EEE 498/598
Overview of Electrical
Engineering
Lecture 3:
Electrostatics: Electrostatic Potential; Charge
Dipole; Visualization of Electric Fields;
Potentials; Gauss’s Law and Applications;
Conductors and Conduction Current
1
Lecture 3 Objectives

To continue our study of electrostatics
with electrostatic potential; charge dipole;
visualization of electric fields and
potentials; Gauss’s law and applications;
conductors and conduction current.
2
Lecture 3
Electrostatic Potential of a Point
Charge at the Origin
r
r
V r     E  d l    aˆ r


Q
40 r 
2
 aˆ r dr 

dr 
Q


2

40 r r 
40 r
Q
P
r
spherically symmetric
Q
3
Lecture 3
Electrostatic Potential Resulting
from Multiple Point Charges
R2
Q2
r 2
r
r 1
P(R,q,f)
R1
Q1
n
V r   
O
k 1
Qk
40 Rk
No longer spherically symmetric!
4
Lecture 3
Electrostatic Potential Resulting from
Continuous Charge Distributions
qel r  dl 
V r  

40 L
R
1
qes r  ds
V r  
40 S
R
1
qev r  dv
V r  
40 V
R
1
5
 line charge
 surface charge
 volume charge
Lecture 3
Charge Dipole

An electric charge dipole consists of a pair
of equal and opposite point charges
separated by a small distance (i.e., much
smaller than the distance at which we observe
the resulting field).
+Q -Q
d
6
Lecture 3
Dipole Moment
• Dipole moment p is a measure of the strength
of the dipole and indicates its direction
+Q
p  Qd
d
p is in the direction from
the negative point charge
to the positive point
charge
-Q
7
Lecture 3
Electrostatic Potential Due to
Charge Dipole
P
observation
point
z
R
+Q
R
r
d/2
d/2
-Q
p  aˆ z Qd
q
8
Lecture 3
Electrostatic Potential Due to
Charge Dipole (Cont’d)
V r   V r ,q  
Q
40 R

Q
40 R
cylindrical symmetry
9
Lecture 3
Electrostatic Potential Due to
Charge Dipole (Cont’d)
P
R
d/2 q
r
R
d/2
R  r 2  (d / 2) 2  rd cos q
R  r 2  (d / 2) 2  rd cos q
10
Lecture 3
Electrostatic Potential Due to
Charge Dipole in the Far-Field
• assume R>>d
• zeroth order approximation:
R  R
R  R
V 0
11
not good
enough!
Lecture 3
Electrostatic Potential Due to Charge
Dipole in the Far-Field (Cont’d)
• first order approximation from geometry:
R
d/2
d/2
d
R  r  cos q
2
d
R  r  cos q
2
q r
R
lines approximately
parallel
12
Lecture 3
Electrostatic Potential Due to Charge
Dipole in the Far-Field (Cont’d)
• Taylor series approximation:
1
1  d
1 d


 r  cos q   1  cos q 
R  2
r  2r


1
d

 1  cos q 
Recall :
r  2r

1
1 1
d

 1  cos q 
R r  2r

x  1
13
1  x n  1  nx,
Lecture 3
Electrostatic Potential Due to Charge
Dipole in the Far-Field (Cont’d)
 d cos q   d cos q 
V r ,q  
1 
  1 


40 r 
2r  
2r  
Qd cos q

2
40 r
Q
14
Lecture 3
Electrostatic Potential Due to Charge
Dipole in the Far-Field (Cont’d)
• In terms of the dipole moment:
V
1
p  aˆ r
40 r
15
2
Lecture 3
Electric Field of Charge Dipole
in the Far-Field
1 V 
 V
E  V   aˆ r
 aˆq

r q 
 r
Qd
ˆ
ˆ



a
2
cos
q

a
sin
q
r
q
3
40 r
16
Lecture 3
Visualization of Electric Fields



An electric field (like any vector field) can be
visualized using flux lines (also called streamlines
or lines of force).
A flux line is drawn such that it is everywhere
tangent to the electric field.
A quiver plot is a plot of the field lines constructed
by making a grid of points. An arrow whose tail is
connected to the point indicates the direction and
magnitude of the field at that point.
17
Lecture 3
Visualization of Electric
Potentials



