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Lecture 4:
Electrostatics: Electrostatic
Shielding; Poisson’s and Laplace’s
Equations; Capacitance; Dielectric
Materials and Permittivity
Lecture 4
1




To continue our study of electrostatics
with electrostatic shielding;
Poisson’s and Laplace’s equations;
capacitance;
dielectric materials and permittivity.
Lecture 4
2

Consider a point charge at the center of a
spherical metallic shell:
b
Electrically
neutral
a
Q
Lecture 4
3

The applied electric field is given by
E app
Q
 aˆ r
2
40 r
Lecture 4
4
 The total electric field can be obtained using Gauss’s law
together with our knowledge of how fields behave in a
conductor.
(1) Assume from symmetry the form of the
field
D  aˆ r Dr r 
(2) Construct a family of Gaussian surfaces
spheres of radius r where
0r
Lecture 4
5

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
Lecture 4
6
(3) Evaluate the total charge within the volume
enclosed by each Gaussian surface
Qencl   qev dv
V
Lecture 4
7
Gaussian surfaces
for which
0ra
Gaussian surfaces
for which
arb
Gaussian surfaces
for which
r b
Lecture 4
8

For

For
0ra
r b
Qencl  Q
Qencl  Q
Shell is electrically neutral:
The net charge carried by shell is zero.
Lecture 4
9

For
arb
Qencl  0
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
Lecture 4
10

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
Lecture 4
11
(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 4
12
(5) Solve for D on each Gaussian surface
Qencl
D
S
(6) Evaluate E as
E
D
0
Lecture 4
13
Q

ˆ
a
,
r
2
 4 r
0

E  0,

Q
aˆ r
,
2
 40 r
0r a
ar b
r b
Lecture 4
14

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
Lecture 4
15
E
total
electric
field
Eapp
a
b
r
Eind
Lecture 4
16


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     E r dr 
40 r

Lecture 4
17

Since the conductor is an equipotential body (and
potential is a continuous function), we have for
arb
Q
V r   V b  
40b
Lecture 4
18

For
0ra
r
V r   V b    Er dr
a
Q 1 1 1 

  

40  b r a 
Lecture 4
19
V
No metallic shell
a b
r
Lecture 4
20


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
Lecture 4
sphere.
21



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
- - Lecture 4
22
E
total
electric
field
Eapp
R
a b
Eind
Lecture 4
23
V
Grounded
metallic shell acts
as a shield.
a b
R
Lecture 4
24

So far, we have studied two approaches
for finding the electric field and
electrostatic potential due to a given
charge distribution.
Lecture 4
25

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

Lecture 4
26

(B)
qev r  dv
V r   
40 R
V
E r   V r 
 Method 1 is valid only for charges in free space.
Lecture 4
27

Method 2: Find the electric field and
electrostatic potential using
Gauss’s Law
 Dds  q
S
ev
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.
Lecture 4
28

Consider the following problem:
 What are E and V in the region?
V  V1
 r 
Conducting
bodies
V  V2
Neither Method 1 nor
Method 2 can be used!
Lecture 4
29


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 
Lecture 4
30

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).
Lecture 4
31
  D  qev    E 
qev
0
E  V    V  
qev
0
V
2
Lecture 4
32
 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.
Lecture 4
33
Lecture 4
34


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
Lecture 4
35

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.
Lecture 4
36

Two conducting coaxial cylinders exist
such that
y
V   a   V0
V   b   0
+
a
V0
x
b
Lecture 4
37

Assume from symmetry that
V  V  

1 d  d 
 
  0
V
 d  d 
2
Lecture 4
38

Two successive integrations yield
V     C1 ln     C 2

The two constants are obtained from the
two BCs:
V   a   V0  C1 ln a  C 2
V   b   0  C1 ln b  C 2
Lecture 4
39

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 
Lecture 4
40


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
byq   aˆ  E   a    0V0
esa
qesb
a ln b / a 
  0V0
  0  aˆ   E   b  
b ln b / a 
0 
Lecture 4
41
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.
+
Q
V2
V1
+

C
V12
V12 = V2-V1
Lecture 4
42


Capacitance is a positive quantity
measured in units of F = Farads.
Capacitance is a measure of the ability of
a conductor configuration to store
charge.
Lecture 4
43

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
Lecture 4
44


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.
Lecture 4
45


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.
Lecture 4
46

Determine an approximate expression
for the capacitance of a parallel-plate
capacitor by neglecting fringing.
Conductor 2
d
A
Conductor 1
Lecture 4
47

“Neglecting fringing” means to assume
that the field that exists in the real
problem is the same as for the infinite
z
problem.
z=d
V = V12
z=0
V=0
Lecture 4
48

Determine the potential between the
plates by solving Laplace’s equation.
2
dV
 V  2 0
dz
V z  0  0
V  z  d   V12
2
Lecture 4
49
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
Lecture 4
50
• Evaluate the electric field between the plates
dV
V12
E  V   aˆ z
  aˆ z
dz
d
Lecture 4
51
• 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
 0V12 A
Q  qes 2 A 
d
Lecture 4
52
• Evaluate the capacitance
Q 0 A
C

V12
d
Lecture 4
53


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.
Lecture 4
54



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
Lecture 4
to be modified.
55


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.
Lecture 4
56
electron
cloud
nucleus


57
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
Eapp

58
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
Lecture 4
e

e
p   e E loc
dipole
moment
(C-m)
polarizability
(F-m2)
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
59
negative
ion
positive
ion

60
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
Eapp

In the presence of
an applied electric
field, the dipoles
tend to align
themselves with
the applied
electric field.
Lecture 4
61
e

e
p   i E loc
dipole
moment
(C-m)
polarizability
(F-m2)
62
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
Lecture 4field

63
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
Eapp

In the presence of
an applied electric
field, the dipoles
tend to align
themselves with
the applied
electric field.
Lecture 4
64
e

e
p   o E loc
dipole
moment
(C-m)
polarizability
(F-m2)
65
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
Lecture 4field


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
Lecture 4
66





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
Lecture 4
67



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
E loc
 N T
 E 1 
3 0




1
Lecture 4
68

From the macroscopic point of view, it suffices
to use
P  0e E
electron
susceptibility
(dimensionless)
Lecture 4
69


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
Lecture 4
70


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
Lecture 4
71

Gauss’s law in differential form in free
space:
 0   E  q ev

Gauss’s law in differential form in
dielectric:
 0   E  q ev  q evb
Lecture 4
72
 0  E  qev  qevb  qev    P
   0 E  P   qev
• Hence, the displacement flux density vector
is given by
D  0 E  P
Lecture 4
73

Gauss’s law in differential form:
  D  qev

Gauss’s law in integral form:
D

d
s

Q
encl

S
Lecture 4
74

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.
Lecture 4
75



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.
Lecture 4
76

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
Lecture 4
77

The relative permittivity of a dielectric is the ratio
of the permittivity of the dielectric to the
permittivity of free space

r 
0
Lecture 4
78