Electrostatic Boundary Conditions
EE 141 Lecture Notes
Topic 13
(Shades of Triskaidekaphobia!)
Professor K. E. Oughstun
School of Engineering
College of Engineering & Mathematical Sciences
University of Vermont
2014
Motivation
Dielectric Boundary Conditions
Dielectric Boundary Conditions
Let the interface separating two dielectrics with permittivities ǫ1 & ǫ2
be denoted by S with surface normal n̂ directed from medium 2 into
medium 1.
At any point r ∈ S the electric field vector Ej (r), j = 1, 2, on either
side of the interface S may be decomposed into tangential Etj (r) and
normal Enj (r) components with respect to the surface S at that
point as
E1 (r) = Et1 (r) + En1 (r),
E2 (r) = Et2 (r) + En2 (r),
(1)
(2)
for all r ∈ S.
Notice that n̂ changes as r ∈ S varies over the interface surface S
and that this field decomposition will then also vary as this direction
changes.
Dielectric Boundary Conditions
Application of the integral form of Faraday’s law to an infinitesimally
small loop C in the plane of the incident and transmitted electric field
vectors about the point r ∈ S, the upper side tangent to S in
medium 1 and the lower side tangent to S in medium 2, gives
I
Z b
Z d
~
~
E · dℓ =
E2 · d ℓ +
E1 · d ~ℓ = 0
C
a
c
where the contributions from the sides vanish as ∆h → 0 about S.
In addition,
E2 · d ~ℓ = Et2 ∆ℓ,
& E1 · d ~ℓ = −Et1 ∆ℓ
in the limit as ∆ℓ → 0 on S. Hence, in this limit, Faraday’s law gives
Et2 ∆ℓ − Et1 ∆ℓ = 0, or
Et1 (r) = Et2 (r),
r∈S
(3)
The tangential component of E is continuous across the interface S.
Dielectric Boundary Conditions
Application of the integral form of Gauss’ law to an infinitesimally
small “pillbox” of thickness ∆h → 0 with upper surface S1 parallel to
S in medium 1 with outward normal n̂1 = n̂ and lower surface S2
parallel to S in medium 2 with outward normal n̂2 = −n̂, gives
I
ZZ
ZZ
D · ds =
D1 · n̂1 ds +
D2 · n̂2 ds = ̺s ∆s
SG
S1
S2
where the contributions from the sides vanish as ∆h → 0 about S.
Here ̺s = ̺s (r), r ∈ S, denotes the surface charge density residing
on the interface S. In the limit as ∆s → 0, one obtains
n̂ · D1 (r) − D2 (r) = ̺s (r), r ∈ S
(4)
or
Dn1 (r) − Dn2 (r) = ̺s (r),
r∈S
(5)
The normal component of D changes discontinuously across the
interface S by an amount given by the surface charge density ̺s at
that point.
Dielectric Boundary Conditions
At the interface between two dielectrics with ̺s = 0, the boundary
conditions are
ǫ1 En1 = ǫ2 En2 ,
Et1 = Et2 .
Let E1 be at the angle θ1 with respect to the surface normal n̂ and
E2 be at the angle θ2 with respect to the surface normal −n̂, where
Et1
Et2
θ1 = arctan
, θ2 = arctan
.
En1
En2
Dielectric Boundary Conditions
Then
tan θ2 =
so that θ2 = Tan−1
ǫ2
ǫ1
ǫ2 Et1
ǫ2
Et2
=
= tan θ1 ,
En2
ǫ1 En1
ǫ1
ǫ1 tan θ2 = ǫ2 tan θ1
tan θ1 .
Notice that
ǫ1 > ǫ2
ǫ2 > ǫ1
=⇒
=⇒
tan θ1 > tan θ2
tan θ2 > tan θ1
(6)
Boundary Conditions at the Surface of a
Perfect Conductor (σ = ∞)
A perfect conductor may be defined as a material inside which
electric charge can freely flow.
In electrostatics one assumes that the charges have all reached their
equilibrium positions and are now fixed in space. Hence, inside a
conductor the electrostatic field intensity E vanishes and all points
are at the same potential; that is, a conductor forms an equipotential.
When a conductor is charged, the charges arrange themselves so that
the net electric field due to all the charges is zero inside the
conductor. If a conductor is placed in an electrostatic field, the
charges temporarily flow within it in such a manner to produce a
second field that, added to the first, results in a net zero field inside
the conductor. The field outside the conductor is then distorted by
these charges, resulting in an altered static configuration.
Boundary Conditions at the Surface of a
Perfect Conductor (σ = ∞)
By Gauss’ law for an electrostatic field, ̺ = ǫ∇ · E = 0 within a
conductor. Hence, any net static charge on a conductor must reside
on its surface.
At the surface S of a conductor, the electrostatic field intensity E(r)
must be normal to S, for if it were not, there would be a tangential
component of E that would cause the surface charge to flow along
the surface. Hence, by Gauss’ law
Eext (r) =
̺s (r)
n̂,
ǫ
r∈S
(7)
where n̂ denotes the unit outward normal vector to S at the point
r ∈ S, ̺s (r) denotes the surface charge density at that point, and
where Eext (r) denotes the electrostatic field just above the conductor
surface S at that point in a medium with dielectric permittivity ǫ.
Boundary Conditions at the Surface of a
Perfect Conductor (σ = ∞)
Eext(r)
σ=0
σ>0
ε
E(r) = 0
ρs(r)
Illustration of the external electrostatic field lines Eext (r) terminating
on the induced surface charge with density ̺s (r) on a conductor
surface S that is embedded in a dielectric medium with permittivity ǫ.
Problems
Problem 17. Charge Q1 is uniformly distributed over a spherical
surface of radius a surrounding a dielectric with permittivity ǫ1 , and
charge Q2 is uniformly distributed over a spherical surface of radius b
where the dielectric permittivity is ǫ2 for a < r < b and ǫ3 for r > b,
as illustrated. Apply Gauss’ law to determine the electrostatic field in
each of the spherical regions 0 ≤ r < a, a < r < b, and r > b. Show
that the appropriate boundary conditions are satisfied at both r = a
and r = b.
Q2
Q1
ε1
ε2
ε3