Lecture 10
Current Density
Ohm’s Law in Differential Form
Sections: 5.1, 5.2, 5.3
Homework: See homework file
LECTURE 10
slide
1
Electric Direct Current – Review
• DC is the flow of charge under Coulomb (electrostatic) forces in
conductors
• Georg Simon Ohm was the 1st to observe and explain the lack of
charge acceleration in metals – electrons move with uniform
averaged speed (drift velocity)
• the electrostatic force is provided by external sources: battery,
charged capacitor
LECTURE 10
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2
Current Density
the current flowing through the cross-section Δs of a conductor
is the amount of transferred charge ΔQ per unit time
∆Q
=
, A=C × s −1
I ( ∆s )
∆t
s∆L = ρv ∆s v
∆Q = ρv ∆V = ρv ∆
∆t , C ⇒ I ( ∆s )

∆V
∆L
n
I ( ∆s=
J n ∆s where J n = ρv v, A/m 2
)
∆L = v∆t
∆s
e
e
e
e ∆Q
e
e
e
e
dQ
I=
dt
∆Q
=
=ρ
v v ∆s , A

∆t
J
current density,
normal component
the current density is a vector
I
∆V
J n= J ⋅ a n
LECTURE 10
J = ρv v, A/m 2
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3
Current and Current Density
the current I is the flux of the current density J
∆I = I ( ∆s ) = J n ∆s = J ⋅ ∆s ⇒ dI = J ⋅ ds ⇒ =
I
∫∫ J ⋅ ds, A
S
Two cylindrical wires are connected in series. Current I = 10
A flows through the junction. The radii of the wires are: r1 =
1 mm, r2 = 2 mm. Find the current densities J1 and J2 in the
two wires.
LECTURE 10
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4
Charge Mobility
• charge velocity in a conductor depends on the charge mobility
ve =
− µe E, v h =
µh E, v i =
µi E, m/s
 metals support electron current
drift electron velocity in metals: vd = −μeE
 semiconductors support both electron and hole currents
 most electrolytes support both electron and ion currents
 in general plasmas support both electron and ion currents
• mobility may in general depend on E (nonlinear conductors)
LECTURE 10
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5
Specific Conductivity – 1
(− ρe µe + ρ p µ p ) E, p ≡ h, i
J=
ρe v e + ρ p v p =

σ
note: ρe < 0
Ohm’s law
• the specific conductivity σ depends on the free-charge density and
its mobility
σ = − ρe µe + ρ p µ p , S/m=(Ω × m) −1
• the charge density depends on the number of charge carriers per unit
volume (number density), e.g., ρe = −eNe
e ≈ 1.6022 × 10−19 , C
=
σ semi ( N e µe + N h µ p )e
in pure semiconductors Ne = Nh
σ metal = N e µe e
LECTURE 10
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6
Specific Conductivity – 2
typical carrier number densities, mobilities, conductivities (below THz)
μe
μh
Ne (m−3)
Nh (m−3)
σ (S/m)
pure Ge
0.39
0.19
2.4x1019
2.4x1019 2.2
pure Si
0.14
0.05
1.4x1016
1.4x1016 4.4x10 − 4
Cu
0.0032
―
1.13x1029
―
5.8x107
Al
0.0015
―
1.46x1029
―
3.5x107
Ag
0.005
―
7.74x1028
―
6.2x107
σ Au
= 4.5 ×107 S/m
Homework: What is the drift velocity of electrons in a Cu wire of
length 10 cm if the voltage applied to both ends of the wire is 1 V.
(Ans.: 3.2 cm/s. Wire may melt if too thin!)
LECTURE 10
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7
Ohm’s Law in Point (Differential) Form
J =σE
• Ohm’s law in circuits
=
I GV
= V / R, A
• assume uniform current distribution in the cross-section of the
conductor between points A and B
V
VAB = E ⋅ L AB ⇒ E = , l =| L AB |
I= J ⋅ s
l
• use Ohm’s law in point form to arrive at Ohm’s law for resistors
s
s
1 l
⇒ I = Js = σ Es = σ ⋅V
⇒G =σ , R = ⋅
l
σ s
l
G = R −1
conductance/resistance of a conductor of length l, constant cross-section s, and
constant current density distribution in s
LECTURE 10
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General Expression for Resistance
B
B
V
R= =
I
∫A E ⋅ dl , Ω
∫∫s J ⋅ ds
=
⇒ R
∫A E ⋅ dl , Ω
∫∫s σ E ⋅ ds
• in homogeneous medium
B
1 ∫A E ⋅ dl
,Ω
R=
⋅
σ ∫∫ E ⋅ ds
s
G= R −1=
E ⋅ ds
∫∫
,S
σ ⋅ Bs
∫A E ⋅ dl
LECTURE 10
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DC Resistance per Unit Length
∆L
• twin-lead line
1 ∆L
2×
,Ω
R=
σ A
1
2
, Ω/m
⇒ R′ =×
σA
I
A
I
A
• coaxial line
1 ∆L 1
∆L
R=
+
,Ω
2
2
2
σ π a σ π (c − b )
1  1
1 
⇒=
R′
 2 + 2 2  , Ω/m
σπ  a c − b 
LECTURE 10
I c b
I
a
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10
Homework: Resistance per Unit Length
Find the resistance per unit length of a coaxial cable whose inner
wire is of radius a = 0.5 mm and whose shield has inner radius b
= 4 mm and outer radius c = 4.5 mm. Both the inner wire and
the shield are made of copper (σCu = 5.8x107 S/m).
ANS: 21 mΩ/m
LECTURE 10
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11
Conservation of Charge/Continuity of Current Law – 1
consider the current flowing through a closed surface
=
I
∫∫ s

