Electrostatic Field Problems:
Planar Symmetry
EE 141 Lecture Notes
Topic 5
Professor K. E. Oughstun
School of Engineering
College of Engineering & Mathematical Sciences
University of Vermont
2014
Motivation
E - Field of a Uniform, Infinitely Extended Planar
Charge Distribution (Symmetry about a Plane)
Let the uniform, infinitely extended planar charge distribution with
surface charge density %s be situated in the yz-plane.
Because of the invariance of the problem in any plane parallel to the
yz-plane, the electrostatic field vector has the form
E(r) = +1̂x E (x),
E(r) = −1̂x E (x),
x > 0,
x < 0.
E- Field of a Uniform, Infinitely Extended Planar
Charge Distribution (Symmetry about a Plane)
Application of Gauss’ law to a cylinder whose faces S1 and S2 on
opposite sides of the yz-plane are parallel to and at equal distances
from that plane and whose side is perpendicular to the yz-plane gives
I
ZZ
ZZ
E · n̂da = E (x)
1̂x · 1̂x da + E (−x)
(−1̂x ) · (−1̂x )da
S
S1
S2
= 2AE (x)
for the left-hand side, where the obvious symmetry E (−x) = E (x)
has been applied, and
ZZZ
ZZ
1
1
1
3
%(r)d r =
%s (y , z)d 2 r = %s A
0
0
0
R
S1 ∪S2
for the right-hand side, where A denotes the surface area of either
(identical) cylinder end S1 or S2 .
E - Field of a Uniform, Infinitely Extended Planar
Charge Distribution (Symmetry about a Plane)
Equating these two expressions then yields the electrostatic field
behavior
%s
E = ±1̂x
(1)
20
where the plus sign is used for x > 0 and the minus sign for x < 0.
A uniform planar surface charge then produces a spatially uniform
electrostatic field in each half-space.
Notice that the E - field is discontinuous across the charge surface,
where
%s
E(x > 0) − E(x < 0) = 1̂x
(2)
0
the magnitude of the discontinuity in direct proportion to the surface
charge density.
E - Field of a Uniform, Infinitely Extended Planar
Charge Distribution (Symmetry about a Plane)
The absolute potential for x > 0 is then given by, with d ~` = 1̂x dx,
Z x
Z ∞
%s
V (x) = −
E · 1̂x dx =
dx,
20 x
∞
which diverges.
This divergence is a consequence of the nonphysical nature of the
charge distribution which extends to infinity in both the y - and
z-directions.
Nevertheless, the equipotential surfaces are seen to be planes parallel
to the yz-plane so that
%s
V (x) = V0 +
x; x < 0,
20
%s
V (x) = V0 −
x; x > 0,
(3)
20
where V0 is some constant dependent upon the reference potential.
Parallel Plate Capacitor I
Problem 13. Consider a pair of uniform, infinitely extended planar
charge distributions parallel to the yz-plane and separated by the
distance d > 0. Let the plane at x = −d/2 have the negative surface
charge density −%s and the plane at x = +d/2 have the positive
surface charge density +%s , where %s is a constant.
1
Use Eq. (1) and superposition to determine the electrostatic
field E(x) in each of the regions x < −d/2, −d/2 < x < d/2,
and x > d/2.
2
Using this expression for the electrostatic field, determine the
behavior of the electrostatic potential V (x) in the interior region
between the two planar surfaces with the negative planar surface
as reference, viz. V (−d/2) ≡ 0.
3
What is the electrostatic potential in each of the exterior regions
x < −d/2 and x > d/2?
The Method of Images
At its most fundamental level, the method of images concerns itself
with the problem of determining the electrostatic potential that is
due to a set of discrete point charges in the presence of a system of
ideal conducting boundary surfaces.
This approach is simply the physical equivalent of the determination
of the appropriate form of the function F (r, r0 ) that appears in the
definition of the Green’s function
1
+ F (r, r0 ),
|r − r0 |
(4)
∇02 G (r, r0 ) = −4πδ(r − r0 ),
(5)
G (r, r0 ) =
where
∇02 F (r, r0 ) = 0;
r0 ∈ R,
(6)
such that G (r, r0 ) satisfies the imposed boundary conditions for either
the Dirichlet or Neumann problem.
