#### Transcript Document

```EEE 498/598
Overview of Electrical
Engineering
Lecture 4:
Electrostatics: Electrostatic Shielding;
Poisson’s and Laplace’s Equations;
Capacitance; Dielectric Materials and
Permittivity
1
Lecture 4 Objectives

To continue our study of electrostatics
with electrostatic shielding; Poisson’s and
Laplace’s equations; capacitance; and
dielectric materials and permittivity.
2
Lecture 4
Ungrounded Spherical Metallic
Shell

Consider a point charge at the center of a
spherical metallic shell:
b
Electrically
neutral
a
Q
3
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)

The applied electric field is given by
E app  aˆ r
Q
40 r
4
2
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)

The total electric field can be obtained
using Gauss’s law together with our
knowledge of how fields behave in a
conductor.
5
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)
(1) Assume from symmetry the form of the
field
ˆ
D  ar Dr r 
(2) Construct a family of Gaussian surfaces
0r 
6
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)

Here, we shall need to treat separately 3 subfamilies of Gaussian surfaces:
1)
a
b
2)
ar b
3)
r b
Dr (r )  0
7
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)
(3) Evaluate the total charge within the volume
enclosed by each Gaussian surface
Qencl   qev dv
V
8
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)
Gaussian surfaces
for which
0r a
Gaussian surfaces
for which
ar b
Gaussian surfaces
for which
r b
9
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)

For

For
0r a
r b
Qencl  Q
Qencl  Q
Shell is electrically neutral:
The net charge carried by shell is zero.
10
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)
Qencl  0
For a  r  b
since the electric field is zero inside
conductor.
 A surface charge must exist on the inner
surface and be given by

qesa
Q

2
4a
11
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)

Since the conducting shell is initially
neutral, a surface charge must also exist on
the outer surface and be given by
qesb
Q

2
4b
12
Lecture 4
Ungrounded Spherical Metallic
Shell (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
13
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)
(5) Solve for D on each Gaussian surface
Qencl
D
S
(6) Evaluate E as
E
D
0
14
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)
Q

ˆ
a
,
r
2
 4 r
0

E  0,

Q
aˆ r
,
2
 40 r
15
0r a
ar b
r b
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)

The induced field is given by
E ind  E  E app
0,
0r a

Q

  aˆ r
, ar b
2
40 r

0,
r b
16
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)
E
total
electric
field
Eapp
a
b
r
Eind
17
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)


The electrostatic potential is obtained by taking
the line integral of E. To do this correctly, we
must start at infinity (the reference point or
ground) and “move in” back toward the point
charge.
For r > b
r
Q
V r     Er dr 
40 r

18
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)

Since the conductor is an equipotential
body (and potential is a continuous
function), we have for
ar b
V r   V b  
19
Q
40b
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)

For 0  r  a
r
V r   V b    Er dr
a
Q 1 1 1 

  

40  b r a 
20
Lecture 4
Ungrounded Spherical Metallic
Shell (Cont’d)
V
No metallic shell
a b
21
r
Lecture 4
Grounded Spherical Metallic
Shell


When the conducting sphere is grounded, we
can consider it and ground to be one huge
conducting body at ground (zero) potential.
Electrons migrate from the ground, so that the
conducting sphere now has an excess charge
exactly equal to -Q. This charge appears in the
form of a surface charge density on the inner
surface of the sphere.
22
Lecture 4



Grounded Spherical Metallic
Shell
There is no longer a surface charge on the
outer surface of the sphere.
The total field outside the sphere is zero.
The electrostatic potential of the sphere is
zero.
- - - b
a
Q
- - 23
Lecture 4
Grounded Spherical Metallic
Shell (Cont’d)
E
total
electric
field
Eapp
R
a b
Eind
24
Lecture 4
Grounded Spherical Metallic
Shell (Cont’d)
V
Grounded
metallic shell acts
as a shield.
a b
25
R
Lecture 4
The Need for Poisson’s and
Laplace’s Equations

So far, we have studied two approaches for
finding the electric field and electrostatic
potential due to a given charge distribution.
