Chapter 1Chapter 1Chapter 1Chapter 1Chapter 1Chapter 1Chapter 1Chapter 1Chapter 1Chapter 1 Introduction to Electrostatics

Electrostatics � phenomena involving time-independent distributions of charge

and fields.

Electrostatics developed as a science of macroscopic phenomena. Some

idealizations like point charge may fail to have meaning microscopically.

the force between two small charged bodies separated in a distance - varies directly as the magnitude of each charge, - varies inversely as the square of the distance, - is directed along the line joining the charges, - attractive if oppositely charged and repulsive if the same type of charge, - the vector summation rule applies.

Electric field: force/(unit charge) at a given point in a limiting process

Coulomb's Law:

the electric field:

1.2 Electric Field

1.1 Coulomb's Law

E �x�=k q1

x�x1

�x�x1�3

F=q E

F �x1,x

2�= k q

1q

2

x1� x2

�x1�x

2�3

In the SI system:

The linear superposition law:

Dirac delta functionDirac delta function: a mathematically improper function with the properties

A discrete set of point charges can be described with delta functions

E �x �=1

4 ��0

�i=1

n

qi

x�x1

�x�x1�3

� E �x �=1

4 � �0

�x ' �x�x '

�x�x '�3d

3x '

where � q= �x ' � � x � y � z , d3

x ' =d x ' d y ' d z '

1. � � x�a �=0 for x a in 1d ,

2. � � x�a� d x=1 if the region of integration includes x=a , and is zero otherwise

3. f � x � � � x�a� d x= f �a�

4. f � x � � ' � x�a� d x=� f ' �a �

5. � � f � x ��=�i�d f

d x� x

i���1

� � x� xi�

6. � �x�X�=� �x1�X

1� � � x

2� X

2� � � x

3�X

3� with Cartesian coordinates in 3d

7. �V� �x�X � d

3x={1 if �V constains x=X

0 if � V does not constain x=X� [� �x�X�]=

1

V

�x �=�i=1

n

qi� �x�x

i�

k= �4 � �0��1=10

�7c

2 � free space permittivity

�0=8.854×10

�12F /M

Gauss's law is sometimes more useful

and leads to a differential eqn for E.

For a set of charges,

the sum is over

only those charges inside the

surface S.

For a continuous charge density:

The eqn is one of the basic eqns of electrostatics. It depends upon (a) the inverse square law for the force between charges; (b) the central nature of the force; (c) the linear superposition of the effects of different charges.

�SE�n d a=

1

�0

V �x � d

3x

1.3 Gauss's Law

E�n d a=q

4 � �0

cos �

r2

d a

=q

4 � �0

d � � r2

d �=cos � d a

� �SE�n d a={q /�0

if q lies inside S

0 if q lies outside S

�SE�n d a=

1

�0

�i

qi

The divergence theorem: for any well-behaved vector field defined within a

volume surrounded by the closed surface

apply the divergence theorem

the differential form of Gauss's law of electrostatics.

�SA�n d a=

V��A d

3x

1.4 Differential Form of Gauss's Law

� V���E� /�

0� d

3x for an arbitrary volume V � ��E�

�0

=0

�SE�n d a=

V��E d

3x=

1

�0

V �x � d

3x

A vector field can be specified almost completely if its divergence and curl are given everywhere in space.

look for an equation specifying curl E as a function of position,

Note that � E=0 depends on the central nature of the force, and on the fact that the force is a function of relative distances only, but does not depend on the inverse square nature.

Since a scalar is easier to deal with than a vector, define the scalar potential

�the work done in moving the charge from A to is

�� �=0 for all �

1.5 Another Equation of Electrostatics and the Scalar Potential

W =�A

B

F�d �=�qA

B

E�d � � F=q E

=qA

B

� ��d �=qA

B

d �=q ��B��

A�

�×E=0 � E �x �=�1

4 � �0

� �x ' �

�x�x '�d

3x ' �

1

4 � �0

�x ' �x�x '

�x�x '�3d

3x '

E=�� � � � �x�=1

4 � �0

�x ' �

�x�x '�d

3x '

q� can be interpreted as the potential energy of the test charge in the

electrostatic field.

the line integral of the electric field between two points is independent of the path and is the negative of the potential difference between the points:

Stokes's theorem:

With the line integral of the electric field being independent of the path and the application of the Stokes�s theorem � �×E=0

A

B

E�d �=���B��

A� � �E�d �=0 � �� � If the path is closed,

the line integral is zero

�CA�d �=

S��×A��n d a

Gauss's law tells

This does not determine E1 and E2. The eqn

means that there is a discontinuity in the normal component of E in crossing a surface.

