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Lecture 3:
Electrostatics: Electrostatic Potential;
Charge Dipole; Visualization of Electric
Fields; Potentials; Gauss’s Law and
Applications; Conductors and
Conduction Current
Lecture 3
1






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.
Lecture 3
2
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
Lecture 3
3
R2
Q2
r 2
r
r 1
P(R,q,f)
R1
Q1
n
Qk
V r   
k 1 40 Rk
O
No longer spherically symmetric!
Lecture 3
4
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
 line charge
 surface charge
 volume charge
Lecture 3
5

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
Lecture 3
6
• 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
Lecture 3
7
P
observation
point
z
R
+Q
R
r
d/2
d/2
-Q
p  aˆ z Qd
q
Lecture 3
8
Q
Q
V r   V r , q  

4 0 R 4 0 R
cylindrical symmetry
Lecture 3
9
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
Lecture 3
10
• assume R>>d
• zeroth order approximation:
R  R
R  R
V 0
not good
enough!
Lecture 3
11
• 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
Lecture 3
12
• 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
1  x n  1  nx,
Lecture 3
13

d cos q
V r , q  
1 

40 r 
2r
Qd cos q

2
40 r
Q
• In
d cos q
 
  1 
2r
 
terms of the dipole moment:
1 p  aˆ r
V
40 r 2
Lecture 3
14



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
Lecture 3
15



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.
Lecture 3
16



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.
Lecture 3
17


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.
Lecture 3
18
charged sphere
(+Q)
+
+
+
+
metal
insulator
Lecture 3
19




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.
Lecture 3
20


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.
Lecture 3
21
+Q
-Q
Lecture 3
22


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
Lecture 3
23

The electric (displacement) flux density for a
point charge centered at the origin is
Lecture 3
24

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
Lecture 3
25
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.
Lecture 3
26

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
Lecture 3
27


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.
Lecture 3
28


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.
Lecture 3
29
Consider a point charge at the origin:
Q
Lecture 3
30
(1) Assume from symmetry the form of the
field
D  aˆ r D r r 
spherical
symmetry
(2) Construct a family of Gaussian surfaces
spheres of radius r where
0r
Lecture 3
31
(3) Evaluate the total charge within the
volume enclosed by each Gaussian
surface
Qencl   qev dv
V
Lecture 3
32
Gaussian surface
R
Q
Qencl  Q
Lecture 3
33
(4) For each Gaussian surface, evaluate the
integral
surface area
D

d
s

DS

S
magnitude of D
on Gaussian
surface.
of Gaussian
surface.
 D  d s  D r  4 r
2
r
S
Lecture 3
34
(5) Solve for D on each Gaussian surface
Qencl
D
S
Q
D  aˆ r
4 r 2

D
Q
E
 aˆ r
0
4 0 r 2
Lecture 3
35
Consider a spherical shell of uniform
charge density:
q0 , a  r  b
qev  
0, otherwise
a
b
Lecture 3
36
(1) Assume from symmetry the form of the
field
D  aˆ r D r  R 
(2) Construct a family of Gaussian surfaces
spheres of radius r where
0r
Lecture 3
37

Here, we shall need to treat separately 3
sub-families of Gaussian surfaces:
1)
0ra
2)
arb
3)
rb
a
b
Lecture 3
38
Gaussian surfaces
for which
0ra
Gaussian surfaces
for which
arb
Gaussian surfaces
for which
rb
Lecture 3
39
(3) Evaluate the total charge within the
volume enclosed by each Gaussian
surface
Qencl   qev dv
V
Lecture 3
40

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

Lecture 3
41

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

Lecture 3
42
(4) For each Gaussian surface, evaluate the
integral
surface area
D

d
s

DS

S
magnitude of D
on Gaussian
surface.
of Gaussian
surface.
 D  d s  D r  4 r
2
r
S
Lecture 3
43
(5) Solve for D on each Gaussian surface
Qencl
D
S
Lecture 3
44


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






Lecture 3
45

Notice that for r > b
Total charge contained
in spherical shell
Qtot
D  aˆ r
2
4 r
Lecture 3
46
0.7
0.6
q0  1 C/m 3
0.5
a 1m
0.3
b2m
Dr (C/m)
0.4
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
10
R
Lecture 3
47
Consider a infinite line charge carrying
charge per
unit length of qel:
qel
z
Lecture 3
48
(1) Assume from symmetry the form of the
field
D  aˆ  D    
(2) Construct a family of Gaussian surfaces
cylinders of radius  where
0 
Lecture 3
49
(3) Evaluate the total charge within the volume
enclosed by each Gaussian surface
Qencl   qel dl
L
cylinder is
infinitely long!
Q encl  q el l
Lecture 3
50
(4) For each Gaussian surface, evaluate the
integral
surface area
D

d
s

DS

S
magnitude of D
on Gaussian
surface.
of Gaussian
surface.
 D  d s  D   2  l
S
Lecture 3
51
(5) Solve for D on each Gaussian surface
Qencl
D
S
qel
D  aˆ 
2 
Lecture 3
52
D

d
s

Q

q
dv
encl
ev


S
V
ds
V
S
Lecture 3
53


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
Lecture 3
54
 D  d s     D dv   q
S
V
ev
dv
V
 Because the above must hold for any
volume V, we must have
  D  q ev
Differential form
of Gauss’s Law
Lecture 3
55


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.
Lecture 3
56




Conductors
Semiconductors
Dielectrics
Magnetic materials
Lecture 3
57


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.
Lecture 3
58

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
59
Lecture 3


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.
vd  e E
Electron mobility
60
Lecture 3


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.)
Lecture 3
61


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.
Lecture 3
62

•
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.
Lecture 3
63
-
-
-
-
+
+
+
+
+
Lecture 3
64


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.
Lecture 3
65
Et  0
Dn  aˆ n  D  qes
ân
-
- - E=0
+
+ + +
+
Lecture 3
66

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
Lecture 3
67



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
Lecture 3
68