The scalar electric potential can be visualized using
equipotential surfaces.
An equipotential surface is a surface over which V
is a constant.
Because the electric field is the negative of the
gradient of the electric scalar potential, the electric
field lines are everywhere normal to the
equipotential surfaces and point in the direction of
decreasing potential.
18
Lecture 3
Visualization of Electric Fields

Flux lines are suggestive of the flow of

some fluid emanating from positive charges
(source) and terminating at negative charges
(sink).
Although electric field lines do NOT
represent fluid flow, it is useful to think of
them as describing the flux of something
that, like fluid flow, is conserved.
19
Lecture 3
Faraday’s Experiment
charged sphere
(+Q)
+
+
+
+
metal
insulator
20
Lecture 3
Faraday’s Experiment (Cont’d)




Two concentric conducting spheres are
separated by an insulating material.
The inner sphere is charged to +Q. The
outer sphere is initially uncharged.
The outer sphere is grounded momentarily.
The charge on the outer sphere is found to
be -Q.
21
Lecture 3
Faraday’s Experiment (Cont’d)


Faraday concluded there was a
“displacement” from the charge on the inner
sphere through the inner sphere through the
insulator to the outer sphere.
The electric displacement (or electric flux)
is equal in magnitude to the charge that
produces it, independent of the insulating
material and the size of the spheres.
22
Lecture 3
Electric Displacement (Electric
Flux)
+Q
-Q
23
Lecture 3
Electric (Displacement) Flux
Density


The density of electric displacement is the
electric (displacement) flux density, D.
In free space the relationship between flux density
and electric field is
D  0 E
24
Lecture 3
Electric (Displacement) Flux
Density (Cont’d)

The electric (displacement) flux density for
a point charge centered at the origin is
25
Lecture 3
Gauss’s Law

Gauss’s law states that “the net electric
flux emanating from a close surface S is
equal to the total charge contained
within the volume V bounded by that
surface.”
D

d
s

Q
encl

S
26
Lecture 3
Gauss’s Law (Cont’d)
S
By convention, ds
is taken to be outward
from the volume V.
ds
V
Qencl   qev dv
V
Since volume charge
density is the most
general, we can always write
Qencl in this way.
27
Lecture 3
Applications of Gauss’s Law

Gauss’s law is an integral equation for the
unknown electric flux density resulting
from a given charge distribution.
D

d
s

Q
encl

S
known
unknown
28
Lecture 3
Applications of Gauss’s Law
(Cont’d)
In general, solutions to integral equations
must be obtained using numerical
techniques.
 However, for certain symmetric charge
distributions closed form solutions to
Gauss’s law can be obtained.

29
Lecture 3
Applications of Gauss’s Law
(Cont’d)
Closed form solution to Gauss’s law relies
on our ability to construct a suitable family
of Gaussian surfaces.
 A Gaussian surface is a surface to which
the electric flux density is normal and over
which equal to a constant value.

