Lecture3
September 7, 2005
Chapter 3 Static Electric Fields
3-1 Introduction
Charles-Augustin de Coulomb
born June 14, 1736, Angoulême, Fr.
died Aug. 23, 1806, Paris
French physicist best known for the formulation of Coulomb's law,
which states that the force between two electrical charges is
proportional to the product of the charges and inversely
proportional to the square of the distance between them.
Coulombic force is one of the principal forces involved in atomic
reactions.
The experimental Coulomb’s law: the force between two charges
is found to be
qq
F12 = Rˆ12 k 1 2 2
R12
,
where F12 is the vector force exerted by q1 on q2, R̂12 is a unit vector
in the direction from q1 to q2, and k is a proportionality constant
depending on the medium and the system of units.
Torsion balance to measure force between two charges.
3-2 Fundamental Postulates of Electrostatics in Free Space
Electric field intensity is defined as the force per unit charge
F
E = lim
( V/m ) .
q →0
q
where
F = qE
( N ).
The two fundamental postulates of electrostatics in free space
specify the divergence and curl of E.
∇⋅E =
ρ
ε0
and
∇×E = 0.
=> static electric fields are irrotational(conservative).
The two postulates are in the form of point relations/differential
form.
Taking volume integral of the first equation ∇ ⋅ E =
arbitrary volume V, we have
1
∫ ∇ ⋅ Edv = ε ∫
V
0
V
ρ
ε0
over an
ρ dv .
In view of the divergence theorem
v∫
S
E ⋅ ds =
Q
ε0
Gauss’s law: the total outward flux of the electric field intensity
over any closed surface in free space is equal to the total charge
enclosed in the surface divided by .
An integral form is obtained by integrating equation
an open surface and invoking Stokes theorem
v∫
C
over
G
E ⋅ d A = 0.
The scalar line integral of the static electric field intensity around
any closed path vanishes.
E field is irrotational
G
G
E ⋅ d A +∫ E ⋅ d A = 0
C1
C2
G
p2
p1
E ⋅ d A = −∫
∫
∫
∫
p1alongC 1
p2
p1alongC 1
p 2 alongC 2
G
p2
E⋅dA = ∫
p1alongC 2
G
E⋅dA
G
E⋅dA
The scalar line integral of the irrotational E field is independent of
the path; it depends only on the end points.
3-3 Coulomb’s law
Consider a single point charge, q, at rest in a boundless free space.
We draw a hypothetical spherical surface of a radius R centered at
q.
Since a point charge has no preferred directions, its electric field
must be everywhere radial and has the same intensity at all points
on the spherical surface.
v∫
S
(
)
q
ˆ
ˆ
E ⋅ ds = v∫ RE
R ⋅ Rds =
S
ε0
ER v∫ ds = ER ( 4π R 2 ) =
S
q
ε0
Therefore
ˆ = Rˆ
E = RE
R
q
4πε 0 R 2
( V/m ) .
If the charge q is not located at the origin of a chosen coordinate
system
E P = aˆqP
Where
aˆqP
q
4πε 0 R − R '
2
,
is the unit vector drawn from q to P. Since
aˆqP =
We have
EP =
R −R'
R −R'
q ( R − R ')
4πε 0 R − R '
3
( V/m ) .
Example 3-1 Determine the electric field intensity at P(-0.2, 0, 2.3) due to a point charge of 5 (nC) at Q(0.2, 0.1, -2.5) in air.
Solution
JJJG
R = OP = − xˆ 0.2 − zˆ 2.3.
JJJG
R ' = OQ = xˆ 0.2 + yˆ 0.1 − zˆ 2.5.
R − R ' = − x0.4 − y 0.1 + z 0.2,
2
2
2
R − R ' = ( −0.4 ) + ( −0.1) + ( 0.2 ) 


We obtain
1/ 2
= 0.458
( m ).
 1  q ( R − R ')
EP = 

3
 4πε 0  R − R '
5 × 10−9 ( − xˆ 0.4 − yˆ 0.1 + zˆ 0.2 )
0.4583
=214.5 ( − xˆ 0.873 − yˆ 0.218 + zˆ 0.437 ) ( V/m ) .
= ( 9 ×109 )
The quantity within the parentheses is the unit vector
aˆQP = ( R − R ') / R − R ' , and EP has a magnitude of 214.5 (V/m).
Note that
1
4πε 0
=
µ0c 2
= 10−7 c 2 ≅ 10−7 ( 3 × 108 ) = 9 × 109 (m/F). .
4π
When a point charge q2 is placed in the field of another point
charge q1 at the origin, a force F12 is experienced by q2 due to q1.
qq
F12 = q2E12 = Rˆ 1 2 2 ( N ) .
4πε 0 R
This equation is a mathematical form of Coulomb’s law.
Example 3-2 A total charge Q is uniformly put on a thin spherical
shell of radius b. Determine the electric field intensity at an
arbitrary point inside the shell.
Solution
a) The statement in a) is wrong! Why?
b) Uniform surface charge density
ρs =
Q
.
4π b 2
The magnitude of the electric field intensity at P due to charges on
the elementary surfaces ds1 and ds2 is
dE =
ρ s  ds1 ds2 
−

.
4πε 0  r12 r22 
But the solid angle equals
dΩ =
ds1
ds
cos α = 22 cos α .
2
r1
r2
Combining the expressions of dE and dΩ, we find
dE =
ρs  d Ω
dΩ 
−

 = 0.
4πε 0  cos α cos α 
Since the above result applies to every pair of elementary cones,
we conclude that E = 0 everywhere inside the conducting shell.