Chapter 4 : Solution of Electrostatic ProblemsLecture 8-1 Static Electromagnetics, 2007 SpringProf....

Chapter 4 : Solution of Electrostatic Problems Lecture 8-1

Static Electromagnetics, 2007 Spring Prof. Chang-Wook Baek

Chapter 4. Solution of Electrostatic Problems

• Poisson's equation

– Two governing differential equations for electrostatics :

In a linear, isotropic medium (DD = EE) :

For a simple medium (linear, isotropic, and homogeneous) :

– Poisson's equation in coordinate systems RCS :

CCS :

SCS :

, 0D E ��

( )E V ��

V2

)(V/m

ˆˆˆˆˆˆ

22

2

2

2

2

2

2

z

V

y

V

x

V

z

Vz

y

Vy

x

Vx

zz

yy

xxVV

2

2

2

2

22 11

z

VV

rr

Vr

rrV

2

2

2222

22

sin

1sin

sin

11

V

R

V

RR

VR

RRV

: Poisson's equations'

1'

4 VV dv

R

Laplacian operator



• Laplace's equation

– If the medium under consideration contains no free charges, Poisson's equation reduces

to

It is the governing equation for problems involving a set of conductors, such as capacitors,

maintained at different potentials.

• Example 4-1 : Parallel-plate capacitor

(a) The potential at any point between the plates

1. Ignoring the fringing field Electric field is

uniform between the plates (as if the plates were infinitely

large) No variation of V in the x and z directions.

2. No free charges in the dielectric : Using

Laplace's eq.

(b) The surface charge densities on the plate

2 0V : Laplace's equations

02

2

dy

Vd211 , CyCVC

dy

dV

0 0, 0

y d yV V V

B.C. : y

d

VV 0

d

Vy

dy

dVyVE 0ˆˆ

nsn ED

0su

V

d

At the upper plate :0

sl

V

d

, At the lower plate :

V2



• Uniqueness of electrostatic solutions

– Uniqueness theorem : A solution of Poisson's equation that satisfies the given

boundary conditions is a unique solution. A solution of an electrostatic problem satisfying its boundary solution is the only possible

solution, irrespective of the method by which the solution is obtained.

A solution obtained by intelligent guessing is the only correct solution.

(Proof) Assume that there are two solutions V1 and V2 satisfying Poisson's equation in .

0 boundaries conductingOn ,0

.in equation sLaplace' satisfies then , Define

,

2

21

22

12

dd

dd

VV

VVVV

VV

))(( )(22 fAAfAfVVVVV ddddd

)(2dvVSdVV dS dd

3/1 RConsidering very large S02R

02 dvVd

Since is nonnegative everywhere, should be zero. 2

dV dV

Vd is constant at all points in volume including bounding surfaces S1, S2, …. Sn where Vd = 0.

Vd = 0 throughout the volume V1 = V2

Integrating both sides over volume :



• Method of images

– Image charge method : In solving some electrostatic problems, the bounding surfaces

can be replaced with appropriate "fictitious" charges (image charges) to satisfy the

boundary conditions This must be the only solution according to uniqueness

theorem.

– Example : A point charge Q above a very large grounded conducting plane What is

the potential at every point above the conducting plane (y > 0)?

The potential above the conducting plane

satisfies Laplace's equation except at the

point charge.

Boundary conditions :

(1) V(x, 0, z) = 0, (2) At far point, V 0

What really happens is :

(1) The point charge Q produces an electric field.

(2) The field induces surface charges s on the grounded conducting plane.

In order to calculate V, we must find out s, but it is hard to determine s.

02

2

2

2

2

22

z

V

y

V

x

VV



(Cont'd)

(1) Potential above the conducting plate (y > 0)

Another point charge –Q (image charge) is

located at

y = – d and the conducting plate is removed.

This situation satisfies the original

boundary conditions.

This potential is the only solution of the original

problem by uniqueness theorem.

(2) Surface charge density on the conducting plate

(0, d, 0)

(0, -d, 0)

region 0 in the ),,(point aat 11

4),,(

0

yzyxPRR

QzyxV

1/ 2 1/ 22 22 2 2 20

1 1( , , )

4

QV x y z

x y d z x y d z

2/32222/32220 )(

)(

)(

)(

4 zdyx

dy

zdyx

dyQ

y

VEy

3/ 2 32 2 20 0 0 0

2( ,0, )

4 2s

n y

Q d QdE E x z

Rx d z

222 dzx

302

s

induced sS

Qd

R

Q ds Q

[HW] Image charges must be located outsidethe region in which the filed is to be determined. Explain why.



• Boundary-Value Problems (BVP)

– BVP in electrostatics : the problem consisting of conductors maintained at specified

potentials with no isolated free charges This kind of problems cannot be solved

by the image method We must solve these problems by solving Laplace's

equation (it is a differential equation!) directly.

– Types of BVP Dirichlet problems : The value of potential is specified everywhere on the boundaries.

Neumann problems : The normal derivative of the potential is specified everywhere on the

boundaries

Mixed BVP : The potential is specified over some boundaries, and the normal derivative

of the potential is specified over the remaining ones.