The Method of Images
As an illustration of the general approach taken in the method of
images, consider the electrostatic potential produced by a system of
discrete point charges in the presence of a given configuration of
perfectly conducting boundary surfaces S, which may be written in
the form
ZZ
ρs (r0 ) 2 0
1
d r,
(7)
V (r) = V1 (r) +
0
4π0
S |r − r |
where
1
V1 (r) is the potential due to the system of point charges alone,
2
and where the surface integral represents the contribution to the
total potential from the (unknown) surface charge density ρs (r0 )
residing on the system of conductor surfaces.
Comparison of (7) with (4) shows that F (r, r0 ) is associated with the
surface integral. Notice further that the Laplacian of the surface
integral in Eq. (7) vanishes in the region between the conductors
where the potential V (r) is to be determined, as required by Eq. (6).
The Method of Images
It may then occur that this surface integral contribution to the total
potential V (r) can be replaced by a potential V2 (r) that is produced
by another specified distribution of charges; this is the essence of the
method of images.
This replacement is possible provided that the surfaces of the entire
system of conductors coincide with specific equipotential surfaces of
the total potential V (r) = V1 (r) + V2 (r). Uniqueness then
guarantees that this solution is valid.
The specified fictitious charge sources that produce V2 (r) are called
image charges. Their apparent location is in the interior region of the
various conductors comprising the system and the potential
V (r) = V1 (r) + V2 (r) is a valid solution to the problem only in the
exterior region between the conductor bodies.
Boundary Value Problems Rectangular Coordinates
Laplace’s equation in three-dimensional rectangular coordinates
∂ 2φ ∂ 2φ ∂ 2φ
+
+
=0
(8)
∂x 2 ∂y 2 ∂z 2
admits a separated solution of the form φ(x, y , z) = X (x)Y (y )Z (z).
With this substitution, Laplace’s equation (8) becomes, after division
by φ = XYZ ,
1 d 2X
1 d 2Y
1 d 2Z
+
+
= 0.
(9)
X (x) dx 2
Y (y ) dy 2
Z (z) dz 2
This equation can be satisfied for arbitrary values of the independent
coordinates x, y , z only if each term is separately constant, so that
1 d 2X
1 d 2Y
1 d 2Z
2
2
=
−α
,
=
−β
,
= γ 2 , (10)
2
2
2
X (x) dx
Y (y ) dy
Z (z) dz
where the separation constants α, β, and γ satisfy the relation
α2 + β 2 = γ 2 .
(11)
Boundary Value Problems Rectangular Coordinates
If both α and β are real-valued, then so also is γ = ± (α2 + β 2 )
and the elementary solutions of the three ordinary differential
equations appearing in Eq. (10) are then e ±iαx , e ±iβy , and
2
2 1/2
e ±iγz = e ±i (α +β ) z .
1/2
The potential function φ(x, y , z) can then be constructed from the
elementary product solutions
φαβ = e ±iαx e ±iβy e ±i (α
2 +β 2 1/2 z
)
(12)
through linear superposition.
Allowed values of the separation constants α and β are determined
by imposing specific boundary conditions on the potential. In doing
this, notice that more appropriate forms of the elementary solutions
(e.g., as sine and cosine functions) may first be inferred from the
imposed symmetry properties of the boundary conditions.
Fourier Series Representation: Example I
Consider determining the electrostatic potential φ(r) = φ(x, y , z)
interior to a rectangular box with dimensions a, b, c along the x, y , z
coordinate axes, as illustrated.
The top surface z = c is held at the potential φ(x, y , c) = V (x, y )
and all of the remaining surfaces are held at zero potential.
Fourier Series Representation: Example I
Because φ(0, 0, 0) = 0, φ(a, y , z) = 0, and φ(x, b, z) = 0, the
required forms of the elementary solutions (12) are
p
α2 + β 2 z .
X (x) = sin (αx), Y (y ) = sin (βy ), Z (z) = sinh
Since φ(a, y , z) = 0, then sin (αa) = 0 ⇒ αa = mπ, m = 1, 2, 3, . . . .
Since φ(x, b, z) = 0, then sin (βb) = 0 ⇒ βb = nπ, n = 1, 2, 3, . . . .
Define:
mπ
αm ≡
,
a
nπ
βn ≡
,
b
γmn ≡ π
m2 n2
+ 2
a2
b
1/2
.
