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6.641 Electromagnetic Fields, Forces, and Motion
Spring 2009
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6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7: Polarization and Conduction
I. Experimental Observation
A. Fixed Voltage - Switch Closed ( v = Vo )
As an insulating material enters a free-space capacitor at constant voltage
more charge flows onto the electrodes; i.e. as x increases, i increases.
B. Fixed Charge - Switch open (i=0)
As an insulating material enters a free space capacitor at constant charge,
the voltage decreases; i.e. as x increases, v decreases.
II. Dipole Model of Polarization
A. Polarization Vector P = N p = N q d
N dipoles/Volume ( P is dipole density)
( p = q d dipole moment)
d
+q
−q
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 1 of 27
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7 Page 2 of 27 Q inside V = −
∫ q N d i da = ∫ ρ
S
P
dV
V
paired charge or
equivalently
polarization
charge density
Q inside V = − ∫ P i da = − ∫ ∇ i P dV =
S
V
∫ρ
P
dV
(Divergence Theorem)
V
P = qN d
∇ i P = −ρP
B. Gauss’ Law
∇i
( ε E) = ρ
o
total
= ρu + ρP = ρu − ∇ i P
unpaired charge
density; also
called free charge
density
∇i
(ε
o
)
E + P = ρu
D = εo E + P
Displacement Flux Density
∇ i D = ρu
C. Boundary Conditions
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 3 of 27
∫ D i da = ∫ ρ
∇ i D = ρu ⇒
S
∇ i P = −ρP ⇒
dV ⇒ n i ⎡⎣D a − Db ⎤⎦ = σ su
∫ P i da = − ∫ ρ
V
S
(
u
V
)
∇ i ε o E = ρu + ρP ⇒
∫ε
o
P
dV ⇒ n i ⎡⎣P a − P b ⎤⎦ = −σ sp
E i da =
S
∫ (ρ
V
u
+ ρ P ) dV ⇒ n i ε o ⎡⎣Ea − Eb ⎤⎦ = σ su + σ sp
D. Polarization Current Density
∆Q = q N dV = q N d i da = P i da
[Amount of Charge passing through
surface area element da ]
d ip =
∂∆Q
∂P
=
i da
∂t
∂t
[Current passing through surface
area element da ]
= Jp i da
polarization current density Jp =
∂P
∂t
Ampere’s law:
∇ x H = Ju + Jp +
εo
∂E
∂t
= Ju +
∂P
+
∂t
= Ju +
∂
ε o
E + P
∂t
= Ju +
∂ D
∂t
(
εo
∂E
∂t
)
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 4 of 27
III. Equipotential Sphere in a Uniform Electric Field
lim Φ (r, θ ) = −E or cos θ
r→∞
⎣⎡Φ = −E o z = −E or cos θ⎤⎦
Φ (r = R, θ ) = 0
⎡
R3 ⎤
Φ (r, θ ) = −Eo ⎢r − 2 ⎥ cos θ
r ⎦
⎣
This solution is composed of the superposition of a uniform electric field
plus the field due to a point electric dipole at the center of the sphere:
Φ dipole =
p cos θ
4πε or 2
with p = 4 πε oE oR 3
This dipole is due to the surface charge distribution on the sphere.