J ⋅ ds
[v]
total positive flux corresponds to an outflow of charge (charge
inside volume decreases) – continuity of current (conservation
dQencl of charge) in integral form
I=
−

∫∫ J ⋅ ds =
NOTE THE NEGATIVE SIGN!
dt
s[ v ]
in circuits we assume that no charge accumulates at nodes
I= 
∫∫ J ⋅ ds = ∫∫ J ⋅ ds + ∫∫ J ⋅ ds + ∫∫ J ⋅ ds = 0
s1
s2
s3



s[ v ]
− I1
I2
Kirchhoff’s current law follows
from conservation of charge
∑ In = 0
n
s3
I3
I1
LECTURE 10
I3
s1
s2
s
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12
I2
Conservation of Charge/Continuity of Current – 2
apply Gauss’ (divergence) theorem to conservation-of-charge law
I=
∫∫ J ⋅ ds =

s[ v ]
dQencl
d
= − ∫∫∫ ρ v dv
dt
dt v
∫∫∫ ∇ ⋅ Jdv = −
v
∂ρv
⇒ ∇⋅J = −
∂t
continuity of current (conservation
of charge) in point form
the equation of charge relaxation
ρv
∂ρv hm
1 ∂ρv
ε E) = ρv ⇒∇ ⋅ E =
, also ∇ ⋅ (
σ E) = −
∇ ⋅ (
⇒∇ ⋅ E = −
ε
∂t
σ ∂t
D
J
hm
σ
∂ρv σ
− ⋅t
−t / τ
⇒
+ ρv =
0 =
ε
ρ
(
t
)
ρ
=
e
ρ
e
, τ ε /σ
0
0 =
∂t ε
charge relaxation constant
LECTURE 10
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Charge Relaxation
• consider an isolated conductor into which some charge Q0 is
injected initially
• Coulomb forces push the charge carriers apart until they redistribute and settle on the surface
• the process continues until no free charge is left inside the
conductor
• the time for this to happen is about 3τ where τ = ε / σ
• this is also the time required to discharge a charged capacitor
through a shorting conductor
LECTURE 10
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Charge Relaxation Illustrated
1
τ =3s
ρ0 = 1
exp(-t/τ)
0.8
curve tangent at t = 0,
intersects time axis at t = τ
0.6
(
d
ρ0e −t /τ
dt
0.4
1/e
)
= −
t =0
ρ0
τ
0.2
0
0
τ = 35
10
15
20
time (s)
Example: Calculate the time required to restore charge neutrality
in Cu where ε = ε0 and σ = 5.8x107 S/m.
3 × 8.8542 × 10−12
−19 , s
T ≈=
3τ 3ε 0 /=
σ
≈
4.6
×
10
5.8 × 107
LECTURE 10
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Joule’s Law in Differential Form
• consider sufficiently small volume Δv = Δs·ΔL where the E-field
and the charge density ρv are constant
• since charge is moving with uniform drift velocity ud, the E-field
does work on the charge (this work is converted into heat)
∆We= Q
E ⋅ ∆L= we ∆v, J

F
• power is work done per unit time
w ∆v QE ⋅ ∆L
∆We
∆P =
= p∆v = e =
= QE ⋅ u d , W
∆t
∆t
∆t
power density
∆P Q E ⋅ u d
⇒ p=
=
= ρ v E ⋅ u d = E ⋅ J, W/m3
∆v
∆v
Joule’s law in differential form: dissipated power per unit volume
p = E ⋅ J = σ (E ⋅ E) = σ | E |2 , W/m3
LECTURE 10
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Joule’s Law in Integral Form
• power dissipated in conductors
P=
∫∫∫ pdv =
v
∫∫∫ E ⋅ Jdv =
v
2 dv, W
|
|
σ
E
∫∫∫
v
• Joule’s law in circuit theory
o assume that in a piece of conductor, E does not depend on
the cross-section, only J (or σ) does, while J (or σ) does not
depend on the length
o assume that E and J are collinear
P =∫∫∫ E
⋅ J dsdL

 =∫∫ ∫ EJdLds
v
⇒P=
p
∫L
dv
S L
EdL × ∫∫ Jds = V ⋅ I = RI 2 = V 2 / R, W
S
LECTURE 10
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You have learned:
what current density is and how it relates to the total current
that drift velocity of charge in conductors is proportional to the
strength of E and the coefficient of proportionality is the mobility
what specific conductivity is and how it relates J to E through
Ohm’s law in differential (point) form
how to compute the resistance/conductance of conducting bodies
that charge is preserved and the rate of change of the charge
density determines the divergence of the current density
(continuity of current in point form)
what charge relaxation is and how it depends on the permittivity
and conductivity of the material
how to find the dissipated power from the E field using Joule’s law
LECTURE 10
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