(13)
The partial potential that satisfies all of the φ(r) = 0 boundary
conditions is then given by
φmn (x, y , z) = sin (αm x) sin (βn y ) sinh (γmn z).
(14)
Fourier Series Representation: Example I
In order to satisfy the final boundary condition φ(x, y , c) = V (x, y ),
it is necessary to construct the potential as a linear superposition of
the partial potentials φmn (x, y , z) as
φ(x, y , z) =
∞ X
∞
X
Amn sin (αm x) sin (βn y ) sinh (γmn z).
(15)
m=1 n=1
Application of this expression so as to satisfy the bc at z = c gives
V (x, y ) =
∞ X
∞
X
Amn sinh (γmn c) sin (αm x) sin (βn y ),
(16)
m=1 n=1
which is simply a double Fourier series of the function V (x, y ) with
coefficients
Z a Z b
4
dx
dy · V (x, y ) sin (αm x) sin (βn y ).
Amn =
ab sinh (γmn c) 0
0
(17)
Fourier Series Representation: Example I
In the more general case where the rectangular box has surface
potentials that are, in general, different from zero on either some or
all six sides of the box, then the required solution for the interior
potential can be directly obtained by a linear superposition of the
appropriate solutions equivalent to the above solution (17) for each
face.
Fourier Series Representation: Example II
As an elementary yet illustrative example of a two-dimensional
boundary value problem, consider the electrostatic potential φ(x, y )
in the semi-infinite region 0 ≤ x ≤ a, y ≥ 0 subject to the bcs
φ(0, y ) = φ(a, y ) = 0; y ≥ 0,
φ(x, 0) = V ; 0 ≤ x ≤ a,
with asymptotic behavior
φ(x, y ) → 0 as y → ∞ with 0 ≤ x ≤ a.
Fourier Series Representation: Example II
The appropriate forms of the elementary solutions (12) are then
X (x) = sin (αx),
Y (y ) = e −αy .
(18)
Since φ(a, y ) = 0, then sin (αa) = 0 ⇒ αa = nπ, n = 1, 2, 3, . . . .
Hence, the elementary solutions that satisfy the boundary conditions
φ(0, y ) = φ(a, y ) = 0 and are such that they approach zero as
y → ∞ are given by
φn (x, y ) = e −nπy /a sin (nπx/a).
(19)
In order to satisfy the final bc that φ(x, 0) = V for 0 ≤ x ≤ a, it is
necessary to construct the potential φ(x, y ) as a linear superposition
of the partial potentials as
φ(x, y ) =
∞
X
n=1
An e −nπy /a sin (nπx/a).
(20)
Fourier Series Representation: Example II
Application of the final bc then gives the Fourier coefficients An as
Z
2 a
φ(x, 0) sin (nπx/a)dx,
(21)
An =
a 0
which, for φ(x, 0) = V yields
An =
4V
, for n odd,
nπ
(22)
and An = 0 for n even. The Fourier series representation of the
electrostatic potential in the charge-free, semi-infinite region
0 ≤ x ≤ a, y ≥ 0 is then given by
∞
4V X 1 −nπy /a
φ(x, y ) =
e
sin (nπx/a),
π n=1 n
the summation taken over odd n.
(23)
Fourier Series Representation: Example II
For y ∼ a/π, only the first few terms in (23) are appreciable, and as
y increases above a/π the potential rapidly approaches its asymptotic
form that is given by the first term in the series as
φ(x, y ) ∼
4V −πy /a
e
sin (πx/a); as y → ∞.
π
(24)
This rapid trend to its asymptotic form is evident in the graph on the
following page where the solid curves describe the exact potential at
a fixed value of y /a as given by Eq. (23), while the dashed curves
describe the corresponding asymptotic behavior given by Eq. (24).
Fourier Series Representation: Example II
Boundary Value Problems in 2-D Angular Regions
A problem of considerable practical interest to electrostatics is that in
which two conducting surfaces come together in such a way that the
local geometry about the line of intersection can be approximated as
the intersection of two half-planes.
y
2
φ=0
∆
P(r,ϑ)
r
β
ϑ
x
The general solution of this problem provides insight into the local
behavior of the electrostatic potential and field, as well as the surface
charge density, in the neighborhood of sharp corners and wedges.
