7 Poisson’s and Laplace’s equations
Summarizing the properties of electrostatic fields we have learned so far, they
satisfy the constraints
∇
·
D
=
ρ
and
∇ ×
E
= 0
where
D
=
o
E
;
in addition
E
=
∇
V
as a consequence of
∇ ×
E
= 0
.
•
Combining the equations above, we can rewrite Gauss’s law as
∇
·
D
=
o
∇
·
E
=

o
∇
·
(
∇
V
) =
ρ
,
from which it follows that
∇
2
V
=

ρ
o
,
Poisson’s eqn
where
∇
2
V
≡
∂
2
V
∂
x
2
+
∂
2
V
∂
y
2
+
∂
2
V
∂
z
2
is known as
Laplacian
of
V
.
–
A special case of Poisson’s equation corresponding to having
ρ
(
x, y, z
) =
0
everywhere in the region of interest is
∇
2
V
= 0
Laplace’s eqn
.
1
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Focusing our attention first on Laplace’s equation, we note that the equation
can be used in charge freeregions to determine the electrostatic potential
V
(
x, y, z
)
by matching it to specified potentials at boundaries as illustrated
in the following examples:
z
x
y
z
=
d
= 2 m
V
(
d
) =

3 V
V
(0) = 0
z
= 0
V
(
z
) =?
z
V
(
z
)
V
(
z
) =
Az
+
B
Example 1:
Consider a pair of parallel conducting metallic plates of infinite extents
in
x
and
y
directions but separated in
z
direction by a finite distance of
d
= 2
m (as shown in the margin). The conducting plates have nonzero surface charge
densities (to be determined in Example 2), which are known to be responsible for
an electrostatic field
E
= ˆ
zE
z
measured in between the plates. Each plate has
some unique and constant electrostatic potential
V
since neither
E
(
r
)
nor
V
(
r
)
can dependent the coordinates
x
or
y
given the geometry of the problem.
Using Laplace’s equation determine
V
(
z
)
and
E
(
z
)
between the plates if the potential
of the plate at
z
= 0
is 0 (the ground), while the potential of the plate at
z
=
d
is

3
V.
Solution:
Since the potential function
V
=
V
(
z
)
between the plates is only dependent
on
z
, it follows that Laplace’s equation simplifies as
∇
2
V
=
∂
2
V
∂
x
2
+
∂
2
V
∂
y
2
+
∂
2
V
∂
z
2
=
∂
2
V
∂
z
2
= 0
.
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 Summer '10
 KUDEKI
 Electrostatics, Electric charge, Poisson, surface charge densities, Poisson’s Equation

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