take limit when Δh → 0 (1) (2) (1) (2) ⇒ − Etan1 ∆w + Etan1 ∆w = 0 ⇒ Etan1 = E tan1
•
the same is proven for the other pair of tangential field components with a contour along atan2 and an (1) (2) = E tan2 Etan2
• boundary
E(1,2) tan
condition for Etan in vector form
=
(1,2) Etan1 a tan1
+
(1,2) E tan2 a tan2
(2) ⇒ E(1) = E tan tan
the tangential E component is continuous across dielectric interface LECTURE 13
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3
BCs for the Tangential Field Components – 3 • equivalent vector formulation a n × E = a n × ( E na n + E tan ) = a n × E tan (2) (1) (2) = ⇒ × = × E(1) E a E a E n n tan tan tan tan
a n × (E •
(2)
an E tan
−E ) = 0 (1)
a n × E tan
the tangential components of the flux density
⇒
D(1) tan ε1
=
D(2) tan
D(1) tan
ε 2
D(2) tan
=
ε 1 ε 2
the tangential D component is discontinuous •
when medium 1 is a perfect conductor (particular case) (1) (2) (1) = = 0, = E(2) E D D tan tan tan tan = 0 LECTURE 13
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4
BCs for the Normal Field Components – 1 •
apply Gauss’ law over a closed surface centered around the interface (2)
Dn
an
d s A = l x l y
l y
ε 2
region 2 a tan2 a tan1
region 1
l x
S
(2) D tan
∆h
an
(1)
d s
D tan (1)
Dn
ε 1
lim
∆h → 0
∫∫
(2) (1) ⋅ = − d D D ( D s n n ) ⋅ A = Q f = 0 (2)
S
(2)
Dn
= Dn(1) ⇒ ε 2 En(2) = ε 1E n(1)
E n
(1) E n
=
ε 1 ε 2
the normal D component is continuous while the normal E component is discontinuous across dielectric interfaces LECTURE 13
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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: E x(1)= 4 V/m, E y(1)= 3 V/m, and E z(1)= 1 V/m. What are the field components in medium 2? (2)
=
(2)
=
(2)
=
E x
E y E z
LECTURE 13
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6
BCs for the Normal Field Components – 3
•
when medium 1 is a perfect conductor lim
∆h→0
∫∫
S
0
(2) (1) ⋅ = ( − D s d D D n n ) ⋅ A = Q f = ρ sf ⋅ A
(2) Dn
•
= ρ sf ,
(2) E n
=
ρ sf ε 0ε r 2
Dn(1) = E n(1) = 0
we have already derived this BC in Lecture 11
LECTURE 13
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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 tan α1
=
ε 2 ε 1
(2)
En
=
2 E n(1)
E
(2)
α 2
ε r1 =
(2)
(1)
E tan
ε r 2
Etan
=
(1)
E tan
an
(1)
2ε r 2 E
(1)
α 1
E n
LECTURE 13
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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 tan α1
=
ε 2
E
ε 1
(2)
V =
tan α 2 tan α1
ε r 1
=
=
1 ε r 1
tan α 1
c o n α 2 s t α 2
ε r 2 = 1 ε r 1 = ?
E
(1)
α 1
an α 1
tan α 2 LECTURE 13
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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