11/3/03
Poisson and Laplace's Equations
Recall Gauss's law and Faraday's law (under electrostatic conditions) in differential form:
free charge!
( )
v
==
ε
ρ
⋅⋅
DE
∇∇
=0
=
×
⇒−
EE
Φ
∇
Substituting
E
into Gauss's law yields
=− Φ
( )
v
⋅
−Φ
=
.
Assuming that the permittivity is a constant throughout the given region (i.e. the medium
is homogeneous), then
can be taken out of the divergence operation.
Thus,
free charge!
()
v
⋅Φ
=
−
.
Substituting for the del operator in rectangular coordinates,
222
2
ˆˆˆ ˆ ˆ ˆ
,
x
yz
x y z
xyz
⎛⎞
⎛
∂∂∂ ∂
Φ∂
Φ
∂
Φ
=
+
+
⋅
+
+
⎜⎟
⎜
∂∂∂
∂ ∂ ∂
⎝⎠
⎝
∂Φ
=++
≡∇ Φ
xyz x y z
⎞
⎟
⎠
we obtain
Poisson's equation
:
∇=
−
2
Φ
v
.
Poisson's Equation
The operator
22
2
2
2
x
+
+
in rectangular coordinates and is referred to as the
Laplacian operator
.
In cylindrical coordinates, it is
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2
2
11
z
ρ
ρρ
φ
⎛⎞
∂
∂Φ
∇Φ=
+
+
⎜⎟
∂
∂∂
⎝⎠
∂
and in spherical coordinates is
2
2
2
1
sin
sin
sin
r
rr
r r
r
θ
2
θθ
Φ
∂
∂
Φ
∂
⎛
⎞
+
+
⎜
⎟
∂
∂
∂
⎝
⎠
Φ
.
Poisson’s equation reduces to
Laplace's equation
in a region in which there are no
volume sources.
That is, letting
v
=
0 in Poisson's equation yields
∇=
2
0
Φ
Laplace's Equation
It is important to recognize that the
potential may be obtained by solving these equations
with the correct sources and subject to the appropriate boundary conditions.
EXAMPLE:
ParallelPlate Capacitor
o
V
Φ =
z
0
Φ =
ρ
s

ρ
s
ε
x
h
V
o
Note a reference potential
Φ=
has been
chosen
on the bottom plate––this choice is
arbitrary.
But then the potential on the top conductor
must be
0
o
V
Φ =
to have the correct
potential difference as determined by the impressed DC source.
Since the region between
the conductors can be considered source free (no free charge should exist between the
plates, so
0
v
=
there),
Poisson’s equation becomes Laplace's equation,
222
2
0,
xyz
+
+
=
∂∂∂
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 Spring '08
 Staff

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