Download Chapter 2A

Chapter 2 - Electrostatics
Part A
2.0 Introduction

Field theory lays the foundation of electrical,
electronics, telecom and computer engineering.

It is used in understanding the principle of
radar, remote sensing, satellite communication,
TV reception, microwave devices, fiber optics,
electromagnetic compatibility, etc.
2.0 Introduction



A number of phenomena, which cannot be
adequately explained by circuit theory, can
possibly be explained by field theory.
In circuit theory, voltages and currents are the
main system variables, and they are treated as
scalar quantities.
On the other hand, most of the variables in
electromagnetic field theory are vectors having
magnitude as well as direction.
2.0 Introduction




In a simple term, electromagnetic is the study of
the effects of electric charges at rest and in
motion
Both positive and negative charges are sources
of electric field
Moving charges produce a current, which gives
rise to magnetic field
A time varying electric field is accompanied by
a magnetic field, and vice versa
2.1 Nature of Electric Charge



In general, an atom is electrically neutral (total
number of positive charge of protons = total
number of negative charge of electrons).
However, when electrons are removed from
atoms of a body, then there will be more positive
charges than the negative charges in the atoms
and the body is said to be positively charged.
On the other hand, if some electrons are added to
it, the number of negative charges is more than
positive charges, then the body is said to be
negatively charged.
2.1 Nature of Electric Charge

Electric charge properties:
1) Law of conservation of electric charge


The net electric charge can neither be created nor destroyed
If a volume contains np protons and ne electrons, then the
total charge is
q  n p e  ne e  (n p  ne )e
2) Principle of linear superposition

The total vector electric field at a point in space due to a
system of point charges is equal to the vector sum of the
electric fields at that point due to the individual charges.
2.1 Nature of Electric Charge

First Law of ELECTROSTATICS:
“Like charges repel each other, opposite
charges attract each other”
2.2 Concept of Electric Field


The electromagnetic force consists of an
electrical force, Fe and magnetic force, Fm.
The electrical force, Fe, is similar to the
gravitational force, but with two differences:


The source of the gravitational force is mass, whereas
the source of the electrical field is electric charge.
Electric charge may have positive or negative
polarity, whereas mass does not exhibit such a
property.
2.2 Concept of Electric Field




All matter contains a mixture of neutrons, positively
charged protons, and negatively charged electrons.
The fundamental quantity of charge is a single
electron, usually denoted by the letter e.
The unit of electric charge is coulomb (C), named in
honor of the 18th century French Scientist Charles
Augustin de Coulomb.
The magnitude of e is:
e  1.6 1019 (C)
2.2 Concept of Electric Field

The charge of a single electron is:
qe  e  1.6 10 19 (C)

And that of a proton is equal in magnitude but
opposite in polarity:
q p  e  1.6 1019 (C)

Neutrons have no net charge
2.2 Concept of Electric Field




Every charged object sets up an electric field in
the surrounding space.
A second charge "feels" the presence of this field.
The second charge is either attracted toward the
initial charge or repelled from it, depending on the
signs of the charges.
Since the second charge also has an electric field,
the first charge feels its presence and is either
attracted or repelled by the second charge, too.
2.2 Concept of Electric Field


If we place two positively charged particles A and B at
a distance as shown in figure below, there is a
repulsive force between these two particles.
This force is of the action-at-a-distance type and can
be felt without any intermediate medium between A
and B.
2.2 Concept of Electric Field



Let us move particle B away.
Point P is the point where particle B was placed, and
is now in the electric field created by particle A.
When particle B is placed back at P, a force F will be
exerted on the particle B by the electric field.
2.2 Concept of Electric Field



To verify the existence of an electric field at a
point, a test charge is placed at that point.
If the test charge feels an electric force, then the
electric field exists at that point.
Thus, an electric field is said to exist at a particular
point if an electric force is acting on a charged
particle at that point.
2.2 Concept of Electric Field