The tangential component of E is continuousacross a boundary surface from eqn (*).

In this case

For volume or surface distributions of charge, the potential is continuous, even within the charge distribution. E is bounded, even though discontinuous across a surface distribution of charge.

With point or line charges, or dipole layers, the potential is no longer continuous.

The dipole-layer distribution of strength is formed by

letting S' approach infinitesimally close to S while the

surface-charge density becomes infinite such that their product approaches the limit:

�E2�E

1��n=

�

�0

1.6 Surface Distributions of Charges & Dipoles and Discontinuities in the Electric Field & Potential

limd �x � � 0

� �x� d �x �=D �x �

� �x�=1

4 � �0

S

� �x ' �

�x�x '�d a '

For a constant surface-dipole-moment density, the potential is the product of the moment divided by 4��0 and the solid angle

subtended at the observation point by the surface, regardless of its shape.

In the integrand we notice that it is the sum of the potential

of a point dipole with dipole moment dp=nDda'

� �x�=1

4 � �0

S

� �x ' �

�x�x '�d a '�

1

4 � �0

S '

� �x ' �

�x�x '�n d�d a ' '

�1

4 � �0

S

D �x ' � n�� ' � 1

�x�x '� � d a ' �1

�x�a�=

1

x�a�� �

1

x���

=�1

4 ��0

S

D �x ' �cos � d a '

�x�x '�2

=�1

4 ��0

S

D �x ' � d �

����

4 � �0

D if D= const

� �x�=1

4 � �0

S

�x�x ' ��d p

�x�x '�3

There is a discontinuity in potential in crossing a double layer. The total potential jump in crossing the surface is:

The potential has a discontinuity of D/�0 in crossing from the inner to the outer

side, being -D/2�0 on the inner side and +D/2�0 on the outer.

�2��

1=

D

�0

To verify the result directly and avoid being singular in the resulting integrand, we invoke a limiting procedure:

It is well-behaved everywhere for nonvanishing a, but as a � 0

The singular nature of �2(1/r) can be expressed as

��E=

�0

�×E=0 � E=�� �

��2 �=�

�0

� Poisson equation

�2 �= 0 � Laplace equation

� � �x�=1

4 � �0

�x ' �

�x�x '�d

3x '

1.7 Poisson and Laplace Equations

�2 1

�x�x '�=�4 � � �x�x ' �

{��� , r=0

�= 0, r 0

�a�x��

1

4 � �0

�x ' �

�x�x ' �2�a2

d3

x ' , r��x�x '�

� �2 �a�x �=

1

4 ��0

�x ' � � 2 1

r2�a

2d

3x '=�

1

4 ��0

3 a2 �x ' �

�r2�a

2�5/2

d3

x '

�2�

a�r � 0���

3 a2 �x �

4 � �0

d3

x

�r2�a2�5/2

=�3 a

2 �x �

�0

0

�r

2d r

�r 2�a2�5/2

=� �x�

�0

To handle the boundary conditions it is necessary to develop some new mathematical tools. The divergence theorem:

Convert the Poisson differential equation into an integral equation

1.8 Green's Theorem

V

��A d3

x=�S

A�n d a and A=! � � � [���! � ��=! �2 ��� !�� �

! � ��n=!" �

" n

� V�! �2 ��� !�� �� d

3x=�

S!" �

" nd a � �Green's 1st identity� � !#�

� V�! �

2��� �

2!� d

3x=�

S �! " �

" n��

"!

" n � d a � �Green's 2nd identiy

Green theorem ��2 �=�

�0

, �=1

R=

1

�x�x '�� � 2 �=�4 �� �x�x ' �

� V [�4 �� �x ' � � �x�x ' ��

�x ' �

�0

R ] d3

x=�S �� "

" n '

1

R�

1

R

" �

" n ' � d a '

�

� �x�=1

4 � �0

V

�x ' �

Rd

3x�

1

4 ��

S � 1

R

" �

" n '��

"

" n '

1

R � d a ' � x$V �� ' �

0=1

4 � �0

V

�x ' �

Rd

3x�

1

4 ��

S � 1

R

" �

" n '��

"

" n '

1

R � d a ' � x%V

For eqn (&')

This result is not a solution to a boundary-value problem, but only an integral statement, since the arbitrary specification of both � and �' (Cauchy boundary conditions) is an overspecification of the problem.

