the boundaries, over whichthe potential or its normal derivative is specified, coincide with the coordinate surfacesof an orthogonal, curvilinear coordinate system. In such cases, the solution can beexpressed as a product of three one-dimensional functions, each dependingseparately on one coordinate variable only, The procedure is called the method ofseparation of variables.

Classifications of boundary-value problems:

Dirichlet problems,in which the value of the potential is specified everywhere on the boundaries;

Neumann problems,in which the normal derivative of the potential is specified everywhere on the

boundaries;

Mixed boundary-value problems,in which the potential is specified over some boundaries and the normal

derivative of the potential is specified over the remaining ones.

3.6 Method of Separation of Variables


Letʼs investigate Laplaceʼs equation for scalar electric potential V in Cartesiancoordinates system first:

)()()(),,( zZyYxXzyxV =To apply the method of separation of variables, we assume

I order for the above equation to be satisfied for all values of x,y,z, we must have:

Therefore,


02

2

2

2

2

2

=!

!+

!

!+

!

!

z

V

y

V

x

V

0)(

)()()(

)()()(

)()(2

2

2

2

2

2

=++dz

zZdyYxX

dy

yYdzZxX

dx

xXdzZyY

That is 0)(

)(

1)(

)(

1)(

)(

12

2

2

2

2

2

=++dz

zZd

zZdy

yYd

yYdx

xXd

xX

,)(

)(

1 2

2

2

xk

dx

xXd

xX!= ,

)(

)(

1 2

2

2

ykdy

yYd

yY!=

2

2

2 )(

)(

1zk

dz

zZd

zZ!=

,0)()( 2

2

2

=+ xXkdx

xXdx

,0)()( 2

2

2

=+ yYkdy

yYdy 0)(

)( 2

2

2

=+ zZkdz

zZdz

Consequently,

and ,0222=++ zyx kkk


Constant k and the function forms are determinedby the given boundary conditions. For example, ifthe potential V approaches to 0 when x approachesto infinite, the possible solution form is witha positive real k.


kxDe

!


With V independent of z, we have


),,(),,( yxVzyxV =

EXAMPLE Two grounded,semi-infinite,parallel-plane electrodes are separated by adistance b.A third electrode perpendicular to and insulated from both is maintained at aconstant potential V0. Determine the potential distribution in the region enclosed by theelectrodes.

kxeDxX!

= 2)(

Solution:

In the x-direction: 0),( ,),0( 0 =!= yVVyV

In the y-direction: 0),( ,0)0,( == bxVxV

Since , we choose , where k is a positive real number and0),( =! yV

kkjkkkkk yxxy ==!="= , 222

The boundary conditions in y-direction suggest that )sin()( 1 kyAyY =

An appropriate solution of the Laplaceʼs equation satisfying partial boundary conditions(namely, harmonic functions) is

)sin(),( 120 kyeADByxV kx

C

n

n

!=

321

0)( BzZ =


The constant k is determined by the BC aty=b, that is


Therefore, the harmonic function becomes

In order to satisfy the BC at x=0, we use the principle of linear superposition of Vn to findthe specific solution of the given BC:

In order to evaluate the coefficients , we multiply both sides of the equation byAnd integrate the products from y=0 to y=b:

)sin(),( 120 kyeADByxV kx

C

n

n

!=

321

0)sin(),( ==!

kbeCbxVkx

nn 0)sin( =kb ,...3,2,1 ,/ == nbnk !

)sin(),( / yb

neCyxV bxn

nn

!!"=

!!"

=

#"

=

==1

/

1

)sin(),(),(n

bxn

n

n

n yb

neCyxVyxV

$$ byVyb

nCyV

n

nn <<==!"

=

0 ,)sin(),0( 0

1

#

nC )sin( y

b

m!

!"! =#

=

b

n

b

n dyyb

mVdyy

b

my

b

nC

0

0

10

)sin()sin()sin($$$



!"! =#

=

b

n

b

n dyyb

mVdyy

b

my

b

nC

0

0

10

)sin()sin()sin($$$


Method of separation of variables in Cylindrical coordinates


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