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The magnitude and direction of the electric field
vector varies from point to point in the field space.
Electric field can also vary with time.
In this subject we are studying on electric field
produced by static electrical charges which is not
changing with time.
Electrostatic field - field from electric charge at
rest.
2.2 Concept of Electric Field

Coulomb’s experiments demonstrated that:



Two like charges repel one another; whereas two
charges of opposite polarity attract
The force acts along the line joining the charges, and
Its strength is proportional to the product of the
magnitudes of the two charges and inversely
proportional to the square of the distance between
them
2.2 Concept of Electric Field

These properties that constitute Coulomb’s Law can be
expressed mathematically as:
Fe21
q1q2
ˆ
Fe21  R
(N) (in free space)
12
2
40 R12
where
Fe21 is the electrical force acting on
+q2
R̂12
+q1
Fe12
R 12
charge q2 due to charge q1
R12 is the distance between th e two charges
ˆ is a unit vecto r pointing from charge q to charge q
R
12
1
2
 0 is a universal constant called the electrical permittivi ty of free space
 0  8.854 10 12 farad per meter (F/m)
2.2 Concept of Electric Field
Fe21
+q2
R̂12
+q1
Fe12


R 12
The two charges are assumed to be in free space
(vacuum) and isolated from all other charges
The force Fe12 acting on charge q1 due to charge q2 is
equal to force Fe21 in magnitude, but opposite in
direction:
Fe12  Fe21
2.2 Concept of Electric Field

We can express the unit vector R̂12in terms of the
position vectors R1 and R2 of the charges q1 and q2
respectively as:
z
ˆ  R 2  R1
R
12
R 2  R1
Fe21 
1
40
q1q2
q2
R12
q1
R 2 - R1
R 2 - R1
3
(N)
R1
y
since
R 12  R 2 - R 1
and
1
40
R2
x
is called proportion ality constant
2.2 Concept of Electric Field

For other medium,
q1q2 R 2 - R 1
q1q2 R 2 - R 1
Fe 21 

3
4 R 2 - R 1
40 r R 2 - R 1 3

(N)
From the expression given above, the force between
two-point charge is:



directly proportional to the product of the quantities of
both charges
inversely proportional to the square of the distance
between the two charges
dependent on the medium in between the two charges
2.3 Electric Field Intensity, E

A static electric charge sets up an electric field in the
region of space that surrounds it. The quantity we use to
measure the strength of the electric field is its intensity.
The intensity of an electric field (or electric field
strength), E, is the force exerted by the electric field on
a unit test charge.
2.3 Electric Field Intensity, E

Coulomb’s law states that:

An isolated charge, q induces an electric field, E, at every
point in space. At any specific point, P, the electric field
intensity, E can be expressed as:
ˆ
ER

q
4R 2
(V/m)
In the presence of an electric field, E, which may be due to a
single charge or a distribution of many charges, the force
acting on a test charge, q' when it is placed at that point, is
given by
F  qE
(N)
2.4 Streamlines (Flux lines)

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
A convenient method to represent field variations
graphically is the use of streamlines, or flux lines.
These are directed lines or curves that indicate the
direction of the vector field at each point.
These continuous lines of force emanate from a
positive charge and end on a negative charge.
These lines always leave or enter a conducting surface
normally.
The lines indicate the path that a small positive test
charge would take if it were placed in the field.
2.4 Streamlines (Flux lines)

A line tangent to a flux line indicates the direction
of the electric field at that point.
+q
Streamlines surrounding a
positive charge
-q
Streamlines surrounding a
negative charge
2.4 Streamlines (Flux lines)

The arrow shows the direction of the tangential line.
The streamlines do not cross among themselves.
+q
-q
Electric field lines between two
opposite charges
+q
+q
Electric field lines between two
positive charges
2.4 Streamlines (Flux lines)

