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
Exercise: A charge q is spread uniformly on a disc of radius a. What is the electric filed?
Solution: Let us take the coordinate origin at the center of the disc, with the z-axis
perpendicular to it. The symmetry of this case calls for cylindrical coordinates, z ,  , for

q


r , and z ' ,  ' , ' for r ' . The charge density is then  (r ' )  2  ( z ' ) for  '  a . We hence have
a

q
E( , z)  2
a
q
 2
a
2
a
0
0
2
a
0
0
 d ' d '  '
 d ' d '  '
zˆz  ˆ (    ' cos(   ' ))  ˆ ' sin(    ' )
z
2
  2   '2 2  ' cos(   ' )

3/ 2
zˆz  ˆ (    ' cos  ' )  ˆ ' sin  '
z
2
  2   '2 2  ' cos  '

3/ 2
This expression can be put in the form
2
a

q  

1  
1
  d ' d '  '
E (  , z )   2  zˆ  ˆ
 ˆ
1/ 2

   0
a  z
z 2   2   '2 2  ' cos  '
0


A simpler case is when the field is required on the axis, namely, for
a

2q   
1
E (0, z )   2  zˆ  d '  '
a  z  0
z 2   '2

  0 . Then we are left with

1/ 2
which gives, for z>0,



2q   
2q 
z
E (0, z )   2  zˆ  z 2  a 2  | z |  zˆ 2 1 
a  z 
a 
z2  a2




and

2q 
z
E (0, z )  zˆ 2   1 
a 
z2  a2




Fir z<0. We see that at the disc itself, there is a discontinuity in the electric field along the zdirection,




4q
 q 
Ez  0   Ez  0   2  4  2 
a
 a 
We will see later on how to deal with the field in the general case.
2. Gauss’s law
Since we can write
 
 
  r  r'
  
   1
1 
E (r )   dr '  (r ' )   3   dr '  (r ' ) '       dr '  (r ' )  
| r  r '|
| r  r '|
 | r  r '| 
then
 
   2
  
 
 
1 
ˆ
E
(
r
)

n

d
r
d
r
'

(
r
'
)



4

d
r
d
r
'

(
r
'
)

(
r

r
'
)

4

d
r
 (r )




 | r  r '| 
S
 





2
n
 
 
ˆ
 E (r )  n  4  dr  (r )  4  qi
and in summary
i 1
S
There are two more results to the expression
 
 
  r  r'
  
   1
1 
E (r )   dr '  (r ' )   3   dr '  (r ' ) '       dr '  (r ' )   .
| r  r '|
| r  r '|
 | r  r '| 
Firstly,
  
 

1
  E (r )    dr '  (r ' ) 2    4(r ) .
| r  r '|
Secondly,
  
  E (r )  0 ,
since (rot grad)=0 .
Exercise: Calculate the electric field of an infinite cylinder, of radius a, charged with a constant
charge density

per unit length.
Solution: 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
3

2
E (  )  ˆ

  a.
,
3. Scalar potential
Since we have found that
 
   (r ' )
E (r )    dr '  
| r  r '|


  (r ' )
(r )   dr '  
| r  r '|
we can define
 
 
E (r )   (r ).
and then
 
 E  dl   B   A ,
B
A
This means that
(work per charge)
 
 E  dl  0
Exercise: Find the potential of a uniformly charged straight wire of length
unit length is
.
2 . The charge per
Hint:

dx
1  x2
 ln( x  1  x 2 ) .
Solution: Let the wire be along the z-axis, such that the x-y plane cuts it at its middle point. In
our case, the charge density is

 (r ' )   ( x' ) ( y ' )
and therefore, from the definition of the scalar potential we find

 ( x, y, z )    dz '

1
x 2  y 2  ( z  z' )2
.
4
r 2  x2  y 2.
It is useful now to define

dz '
( x, y,0)   
r  z'
2

2
Then on the x-y plane the potential is given by
  ln
  2  r 2
   2  r 2
For points in the x-z plane

( x,0, z )   

dz '
x  ( z  z' )
2
2
  ln
z    ( z  ) 2  x 2
z    ( z  ) 2  x 2
By symmetry, the potential at any point is
( x, y, z )   ln
z    ( z  ) 2  r 2
z    ( z  ) 2  r 2
.
Let us now find the surfaces of equal-potential. These are given by
( x, y, z )  const. 
z    ( z  ) 2  r 2
z    ( z  ) 2  r 2

c
c
where c is a constant. The equal-potential surface is given by a function relating z to
z  
r 2 . When
the above equation gives
(c 2   2 ) 2
r 
c2
2
When
r2  0
we find that
this equation gives z=c. Therefore, if we write the function
Ar 2  Bz 2  1
B
1
c2
A
1
c2  2
and the equi-potential surfaces are ellipsoids. The potential there is
 0   ln
c
.
c
5
4. surface charge distributions
Let us consider a surface charged with surface charge density
surface of area A in the x-y plane we have