26
Lecture 4
The Need for Poisson’s and
Laplace’s Equations (Cont’d)

Method 1: given the position of all the charges,
find the electric field and electrostatic potential
using

(A)
qev r  R dv
E r   
3
4

R
0
V
P
V r     E  d l

27
Lecture 4
The Need for Poisson’s and
Laplace’s Equations (Cont’d)

(B)
qev r  dv
V r   
40 R
V
E r   V r 
 Method 1 is valid only for charges in free space.
28
Lecture 4
The Need for Poisson’s and
Laplace’s Equations (Cont’d)

Method 2: Find the electric field and
electrostatic potential using
 Dds  q
ev
S
Gauss’s Law
dv
V
P
V r     E  d l

 Method 2 works only for symmetric charge
distributions, but we can have materials other
than free space present.
29
Lecture 4
The Need for Poisson’s and
Laplace’s Equations (Cont’d)

Consider the following problem:
 What are E and V in the region?
V  V1
 r 
Conducting
bodies
30
V  V2
Neither Method 1 nor
Method 2 can be used!
Lecture 4
The Need for Poisson’s and
Laplace’s Equations (Cont’d)


Poisson’s equation is a differential equation for the
electrostatic potential V. Poisson’s equation and the
boundary conditions applicable to the particular
geometry form a boundary-value problem that can
be solved either analytically for some geometries or
numerically for any geometry.
After the electrostatic potential is evaluated, the
electric field is obtained using
Er   V r 
31
Lecture 4
Derivation of Poisson’s Equation

For now, we shall assume the only
materials present are free space and
conductors on which the electrostatic
potential is specified. However, Poisson’s
equation can be generalized for other
materials (dielectric and magnetic as well).
32
Lecture 4
Derivation of Poisson’s Equation
(Cont’d)
  D  qev    E 
qev
0
E  V    V  
qev
0
V
2
33
Lecture 4
Derivation of Poisson’s Equation
(Cont’d)
 V 
2
qev
0
Poisson’s
equation
 2 is the Laplacian operator. The Laplacian of a scalar
function is a scalar function equal to the divergence of the
gradient of the original scalar function.
34
Lecture 4
Laplacian Operator in Cartesian,
Cylindrical, and Spherical Coordinates
35
Lecture 4
Laplace’s Equation


Laplace’s equation is the homogeneous
form of Poisson’s equation.
We use Laplace’s equation to solve problems
where potentials are specified on conducting
bodies, but no charge exists in the free space
region.
Laplace’s
2
equation
 V 0
36
Lecture 4
Uniqueness Theorem

A solution to Poisson’s or Laplace’s
equation that satisfies the given boundary
conditions is the unique (i.e., the one and
only correct) solution to the problem.
37
Lecture 4
Potential Between Coaxial Cylinders
Using Laplace’s Equation

Two conducting coaxial cylinders exist such that
V   a   V0
y
V   b   0
+
a
V0
x
b
38
Lecture 4
Potential Between Coaxial Cylinders
Using Laplace’s Equation (Cont’d)

Assume from symmetry that
V  V  

1 d  d 
 
  0
V
 d  d 
2
39
Lecture 4
Potential Between Coaxial Cylinders
Using Laplace’s Equation (Cont’d)

Two successive integrations yield
V    C1 ln    C2

The two constants are obtained from the two
BCs:
V   a   V0  C1 ln a  C2
V   b   0  C1 ln b  C2
40
Lecture 4
Potential Between Coaxial Cylinders
Using Laplace’s Equation (Cont’d)

Solving for C1 and C2, we obtain:
V0
C1 
ln a / b 

V0 ln b 
C2  
ln a / b 
The potential is
V0

V   
ln  
ln a / b   b 
41
Lecture 4
Potential Between Coaxial Cylinders
Using Laplace’s Equation (Cont’d)


The electric field between the plates is given
by:
V0
dV
E  V  aˆ 
 aˆ 
d
 ln b / a 
The surface charge densities on the inner and
outer conductors are given by
 0V0
qesa   0 aˆ   E   a  
qesb
a ln b / a 
  0V0
  0  aˆ   E   b  
b ln b / a 
42
Lecture 4
Capacitance of a Two Conductor
System
The capacitance of a two conductor system is
the ratio of the total charge on one of the
conductors to the potential difference between
that conductor and the other conductor.
