chapter 1ocw.nctu.edu.tw/course/classical_electrodynamics/electrodynamics... · classical...

Classical Electrodynamics

Chapter 1Introduction and Survey

A First Look at Quantum Physics

2011 Classical Electrodynamics Prof. Y. F. Chen

A First Look at Quantum Physics


Contents

§1.1 Coulomb’s law and electric field§1.2 Electric field and electric potential§1.3 Discontinuity of electric field and potential§1.4 Poisson and Laplace equation§1.5 Green’s theorem and Green function§1.6 Electrostatic potential energy and energy density

(1) The force between two point charges is given by:

31

11

041

xxxxqqF

(2) The electric field of a point charge can be defined via force:

31

1

0

1

21 4 xx

xxqqFE


§1.1 Coulomb’s law and electric field

Note 1: If there are many point charges:

i

i xxxxqxE 31

1

041)(

Note 2: If the source is a distribution:

''

')'(4

1)( 33

0

xdxx

xxxxE

Compare the two equations above: i

ii xxqx )(

Note 3: The locality of the charge density could not be precisely determined.


§1.1 Coulomb’s law and electric field

§1.2 Electric field and electric potential

(1) For 'xx

0

'3

'3

''

1''

1

''

'1'

'1

33

33

32

xxxx

xxxx

xxxx

xxxx

xxxx


(2) For 'xx

4

ˆˆ1

ˆ''

'1'

'1

22

3

332

dRaaR

danxxxx

xdxx

xdxx

RR

Consequently: '4'

12 xxxx

'x

xRaRxx ˆ



0

3

0

32

0

33

0

33

0

)(

')'(4)'(4

1''

1)'(4

1

''')'(

41'

'')'(

41

x

xdxxxxdxx

x

xdxxxxxxd

xxxxxE

(3)

Note : The locality of the charge density could be precisely determined.



(4) Gauss’s law:

Differential form:0

E

Integral form:00

33 ˆ

QxddanExdE

(5) (i)

''')'(

41 3

30

xdxxxxxE

0'

'''

'3''

353

xx

xxxxxx

xxxxxx

0 E



(5) (ii) The electric field can be expressed as the negative gradient of the electric

potential:

'

'1)'(

41'

'')'(

41)( 3

0

33

0

xdxx

xxdxx

xxxxE

Note the curl of the gradient of any well-behaved scalar function of positionvanishes. As a result, we can obtain:

0

E



(5) (iii) With Stokes’s theorem:

0ˆ dldldEdanE

0 E

(5) (iv) depends on the central nature of the force between charges, and

on the fact that the force is a function of relative distances only, but does

not depend on the inverse square nature.

0 E



§1.3 Discontinuity of electric field and potential(1) The tangential component of the electric field is continuous:

0ˆ ldEdanE

tttt EElEE 2121 0

(2) The discontinuity of the normal component of the electric field means the

existence of the charges at the boundary:

0

3 ˆ

inQdanExdE

0

12 )ˆ(ˆQAnEnE

0012

1

AQEE in

nn


E1

E2( ) x

Side 1

Side 2 n

(3) The discontinuity of the electric potential is due to the existence of the dipolelayer, and it can be analogous to the situation of the capacitance.

For the situation of the single layer, the electric potential is continuous, but theelectric field is not, as shown in the left figure.

For the situation of the dipole layer, if the distance d is limited to zero, then theelectric field can be view to be continuous, but the electric potential is not, asshown in the right figure.

single layer

Position (m)-3 -2 -1 0 1 2 3

Elec

tric

field

(V/m

)

-2

0

2

4

6

8

Elec

tric

pote

ntia

l (V

)

-2

0

2

4

6

8

electric fieldelectric potential

dipole layer

Position (m)-2 -1 0 1 2 3

Elec

tric

field

(V/m

)

-4

-2

0

2

4

6

Elec

tric

pote

ntia

l (V

)

0

2

4

6

8

10

electric fieldelectric potential

d


§1.3 Discontinuity of electric field and potential

§1. 4 Poisson and Laplace equation

(1)

'

'')'(

41)( 3

30

xdxx

xxxxE

This expression is convenient to be used in the situation of free space or the chargedistribution being point charge.