30
Lecture 3
Electric Flux Density of a Point
Charge Using Gauss’s Law
Consider a point charge at the origin:
Q
31
Lecture 3
Electric Flux Density of a Point
Charge Using Gauss’s Law (Cont’d)
(1) Assume from symmetry the form of the field
D  aˆr Dr r 
spherical
symmetry
(2) Construct a family of Gaussian surfaces
spheres of radius r where
0r 
32
Lecture 3
Electric Flux Density of a Point
Charge Using Gauss’s Law (Cont’d)
(3) Evaluate the total charge within the volume
enclosed by each Gaussian surface
Qencl   qev dv
V
33
Lecture 3
Electric Flux Density of a Point
Charge Using Gauss’s Law (Cont’d)
Gaussian surface
R
Q
Qencl  Q
34
Lecture 3
Electric Flux Density of a Point
Charge Using Gauss’s Law (Cont’d)
(4) For each Gaussian surface, evaluate the integral
 D  d s  DS
S
magnitude of D
on Gaussian
surface.
surface area
of Gaussian
surface.
 D  d s  D r  4 r
2
r
S
35
Lecture 3
Electric Flux Density of a Point
Charge Using Gauss’s Law (Cont’d)
(5) Solve for D on each Gaussian surface
Qencl
D
S
Q
D  aˆ r
4 r 2
 E
36
D
0
 aˆ r
Q
40 r 2
Lecture 3
Electric Flux Density of a Spherical
Shell of Charge Using Gauss’s Law
Consider a spherical shell of uniform charge
density:
q0 , a  r  b
qev  
0, otherwise
a
b
37
Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)
(1) Assume from symmetry the form of the field
D  aˆr Dr R 
(2) Construct a family of Gaussian surfaces
spheres of radius r where
0r 
38
Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)

Here, we shall need to treat separately 3 subfamilies of Gaussian surfaces:
1)
0r a
2)
ar b
3)
r b
a
b
39
Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)
Gaussian surfaces
for which
0r a
Gaussian surfaces
for which
ar b
Gaussian surfaces
for which
r b
40
Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)
(3) Evaluate the total charge within the volume
enclosed by each Gaussian surface
Qencl   qev dv
V
41
Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)

For

For
0r a
Qencl  0
ar b
r
Qencl
4 3
4
3
  q0 dv  q0  r  q0  a
3
3
a

4
3
3
 q0  r  a
3
42

Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)

For
r b
b
Qencl
4 3
4
3
  qev dv  q0  b  q0  a
3
3
a

4
3
3
 q0  b  a
3
43

Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)
(4) For each Gaussian surface, evaluate the integral
 D  d s  DS
S
magnitude of D
on Gaussian
surface.
surface area
of Gaussian
surface.
 D  d s  D r  4 r
2
r
S
44
Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)
(5) Solve for D on each Gaussian surface
Qencl
D
S
45
Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)


0r a
0,

4
3
3
a

r

q
3

0


q
a
3
0
ar b
 aˆ r  r  2 ,
D  aˆ r
2
r 
3
4 r


4
3
3
a

b

q

3
3
0
q
a

b
0
aˆ r 3
ˆ
r b
,
a

r
2
2

r
3
4 r





46

Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)

Notice that for r > b
Total charge contained
in spherical shell
Qtot
D  aˆ r
2
4 r
47
Lecture 3
Electric Flux Density of a Spherical Shell
of Charge Using Gauss’s Law (Cont’d)
0.7
0.6
q0  1 C/m3
0.5
Dr (C/m)
a 1m
b2m
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
10
R
48
Lecture 3
Electric Flux Density of an Infinite
Line Charge Using Gauss’s Law
Consider a infinite line charge carrying charge per
unit length of qel:
qel
z
49
Lecture 3
Electric Flux Density of an Infinite Line
Charge Using Gauss’s Law (Cont’d)
(1) Assume from symmetry the form of the field
D  aˆ  D  
(2) Construct a family of Gaussian surfaces
cylinders of radius  where
0  
50
Lecture 3
Electric Flux Density of an Infinite Line
Charge Using Gauss’s Law (Cont’d)
(3) Evaluate the total charge within the volume
enclosed by each Gaussian surface
Qencl   qel dl
L
cylinder is
infinitely long!
Qencl  qel l
51
Lecture 3
Electric Flux Density of an Infinite Line
Charge Using Gauss’s Law (Cont’d)
(4) For each Gaussian surface, evaluate the integral
 D  d s  DS
S
magnitude of D
on Gaussian
surface.
surface area
of Gaussian
surface.
 D  d s  D   2  l
S
52
Lecture 3
Electric Flux Density of an Infinite Line
Charge Using Gauss’s Law (Cont’d)
(5) Solve for D on each Gaussian surface
Qencl
D
S
qel
D  aˆ 
2 
53
Lecture 3
Gauss’s Law in Integral Form
 Dds  Q
encl
S
  qev dv
V
ds
V
S
54
Lecture 3
Recall the Divergence Theorem