The magnitude of the electric field is represented either
by the number of lines (density) passing
perpendicularly (normally) through a surface, or by the
length of the directed lines.
A
B
Stronger field in region A than B
since the density of equal length
directed lines is higher
q
Field strength is strongest at the
region closest to q and decreases
when it is further away
2.5 Electric Flux



In SI units, one line of electric flux is defined as a tube
of lines (known as Faraday tube) of force emanates
from +1 C and terminates on -1 C charge.
Since electric flux is numerically equal to the charge, so
unit of flux is measured in coulomb, which is
represented by .
So = Q coulombs.
2.5 Electric Flux

The number of lines of force per unit area is directly
proportional to the electric field intensity at that point.
They can be related by the equation below:
N
  oE
An
For free space
where
N = number of lines of force,
An = normal surface area to the field direction at that point,
o = the permittivity of the space.
2.5 Electric Flux

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If A1=A2=A and N1>N2, then
N1/A1 > N2/A2
 0E1> 0E2, and
E1>E2
2.6 Electric Flux Density, D


Flux density is given by the normal flux per unit area,
denotes as D.
If a flux of  coulombs passes normally through an area of
A m2, then flux density,

2
D
A
C/m

It is related to electric field intensity by the relation,
D   o r E

Therefore D is a vector field similar to which can be
represented by lines of force or lines of displacement (or
simply "displacement").
2.6 Electric Flux Density, D

D is a vector quantity whose direction at every
point is the same as that of E but whose
magnitude is
 o r  E

One useful property of D is that its surface
integral over any closed surface equals the
enclosed surface charge.
2.7 Maxwell’s Equations

Electromagnetism is based on a set of four fundamental
relations known as Maxwell’s equations:
  D  v
B
E  
t
B  0
D
H  J 
t



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E and D are electric field quantities
interrelated by D = εE
ε is the electrical permittivity of the
material
B and H are magnetic field quantities
interrelated by B = µH,
µ is the magnetic permeability of the
material.
ρv is the electric charge density per
unit volume
J is the current density per unit area
2.7 Maxwell’s Equations
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Maxwell’s equations hold in any material, including free
space (vacuum) and at any spatial location
All the quantities in Maxwell’s equations may be a
function of time, t.
These equations are deduced from experimental
observations reported by Gauss, Ampere, Faraday and
others.
Maxwell’s equations form the fundamental tenets of
electromagnetic theory.
2.7 Maxwell’s Equations

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In the static case, none of the quantities appearing in
Maxwell’s equations are a function of time (  t  0 ).
In this case, all charges are permanently fixed in
space, or if they move, they do so at a steady rate so
that ρv and J are constant in time.
The time derivatives of B and D are zero, and
Maxwell’s equations are reduced to:
  D  v
E  0
Electrosta tics
B  0
H  J
Magnetosta tics
2.7 Maxwell’s Equations
Static and Dynamic Fields
Condition
Field Quantities
Electrostatics
Stationary
charges
Electric Field intensity, E (V/m)
Electric Flux density, D (C/m2)
D = εE
Magnetostatics
Steady
currents (DC)
Magnetic flux density, B (T)
Magnetic field intensity, H (A/m)
B = μH
Dynamics
(Time varying
fields)
Time varying
current
E, D, B and H
2.7 Maxwell’s Equations

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Maxwell’s four equations separate into two uncoupled
pairs.
The first pair involving only the electric field quantities
E and D, while the second pair involving only the
magnetic field quantities B and H.
The electric and magnetic fields are no longer
interconnected in the static case
This allows the study of electricity and magnetism as
two distinct and separate phenomena, as long as the
spatial distributions of charge and current flow remain
constant in time.
2.8 Charge and Current Distribution

There are various forms of electric charge
distributions in electromagnetics.