+
-
V2
V1
+

Q
C
V12
V12 = V2-V1
43
Lecture 4
Capacitance of a Two Conductor
System
Capacitance is a positive quantity
 Capacitance is a measure of the ability of
a conductor configuration to store charge.

44
Lecture 4
Capacitance of a Two Conductor
System

The capacitance of an isolated conductor
can be considered to be equal to the
capacitance of a two conductor system
where the second conductor is an infinite
distance away from the first and at ground
potential.
Q
C
V
45
Lecture 4
Capacitors


A capacitor is an electrical device consisting of two
conductors separated by free space or another
conducting medium.
To evaluate the capacitance of a two conductor
system, we must find either the charge on each
conductor in terms of an assumed potential
difference between the conductors, or the potential
difference between the conductors for an assumed
charge on the conductors.
46
Lecture 4
Capacitors (Cont’d)
The former method is the more general
but requires solution of Laplace’s equation.
 The latter method is useful in cases where
the symmetry of the problem allows us to
use Gauss’s law to find the electric field
from a given charge distribution.

47
Lecture 4
Parallel-Plate Capacitor

Determine an approximate expression for the
capacitance of a parallel-plate capacitor by
neglecting fringing.
Conductor 2
d
A
Conductor 1
48
Lecture 4
Parallel-Plate Capacitor (Cont’d)

“Neglecting fringing” means to assume that
the field that exists in the real problem is the
same as for the infinite problem.
z
z=d
V = V12
z=0
V=0
49
Lecture 4
Parallel-Plate Capacitor (Cont’d)

Determine the potential between the plates
by solving Laplace’s equation.
2
dV
 V  2 0
dz
V z  0  0
2
V  z  d   V12
50
Lecture 4
Parallel-Plate Capacitor (Cont’d)
2
dV



0

V
z
 c1 z  c2
2
dz
V  z  0   0  c2
V12
V  z  d   V12  c1d  c1 
d
V12
V z  
z
d
51
Lecture 4
Parallel-Plate Capacitor (Cont’d)
• Evaluate the electric field between the plates
dV
V12
E  V  aˆ z
 aˆ z
dz
d
52
Lecture 4
Parallel-Plate Capacitor (Cont’d)
• Evaluate the surface charge on conductor 2
qes 2
V12  0V12
  0 aˆ n  E   0  aˆ z   aˆ z

d
d
• Evaluate the total charge on conductor 2
Q  qes 2 A 
 0V12 A
d
53
Lecture 4
Parallel-Plate Capacitor (Cont’d)
• Evaluate the capacitance
Q 0 A
C

V12
d
54
Lecture 4
Dielectric Materials


A dielectric (insulator) is a medium which possess no
(or very few) free electrons to provide currents due to
an impressed electric field.
Although there is no macroscopic migration of charge
when a dielectric is placed in an electric field,
microscopic displacements (on the order of the size of
atoms or molecules) of charge occur resulting in the
appearance of induced electric dipoles.
55
Lecture 4
Dielectric Materials (Cont’d)



A dielectric is said to be polarized when
induced electric dipoles are present.
Although all substances are polarizable to some
extent, the effects of polarization become
important only for insulating materials.
The presence of induced electric dipoles within
the dielectric causes the electric field both inside
and outside the material to be modified.
56
Lecture 4
Polarizability
Polarizability is a measure of the ability
of a material to become polarized in the
presence of an applied electric field.
 Polarization occurs in both polar and
nonpolar materials.

57
Lecture 4
Electronic Polarizability
electron
cloud
nucleus


58
In the absence of an
applied electric field, the
positively charged
nucleus is surrounded by
a spherical electron cloud
with equal and opposite
charge.
Outside the atom, the
electric field is zero.
Lecture 4
Electronic Polarizability (Cont’d)
Eapp

59
In the presence of an
applied electric field,
the electron cloud is
distorted such that it
is displaced in a
direction (w.r.t. the
nucleus) opposite to
that of the applied
electric field.
Lecture 4
Electronic Polarizability (Cont’d)
e

e
p   e E loc
dipole
moment
(C-m)
polarizability
(F-m2)
60
The net effect is that
each atom becomes a
small charge dipole
which affects the total
electric field both
inside and outside the
material.
Lecture 4
negative
ion
Ionic Polarizability
positive
ion

61
In the absence of an
applied electric field,
the ionic molecules
are randomly oriented
such that the net
dipole moment within
any small volume is
zero.