(2)0

E

EE 00

2

: Poisson equation

If = 0: 02

: Laplace equation

When we deal with the problems involving the boundary condition or finite region,using Poisson equation of Laplace equation together with special mathematicaltechniques (for example, Green function) is a convenient way to solve the problem.


§1. 5 Green’s theorem and Green function

(1) Green’s first identity:

2

,let

vfvfvfvf

Integrate the above equation and use the divergence theorem:

xd

dan

danxd

32

3

ˆ

dan

xd 32


(2) Green’s second identity (also known as Green’s theorem):

)1......(2

Interchange and :

)2......(2

(1)-(2) and integrate:

dann

xd 322



(3) To solve the Poisson equation with the boundary condition, firstly we can solvethe impulse response with the same boundary condition:

)'-(4)' ,(2 xxxxG

With the replacement of: G ,

(i)

)(4')'-()'(4

')' ,(')'(''3

3232

xxdxxx

xdxxGxxd

(ii) ')' ,()'(1')'(')' ,('' 3

0

3232 xdxxGxxdxxxGxd

(iii) ''

)' ,()'(''

dan

xxGxdan

(iv) ''

)'()' ,(''

danxxxGda

n



''

)' ,()'(41'

')'()' ,(

41

')' ,()'(4

1)( 3

0

dan

xxGxdanxxxG

xdxxGxx

(4) Advanced discussions:

(i) Free space means no boundary condition:'-

1)' ,(xx

xxG

'

'1)'(

41')' ,()'(

41)( 3

0

3

0

xdxx

xxdxxGxx



(ii) For Neumann boundary condition: the simplest case isSn

GN 4'

SNN danxxxGxdxxGxx

''

)'()' ,(41')' ,()'(

41)( 3

0

whereS

daxS

')'(

Note that the electric potential for a point charge is:'4

1)(0 xx

Qx

If the total charge is expressed as the surface charge: ')'( daxQ

'

')'(

41)(

0

daxx

xx

Compare with the term of the red box with the Green function in thefree space:

'-1)' ,(

xxxxG



As a consequence, the interpretation of the surface integral of the redbox is the potential due to the surface charge density given above.

' )'(

' 0

0 nx

n

The discontinuities in the electric field across the surface then lead tozero field outside the volume V:

'

'

' ,0

0

0

12

012

n

n

nEE

EE

nn

nn

0

'1

n

E n

02 nE

'n̂Surface



(iii) For Dirichlet boundary condition: 0DG

''

)' ,()'(41')' ,()'(

41)( 3

0

dan

xxGxxdxxGxx DD

Note that the electric potential for a dipole is:

dxxQ

xxQx

''4

1)(0

With Taylor expansion: ......'

1'

1'

1

d

xxxxdxx

'1'

4'1

4)(

00 xxdQ

xxdQx

If the total charge is expressed as the surface charge: ')'( daxQ

And define dipole moment: danDddaP 'ˆ



'

'1''ˆ

41)(

0

daxx

nDx

Compare with the term of the blue box with the Green function in thefree space:

'-1)' ,(

xxxxG

00

DD

As a consequence, the interpretation of the surface integral of the bluebox is the potential due to the dipole layer D given above.



The discontinuities in the electric potential across the surface thenlead to zero potential outside the volume V:

01

02

D

'n̂Surface

0

0

12

012

,0

D

D

D



§1. 6 Electrostatic potential energy and energy density

(1) For a point charge, the work done on the charge is given by: )( iii xqW

(2) If the potential is produced by other charges, then the potential is given by:

1

1041)(

N

j ji

ji xx

qx

So that the potential energy of the charge qi is:

1

104

N

j ji

jii xx

qqW

(3) The total potential energy of all the charges due to all the forces acting betweenthem is:

i j ji

jiN

j ij ji

ji

xxqq

xxqq

W 0

1

10 81

41

It is understood that i = j terms (infinite “self-energy” terms) are omittedin the double sum.


(4) For a continuous charge distribution:

xdxxxdxdxxxxW 333

0

)()(21'

')'()(

81

(5)

xdEW

EEE

E

320

0

21

&

The potential energy expressed in (5) is definitely nonnegative. Thisseems to contradict our impression from (3) that the potential energy oftwo charges of opposite sign is negative. The reason for this apparentcontradictions is that (5) contains “self-energy” contributions to theenergy density, whereas the double sum in (3) is not.


§1. 6 Electrostatic potential energy and energy density

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