If the charges are in motion, they constitute current
distribution

Charge may be distributed over
 a volume of space,
 across a surface, or
 along a line
2.8 Charge and Current Distribution
Charge Densities
 In electromagnetics, we treat the net charge contained in
an elemental volume, ∆v as if it were uniformly
distributed within it.
 Therefore, the volume charge density, ρv can be
expressed as
v  lim
v 0

q dq

v dv
(C/m 3 )
where ∆q is the charge contained in ∆v.
In general, ρv is defined at a given point in space (x, y, z),
and at a given time, t.
2.8 Charge and Current Distribution
Charge Densities
 Physically, ρv is the average charge per unit volume
for a volume ∆v centered at (x, y, z), with ∆v large
enough to contain a large number of atoms and at the
same time, small enough to be regarded as a point at
the macroscopic scale under consideration.
 The variation of ρv with spatial location is called
spatial distribution, or simply distribution.
 The total charge contained in a given volume, v can
be expressed as
Q   v dv (C)
v
2.8 Charge and Current Distribution
Charge Densities
 In conductors, electric charge may be distributed
across the surface of a material.
 The quantity of interest is the surface charge density,
ρs which can be defined as
q dq
 s  lim

ds
s 0 s
(C/m 2 )
where ∆q is the charge across an elemental surface
area, ∆s.
2.8 Charge and Current Distribution
Charge Densities
 Similarly, if the charge is distributed along a line,
which need not be straight, the distribution can be
characterized as line charge density, ρl , defined as
q dq
l  lim

dl
l 0 l
(C/m)
where ∆q is the charge across an elemental length, ∆l.
2.8 Charge and Current Distribution

Example 1 – Line charge Distribution
Calculate the total charge Q contained in
a cylindrical tube of charge oriented
along the z-axis as shown. The line
charge density is l = 4z, where z is the
distance in meters from the bottom end
of the tube. The tube length is 10cm.
Solution:
The total charge Q is
0.1
0.1
0
0
Q   l dz   4 zdz  2 z
2 0.1
0
 2 102 C
2.8 Charge and Current Distribution

Example 2 – Surface Charge Distribution
The circular disk of electric charge
shown is characterized by a azimuthally
symmetric surface charge density that
increases linearly with r from zero at the
center to 6 C/m2 at r = 3 cm.
Find the total charge present on the disk
surface.
Solution:
ρs is symmetrical with respect to the
azimuth angle,  and can be expressed
as a function of r.
2.8 Charge and Current Distribution
6r
2
2

2

10
r
(C/m
)
2
3 10
where r is in meters.
s 
In polar coordinates, an element of area is ds = r dr d.
For the disk shown, the limits of integration are from 0 to 2 for 
and from 0 to 3  10-2 m for r. Therefore,
Q    s ds  
S
2

3102
 0 r 0
2 10 r r drd
2
3 310
r
 2  2 10
3
2
0
2
 11.31 (mC)
2.8 Charge and Current Distribution

Example 3 – Volume Charge Distribution
For a spherical shell centered at the origin and extends between
R = 3cm and R = 4cm, find the total charge contained in the shell
if the volume charge density is 5R×10-4 (C/m3).
Solution:  v  5R 10 4 (C/m 3 )
Q    v dv
v

0.04
0.03

2
 
0
5R 4

10 4
4

5 R 10 4 R 2 sin  dR d d
0
0.04
 2  2
0.03
 5 10 4 0.04   0.03
 2.75 (nC)
4
4

2.8 Charge and Current Distribution
Current Density
 Consider a tube of charge with volume charge density
ρv moving with a mean velocity, u along the axis of a
tube, as shown in the figure below.
 The charges move a distance ∆l = u ∆t over a period
of ∆t.
 The amount of charge that crosses the tube’s crosssectional surface, ∆s' in time ∆t is therefore
2.8 Charge and Current Distribution
Current Density
 Now consider the more general case where the charges
are flowing through a surface ∆s whose surface normal n̂
is not necessarily parallel to u, as shown in figure

