Electromagnetism
Chapter 7: Laplace Equation
Pablo Laguna
Georgia Institute of Technology, USA
Fall 2012
Notes based on textbook: Modern Electrodynamics by A. Zangwill
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Electromagnetism
Introduction
Recall that:
In vacuum
∇ · E = −∇2 ϕ =
ρ
0
Inside a conductor E = 0 and thus ϕ = constant
Inside a dielectric
∇ · E = −∇2 ϕ =
The equation
∇2 ϕ = −
ρ
0
or
ρf
0 κ
∇2 ϕ = −
ρf
is called Poisson Equation
The source-free Poisson equation ∇2 ϕ = 0 is called Laplace Equation
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Boundary Conditions
∇2 ϕ = 0 does not necessarily implies that the only solutions are ϕ = 0
or ϕ = constant
Most of the solutions of interest are determined by boundary conditions
The following types of boundary conditions apply to both, the Laplace
and Poisson equations.
Dirichlet:
ϕ|∂Ω = ϕ0
Neumann:
n · ∇ϕ|∂Ω = g
Mixed:
ϕ|Γ1 = ϕ0
and
n · ∇ϕ|Γ2 = g
such that Γ1 ∪ Γ2 = ∂Ω
Robin:
(a ϕ + b n · ∇ϕ)|∂Ω = f
with a and b constants.
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Boundaries with Cartesian Symmetry
Ansatz ϕ(x, y , z) = X (x)Y (y )Z (z), thus
∇2 ϕ =
becomes
∂2ϕ
∂2ϕ
∂2ϕ
+
+
=0
∂x 2
∂y 2
∂z 2
X 00 (x)
Y 00 (y )
Z 00 (z)
+
+
=0
X (x)
Y (y )
Z (z)
Therefore
d 2X
= α2 X ,
dx 2
d 2Y
= β2Y ,
dy 2
and
d 2Z
= γ2Z
dz 2
with α2 , β 2 , and γ 2 separation constants satisfying α2 + β 2 + γ 2 = 0
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The solutions to these equations are

 A0 + B0 x
X (x) =

Aα eαx + Bα e−αx

 C0 + D0 y
Y (y ) =

Cβ eβy + Dβ e−βy

 E0 + F0 z
Z (z) =

Eγ eγz + Fγ e−γz
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α = 0,
α 6= 0.
β = 0,
β 6= 0.
γ = 0,
γ 6= 0.
Example
Let’s find the electrostatic potential inside a rectangular box assuming that all
the walls are fixed at V = 0 except for the z = 0 wall.
Solution: Since at x = 0 and x = a we need X (0) = X (a) = 0 then we set α
and the constants Aα and Bα in such a way that
m π x X (x) ∝ sin
a
similarly
Y (y ) ∝ sin
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n π y b
Electromagnetism
This implies that
ϕ(x, y, z) =
∞ X
∞
X
sin
mπx m=1 n=1
a
sin
where
2
γmn
=
nπy b
mπ 2
a
[Emn exp(γmn z) + Fmn exp(−γmn z)] ,
+
nπ 2
b
Finally, we need to select the constants in such a way that ϕ(x, y , c) = 0 and
ϕ(x, y, 0) = V (x, y )
The solution is
ϕ(x, y , z) =
∞ X
∞
X
Vmn sin
mπx m=1 n=1
a
recall that
sinh x =
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sin
nπy sinh[γ (c − z)]
mn
.
b
sinh(γmn c)
ex − e−x
2
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The coefficients Vmn are obtained from
Vmn
4
=
ab
Za Zb
dx dy V (x, y ) sin
0
mπx a
0
where one makes use of
Zπ
ds sin(ms) sin(ns) =
π
δmn
2
0
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Electromagnetism
sin
nπy b
Application 7.1
y
V'
L
V
0
V
0
0
L
x
Consider an infinitely long, hollow, conducting duct. The walls are maintained
at the constant potentials indicated in the figure. Find the electrostatic
potential everywhere inside the duct.
Solution:
∞
ϕ(x, y ) =
X
V0
mπy
y+
(am emπx/L + bm e−mπx/L ) sin
.
L
L
m=1
where
am =
Ωm emπ − 1
2 sinh mπ
with
Ωm =
and
bm =
Ωm e−mπ − 1
2 sinh mπ
2 V + (−1)m (V 0 − V )
mπ
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Laplace Equation in Spherical Coordinates
In spherical coordinates the Laplace equations reads:
1
∂ϕ
1
1 ∂
∂ϕ
∂
∂2ϕ
r2
+ 2
sin θ
+
=0
∇2 ϕ = 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r 2 sin2 θ ∂φ2
Let’s restrict ourselves first to axi-symmetric cases; that is,
ϕ(r , θ, φ) = ϕ(r , θ). Thus
1 ∂
∂
1
∂ϕ
2 ∂ϕ
r
+
sin
θ
=0
r 2 ∂r
∂r
r 2 sin θ ∂θ
∂θ
Next, introduce the change of variable x = cos θ. Then,
∂
∂ϕ
∂
∂ϕ
r2
+
(1 − x 2 )
=0
∂r
∂r
∂x
∂x
The equation written in this way calls for a separation of variables of the form
ϕ(r , x) = R(r ) M(x)
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Thus, ϕ(r , x) = R(r ) M(x) in
∂
∂
2 ∂ϕ
2 ∂ϕ
r
+
(1 − x )
=0
∂r
∂r
∂x
∂x
yields
d
dr
and
r2
dR
dr
− ν(ν + 1)R = 0
dM
d
(1 − x 2 )
+ ν(ν + 1)M = 0.
dx
dx
with ν real.
The first equation has as a solution
Rν (r ) = Aν r ν + Bν r −(ν+1)
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The equation for x
dM
d
(1 − x 2 )
+ ν(ν + 1)M = 0.
dx
dx
can be rewritten as
(x 2 − 1)
d 2M
dM
+ 2x
− ν(ν + 1)M = 0
dx 2
dx
which is Legendre’s equation.
Legendre’s equation has two linearly independent solutions called Legendre
functions of the first and second kind, denote by Pν (x) and Qν (x),
respectively.
The solution to Laplace’s equation with azimuthal symmetry is then given by
i
Xh
ϕ(r , θ) =
Aν r ν + Bν r −(ν+1) [Cν Pν (cos θ) + Dν Qν (cos θ)]
ν
with Cν and Dν constant to be determined by boundary conditions.
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Laplace Equation in Spherical Coordinates
In spherical coordinates the Laplace equations reads:
1 ∂
1
∂ϕ
1
∂2ϕ
∂ϕ
∂
∇2 ϕ = 2
r2
+ 2
sin θ
+
=0
2
2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
If ϕ(r , θ, φ) = R(r )Y (θ, φ)
d
dr
dR`
r2
= `(` + 1)R`
dr
with solutions
R` (r ) = A` r ` + B` r −(`+1) .
and
1 ∂ 2 Y`m
= `(` + 1)Y`m
sin2 θ ∂φ2
with solutions Y`m , the spherical harmonic functions. The general solution is
given by
1 ∂
−
sin θ ∂θ
ϕ(r , θ, φ) =
∂Y`m
sin θ
∂θ
∞
`
h
X
X
−
i
A`m r ` + B`m r −(`+1) Y`m (θ, φ)
`=0 m=−`
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Laplace Equation in Cylindrical Coordinates
∇2 ϕ =
1 ∂
ρ ∂ρ
ρ
∂ϕ
∂ρ
+
1 ∂2ϕ
∂2ϕ
+
= 0.
ρ2 ∂φ2
∂z 2
If ϕ(ρ, φ, z) = R(ρ)G(φ)Z (z)
ρ
d
dρ
dR
ρ
+ (k 2 ρ2 − α2 )R
dρ
=
0
d 2G
+ α2 G
dφ2
=
0
d 2Z
− k 2Z
dz 2
=
0
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Electromagnetism
Laplace Equation Solutions in Cylindrical Coordinates
d 2G
+ α2 G
dφ2
=
0
d 2Z
− k 2Z
dz 2
=
0
have solutions
Gα (φ) =