Lecture 4
Ionic Polarizability (Cont’d)
Eapp

62
In the presence of an
applied electric field,
the dipoles tend to
align themselves with
the applied electric
field.
Lecture 4
Ionic Polarizability (Cont’d)
e

e
p   i E loc
dipole
moment
(C-m)
polarizability
(F-m2)
63
The net effect is that
each ionic molecule is
a small charge dipole
which aligns with the
applied electric field
and influences the
total electric field
both inside and
outside the material.
Lecture 4
Orientational Polarizability

64
In the absence of an
applied electric field,
the polar molecules
are randomly oriented
such that the net
dipole moment within
any small volume is
zero.
Lecture 4
Orientational Polarizability (Cont’d)
Eapp

65
In the presence of an
applied electric field,
the dipoles tend to
align themselves with
the applied electric
field.
Lecture 4
Orientational Polarizability (Cont’d)
e

e
p   o E loc
dipole
moment
(C-m)
polarizability
(F-m2)
66
The net effect is that
each polar molecule is
a small charge dipole
which aligns with the
applied electric field
and influences the
total electric field
both inside and
outside the material.
Lecture 4
Polarization Per Unit Volume
The total polarization of a given material
may arise from a combination of
electronic, ionic, and orientational
polarizability.
 The polarization per unit volume is given
by

P  N p  N T E loc
67
Lecture 4
Polarization Per Unit Volume (Cont’d)

P is the polarization per unit volume. (C/m2)

N is the number of dipoles per unit volume. (m-3)
p is the average dipole moment of the dipoles in the
medium. (C-m)
T is the average polarizability of the dipoles in the
medium. (F-m2)


T   e   i   o
68
Lecture 4
Polarization Per Unit Volume (Cont’d)

Eloc is the total electric field that actually
exists at each dipole location.
 For gases Eloc = E where E is the total
macroscopic field.
 For solids
1
 N T
E loc  E 1 
3 0

69



Lecture 4
Polarization Per Unit Volume (Cont’d)

From the macroscopic point of view, it
suffices to use
P   0 e E
electron
susceptibility
(dimensionless)
70
Lecture 4
Dielectric Materials


The effect of an applied electric field on a
dielectric material is to create a net dipole
moment per unit volume P.
The dipole moment distribution sets up induced
secondary fields:
E  E app  E ind
71
Lecture 4
Volume and Surface Bound
Charge Densities


A volume distribution of dipoles may be
represented as an equivalent volume (qevb) and
surface (qesb) distribution of bound charge.
These charge distributions are related to the
dipole moment distribution:
qevb    P
qesb  P  nˆ
72
Lecture 4
Gauss’s Law in Dielectrics

Gauss’s law in differential form in free space:
 0  E  qev

Gauss’s law in differential form in dielectric:
 0  E  qev  qevb
73
Lecture 4
Displacement Flux Density
 0  E  qev  qevb  qev    P
   0 E  P   qev
• Hence, the displacement flux density vector
is given by
D  0 E  P
74
Lecture 4
General Forms of Gauss’s Law

Gauss’s law in differential form:
  D  qev

Gauss’s law in integral form:
D

d
s

Q
encl

S
75
Lecture 4
Permittivity Concept

Assuming that
we have

P   0 e E
D   0 1   e E   E
The parameter  is the electric
permittivity or the dielectric constant of
the material.
76
Lecture 4
Permittivity Concept (Cont’d)



The concepts of polarizability and dipole moment
distribution are introduced to relate microscopic
phenomena to the macroscopic fields.
The introduction of permittivity eliminates the need
for us to explicitly consider microscopic effects.
Knowing the permittivity of a dielectric tells us all
we need to know from the point of view of
macroscopic electromagnetics.
77
Lecture 4
Permittivity Concept (Cont’d)

For the most part in macroscopic
electromagnetics, we specify the permittivity of
the material and if necessary calculate the dipole
moment distribution within the medium by
using
P  D   0 E     0 E
78
Lecture 4
Relative Permittivity

The relative permittivity of a dielectric is
the ratio of the permittivity of the
dielectric to the permittivity of free space

r 
0
79
Lecture 4
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