In this case, the amount of charge ∆q flowing through ∆s
is given by
q = v u (s cos Ө )t
q = vus t
2.8 Charge and Current Distribution
Current Density
 The corresponding current is therefore
I = q/t = vus = Js
where
J = vu (A/m2)
J is defined as the current density in ampere per square
meter. For an arbitrary surface S, the total current
flowing through it is then given by
I   J  ds (A)
s
2.8 Charge and Current Distribution
Current Density
 When the current is generated by the actual movement
of electrically charged matter, it is called a convection
current, and J is called the convection current density.
 This is distinct from a conduction current, where atoms
of the conducting medium do not move.
 Because the two types of current are generated by
different physical mechanism, conduction current
obeys Ohm’s law, whereas convection current does not.
2.9 Coulomb’s Law

Imagine an electric charge, q located at the origin of a coordinate system. A test charge q’ at a distance R from q
experiences a force, which according to Coulomb’s law is:
qq
ˆ
F R
(N)
2
4R
ˆ is the unit vecto r directing radially from q to q
where R

Using the above definition the intensity of the electric field
surrounding the charge q is therefore:
E
F ˆ q
q
ˆ
R

R
(V/m)
2
2

q
4R
40 r R
2.9 Coulomb’s Law




This result says that an electric charge located at the
origin generates an electric field (E-field) which
points radially out from the origin.
The strength of the electric field is inverse
proportional to the square of the distance from the
origin.
With F measured in newtons (N) and q' in coulombs
(C), the unit of E is (N/C).
This is the same as volt per meter (V/m).
2.9 Coulomb’s Law

For a material with electrical permittivity ε, the
electrical field quantities D and E are related by:
D  E
with
   r 0
where
 0  8.85 10 12  (1 / 36 ) 10 9 (F/m)
ε0 is the electrical permittivity of free space, and
εr=ε/ε0 is called the relative permittivity (or dielectric
constant) of the material.
2.9 Coulomb’s Law


For most materials and under most conditions, ε of the
material has a constant value independent of both the
magnitude and direction of E.
If ε is independent of the magnitude of E, then the
material is said to be linear because D and E are related
linearly, and if it is independent of the direction of E,
the material is said to be isotropic.
2.9 Coulomb’s Law
Electric field due to multiple point charges


The equation for field E due to a
single charge can be extended to
find the field due to multiple
point charges.
Consider two point charge, q1
and q2 located at position vector
R1 and R2 from the origin of a
coordinate system as shown in
the Figure
x
z
E
E2
E1
R-R1
q1
R1
P
R-R2
R
q2
R2
y
2.9 Coulomb’s Law
Electric field due to multiple point charges

At point P with position
vector R, the electric field, E1
due to q1 alone is given by
E1 
q1 R  R 1 
4 R  R 1
3
z
E
E2
E1
R-R1
(V/m)
q1
R1
P
R-R2
R
q2

Similarly, the electric field
due to q2 alone is
E2 
q2 R  R 2 
4 R  R 2
3
(V/m)
x
R2
y
2.9 Coulomb’s Law
Electric field due to multiple point charges
 The electric field obeys the principle of linear
superposition.
 The total electric field, E at any point in space is equal
to the vector sum of the electric fields induced by all the
individual charges. In this case,
E  E1  E2
1  q1 R  R1  q2 R  R 2 




3
3
4  R  R1
R  R 2 
2.9 Coulomb’s Law
Electric field due to multiple point charges

Generalizing the preceding result to the case of N
point charges, the electric field E at position vector R
caused by charges q1, q2, …., qN located at points
with position vectors R1, R2, …, RN, is given by
E
1
4
N

i 1
qi R  R i 
R  Ri
3
2.9 Coulomb’s Law

Example 4 – Electric Field due to two point charges
Two point charges with q1 = 310-5 C and q2 = -510-5 C are
located in free space at (2,2,-1) and (0,-2,1), respectively,
in a Cartesian coordinate system.
Find:
(a) the electric field E at (4,2,-1) and
(b) the force on a 610-5 charge located at that point.
All distances are in meters.
Solution:
(a) The electric field, E is given by
1  q1 R  R1  q2 R  R 2 
E