x0 + y0 φ
α = 0,

xα exp(iαφ) + yα exp(−iαφ)
α 6= 0.
and
Zk (z) =


s0 + t0 z
k = 0,

sk exp(kz) + tk exp(−kz)
k 6= 0.
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ρ
d
dρ
dR
ρ
+ (k 2 ρ2 − α2 )R
dρ
=
0
has solutions
Rαk (ρ)
=

A0 + B0 ln ρ








Aα ρα + Bα ρ−α



Akα Jα (kρ) + Bαk Nα (k ρ)









Ak I (κρ) + Bαk Kα (κρ)

 αα
k = 0,
α=0
k = 0,
α 6= 0
k2 > 0
k 2 < 0.
with Jα (x) and Nα (x) are called Bessel functions of the first and second kind,
respectively. The modified Bessel functions Iα (x) and Kα (x) are Bessel
functions with pure imaginary arguments:
Iα (κρ) = i −α Jα (ikρ)
Kα (κρ) =
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π α+1
i
[Jα (ik ρ) + iNα (ikρ)]
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Electromagnetism
There are situations with axi-symmetry G(φ) and α = 0 or polar-symmetry
Z (z) = 1 and k = 0. For polar symmetry

x0 + y0 φ
α = 0,

Gα (φ) =

xα exp(iαφ) + yα exp(−iαφ)
α 6= 0.
and
Rαk (ρ) =

 A0 + B0 ln ρ

Aα ρα + Bα ρ−α
k = 0,
α=0
k = 0,
α 6= 0
thus
ϕ(ρ, φ) = (A0 + B0 ln ρ)(x0 + y0 φ) +
X
[Aα ρα + Bα ρ−α ][xα sin αφ + yα cos αφ].
α6=0
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The Electric Field Near a Sharp Corner or Edge
Consider the electric field E(ρ, φ) inside a two-dimensional wedge
(0 ≤ φ ≤ β) bounded by a grounded perfect conductor Since the potential
cannot be singular as ρ → 0
ϕ(ρ, φ) = (A0 + B0 ln ρ)(x0 + y0 φ) +
X
[Aα ρα + Bα ρ−α ][xα sin αφ + yα cos αφ].
α6=0
simplifies to
ϕ(ρ, φ) = A + Bφ +
X
Cα ρα sin(αφ + δα ).
α>0
Since ϕ(0, φ) = 0 we also get A = B = 0. Moreover, ϕ(ρ, 0) = ϕ(ρ, β) = 0
implies that δα = 0 and α = mπ/β where m is a positive integer.
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Thus
ϕ(ρ, φ) =
X
Cm ρ(m π/β) sin
m
mπφ
β
Notice that the m = 1 term dominates as ρ → 0. Therefore,
πφ
ϕ(ρ, φ) ≈ C1 ρ(π/β) sin
β
Then
E = −∇ϕ ≈ −
o
π π/β−1 n
ρ̂ sin(πφ/β) + φ̂ cos(πφ/β) .
ρ
β
and
σ(ρ) = 0 Eφ (ρ, 0) ≈ −
π 0 π/β−1
ρ
β
Notice that |E | → ∞ and σ → ∞ as ρ → 0 when β > π.
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