�x ' �=0 � � �x�=1

4 ��

S � 1

R

" �

" n '��

"

" n '

1

R � d a ' =�S

f �� ," �

" n '� d a '

if S � � , E �S �' R�1�� � � �x �=

1

4 � �0

�x ' �

�x�x '�d

3x '

Dirichlet problem/Dirichlet boundary conditionDirichlet problem/Dirichlet boundary condition: specification of the potential on a

closed surface defines a unique potential problem.

Neumann boundary conditionNeumann boundary condition:specification of the electric field (normal derivative

of the potential) everywhere on the surface also defines a unique problem.

to show the uniqueness of the solution of the Poisson equation inside a volume subject to either Dirichlet or Neumann boundary conditions on the closed bounding surface.

Assume 2 solutions satisfy the same boundary conditions and

there is also a unique solution to a problem with mixed boundary conditions.

A solution to the Poisson equation doesn't necessarily exist with arbitrary � & �'

1.9 Uniqueness of the Solution with Dirichlet or Neumann Boundary Conditions

� �2

U=0 in V , and U�S=0 �Dirichlet � , or" U

" n �S=0 �Neumann �

� V

�U �2U �� U�� U � d

3x=�

S

U" U

" nd a �!=�=U �

V

�� U�2 d3

x=0

� � U =0 inside V � U =const �U=0 �Dirichlet � � �

1=�

2

U=const �Neumann � � �1=�

2� const

U=�1��

2

The function is only one of a class of functions depending on the variables x and

x', and called Green functions, which satisfy

With the generalized concept of a Green function and its additional freedom

F(x, x'), there arises the possibility that we can use Green's theorem and choose

F(x, x') to eliminate one or the other of the two surface integrals, obtaining a

result that involves only Dirichlet or Neumann boundary conditions.

for Dirichlet boundary conditions we demand:

1.10 Formal Solution of Electrostatic Boundary-Value Problem with Green Function

� � �x�=1

4 � �0

V �x ' �G

D�x , x ' � d

3x '�

1

4 ��

S� �x ' �

" GD

" n 'd a '

� '2

G �x , x ' �=�4 � � �x�x ' � � G �x , x ' �=1

�x�x '��F �x , x ' � � � '

2F �x , x ' �=0

GD�x , x ' �=0 for x ' on S

� '2�x�x '��1=�4 � � �x�x ' � � the potential of a unit point source

!=� ,

�=G �x , x ' ��

� �x �=1

4 � �0

V �x ' �G �x , x ' � d

3x '

�1

4 ��

S [G �x , x ' �" �

" n '�� �x ' �

" G �x , x ' �

" n ' ] d a '

Gauss's theorem gives

For Neumann boundary conditions, the simplest allowable one is

where S is the total area of the boundary surface

The customary Neumann problem is the so-called exterior problem in which the volume is bounded by two surfaces, one closed and finite, the other at infinity. Then the surface area is infinite; the average value vanishes.

The mathematical symmetry property G(x, x') = G(x', x) merely represents the

physical interchangeability of the source and the observation points.

For the physical meaning of F(x, �')/4��0, it is a solution of the Laplace equation

inside V and represents the potential of charges external to the volume V.

�4 �=V�2

G d3

x ' =�S� G�d a '=�

S

" G

" n 'd a '

" GN

" n '�x , x ' �=�

4 �

Sfor x ' on S instead of

" GN

" n '�x , x ' �=0 for x ' on S

� � �x�= (�)S�

1

4 � �0

V �x ' �G

N�x , x ' � d

3x '�

1

4 ��

SG

N

" �

" n 'd a '

If a point charge is brought from infinity to a point in a scalar potential (which vanishes at infinity), the work done on the charge is

expresses the electrostatic potential energy in terms of the positions of the charges and emphasize the interactions between charges via Coulomb forces.

An alternative approach is to emphasize the electric field and to interpret the energy as being stored in the electric field surrounding the charges

1.11 Electrostatic Potential Energy & Energy Density; Capacitance

Wi=q

i� �x

i�=

qi

4 � �0

�j=1

n�1q

j

�xi�x

j�

� � �xi�=

1

4 � �0

�j=1

n�1q

j

�xi�x

j�

� W=1

4 � �0

�i=1

n

�j *i

qiq

j

�xi�x

j�=

1

8� �0

�i

�j

qiq

j

�xi�x

j�

for i j �no self-energy term �

� W=1

8 ��0

�x� �x ' �

�x�x '�d

3x d

3x '=

1

2 �x � � �x� d

3x for a continuous

charge distribution

� W =1

2 �x� � �x� d

3x=�

�0

2� �2 � d

3x=

�0

2�� ��2 d

3x=

�0

2�E�2 d

3x

� w=�0

2�E�

2+0 � energy density

The energy density is positive definite. This contradicts our impression that the potential energy of two charges of opposite sign is negative. This apparent contradiction comes from "self-energy" contributions to the energy density. Ex:

E=q1

4 � �0

x�x1

�x�x1�3�

q2

4 � �0

x�x2

�x�x2�3

� 32 �2 �0

w=q

1

2

�x�x1�4�

q2

2

�x�x2�4

,self energy

�2q

1q

2�x�x

1���x�x

2�

�x�x1�3�x�x

2�3

,w

int

define �x�x

1

�x1�x

2�

, n�x

1�x

2

�x1�x

2�

� 16 �2 �0

Wint=q

1q

2�x�x

1���x�x

2�

�x�x1�3�x�x

2�3

d3

x=q

1q

2

�x1�x

2� ���n�

3��n�3d

3

=q

1q

2

�x1�x

2�� �1����n��1

d3 =�

q1

q2

�x1�x

2��2 �1

� �n�d

3

� Wint=

1

4 ��0

q1

q2

�x1�x

2� �

3��

��n�d

3 =1

4 ��0

q1

q2

�x1�x

2�

� as expected

Forces acting between charged bodies can be obtained by calculating the change in the total electrostatic energy of the system under small virtual displacements.

To calculate the force per unit area on the surface of a conductor with a surface-charge density, imagine a small outward displacement of an elemental area of the surface

an outward force per unit area equal to w at the surface of the conductor.

For a system of n conductors, the electrostatic potential energy can be expressed in terms of the potentials alone and certain geometrical quantities called coefficients of capacity

The capacitance of a conductor is the total charge on the conductor when it is maintained at unit potential, all other conductors being held at zero potential.

the capacitance of 2 conductors carrying equal and opposite charges in the presence of other grounded conductors is defined as the ratio of the charge on one conductor to the potential difference between them.

The potential energy for the system of conductors

Vi= �

j=1

n

pi j

Qj

�i=1, 2,� , n� � Qi= �

j=1

n

Ci j

Vj

�i=1, 2,� , n �

Ci i

: capacities or capacitances , Ci j�i j � : coefficients of induction

W =1

2�i=1

n

QiV

i=

1

2�

i , j=1

n

Ci j

ViV

j

w=�

0

2�E�2=

�2

2 �0

� � W =��2

2 �0

� a � x

The expression of the energy in terms of the potentials and the Cij, or in terms of

the charges and the coefficients pij permits the application of variational methods

to obtain approximate values of capacitances.

Selected problems: 1.3, 1.6, 1.9, 1.10, 1.14, 1.15, 1.17, 1.21

Variational methods provide formal techniques for the derivation of "equations of motion" and also practical methods for obtaining approximate, but often accurate, solutions to problems not amenable to other approaches.

consider the functional

I is a stationary minimum if � satisfies a Poisson-like equation within the volume

and the departures �� vanish on the boundary.

the minimization of the functional yields the "eqn of motion" of the electrostatic potential in the presence of a charge density and Dirichlet boundary conditions.

the stationary nature of the extremum of I permits a practical approach to an

approximate solution for �(�).

1.12 Variational Approach to the Solution of the Laplace and Poisson Equations

� I = I [��� �]� I [�]=V

[� ��� � ��1

2�� � ��2�g � �] d

3x

�V[��

2�� g] � � d

3x��

S� �

" �

" nd a � !=��

�=�in Green's 1st identity

� ��S=0 & �2 ��g=0 � � I1�1st-order of � I =0

I [�]=1

2

V�� ��� ��2 g �� d

3x � g �x � : source function

a: choose a flexible "trial" function �(�) = А-(�,.,/,...) to satisfy the boundary

conditions on the surface.

b: calculation of I[�] gives I(А,.,/,...).

c: vary the parameters to locate the extremum of I(А,.,/,...).

With the optimum parameters, the trial solution is the best possible approximation to the true solution with the particular functional form chosen.

For the Laplace eqn, the normalization constant А is determined by the Dirichlet

boundary condition. For the Poisson eqn, it is determined by the source strength

g(x), as well as the boundary values.

For Neumann boundary conditions

I [�]=V

�1

2� ��� ��g �� d

3x��

S

f � d a � f �s� is defined on S

� � I1=

V[�� 2 ��g ]� � d

3x��

S[" �

" n� f �x�] � � d a

�2 ��g=0 &" �

" n �S= f �s� � � I1=0 independent of � �

consider a hollow circular cylinder of unit radius centered on the z-axis with an

interior source density g(x)=g(),

Two possible trial functions:

-1 is a less accurate representation of � because � should have a maximum or

minimum at the origin with vanishing slope.