3 
40  R  R1 3
R  R 2 
2.9 Coulomb’s Law
The vectors R1, R2 and R can be
R 1  2xˆ  2yˆ  zˆ
R 2  2yˆ  zˆ
expressed as:
R  4xˆ  2yˆ  zˆ
Therefore,
(b)
F  q3E
1  32xˆ  54xˆ  4yˆ  2zˆ  
5


10

40  8
216
71xˆ  10yˆ  5zˆ

10 -5
(V/m)
4320
E
71xˆ  10yˆ  5zˆ
 10 -5
4320
71xˆ  10yˆ  5zˆ

 10 -10
(N)
720
 6  10 5 
2.9 Coulomb’s Law
Electric field due to a charge distribution

It is possible to have continuous charge distribution along a
line, on a surface, or in a volume as shown in the figure below.
ρL
+
+
+
+
+
ρs
+
+
+
+
+
+
+
where
+
+
+
Surface charge
+
+
+
+
+
Line charge
+
+
+
+
+
+
+
ρv
+
+
+
+
Volume charge
L = line charge density, C/m ,
S = surface charge density,C/m2 ,
v = volume charge density , C/m3
2.9 Coulomb’s Law
Electric field due to a charge distribution
 Replacing Q with charge element, the electric field
intensity :
ρL
+
+
+
+
+
+
+
+
ρs
+
+
+
+
+
+
Line charge
 dl ˆ
E L 2 R
4o R
+
+
+
+
+
Surface charge
 ds ˆ
E S 2 R
4o R
+
+
+
+
+
+
+
+
ρv
+
+
+
+
Volume charge
E
 v dv ˆ
R
2
4o R
where R is the distance of a field point from a
differential elements dl, ds or dv contributed by an
element dQ of the charge body.
2.9 Coulomb’s Law

Example 5 – Line Charge using Coulomb’s law
A ring of charge in free space is characterized by a
uniform charge density of ρl. The ring has a radius b
and positioned in the x-y plane as shown in the figure.
Find the electric field intensity, E at a point P(0, 0, h)
along the axis of the ring at a distance h from its
center.
Solution:
Consider the electric field generated by a differential
segment of the ring, for example, at (b, , 0) as indicated by
segment 1 in the figure. This segment has a length,
dl = b d and contains charge dq = ρl dl = ρl b d. The
vector, R'1 from segment 1 to point P(0, 0, h) is
R1  brˆ  hzˆ
2.9 Coulomb’s Law
R1  R 1  b 2  h 2
Therefore,
ˆ   R 1   brˆ  hzˆ
R
1
R 1
b2  h2
The electric field at P(0, 0, h) due to charge of segment 1 is
dE1 
ˆ  l dl  l b  brˆ  hzˆ d
R
1
40
R12
40 b2  h2 3 2
1


The field, dE1 has 2 components: (1) dE1r along  r̂ and (2)
dE1z along ẑ .
For symmetry consideration, a field dE2 generated by segment
2 is located diametrically opposite the location of segment 1.
This field is identical to dE1, except it has an opposite
component r̂ . Hence, the sum of the r̂ components
cancelled each other, while the ẑ components add. The
sum of the two contributions can be expressed as
dE  dE1  dE 2  zˆ
 l bh
1
d
32
2
2
20 b  h 
2.9 Coulomb’s Law
For every ring segment in the semicircle defined over the range 0
≤  ≤ π, there is a corresponding segment located diametrically
opposite at ( + π). The total field generated by the ring can be
obtained by integrating the above equation over a semicircle as:

l bh
1
d
20 b 2  h 2 3 2 0
 bh
1
 zˆ l
2 0 b 2  h 2 3 2
E  zˆ
 zˆ

h
40 b  h
2

2 32
Q
where
Q  2bl is the total charge contained in the ring.
2.9 Coulomb’s Law

Example 6 – Surface Charge using Coulomb’s law
z
Find the expression of E for an infinite sheet of charge
in the x-y plane with uniform charge density, ρs.
P (0,0,h)
2
Solution:
The charge associated with an elemental area, ds is
the total charge is therefore

dQ   s ds
Q    s ds
x
The contribution to E field at point P(0,0,h) by the
elemental surface shown in the figure is
dQ
dE 
From the figure, we can write
h
40 R
R   ppˆ  hzˆ , R  R 
dQ   s ds   s p d dp
2
ˆ
R
ˆ  R
p 2  h2 , R
R
ρs
R
p
1
y
2.9 Coulomb’s Law
Substituting these equations gives:
dE 
 s p d dp  ppˆ  hzˆ
3
40
p 2  h2


Due to the symmetry of charge distribution, there is a corresponding element 2 whose contribution
along p̂ cancels that of element 1, as illustrated in the figure. Therefore, the contribution to E p adds up to zero so that E
has only z component. Therefore,

E   dE z  s
40
2
 
0

p 0
hp d dp
p
2
h

2 32
zˆ


sh
sh 
s
2
2 3 2
2
2
2 1 / 2 
ˆ



(2 )   p  h  pdp zˆ 

p

h
z

zˆ


0
0
40
2 0 
2 0

From this expression, we know that an infinite sheet of charge in the x-y plane has only z component. In other words, the
electric field is normal to the sheet and it can be expressed generally as
E
s
nˆ
2 0
Note that the electric field is surprisingly independent of the distance between the sheet and the point of observation, P.
In a parallel plate capacitor, the electric field between the two plates having equal and opposite charge is given by
E
s
 s

nˆ 
(nˆ )  s nˆ
2 0
2 0
0
2.10 Gauss law

Gauss’s law states that:
The total electric flux passing through any closed
surface is equal to the total charge enclosed by
that surface.
2.10 Gauss law




The surface, S referred here need not necessarily be a
physical surface, but can be any hypothetical surface,
which encloses the charge.
From the figure, D·ds is the electric flux flowing
outward through a differential surface element, ds.
The total flux through the surface S is equal to the
enclosed charge, Q.
The surface, S, is called a Gaussian surface.
2.10 Gauss law



From the definition, we can write
Ψ = charge enclosed = Q
That is,
   d   D  ds   v dv
Or
s
Q   D  ds    v dv
s

Applying divergence theorem,
 D  ds     D dv
s

v
Comparing the two volume integrals results in
  D  v
which is the first of the four Maxwell’s equations.
2.10 Gauss law


The first Maxwell’s equation states that the volume
charge density is the same as the divergence of the
electric flux density.
Note that Gauss’s law may be stated in different ways:




In the integral form
In the differential form or point form
Gauss’s law is an alternative statement of Coulomb’s
law – proper application of the divergence theorem to
Coulomb’s law leads to Gauss’s law.
Gauss’s law provides an easy means of finding E or D
for symmetrical charge distributions.
2.10 Gauss law

A continuous charge distribution has




rectangular symmetry if it depends only on x
(or y or z),
cylindrical symmetry if it depends only on ρ,
or
spherical symmetry if it depends only on r
(independent of θ and )
Gauss’s law always hold no matter the
charge distribution is symmetry or not.
2.10 Gauss law



For an isolated point charge, it does not matter which law
is used to determined the expression for E. However, it
does matter when we deal with multiple charges or
continuous charge distribution.
Coulomb’s law can be used to find E for any specified
distribution of charge.
However, Gauss’s law is easier to apply than Coulomb’s
law, but its usage is limited to symmetrical charge
distribution that allows assumption on the variations of
magnitude and direction of D as a function of spatial
location.
2.10 Gauss law

At every point on the surface, only the normal
component of D at the surface contribute to the
integral form of Gauss’s law.

