Lecture 13
Boundary Conditions at
Dielectric Interfaces
Sections: 6.2 (in 8th ed.: 5.8)
Homework: See homework file
LECTURE 13
slide
1
BCs for the Tangential Field Components – 1
• we consider interfaces between two perfect (σ = 0) dielectric regions
• use conservative property of field
• choose contour across interface
0
∫ E ⋅ dL =
C
• contour is small enough to consider field constant along its line
segments
En(2)
En(1)
E(2)
E
(2)
Etan1
(1) l2
E
(1) tan1
∆w
l3
∆h l4
C
l1
region 2
an ε 2
a tan1ε1
region 1
(1)
(2)
− Etan1
l1 + 0.5( En(1) + En(2) )l2 + Etan1
l3 − 0.5( En(1) + En(2) )l4 =
0
LECTURE 13
slide
2
BCs for the Tangential Field Components – 2
• take limit when Δh → 0
(1)
(2)
(1)
(2)
⇒ − Etan1
∆w + Etan1
∆w = 0 ⇒ Etan1
= Etan1
• the same is proven for the other pair of tangential field components
with a contour along atan2 and an
(1)
(2)
Etan2
= Etan2
• boundary condition for Etan in vector form
=
E(1,2)
tan
(1,2)
Etan1
a tan1
+
(1,2)
Etan2
a tan2
(2)
⇒ E(1)
=
E
tan
tan
the tangential E component is continuous across dielectric interface
LECTURE 13
slide
3
BCs for the Tangential Field Components – 3
• equivalent vector formulation
a n × E = a n × ( Ena n + E tan ) = a n × E tan
an
(2)
(1)
(2)
E(1)
=
E
⇒
a
×
E
=
a
×
E
n
n
tan
tan
tan
tan
a n × (E
(2)
E tan
−E ) =
0
(1)
a n × E tan
• the tangential components of the flux density
⇒
D(1)
tan
ε1
(1)
D
ε1
D(2)
tan
tan
=
=
(2)
Dtan ε 2
ε2
the tangential D component is discontinuous
• when medium 1 is a perfect conductor (particular case)
(2)
(1)
(2)
(1)
E=
E
=
0,
D
=
D
=
0
tan
tan
tan
tan
LECTURE 13
slide
4
BCs for the Normal Field Components – 1
• apply Gauss’ law over a closed surface centered around the interface
an
Dn(2) lx
an
ds
∆h
(2)
A = lx l y Dtan
ly
ε2
region 2
a tan2
a tan1
region 1
ε1
lim
∆h →0
S
ds
D(1)
tan
Dn(1)
(2)
(1)
(
⋅
d
=
D
−
D
D
s
n
n ) ⋅ A= Qf = 0
∫∫

S
Dn(2) =
Dn(1) ⇒ ε 2 En(2) =
ε1En(1)
En(2) ε1
=
(1)
ε2
En
the normal D component is continuous while the normal E
component is discontinuous across dielectric interfaces
LECTURE 13
slide
5
BCs for the Normal Field Components – 2
Medium 1 of ε r(1) = 4 is located in the region x < 0. Medium 2 of ε r(2)
= 1 is located in the region x ≥ 0. The field components in medium
1 are: Ex(1)= 4 V/m, E y(1)= 3 V/m, and Ez(1)= 1 V/m. What are the
field components in medium 2?
Ex(2) =
E y(2) =
Ez(2) =
LECTURE 13
slide
6
BCs for the Normal Field Components – 3
• when medium 1 is a perfect conductor
(2)
(1)
⋅
=
−
D
d
s
D
D
(
n
n ) ⋅ A = Q f = ρ sf ⋅ A
∫∫

∆h →0 
lim
S
0
Dn(2)
=
(2)
ρ=
,
E
sf
n
ρ sf
ε 0ε r 2
(1)
(1)
D=
E
=
0
n
n
• we have already derived this BC in Lecture 11
LECTURE 13
slide
7
Dielectric Interfaces: Orientation of Field Vectors (Homework)
• the tangential E component is continuous while the normal one is not
• the E field vector changes its orientation abruptly
Prove that:
tan α 2 ε 2
=
tan α1 ε1
En(2) = 2 En(1)
E(2)
α2
εr2
ε r1 = 2ε r 2
(1)
Etan
E
(1)
α1
(2)
Etan
En(1)
LECTURE 13
=
(1)
Etan
an
slide
8
Field Map at Dielectric Interfaces
Example: Determine approximately the permittivity of a
dielectric slab from the field map at the air-dielectric interface
tan α 2 ε 2
=
tan α1 ε1
E(2)
V
=
t
ns
tan α1
ε r1 =
tan α 2
co
tan α 2
1
=
tan α1 ε r1
ε r2 = 1
ε r1 = ?
α2
an
α2
E(1)
α1
LECTURE 13
α1
slide
9
You have learned:
that the tangential E component is continuous across interfaces,
both dielectric-to-dielectric and PEC-to-dielectric
that the normal D component is continuous across dielectric
interfaces; it is discontinuous across PEC-to-dielectric interfaces
due to the presence of free surface charge
how to interpret field maps at interfaces
LECTURE 13
slide
10