-1=.

1�1���/

1�1��2�0

1�1��3

-2=. �2�1��/ �3�1��0 �4�1�

1

"

" �

" �

" �=�g �� & � �=1�=0

1

2�I [-

2]=

1

2.2�

6

5. /�

4

3. 0�

3

4/2�

12

7/ 0�02�e

2.�e

3/�e

40

where en=

0

1g �� �n�1� d

" I

" .=0

" I

" /=0

" I

" 0=0

�

.=225 e2�420 e

3�210 e

4

/=�420 e2�

2450

3e

3�420 e

4

0=210 e2�420 e

3�

441

2e

4

g ��=5 ��1��104 5 ��1�5

-1 fails badly for <0.3 because it does not respect the vanishing slope at =0.

The insensitivity of I[-] to errors in the trial function shows both a strength and

a weakness of the variational method. If the principle is used to estimate eigenvalues, it does well. Used as a method of estimating a solution, it can fail badly, at least in parts of the configuration space.

The relaxation methodThe relaxation method (iterative finite difference method) is an iterative numerical scheme for the solution of the Laplace or Poisson equation.

First consider the Laplace equation with Dirichlet boundary conditions within a region with a boundary contour. The region is spanned by a square lattice with

lattice spacing h. The potential values on the boundary sites are assumed given.

imagine the functional integral I[�] over S as a sum over small domains.

Consider the neighboring trial values of the potential as fixed, while the value at the center of the sub-area is a variational quantity to be optimized.

S

1.13 Relaxation Method for 2D Electrostatic Problems

approximate the derivatives in the northeast quarter

similarly for the other three quarters

I is minimized if �0 is equal to the average of the values at the "cross" points.

consider the whole functional integral. guess a set of �(i,j) initially and

approximate I[�] by the sum of terms of the form of (**). go over the lattice and

replace half the values by the average of the points (crosses) around them.

Repeat the process to do the other half points.

We could take the average of the values at the corners of the large square instead of the "cross" values, or we could take some linear combination of the two to improve its accuracy.

�" �

" x�NE

��

E��0

h, �

" �

" y�NE

��

N��0

h

� INE=

1

2

0

h /2

d x0

h /2

d y [� " �" x�2

��" �

" y�2

]� 1

8[��

E��

0�2��!

N��

0�2] �**�

� I�1

4[��

E��

0�2���

S��

0�

2���

W��

0�2��!

N��

0�2]

� ��0�

optimum=

1

4��

E��

S��

W��

N� � Minimizing I with respect to

�0

gives the optimum value

Choose

With either the "cross" or "square" averaging separately, the error is O(h4).

The increase in accuracy is at the expense of twice as much computation for each lattice site, but for the same accuracy, far fewer lattice sites are needed

For the Poisson equation:

((F � x , y�))�4

5(F )

c�

1

5(F )

s

where

(F )c=

1

4[F � x�h , y��F � x , y�h��F � x�h , y ��F � x , y�h�]

(F )s=

1

4[F � x�h , y�h��F � x�h , y�h��F �x�h , y�h ��F � x�h , y�h �]

� ((F � x , y�))=F �x , y��3

10h

2 � 2F�

1

40h

4 �2 � 2F�O �h6� � problem 1.22

� �new

�i , j �= ( (� �i , j �) )�O �h6� � � 2 �=0 �Lapalce eqn�

�2 �=�g

�new

�i , j �= ( (� �i , j �) )�h

2

5g �i , j ��

h2

10(g �i , j �)

c�O �h6�

( (N ))=O �(N )2 /3� � ((N )) : the number of sites needed with ((�))

(N ) : the number with the cross or square average

A basic procedure (JacobianJacobian iteration) for the iterative numerical solution of the Laplace or Poisson equation with Dirichlet boundary conditions

1: A square lattice spacing is chosen and the lattice sites, including the sites on the boundary.

2: The values of the potential at the boundary sites are entered in a table of the potential at all sites.

3: A guess is made for the values, �old(i,j), at all interior sites.

4: The 1st iteration cycle begins by systematically going over the lattice sites and

computing ((�(i,j))) or any of the averages. This quantity is entered as �new(i,j) in a table of "new" values of the potential at each site.

5: replace the set of �old by the set of �new and the iteration cycle begins again.

6: Iterations continue until some desired level of accuracy is achieved.

The Gauss-SeidelGauss-Seidel iteration, a better one, replaces �old with �new as soon as the

latter is determined.

consult Numerical RecipesNumerical Recipes for other possible improvements.