Lecture -- Electrostatic Boundary Conditions · 9/21/2019 3 Boundary Conditions for...
Transcript of Lecture -- Electrostatic Boundary Conditions · 9/21/2019 3 Boundary Conditions for...
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Electromagnetics:Electromagnetic Field Theory
Electrostatic Boundary Conditions
Outline
•General classes of electromagnetic materials•Boundary conditions for dielectric‐dielectric interface•Refraction of static fields at a dielectric‐dielectric interface•Boundary conditions for dielectric‐conductor interface• Examples
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General Classes of Electromagnetic
Materials
Slide 3
Classification by Conductivity
Slide 4
Insulator
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• No free charges• Opposes current• Most dielectrics are insulating
Conductor
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• Many free charges• Easily conducts current• Most metals are conducting
Semiconductor
• Often switchable and tunable conductivity
• Silicon, gallium arsenide, etc.
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Boundary Conditions for Dielectric‐Dielectric
Interface
Slide 5
What Are Boundary Conditions?
Slide 6
We often solve electromagnetic problems using differential equations.
22
2 0d E k Edz
2 2 2
2 2 2 0V V Vx y z
The problem is that derivatives are infinite at discontinuities.
x
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What Are Boundary Conditions?
Slide 7
We are forced to solve our differential equations in each homogeneous region separately.
…and then connect our solutions via boundary conditions.
x
221
12 0d E k Edz
2 2 21 1 1
2 2 2 0V V Vx y z
222
22 0d E k Edz
2 2 22 2 2
2 2 2 0V V Vx y z
1 2
1 2
Boundary Conditions0 0
0 0
E E
V V
Deriving Boundary Conditions
Slide 8
Integral equations do not require boundary conditions as long as they do not contain derivatives.
For this reason, we will derive our boundary conditions using Maxwell’s equations in integral form.
0L
E d
S
Q D ds
Boundary conditions for tangential electric fields.
Boundary conditions for normal electric fields.
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Analysis Setup
Slide 9
Let’s examine the interface between two different dielectrics.
1
2
1
2
Analysis Setup
Slide 10
1
2
1
2
We wish to examine the relation between electric fields on either side of the interface, so that if one is known the other can be calculated.
1E
2E
1,tE
1,nE
2,tE
2,nE
It will be useful to separate the field on either side of the interface into tangential and normal components.
Let’s examine the interface between two different dielectrics.
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Derivation of Tangential BCs
Slide 11
Apply the following integral to a closed path spanning some section of the interface.
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2
1
2
1E
2E
1,tE
1,nE
2,tE
2,nE
0L
E d
0
00
0
b c
a bd a
c d
E d E d E d
E d E d E d
wh
a
b
c
d
1,t 1,n 2,n
2,t 2,n 1,n
2 2
2 2
h hE w E E
h hE w E E
Derivation of Tangential BCs
Slide 12
Cancel like terms with opposite sign.1
2
1
2
1E
2E
1,tE
1,nE
2,tE
2,nE
wh
a
b
c
d
1,t 1,n0 2hE w E 2,n 2
hE
2,t 2,n 2hE w E 1,n 2
hE
1,t 2,tE w E w
From this, it is concluded that the tangential component of 𝐸 is continuous across the interface.
1,t 2,tE E
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Derivation of Tangential BCs
Slide 13
Apply the constitutive relation to get the boundary condition for 𝐷.1
2
1
2
1E
2E
1,tE
1,nE
2,tE
2,nE
wh
a
b
c
d
1,t 2,tE E
1,t 2,t
1 2
D D
The tangential component of 𝐷 is NOT continuous across the interface, but the ratio of 𝐷 /𝜀 is.
Derivation of Normal BCs
Slide 14
Place some charge density s on the surface.
1
2
1
2
1D
2D
1,tD
1,nD
2,tD
2,nD
S
Q D ds
sApply the following surface integral to a pillbox spanning the interface.
S
h
top bottom sides
Q D ds D ds D ds
Separate the closed‐surface integral into three separate surface integrals.
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Derivation of Normal BCs
Slide 15
In the limit as h 01
2
1
2
1D
2D
1,tD
1,nD
2,tD
2,nD
sS
h
top bottom sides
Q D ds D ds D ds
1,n 2,n D S D S
The total charge encompassed within the pillbox is
sQ S
Putting these together gives
s 1,n 2,nS D S D S
Derivation of Normal BCs
Slide 16
The final boundary condition is then1
2
1
2
1D
2D
1,tD
1,nD
2,tD
2,nD
sS
h
s 1,n 2,nS D S D S
1,n 2,n sD D
In the absence of charge (i.e. s = 0)
1,n 2,n s 0D D
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Derivation of Normal BCs
Slide 17
Apply the constitutive relation to get the boundary condition for 𝐸.1
2
1
2
1D
2D
1,tD
1,nD
2,tD
2,nD
sS
h
1,n 2,n sD D
1 1,n 2 2,n sE E
In the absence of charge (i.e. s = 0)
1 1,n 2 2,n s 0E E
The normal component of 𝐸 is NOT continuous across the interface, but the product of 𝜀𝐸 is.