– How to solve the Laplace's equation in general? One-dimensional problem (e.g. V is a function of one variable) : direct integration with

the given boundary condition

Two- or three-dimensional (e.g. V is a function of more than one variables) : separation of

variable method with the given boundary conditions



• Example 4-6 : Two grounded, semi-infinite parallel-plate electrodes by a

distance and maintained at a specified potential at one end

Sol) This is a BVP in Cartesian coordinate system Finding V

between the plates.

Assume that ( , , ) ( ) ( ) ( ) (i.e. separable). Inserting into the Laplace equation,V x y z X x Y y Z z V

2 2 2

2 2 20

V V V

x y z

2 2 2

2 2 2

2 2 2

2 2 2

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 0

1 ( ) 1 ( ) 1 ( )0 : Each of three terms should be constant.

( ) ( ) ( )

d X x d Y y d Z zY y Z z X x Z z X x Y y

dx dy dz

d X x d Y y d Z z

X x dx Y y dy Z z dz

2 2 22 2

2 2 2

2 22 2 2 2 2

2 2

1 ( ) 1 ( ) ( )0 ( ) 0

( ) ( )

( ) ( )Similarly ( ) 0, ( ) 0 , 0

x x

y z x y z

d d X x d X x d X xk k X x

dx X x dx X x dx dx

d Y y d Z zk Y y k Z z k k k

dy dz

separation constant

: X(x), Y(y), and Z(z) can be expressed as a second-order ordinary differential equation.



(Cont'd) Because of the infinity in the z-direction,

( , , ) ( , ) (1)V x y z V x y

0

2 2 2

From (1) : ( ) 0

In this case, must be imaginary and real. ( 0, real)

z

x y y x

Z z B k

k k k k k k

2

222

22

12

( )From (3) : ( ) 0 ( )

( )From (4) : ( ) 0 ( ) sin

kxd X xk X x X x D e

dx

d Y yk Y y Y y A ky

dy

0 2 1( , ) ( ) ( ) ( ) sin sinkx kxn nV x y X x Y y Z z B D A e ky C e ky

From (5) : ( , ) sin 0 sin 0 ( 1 2 3 )

( , ) sin

kxn n

n x

bn n

nV x b C e kb kb k n , , , ....

b

nV x y C e y

b

(why?)

Boundary conditions in the x, y directions are

0(0, ) (2), ( , ) 0 (3)

( ,0) 0 (4), ( , ) 0 (5)

V y V V y

V x V x b



(Cont'd)

The solution should also satisfy the boundary condition (2)

V(0, y) = V0 for all values of y from 0 to b.

: Since Laplace's equation is a linear partial differential equation, the superposition

of Vn(x, y) for the different values of n is also a solution Fourier series

expansion.

( , ) sinn x

bn n

nV x y C e y

b

01 1

(0, ) (0, ) sin , 0n nn n

nV y V V y C y y b

b

1

0 00 sin sin sin

n

bb

n dyyb

mVdyy

b

my

b

nC

02

if is odd,

0 if is even.

bVm

mm

dyyb

mny

b

mnC bn

)(cos

)(cos

2 0

if , 2

0 if .

nCb m n

m n

.even is if 0

odd, is if 4 0

n

nn

VCn

0

1

4 1( , ) sin sin

1, 3, 5, ...., 0 and 0

n x n x

b bn

n n odd

Vn nV x y C e y e y

b n b

n x y b



• Example 4-8 : The potential distribution between a long coaxial cable

Sol) This is a BVP in cylindrical coordinate system.

Laplace's equation in cylindrical coordinate system :

General solution of the above equation is known as

Bessel functions. (solution of Laplace’s equation in CCS)

Because the given structure is very long and has a symmetry in -direction, the

Laplace's equation in this problem must have only a r-dependence.

Integrating both sides yields

Boundary conditions : V(a) = 0, V(b) = V0

011

2

2

2

2

2

z

VV

rr

Vr

rr

10 0

V d dV dVr r r C

r r r dr dr dr

( )dV r C

dr r

( ) lnV r C r D

0

( ) ln 0

( ) ln

V a C a D

V b C b D V

0 0 0ln, ( ) ln

ln( / ) ln( / ) ln( / )

V V b V bC D V r

b a b a b a r



• Example 4-2 : Uniform spherical clouds of electrons

Sol) Poisson's equation in spherical coordinate system :

There are symmetries in the and directions :

(a) Inside the clouds : Solve Poisson's equation.

(b) Outside the clouds : Solve Laplace's equation .

At the surface R = b, the EE-field should be continuous

0 (0 ), 0 ( ) R b R b

22

2 2 2 2 20

1 1 1sin

sin sin

V V VR

R R R R R

2

20

1 VR

R R R

2 2 2 30 0 0 11 2

0 0 0

0 12

0

3 3

ˆ ˆ3

Cd dV dV dVR R R R C R

dR dR dR dR R

CdVE V R R R

dR R

�� , Since EE must be finite at R = 0, C1 must be 0 0

0

ˆ3

E R R

��

2 2 2 22 2 2

ˆ ˆ0 C Cd dV dV dV dV

R R C E V R RdR dR dR dR R dR R

��

0

30

20

022

3

3

b

Cbb

C

30

20

ˆ3

bE R

R

��

Legendre functions. (solution of Laplace’s equation in SCS)

Chapter 4 : Solution of Electrostatic ProblemsLecture 8-1 Static Electromagnetics, 2007 SpringProf....

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