σ s (r = R, θ ) =
ε oEr (r = R, θ ) = − ε o
∂Φ
∂r
=
r =R
⎡
ε oE o ⎢1 +
⎣
2R 3
r3
⎤
⎥ cos θ
r =R ⎦
= 3εoEo cos θ
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 5 of 27
IV. Artificial Dielectric
E =
v
,
d
σs =
q = σs A =
C =
εA
d
εE =
εv
d
v
q
εA
=
v
d
_
υ
E
d
ε
+
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
For spherical array of non-interacting spheres (s >> R)
_
P = 4 π ε o R 3 E o i z ⇒ Pz = N p z = 4 π ε o R 3 E o N N= 1
3
s
3
⎡
⎛R ⎞ ⎤
P = εo ⎢4 π ⎜ ⎟ ⎥ E = ψ e εo E
⎝ s ⎠ ⎥⎦
⎣⎢
3
⎛
⎛R ⎞ ⎞
⎜ ψe = 4 π ⎜ ⎟ ⎟
⎜
⎟
⎝ s ⎠ ⎠
⎝
ψ e (electric susceptibility) D=
ε o E + P = ε o ⎡⎣1 + ψ e ⎤⎦ E = ε E
εr (relative dielectric constant) ⎛
R
⎞
ε = εr ε o = ε o ⎣⎡1 + ψ e ⎦⎤ = ε o ⎜⎜1 + 4 π ⎛⎜ ⎞⎟ ⎟⎟
3
⎝
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
⎝s⎠ ⎠
Lecture 7
Page 6 of 27
V. Demonstration: Artificial Dielectric
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7 Page 7 of 27 Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
E=
v
εv
⇒ σs = ε E =
d
d
εA
q = σs A =
d
∆i = ω ∆C
∆C =
v ⇒ C=
V =
(ε − εo ) A
d
q
ε A
=
v
d
v
o
Rs
3
⎛R ⎞ A
= 4 π εo ⎜ ⎟
⎝ s ⎠ d
R=1.87 cm, s=8 cm, A= (0.4)2 m2, d=0.15m ω =2π(250 Hz), Rs=100 k Ω , V=566 volts peak ∆ C=1.5 pf
v 0 = ω ∆ C R s V
=(2π) (250) (1.5 x 10-12) (105) 566 = 0.135 volts
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
peak
Lecture 7
Page 8 of 27
VI. Plasma Conduction Model (Classical)
m+
∇p +
dv +
= q + E − m + ν + v+ −
dt
n+
m−
∇p −
dv−
= −q− E − m − ν − v− −
dt
n−
p + = n + kT , p − = n −kT
k=1.38 x 10-23 joules/oK Boltzmann Constant
A. London Model of Superconductivity [ T → 0 , ν ± → 0 ] m+
dv +
=
q+ E
dt
J+ = q+ n + v+
,
m−
,
dv −
= − q− E
dt
J− = − q− n − v−
(
)
q+ E
q 2n
dJ+
d
dv +
q+ n + v+ = q+ n +
= q+ n +
= + + E
=
dt
dt
dt
m+
m+
(
)
ωp2 +
(
ε
)
− q− E
q 2n
dJ−
d
dv−
=−
q− n − v− = − q− n −
= − q− n −
= − − E
dt
dt
dt
m−
m−
(
)
ω2p− ε
ωp2+ =
q + 2 n +
,
m+ ε
ωp2− =
q− 2 n−
m− ε
( ωp = plasma frequency)
For electrons: q-=1.6 x 10-19 Coulombs, m-=9.1 x 10-31 kg
n-=1020/m3 , ε = ε o ≈ 8.854 x 10 −12 farads/m
ωp− =
fp − =
ωp −
2π
q−2 n−
≈ 5.6 x 1011 rad/s
m− ε
≈ 9 x 1010 Hz
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 9 of 27
B. Drift-Diffusion Conduction [Neglect inertia]
0
m+
∇ (n + k T )
q+
dv +
kT
= q + E − m + ν + v+ −
E−
⇒ v+ =
∇n +
dt
n+
m+ ν +
m + ν +n+
0
∇ (n − k T ) −q −
dv −
kT
m−
= −q− E − m − ν − v− −
⇒ v− =
E−
∇n −
dt
n−
m−ν −
m − ν −n−
J+ = q + n +
v+ =
q + 2n +
qkT
∇n +
E− +
m+ ν +
m + ν +
J− = −q − n − v− =
ρ + = q+ n + ,
q − 2n −