Refraction of Static Fields at a Dielectric‐Dielectric Interface
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Analysis Setup
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We want a single equation that relates 1, 2, 1, and 2 without any field quantities in the equation.
1
2
1
2
1E
1,tE
1,nE
2,tE
2,nE
1,tD
1,nD
2,tD 2,nD
1D
2E
2D
1
2Given the angles 1 and 2, the field components can be written as
1 1,t t 1,n nˆ ˆE E a E a
1 1 t 1 1 nˆ ˆsin cosE a E a
2 2,t t 2,n nˆ ˆE E a E a
2 2 t 2 2 nˆ ˆsin cosE a E a
Derivation of Refraction Law
Slide 20
Apply the boundary conditions for tangential components.
1
2
1
2
1E
1,tE
1,nE
2,tE
2,nE
1,tD
1,nD
2,tD 2,nD
1D
2E
2D
1
2
1,t 2,tE E
1 1 2 2sin sinE E
Apply the boundary conditions for normal components.
1 1,n 2 2,nE E
1 1 1 2 2 2cos cosE E
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Derivation of Refraction Law
Slide 21
We now have1
2
1
2
1E
1,tE
1,nE
2,tE
2,nE
1,tD
1,nD
2,tD 2,nD
1D
2E
2D
1
2
1 1 2 2sin sinE E
Divide these equations to get
1 1 1 2 2 2cos cosE E
1 1 2 2
1 1 1 2 2 2
sin sincos cos
E EE E
Simplify
1 2
1 2
tan tan
This is NOT Snell’s law.
Boundary Conditions for Dielectric‐
Conductor Interface
Slide 22
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Analysis Setup
Slide 23
We start like we did for the dielectric‐dielectric interface.
1
2
1
2
1E
2E
1,tE
1,nE
2,tE
2,nE
Assume the conductor is perfect.
Recall Ohm’s law
J E
In order for 𝐽 not to be infinite, 𝐸 0inside the conductor.
Analysis Setup
Slide 24
If E2,t is zero, then1
2
1
2
1E
1,tE
1,nE
1,t 0E
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Analysis Setup
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There can only be a normal component for the electric field at the interface with a perfect conductor.
1
2
1
2
1 1,nE E
1 1,n nˆE E a
Notes About Perfect Conductors
•No electric field can exist inside of a perfect conductor (i.e. 𝐸 0).• Electric potential V is constant throughout a perfect conductor (i.e. 2V = 0).• The electric field at the boundary has no tangential component. The electric field can only be normal at the interface to a metal.
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Examples
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Example #1
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Let there be an interface between two semi‐infinite media in the x‐y plane. The dielectric constant of the first medium is 2.0 and the second medium is 4.4.
1. Given that the electric flux density in medium 1 is 𝐷 2.1𝑎 0.7𝑎 1.5𝑎 , calculate the electric flux density in medium 2, 𝐷 .
2. Calculate the angle 𝐷 makes with the interface.3. Using the law of refraction, calculate the angle 𝐷 makes with the interface.
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Example #1 – Problem Setup
Slide 29
Start by visualizing the problem and setting up the coordinates.
Example #1 – Problem Setup
Slide 30
We start by visualizing the problem and setting up the coordinates.
Plot .1D
1 ˆ ˆ ˆ2.1 0.7 1.5x y zD a a a
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Example #1 – Part 1
Slide 31
Separate 𝐷 into tangential and normal components.
1,t ˆ ˆ2.1 0.7x yD a a
1,n ˆ1.5 zD aTangential Normal
1 ˆ ˆ ˆ2.1 0.7 1.5x y zD a a a
Example #1 – Part 1
Slide 32
Apply boundary condition for normal component.
2,nˆ1.5 za D
1,n 2,nD D
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Example #1 – Part 1
Slide 33
Apply boundary condition for tangential component.
2,t 4.4 ˆ ˆ2.1 0.72.0 x yD a a
1,t 2,t
1 2
D D
22,t 1,t
1
D D
2,t ˆ ˆ4.62 1.54x yD a a
Example #1 – Part 1
Slide 34
Gather both components to get overall 𝐷 .2 2,t 2,nD D D
2 ˆ ˆ ˆ4.62 1.54 1.5x y zD a a a
2 ˆ ˆ ˆ4.62 1.54 1.5x y zD a a a
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Example #1 – Part 2
Slide 35
Calculate the angle 1 of 𝐷 .
cos ABA B A B
Recall the property of dot products.
1
1
ˆzAB
A D
B a
We can calculate 1 by letting
Our dot product becomes
1 1 1ˆ ˆ cosz zD a D a
Example #1 – Part 2
Slide 36
Continued…Solve the dot product equation for 1.
1 1 1ˆ ˆ cosz zD a D a
1 1coszD D
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1
cos zDD
1
1 2 2 2
1.5cos2.1 0.7 1.5
1 55.9
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Example #1 – Part 2
Slide 37
Calculate the angle 2 of 𝐷 .
1 2
1 2
tan tan
The law of refraction is
Solving this for 2 gives
1 22 1
1
tan tan
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4.4tan tan 55.92.0
2 72.9
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