Boundary Value Problems in 2-D Angular Regions
Laplace’s equation in two-dimensional polar coordinates (r , ϕ)
∂φ
1 ∂ 2φ
1 ∂
r
+ 2 2 = 0.
r ∂r
∂r
r ∂ϕ
(25)
admits a separated solution of the form φ(r , ϕ) = R(r )Ψ(ϕ) which,
when substituted in Eq. (13), leads to the equation
r d
dR
1 d 2Ψ
r
+
= 0.
(26)
R(r ) dr
dr
Ψ(ϕ) dϕ2
This equation can be satisfied for arbitrary values of the independent
coordinates r and ϕ only if each term is separately constant, so that
r d
dR
1 d 2Ψ
r
=−
= ν 2,
(27)
R(r ) dr
dr
Ψ(ϕ) dϕ2
where ν 2 is the separation constant.
Boundary Value Problems in 2-D Angular Regions
The elementary solutions of these equations are:
R(r ) = ar ν + br −ν , Ψ(ϕ) = A cos (νϕ) + B sin (νϕ) when ν 6= 0,
R(r ) = a0 + b0 ln (r ), Ψ(ϕ) = A0 + B0 ϕ when ν = 0.
For boundary value problems in which there is no restriction on the
azimuthal angle ϕ, the separation constant ν must be a positive
integer or zero in order for the potential to be single-valued.
Single-valuedness also requires that B0 = 0. The general solution is
then of the form
∞
X
n
φ(r , ϕ) = a0 +b0 ln (r )+
an r sin (nϕ + αn ) + bn r −n sin (nϕ + βn ) .
n=1
(28)
If the origin is included in the domain of φ(r , ϕ) and there is no
charge at r = 0, then all of the coefficients bn are zero.
Boundary Value Problems in 2-D Angular Regions
For a two-dimensional angular region, the azimuthal angle ϕ is
restricted to the domain 0 ≤ ϕ ≤ β with β ≤ 2π, in which case the
angle ϕ is not periodic in its value. The boundary condition that
φ(r , 0) = φ(r , β) = V
∀r ≥ 0
(29)
then requires that b0 = B0 = 0 and that b = A = 0 in the elementary
solutions of the separated equations, so that
X
φ(r , ϕ) = a0 +
aν r ν sin (νϕ).
(30)
ν>0
Application of the boundary condition given in Eq. (17) then shows
that a0 = V and that ν = mπ/β, m = 1, 2, 3, . . . .
Boundary Value Problems in 2-D Angular Regions
The general solution for the electrostatic potential in this case is then
given by
∞
X
φ(r , ϕ) = V +
am r mπ/β sin (mπϕ/β).
(31)
m=1
The remaining coefficients am are specified by the imposed behavior
of the potential in the region r 0 removed from the corner.
In the immediate neighborhood of the corner (r = 0) the potential is
approximately given by the first term in the above series as
φ(r , ϕ) ' V + a1 r π/β sin (πϕ/β) as r → 0,
with associated electric field components
π
∂φ
π
Er (r , ϕ) = −
' − a1 r ( β −1) sin (πϕ/β),
∂r
β
π
1 ∂φ
π
Eϕ (r , ϕ) = −
' − a1 r ( β −1) cos (πϕ/β),
r ∂ϕ
β
as r → 0.
(32)
(33)
(34)
Boundary Value Problems in 2-D Angular Regions
The surface charge densities on the conductors at ϕ = 0 and ϕ = β
are equal and are given by
π
π
%s (r ) = 0 Eϕ (r , 0) ' −0 a1 r ( β −1) as r → 0.
(35)
β
Both electric
field components and the surface charge density then
π
vary as r β −1 as r → 0.
2
β = 2π
(β|ρs|)/(πε0a1)
β = 3π/2
β = 5π/4
1
β=π
β = 3π/4
β = π/2
β = π/4
0
0
0.5
r
1.0
Boundary Value Problems in 2-D Angular Regions
For small β one has a very deep corner, the power of r becomes
very large, and there is no charge accumulation in the corner.
For a flat surface (β = π), both the surface charge density and
the electrostatic field become independent of r .
When β > π the corner becomes a two-dimensional wedge and
both the field and the surface charge density become singular as
r → 0. This singularity in %s is integrable so that the total
surface charge within a finite distance from the edge remains
finite.