(charge per area). Then for a
 
d
r
  (r )   dxdydz ( z )  A
Using this result in Gauss's theorem, we find that there is discontinuity of the z-component of
the electric field,




Ez 0   Ez 0   4
For a general surface, there is a discontinuity in the normal component of the electric field
across the surface,
E

2


 E1  nˆ  4
while the tangential components are continuous.
Exercise: Find the electric field of a uniformly charged (infinite) plane perpendicular to the zaxis, of charge
 per unit area.
Solution: By symmetry, the field must be along z, and it may depend only on z,
   zˆE ( z ),
E (r )  
 zˆE (| z |),
z  0,
z  0.
It follows that
E ( z )  2
5. Poisson and Laplace equations
6
From Gauss's law the scalar potential satisfies the Poisson equation,
 2   4
Excercise: Given the scalar potential
where

e r  r 
(r )  q
1  
r 
2 
is a constant, let us find the charge density that gives rise to this potential.
Solution: We solve the Poisson equation. Firstly, we write it explicitly in spherical coordinates,
1 2
(r(r ))  4(r )
r r 2
Then we insert in the left hand side the explicit expression of the scalar potential. This gives
 (r )  2qe r 3
We will encounter mostly two situations: either one is given the charge density, or one is given
the electric field at certain locations in space. In both cases we will need to find the scalar
potential, from which the electric field in the entire space can be found.
6. Electrostatic potential energy
The potential energy of a collection of point charges is:
W 
qi q j
1
  
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
By integrating by parts, using
7

 


 
 


  
(r ) 2 (r )    (r )(r )  (r )  (r )

we obtain


  
  2
1
1
ˆ
W 
dan  (r )(r ) 
dr (r )
8 S
8 
Since the first term—the surface term vanishes (it goes like
which is
r 3 times the area of the surface,
r 2 ), we find
W
   2 1
 2
1
d
r
|


(
r
)
|

d
r
E
8 
8 
Exercise: Find the electrostatic energy of a uniformly charged sphere of radius a.
Solution: Let us first find the electric field, using Gauss’s law. By symmetry,
 
E (r )  rˆE (r ).
Then,

4a 3
ra
 4 3   4Q,
2
E (r )4r  
3
3
4

r
r
4
  4Q 3 , r  a

3
a
which gives
The energy is then
Q
2
r
E (r ) 
Qr
a3
ra
r, a
a

2
 3 Q2
4 2 
2 r
2 1
W
Q   drr 6   drr 4  
8
a
r  5 a
a
0
8
7. Conductors
a.
b.
c.
d.
The
The
The
The
electric field is zero inside a conductor (at equilibrium).
potential inside a conductor is constant.
charge density inside a conductor is zero.
electric field on the surface of a conductor satisfies:
where

Et  0

En  4
,
is the charge density on the surface.
Exercise: Find the electric field of a conducting sphere, of radius a, carrying a charge q.
Solution: Since there is no charge density inside the conducting sphere, the charge must be only
on the surface. Then the Poisson equation becomes the Laplace equation,
 2  0
everywhere, except on the surface of the sphere. By symmetry, the scalar potential (and as a
result, the magnitude of the electric field) depends only on r,
 
E ( r )  rˆE ( r ) . Let us first solve
the Laplace equation everywhere, except on the surface of the sphere. In spherical coordinates,
we have
 2 
1 d  2 d 
r
  0,
r 2 dr  dr 
with the solution
 (r ) 
A
B
r
where A and B are constants. The potential vanishes at infinity (one can always choose it like
that), and therefore B=0. The magnitude of the electric field outside the sphere (inside the
conducting sphere the electric field is zero) is then
E (r ) 
A
r2
We know also the magnitude of the electric field at r=a, on the sphere surface,
E (a )  4
q
A

4a 2 a 2
9
Therefore we find A=q. Our result for the magnitude of the electric field is hence
E (r ) 
q
,
r2
as expected.
Exercise: A spherical conducting shell of inner radius a and outer radius b carries a charge
density  on its outer shell. There is a point charge q placed at the center of the shell. Find
the electric field in the space.
Solution: By symmetry, the electric filed is only along
r̂ . We also know that the field in the
region a<r<b is zero. The field in the region r<a is clearly E (r ) 
q
. Let us now look at what
r2
happens when r=a. We have
E (a  ) 
q
a2
E (a  )  0
There is a discontinuity, and therefore there appears surface charge density at r=a, of
magnitude
4 | a  
q
a2
In other words, the inner surface is charged with –q. As a result, the field at r>b is
4b 2
E (r ) 
r2
Moral: always check the boundary conditions!
10