qkT
∇n −
E+ −
m−ν −
m − ν −
ρ − = −q− n −
J+ = ρ + µ + E − D + ∇ρ +
J− = −ρ − µ − E − D − ∇ρ −
µ+ =
q+
m+ ν +
,
µ− =
q−
m− ν −
,
charge mobilities
D+ =
D− =
kT
m + ν +
kT
m − ν −
Molecular
Diffusion
Coefficients
D+
D
kT
= thermal voltage (25 mV @ T ≈ 300o K)
= − =
µ+
µ−
q
Einstein’s Relation
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 10 of 27
(
C. Drift-Diffusion Conduction Equilibrium J+ = J− = 0
)
J+ = 0 = ρ + µ + E − D + ∇ρ + = −ρ + µ + ∇Φ − D + ∇ρ +
J− = 0 = −ρ − µ − E − D − ∇ρ − = ρ − µ − ∇Φ − D − ∇ρ −
∇Φ = −
∇Φ =
D+
−k T
∇ρ + =
∇ (ln ρ + )
q
ρ+ µ +
D−
kT
∇ρ − =
∇ (ln ρ − )
ρ− µ−
q
ρ + = ρ o e − qΦ / kT
Boltzmann Distributions
ρ − = −ρ o e + qΦ / kT
ρ + ( Φ = 0 ) = −ρ − ( Φ = 0 ) = ρo
∇2 Φ =
−ρ
ε
=−
( ρ+
+ ρ− )
ε
=
−ρo
⎡e
ε ⎣
[Potential is zero when system is charge neutral]
− qΦ / kT
Small Potential Approximation:
sinh
− e+ qΦ / kT ⎤⎦ =
2ρo
ε
qΦ
kT
(Poisson-Boltzmann Equation)
sinh
qΦ
<< 1
kT
qΦ qΦ
≈
kT
kT
∇2 Φ −
2ρ0q
Φ=0
εk T
∇2 Φ −
Φ
=0 ;
L d2
Ld =
εk T
2 ρo q
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Debye Length
Lecture 7
Page 11 of 27
D. Case Studies
1. Planar Sheet
d2 Φ
Φ
− 2 =0
2
dx
Ld
B.C. Φ ( x → ±∞ ) = 0
⇒
Φ = A1 ex / L d + A2 e−x / L d
⇒ Φ (x) =
Φ ( x = 0 ) = Vo
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Vo e−x / L d
x>0
Vo e+x / Ld
x<0
Lecture 7
Page 12 of 27
Vo −x / L d
e
Ld
Ex = −
x>0
dΦ
=
dx
−Vo x / L d
e
Ld
−
ρ=ε
dEx
=
dx
−
ε Vo
L d2
ε V
o
L
2
d
x<0
e−x / Ld
x>0
e+x / Ld
x<0
2 ε
Vo
σs ( x = 0 ) = ε ⎡⎣Ex ( x = 0+ ) − Ex ( x = 0− ) ⎤⎦ =
L d
2. Point Charge (Debye Shielding)
∇2 Φ −
Φ
=0
L d2
⇒
d2
rΦ
(r Φ ) − 2 = 0
dr 2
Ld
0
r Φ = A1e
1 ∂ ⎛ 2 ∂Φ ⎞
⎜r
⎟
r2 ∂r ⎝ ∂r ⎠
=
Φ (r ) =
−r / Ld
+ A2e
+r / L d
Q
e−r / L d
4 π ε r
1 ∂2
(r Φ )
r ∂r2
E. Ohmic Conduction
J+ = ρ + µ + E − D + ∇ρ +
J− = −ρ − µ − E − D − ∇ρ
−
If charge density gradients small, then ∇ρ ± negligible ⇒ ρ + = −ρ − = ρ o
J = J+ + J− = ( ρ + µ + − ρ − µ − ) E = ρ o ( µ + + µ − ) E = σE
σ = ohmic conductivity J = σ E (Ohm’s Law) 6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 13 of 27
F. pn Junction Diode
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7 Page 14 of 27 ∆Φ = Φ n − Φ p =
Φ ( x = 0 )= Φ p +
∆Φ = Φ n − Φ p =
=
k T N A ND
ln
q
ni2
q NA xp2
2ε
= Φn −
q ND x n2
2ε
q NA xp2
q ND x n2
+
2ε
2ε
q ND x n
q ND x n2 ⎛
ND ⎞
xn + xp ) =
(
⎜1 +
⎟
2ε
2ε
NA ⎠
⎝
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 15 of 27
VII. Relationship Between Resistance and Capacitance In Uniform Media Described
by ε and σ .
