Poisson’s and Laplace’s
Equations
1
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.
2
• 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

3
– (B)
qev r  dv
V r   
4 0 R
V
E r   V r 
 Method 1 is valid only for charges in free space.
4
• Method 2: Find the electric field and
electrostatic potential using
 Dds  q
S
Gauss’s Law
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.
5
• Consider the following problem:
 What are E and V in the region?
V  V1
 r 
V  V2
Conducting
bodies
6
Neither Method 1 nor
Method 2 can be used!
• 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 
7
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).
8
Derivation of Poisson’s Equation
(Cont’d)

D    E 
0

E  V    V  
0
2
V
9

 V 
0
2
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.
10
Laplacian Operator in Cartesian, Cylindrical, and
Spherical Coordinates
11
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.
 V 0
2
12
Laplace’s
equation
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.
13
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
14
Potential Between Coaxial Cylinders Using
Laplace’s Equation (Cont’d)
• Assume from symmetry that
V  V  
•
1 d  d 
 
  0
V
 d  d 
2
15
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
16
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 
17
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 
18
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.
+
Q
V2
+
V1
V12 = V2-V1
19
C
V12
Capacitance of a Two Conductor
System
• Capacitance is a positive quantity measured
in units of Farads.
• Capacitance is a measure of the ability of a
conductor configuration to store charge.
20
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
21
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.
22
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.
23
Parallel-Plate Capacitor
• Determine an approximate expression for
the capacitance of a parallel-plate capacitor
by neglecting fringing.
Conductor 2
d
A
Conductor 1
24
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
25
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
V  z  d   V12
2
26
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
27
Parallel-Plate Capacitor (Cont’d)
• Evaluate the electric field between the plates
dV
V12
E  V  aˆ z
 aˆ z
dz
d
28
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
29
Parallel-Plate Capacitor (Cont’d)
• Evaluate the capacitance
Q 0 A
C

V12
d
30
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.
31
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.
32
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.
33
Electronic Polarizability
electron
cloud
• 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.
nucleus
34
Electronic Polarizability (Cont’d)
Eapp
• 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.
35
Electronic Polarizability (Cont’d)
e
• The net effect is that
each atom becomes
a small charge dipole
which affects the
total electric field
both inside and
outside the material.
e
p   e E loc
dipole
moment
(C-m)
polarizability
(F-m2)
36
negative
ion
Ionic Polarizability
positive
ion
• 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.
37
Ionic Polarizability (Cont’d)
Eapp
• In the presence of an
applied electric field,
the dipoles tend to
align themselves
with the applied
electric field.
38
Ionic Polarizability (Cont’d)
e
e
p   i E loc
dipole
moment
(C-m)
polarizability
(F-m2)
39
• 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.
Orientational Polarizability
• 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.
40
Orientational Polarizability (Cont’d)
Eapp
• In the presence of an
applied electric field,
the dipoles tend to
align themselves
with the applied
electric field.
41
Orientational Polarizability (Cont’d)
e
e
p   o E loc
dipole
moment
(C-m)
polarizability
(F-m2)
42
• 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.
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
43
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
44
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
E loc
 N T
 E 1 
3 0

45



1
Polarization Per Unit Volume (Cont’d)
• From the macroscopic point of view, it suffices
to use
P   0 e E
electron
susceptibility
(dimensionless)
46
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
47
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
48
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
49
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
50
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
51
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.
52
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.
53
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
54
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
55