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LECTURE 10 slide 1
Lecture 10
Current Density
Ohms Law in Differential Form
Sections: 5.1, 5.2, 5.3Homework: See homework file
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LECTURE 10 slide 2
Electric Direct CurrentReview
DC is the flow of charge under Coulomb (electrostatic) forces in
conductors
the electrostatic force is provided by external sources: battery,
charged capacitor
Georg Simon Ohm was the 1st to observe and explain the lack of
charge acceleration in metalselectrons move with uniform averaged
speed (drift velocity)
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LECTURE 10 slide 3
Current Density
the current flowing through the cross-section s of a conductor
is the amount of transferred charge Q per unit time
1( ) , A=C ss
QI
t
, Cv v vV L
Q V s L s v t
( ) , A
n
s v
J
Q I v s
t
( ) wheres n I J s 2, A/mn vJ v
the current density is a vector2, A/mvJ vn nJ J a
e
e
e
e
e
e
s
L v t
e
e
Q V
dQI
dt
I
current density,normal component
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LECTURE 10 slide 4
Current and Current Density
the currentIis the flux of the current density J
, A
S
I d J s( ) s n I I J s dI d J s J s
Two cylindrical wires are connected in series. CurrentI= 10
A flows through the junction. The radii of the wires are: r1=1 mm, r2= 2 mm. Find the current densitiesJ1 andJ2 in the
two wires.
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LECTURE 10 slide 6
Specific Conductivity1
( ) , ,e e p p e e p p p h i
J v v E
note: e < 0
specific conductivity depends on the free-charge density and its
mobility
1, S/m=( m)e e p p
charge density depends on the number of charge carriers per unit
volume (number density), e.g., e = eNe
semi ( )e e h p N N e
metal e eN e
191.6022 10 , Ce
in pure semiconductorsNe =Nh
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LECTURE 10 slide 7
Specific Conductivity2
typical carrier number densities, mobilities, conductivities (low
frequency, below THz)
e h Ne (m3) Nh (m
3) (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
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!)
7
Au 4.5 10 S/m
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LECTURE 10 slide 8
Ohms Law in Point (Differential) Form
J E
Ohms law in circuits/ , A I GV V R
assume uniform current distribution in the cross-section of the
conductor between pointsA andB
I J s , | | AB AB ABV
V E ll E L L
1G R
s
I Js sE V l
1
,
s l
G Rl s
use Ohms law in point form to arrive at Ohms law for resistors
conductance/resistance of a conductor of length l, constant cross-section s, and
constant current density distribution in s
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LECTURE 10 slide 9
General Expression for Resistance
,
B
A
s
dVR
I d
E l
J s
,
B
A
s
dR
d
E l
E s
in homogeneous medium
1,
B
A
s
dR
d
E l
E s
1
, Ss
B
A
d
G Rd
E s
E l
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LECTURE 10 slide 10
DC Resistance per Unit Length
twin-lead line
1
2 ,
12 , /m
L
R A
RA
coaxial line
2 2 2
2 2 2
1 1,
( )1 1 1
, /m
L LR
a c b
Ra c b
LI
IA
A
a
bI c
I
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LECTURE 10 slide 11
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 (Cu= 5.8x107 S/m).
ANS: 21 m/m
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LECTURE 10 slide 12
consider the current flowing through a closed surface[ ]vs
I d J s
total positive flux corresponds to an outflow of charge (chargeinside volume decreases)
[ ]
encl
vs
dQI d
dt J s NOTE THE NEGATIVE SIGN!
Conservation of Charge/Continuity of Current1
in circuits we assume that no charge accumulates at nodes
1 2
[ ] 1 2
3
3
0
v
I I I
s s s s
I d d d d
J s J s J s J s
Kirchhoffs current law follows
from conservation of charge
0nn
I s
1s
3s
2s
1I
3I
2I
continuity of current (conservation
of charge) in integral form
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LECTURE 10 slide 13
Conservation of Charge/Continuity of Current2
apply Gauss (divergence) theorem to conservation of charge law
[ ]
inside
v
v
s v v
dQ d I d dv dvdt dt
J s J
v
t
J continuity of current (conservation
of charge) in point form
the equation of charge relaxationhm hm 1
( ) , also ( )v v vv
t t
D J
E E E E
0v vt
/
0 0( ) , / t
tt e e
charge relaxation constant
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LECTURE 10 slide 14
Charge Relaxation
/
consider an isolated conductor into which some charge Q0 is
injected initially
Coulomb forces push the charge carriers apart until they re-
distribute 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 3where
this is also the time required to discharge a charged capacitor
through a shorting conductor
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LECTURE 10 slide 15
Charge Relaxation Illustrated
Example: Calculate the time required to restore charge neutralityin Cu where = 0 and = 5.8x10
7 S/m.12
190
7
3 8.8542 103 3 / 4.6 10 , s
5.8 10T
0 5 10 15 200
0.2
0.4
0.6
0.8
1
time (s)
exp(-
t/
)
= 3
1/e
curve tangent at t = 0,
intersects time axis at t =
3 s
0/00
t
t
de
dt
0 1
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LECTURE 10 slide 16
Joules Law in Differential Form
consider sufficiently small volume v = sL where the E-field
and the charge density v are constant
since charge is moving with uniform drift velocity ud, the E-fielddoes work on the charge (this work is converted into heat)
, Je eW Q w v F
E L
power is work done per unit time, W
e ed
W w v QP v Q
t t tp
E LE u
power density
3, W/mvd dQv
p
E uE u E J
Joules law in differential form: dissipated power per unit volume
2 3( ) | | , W/mp E J E E E
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LECTURE 10 slide 17
Joules Law in Integral Form
2| | , W
v v v
P pdv dv dv
E J E
power dissipated in conductors
, WL S
P EdL Jds V I
p dvv S L
P dsdL EJdLds E J
Joules law in circuit theory
assume that in a piece of conductor, E does not depend on the
cross-section, only J (or ) does, while J (or ) does notdepend on the length
assume that E and J are collinear
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LECTURE 10 slide 18
You have learned:
what current density is and how it relates to the total current
how to compute the resistance/conductance of conducting bodies
that drift velocity of charge in conductors is proportional to thestrength ofE and the coefficient of proportionality is the mobility
what specific conductivity is and how it relates J to E through
Ohms law in differential (point) form
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 from the Efield the dissipated power using Joules law
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