Download ELECTRODYNAMICS—lecture notes second semester 2004 Ora Entin-Wohlman

ELECTRODYNAMICS—lecture notes
second semester 2004
Ora Entin-Wohlman
References:
1. J. D. Jackson, “Classical Electrodynamics”,
Wiley.
2. G. B. Arfken, "Mathematical methods for
Physicists", Academic Press.
1
Summary of vector analysis

F
F
F
F  xˆ
 yˆ
 zˆ
x
y
z


r dF

F
(r
)
The gradient of
: F
r dr
  Vx Vy Vz


V 


2. Divergence,   ,
x
y
z

1. Gradient,  ,
xˆ
 




V

3. Curl (rotor),   ,
x
Vx
4.
5.
6.
7.
yˆ

y
Vy
zˆ

z
Vz
2F 2F 2F


Div.grad=Laplacian,  ,  F 
x 2 y 2 z 2
 
Rot  grad =0,   F  0
  
Div.rot=0,     V  0

 
  

2


(


V
)


(


V
)


V
Rot  rot,
2
2
8. Gauss’s theorem

 
ˆ
V

n
da

d
r

  V
S
V
Important consequence:

  1
  r  4
 dr    r   dr   r 3   0




Exercise: show that when V  A  P, where A is a constant vector,

 
ˆ
n

P
da

d
r
Gauss’s theorem gives 
  P.
S
V
9. Stoke’s theorem
 
 
V

d




  V  nˆda
S



Exercise: show that when V  AF , where A is a constant vector,


ˆ
n


Fda

Fd
Stoke’s theorem gives 
 
S
2
Electrostatics
1. Coulomb’s law
 

The electric field, E (r ) , at point r , due to a collection of point
charges, q i , located at ri , is
 
n
 
r  ri
E ( r )   qi   3
| r  ri |
i 1
When those charges can be described as a charge-density (of

dimension charge per unit volume),  (r ' ) , then
 
 
  r  r'
E (r )   dr '  (r ' )   3
| r  r '|
The two expressions coincide for
n

 
 (r )   qi  (r  ri )
i 1
The delta-function
Dirac delta-function obeys the following in one-dimension:
 ( x  a)  0, x  a
a
 dx ( x  a)  1
a
and has the properties
3
 dxf ( x) ( x  a)  f (a)
  f ( x)   
i
1
df
dx
 dxf ( x) ' ( x)   f ' (a)
 ( x  xi )
xi
In more than one dimension, the delta-function becomes a
product. In cartesian coordinates


 (r  R)   ( x  X ) ( y  Y ) ( z  Z )
In spherical coordinates
 
 (r  r ' ) 
1
 (r  r ' ) (cos   cos ' ) (   ' )
r2
Examples:
1. Gaussian representation of a one-dimensional delta-function,
the limit   0 of
  x 2 
 ( x0 
exp    
 
    
1
2. Gaussian representation of a three-dimensional delta-function,
the limit   0 of
 1 
 (r )   
 2 
3/ 2


 1

exp

x2  y2  z 2 
3
2


 2

1
4
3. Other one-dimensional delta-functions: (a) Lorentzian--the
1
1
limit   0 of  ( x) 
; (b) of
 1  ( x /  ) 2
sin( x /  )
1
 ( x) 

x
2
1 ( x /  )
,x 0
(c) and of  ( x)  e

4. From Gauss’s theorem,
 (r )  
1/ 
dte
ixt
1 /
1 2 1 
  
4
r
2. Gauss’s law
n

 
ˆ
E

n
da

4

q

4

d
r
 i

  (r )
S
i 1
S
V
n
V
q
E
 
  
   r  r'

  E (r )   dr '  (r ' )    3  4(r )
| r  r '|
  
  E (r )  0
explanation:
 r

  3  4 (r )
r
and
 r
 3  0
r
5
Example: Calculate the electric field of an infinite cylinder, of
radius a, charged with a constant charge density  per unit
length. We use cylindrical coordinates,  ,  , z . By symmetry, the
field is only along the ̂ -direction. Consider a volume of length L
around the cylinder. By Gauss’s law,
Q

EL2  4   L 2  4   L 2 ,   a ,
 SL 
S
Q

EL2  4   La 2  4   La 2 ,   a ,
 SL 
S
giving

2
E (  )  ˆ 2  ,   a ,
a

2
E (  )  ˆ
  a.
,

3. scalar potential
 
   (r ' )
 
E (r )    dr '    (r ),
| r  r '|


  (r ' )
(r )   dr '  
| r  r '|

1
 1
 r

r
2 1
   3  4 (r )
since    3      
r
r
r
r
r
 
 E  dl   B   A ,
B
Thus:
A
 
 E  dl  0
(work per charge)
6
4. surface charge distributions
n
E2

 (r )
E1




E 2  E1  nˆ  4
discontinuity in the normal component of
the electric field. The tangential components are continuous.
5. Poisson and Laplace equations
When there is charge density, the potential satisfies the Poisson
equation:
 2   4 . Otherwise, it satisfies the Laplace
equation:
 2  0 .
Example: Let us find the charge distribution giving rise to the
scalar potential
e r  r 
( r )  q
1  
r 
2
Let us the Laplacian of this potential (in Cartesian coordinates)
7

r 
x
(r ) 
 q F (r ),
x
x r
r
 2 
r  1

F (r )  e  2  
r
r
2


 F (r )  x  2 

2
 x



(
r
)

q
F
(
r
)

q

F
(
r
)
 
2


x
x r
 r  r
 r

Therefore,
 2 (r ) 
q r 3
e  and
2
 (r )  2qe r 3
6. Electrostatic potential energy
The potential energy of a collection of point charges is:
qi q j
1
W    
2 i j | ri  r j |
The potential energy of continuous charge distribution is:


   (r ' )  (r ) 1  

 

1
1
W    dr ' dr     dr  (r )(r )    dr (r ) 2 (r )
2
| r  r '|
2
8
W
   2 1
 2
1
d
r
|


(
r
)
|

d
r
E
8 
8 
8