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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