∫ D i da
q
C= u =
v
S
∫ E i ds
ε
=
S
∫ E i ds
L
R=
v
=
i
∫ E i ds
L
∫
J i da
L
∫ E i ds
=
L
σ
S
RC =
∫ E i ds
∫ E i da
S
∫ E i da
S
ε
∫ E i da
L
σ
∫ E i da
L
∫ E i ds
=
ε
σ
L
Check:
Parallel Plate Electrodes: R =
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
l
,
σA
C=
εA
l
⇒ RC =
ε
σ
Lecture 7
Page 16 of 27
Coaxial
R=
ln b
a ,
2 πσl
C=
2 πεl
ln b
⇒ RC = ε
a
σ
Concentric Spherical
1
1
−
R1 R 2
R=
,
4πσ
C=
4πε
1
1
−
R1 R 2
⇒ RC =
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
ε
σ
Lecture 7
Page 17 of 27
VIII. Change Relaxation in Uniform Conductors
∇ i Ju +
∂ ρu
=0
∂ t
∇ iE =
ρu
ε
Ju = σ E
σ ∇ i E
ρ
u
∂ ρu
∂t
+
=0 ⇒
+
σ
ε
ρu = 0
ε
τe =
∂ ρu
∂t
∂ ρu
∂t
+
ρu
τe
ε σ = dielectric relaxation time
=0 ⇒
(
)
ρu = ρ 0
r, t = 0 e − t τ
e
IX. Demonstration 7.7.1 – Relaxation of Charge on Particle in Ohmic Conductor
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 18 of 27
∫ J i da = σ ∫ E i da =
S
S
σ qu
ε
=
−dq
dt
dq
q
+
= 0 ⇒ q = q ( t = 0 ) e − t τe
τe
dt
(τ
e
=
ε
σ
)
Partially Uniformly Charged Sphere
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 19 of 27
ρ0
r < R1
ρu ( t = 0 ) =
QT =
0
ρu ( t ) =
r > R1
ρ 0 e − t τe
( τe
r < R1
0
Er (r, t ) =
4
π R 13 ρ 0
3
= ε σ)
r > R1
ρ 0 r e − t τe
Q r e − t τe
=
3ε
4 π ε R 13
Q e − t τe
4 π ε r2
Q
4 π ε0 r2
0 < r < R1
R1 < r < R 2
r > R2
σ su (r = R 2 ) = ε 0 Er (r = R 2 + ) − ε Er (r = R 2 − )
=
Q
1 − e − t τe
4 π R 22
(
)
X. Self-Excited Water Dynamos
A. DC High Voltage Generation (Self-Excited)
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 20 of 27
Courtesy of Herbert Woodson and James Melcher. Used with permission. Woodson, Herbert H., and James R. Melcher.