The surface, S, should be chosen so that the magnitude
of D is constant and it’s direction is either normal or
tangential at every point of each subsurface of S.
2.10 Gauss law

Example 7 – Point charge using Gauss’s law
For a point charge Q located at the origin, find D on a spherical
surface containing a point, P as shown in the figure.
z
Solution:
A spherical surface centered at the origin will satisfy the symmetry
condition. This surface is called the Gaussian surface. D is always
normal to the Gaussian surface, that is: D  D r
ˆ
D
P
r
r
Applying Gauss’s law (Ψ = Qenclosed) gives
Q   D  ds  Dr  ds , and ds 
2
Q

 
0
0
which is the surface area of the Gaussian surface
therefore,
Q  Dr 4r 2
Thus,
D
Q
r̂
4r 2
y
r 2 sin  d d  4r 2 ,
Gaussian surface
x
2.10 Gauss law

Example 8 – Infinite line charge using Gauss’s law
Consider an infinite line of uniform charge ρl (C/m) which lies along the zaxis. Determine the expression of electric flux density, D at a point, P.

l
Since D has no z component,
surfaces of the cylinder.
Hence, D   l ρˆ or E 
2
 D  ds  0
l
ρˆ
20 
on the top and bottom
ρ
P
D
y
Q   l l   D  ds  D  ds
Q  D 2l
Line charge ρl C/m
Gaussian surface
Solution:
A cylindrical surface containing P is chosen to satisfy the symmetry
condition, as shown in the figure. D is constant and always normal to the
cylindrical Gaussian surface: D  D ρˆ
Applying Gauss’s law to an arbitrary length, l of the line,
ds  2l is the surface area of the Gaussian surface
where
Therefore,
z
x
2.10 Gauss law

Example 9 – Infinite sheet of charge using Gauss’s law
Find the electric flux density, D at a point, P for an infinite sheet of uniform charge ρs (C/m2)
lying on the x-y plane as shown in the figure.
Solution:
To determine D, a rectangular box that cut symmetrically by the
sheet of charge is chosen. Two of its faces are parallel to the sheet. D
will be normal to the sheet:
D  Dz zˆ
z
D
Infinite sheet of charge
Applying Gauss’s law, we obtain:
Q   s  ds   D  ds  Dz  ds  
ds
 top

bottom
D has no component along x̂ and ŷ and D  ds is zero on the
sides of the box. Assuming that the top and bottom area of the box x
have an area, A,
 s A  Dz ( A  A)
Thus,
D
s
2
zˆ
or
E
s
zˆ
2 0
P
y
Gaussian surface
Area A
2.10 Gauss law

Gaussian surface
Example 10 - Uniformly charged sphere
A sphere of radius a has a uniform charge of ρv
C/m3. Find the electric flux density, D by
constructing Gaussian surfaces for cases r ≤ a and r
≥ a separately.
r
a
a
Solution:
The charge is spherically symmetrical and a spherical surface is an appropriate surface.
For r ≤ a, the total charge enclosed by the surface is
Q    v dv   v  dv   v 
2
 0
And

 
Hence, Ψ = Q gives
r 2 sin  dr d d   v
0
r 0
2

   D  ds  Dr  ds  Dr 
 0
r
D   v rˆ ,
3
r

0
4 3
r
3
r 2 sin  d d  Dr 4r 2
for 0  r  a
For r ≥ a, the charge enclosed by the surface is the entire charge, which is,
Q    v dv   v  dv   v 
2
 0
while
Therefore,

   D  ds  Dr 4r 2
a3
D  2  v rˆ ,
3r

for r  a
0
4 3
2
r
sin

dr
d

d



a
v
r 0
3
a
r