Also called Gauss’s
theorem or Green’s
theorem.
Holds for any volume
and corresponding
closed surface.
 D  d s     D dv
S
V
ds
V
S
55
Lecture 3
Applying Divergence Theorem to
Gauss’s Law
 D  d s     D dv   q
ev
S
V
dv
V
 Because the above must hold for any
volume V, we must have
  D  qev
Differential form
of Gauss’s Law
56
Lecture 3
Fields in Materials
Materials contain charged particles that
respond to applied electric and magnetic
fields.
 Materials are classified according to the
nature of their response to the applied
fields.

57
Lecture 3
Classification of Materials
Conductors
 Semiconductors
 Dielectrics
 Magnetic materials

58
Lecture 3
Conductors
A conductor is a material in which
electrons in the outermost shell of the
electron migrate easily from atom to atom.
 Metallic materials are in general good
conductors.

59
Lecture 3
Conduction Current

In an otherwise empty universe, a constant
electric field would cause an electron to
move with constant acceleration.
E
a
-e
e = 1.602  10-19 C
 eE
a
me
magnitude of electron charge
60
Lecture 3
Conduction Current (Cont’d)


In a conductor, electrons are constantly
colliding with each other and with the fixed
nuclei, and losing momentum.
The net macroscopic effect is that the
electrons move with a (constant) drift velocity
vd which is proportional to the electric field.
v d   e E
Electron mobility
61
Lecture 3
Conductor in an Electrostatic
Field
To have an electrostatic field, all charges
must have reached their equilibrium
positions (i.e., they are stationary).
 Under such static conditions, there must be
zero electric field within the conductor.
(Otherwise charges would continue to
flow.)

62
Lecture 3
Conductor in an Electrostatic
Field (Cont’d)


If the electric field in which the conductor is
immersed suddenly changes, charge flows
temporarily until equilibrium is once again
reached with the electric field inside the
conductor becoming zero.
In a metallic conductor, the establishment of
equilibrium takes place in about 10-19 s - an
extraordinarily short amount of time indeed.
63
Lecture 3
Conductor in an Electrostatic
Field (Cont’d)

•
There are two important consequences to the fact
that the electrostatic field inside a metallic conductor
is zero:
 The conductor is an equipotential body.
 The charge on a conductor must reside entirely
on its surface.
A corollary of the above is that the electric field just
outside the conductor must be normal to its surface.
64
Lecture 3
Conductor in an Electrostatic
Field (Cont’d)
-
-
-
-
+
+
+
+
65
+
Lecture 3
Macroscopic versus Microscopic
Fields
In our study of electromagnetics, we use
Maxwell’s equations which are written in
terms of macroscopic quantities.
 The lower limit of the classical domain is
about 10-8 m = 100 angstroms. For smaller
dimensions, quantum mechanics is needed.

66
Lecture 3
Boundary Conditions on the Electric Field at
the Surface of a Metallic Conductor
Et  0
Dn  aˆ n  D  qes
67
aˆ n
-
- - E=0
+
+ + +
+
Lecture 3
Induced Charges on Conductors

The BCs given above imply that if a
conductor is placed in an externally
applied electric field, then
 the
field distribution is distorted so that the
electric field lines are normal to the
conductor surface
 a surface charge is induced on the
conductor to support the electric field
68
Lecture 3
Applied and Induced Electric
Fields



The applied electric field (Eapp) is the field
that exists in the absence of the metallic
conductor (obstacle).
The induced electric field (Eind) is the field
that arises from the induced surface charges.
The total field is the sum of the applied and
induced electric fields.
E  E app  E ind
69
Lecture 3