Electromechanical Dynamics, Part 2: Fields, Forces, and Motion. Malabar, FL: Kreiger Publishing Company, 1968. ISBN: 9780894644597.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 21 of 27
− n C i v1 = C
dv 2
dt
v1 = V1 e st
⇒
−n C i v 2 = C
dv1
dt
− n C i V1 = C s V2
⇒
v 2 = V2 e st
− n C i V2 = C s V1
⎡n Ci
⎢ Cs
⎢
⎢
⎢
⎢ 1
⎣⎢
⎤
1 ⎥ ⎡ V1 ⎤
⎢ ⎥
⎥ ⎢ ⎥
⎥ ⎢ ⎥=0
n Ci ⎥ ⎢ ⎥
⎥ ⎢ ⎥
Cs ⎥⎦ ⎣ V2 ⎦
Det = 0
2
n Ci
⎛ n Ci ⎞
⎜
⎟ =1 ⇒ s = ±
C
⎝ Cs ⎠
⊕ root blows up
n Ci
t
eC
Any perturbation grows exponentially with time
B. AC High Voltage Self – Excited Generation
dv 2
dt
dv 3
−n C i v 2 = C
dt
dv1
−n C i v 3 = C
dt
−n C i v1 = C
⎡n Ci
⎢
⎢
⎢0
⎢
⎢
⎢Cs
⎣
Cs
n Ci
0
;
v1 = V1 e st
v 2 = V2 e st
v 3 = V3 e st
0⎤
⎥
⎥
Cs ⎥
⎥
⎥
n Ci ⎥⎦
⎡ V1 ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ V2 ⎥ = 0
⎢ ⎥
⎢ ⎥
⎢ V3 ⎥
⎣ ⎦
det = 0
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 22 of 27
(n Ci )
3
1
⎛ n Ci ⎞
3
3
+ ( Cs ) = 0 ⇒ s = ⎜
⎟ ( −1)
⎝ C ⎠
s1 = −n Ci C (exponentially decaying solution) ( −1)
13
s2, 3 =
= −1,
1 ± 3j
2
n Ci
⎡1 ± 3 j⎤ (blows up exponentially because sreal >0 ; but also
⎦
2C ⎣
oscillates at frequency simag ≠ 0)
XI. Conservation of Charge Boundary Condition
∇ i Ju +
∫J
u
S
∂ρu
=0
∂t
i da +
d
ρu dV = 0
dt ∫V
n i ⎡⎣ Ja − Jb ⎤⎦ +
d
σsu = 0
dt
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 23 of 27
XII. Maxwell’s Capacitor
A. General Equations
_
Ea ( t ) ix
0<x<a
E=
_
Eb ( t ) ix
a
∫E
x
−b < x < 0
dx = v ( t ) = Eb ( t ) b
+ Ea
( t ) a
−b
n i ⎡⎣ Ja − Jb ⎤⎦ +
Eb ( t ) =
v (t)
b
dσsu
d
⎡εaEa ( t ) − εbEb ( t ) ⎤⎦ = 0
= 0 ⇒ σa Ea ( t ) − σb Eb ( t ) +
dt
dt ⎣
− Ea ( t )
a
b
⎡ v (t)
⎤ d ⎡
⎛ v (t)
⎞⎤
σa Ea ( t ) − σb ⎢
− Ea ( t ) a⎥ +
− Ea ( t ) a ⎟⎥ = 0
⎢εaEa ( t ) − εb ⎜⎜
⎟
⎢⎣ b
⎥⎦ dt ⎢⎣
⎝ b
⎠ ⎦⎥
σ v ( t ) εb dv
σ a⎞
εb a ⎞ dEa ⎛
⎛
+ ⎜ σa + b ⎟ Ea ( t ) = b
+
⎜ εa +
⎟
b ⎠ dt
b ⎠
b
b dt
⎝
⎝
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 24 of 27
B. Step Voltage: v ( t ) = V u ( t )
Then
dv
= V δ ( t ) (an impulse)
dt
At t=0
εb a ⎞ dEa εb dv εb
⎛
=
=
Vδ ( t )
⎜ εa +
⎟
b ⎠ dt
b dt
b
⎝
Integrate from t=0- to t=0+
εb a ⎞ dEa
ε a⎞
⎛
⎛
dt = ⎜ εa + b ⎟ Ea
⎜ εa +
⎟
∫
b ⎠ dt
b ⎠
⎝
t = 0− ⎝
t = 0+
t = 0+
0+
=
t = 0−
∫
t = 0−
εb
b
Vδ ( t ) dt =
εb
b
V
Ea ( t = 0 − ) = 0
εb a ⎞
εb
εb V
⎛
V ⇒ Ea ( t = 0 + ) =
⎜ εa +
⎟ Ea ( t = 0 + ) =
εb b + εb a
b ⎠
b
⎝
For t > 0,
dv
=0
dt
σ a⎞
σ
εb a ⎞ dEa ⎛
⎛
+ ⎜ σ a + b ⎟ Ea ( t ) = b V
⎜ εa +
⎟
b
dt
b
b
⎠
⎠
⎝
⎝
Ea ( t ) =
σb V
ε b + εb a
+ A e − t τ
; τ = a
σ ab + σ b a
σ ab + σb a
Ea ( t = 0 ) =
σb V
+A =
σ ab + σ b a
⎡
⎤
εb V
εb
σb
⇒ A=V⎢
−
⎥
ε ab + εb a
⎣ ε ab + ε b a σ ab + σ b a ⎦
Ea ( t ) =
σb V
εb V
1 − e−t τ +
e−t τ
ε ab + εb a
σ ab + σ b a
Eb ( t ) =
V
a
− Ea ( t )
b
b
(
)
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 25 of 27
⎛V a
⎞
σ su ( t ) = ε a Ea ( t ) − εb Eb ( t ) = ε a E a ( t ) − ε b ⎜ − E a ( t ) ⎟
b
b
⎝
⎠
ε a⎞
V
⎛
= Ea ( t ) ⎜ ε a + b ⎟ − εb
b ⎠
b
⎝
=
V ( σb ε a − σ a εb )
( σ ab + σb a)
(1 − e )
−t τ
C. Sinusoidal Steady State: v ( t ) = R e ⎡ V e jω t ⎤
⎣
⎦
Ea ( t ) = Re ⎡Ea e jωt ⎤
⎣
⎦
Eb ( t ) = Re ⎡Eb e jωt ⎤
⎣
⎦
Conservation of Charge Interfacial Boundary Condition
σ a E a ( t ) − σ b Eb ( t ) +
d
⎡ ε a E a ( t ) − εb Eb ( t ) ⎤⎦ = 0
dt ⎣
Ea ⎡σ
⎣ a + j ω ε a ⎤⎦ − Eb ⎡σ
⎣ b + j ω εb ⎤⎦ = 0
Eb b + Ea a = V
Eb =
V Ea a
−
b
b
⎛ V Ea a ⎞
Ea ⎡σ
⎣ a + j ω ε a ⎤⎦ − ⎜⎜ b − b ⎟⎟ ⎡⎣σ b + j ω εb ⎤
⎦= 0
⎝
⎠
a
V
⎡
⎤
E a ⎢ σ a + j ω ε a + ( σb + j ω εb ) ⎥ = ⎡σ
b + j ω εb ⎤
⎦= 0
b
b⎣
⎣
⎦
Ea
Eb
V
=
=
j ω εb + σb
j ω εa + σa
⎡⎣b ( σ a + j ω ε a ) + a ( σ b + j ω εb ) ⎤⎦
σ su =
=
ε a E a − εb Eb
⎡⎣b ( σ a
( ε a σb − εb σ a ) V
+ j ω ε a ) + a ( σb +
j ω εb ) ⎤⎦
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 26 of 27
D. Equivalent Circuit (Electrode Area A)
I = ( σ a + j ω ε a ) Ea A = ( σ b + j ω ε b ) Eb A
=
V
Ra
Rb
+
R a C a j ω + 1 R b Cb j ω + 1
Ra =
Ca =
a
b
, Rb =
σa A
σb A
εa A
a
, Cb =
εb A
b
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 7
Page 27 of 27