Current- Continuity Equation

Topics:

Conductivity and resistivity

8.022 (E&M) Lecture 7

Electrical currents

Ohms law in microscopic and macroscopic form

2

Electric current I

i

Units: cgs: esu/s C/s=ampere (A) 9

dQI dt

=

G. Sciolla MIT 8.022 Lecture 7

Consider a region in which there is a flow of charges: E.g. cyl ndrical conductor

We define a current: the charge/unit time flowing through a certain surface

SI: Conversion: 1 A = 2.998 x 10 esu/s

1

3Current density J

Velocity of each charge: u

Q / t where

cosprismqnV qnAI qnt t t t

= = = = = = G GGGi i

A G

u t

J qnu u G G G lume charge densiG. Sciolla MIT 8.022 Lecture 7

Number density: n = #charges / unit volume

Current flowing through area A: I = Q= q x number of charges in the prism

Where we defined the current density J as:

Q q N u t u A J A

: =vo ty

4

More realistic case

i i u

Multiple charge carriers:

l i ind of i k i i k

i average velocity

J:

k k k k k k k

J u G G G

1 ( )k ik

u uN

= G G k k k k k

k k J u G G G

S

I = GG i G. Sciolla MIT 8.022 Lecture 7

We made a number of unrealistic assumptions: only 1 k nd of charge carriers: we could have several, e.g.: + and ons

assumed to be the same for all particles: unrealistic! regular surface with J constant on it

E.g.: so ut on with different k ons NB: + ion with velocity u is equivalent to on with veloc ty -u

Velocity: Not all charges have the same veloc ty

Arbitrary surface S, arbitrary

q n u

k i

q n u

J dA

2

5

Non standard currents

Other kinds of currents (Demo F5)

+ + + +

----

Na+

Cl-


We usually think of currents as electrons moving inside a conductor This is only one of the many examples!

Ions in solution such as Salt (NaCl) in water

6

The continuity equation

Some charge exits

S

J

J

J J

J

JS

Q t

= GG iv

00S V

V

V

J tQ

t t

J t

= + = +

= =

GG G i i GG i i

v v


A current I flows through the closed surface S: Some charge enters

What happens to the charge after it enters? Piles up inside Leaves the surface

NB: - because dA points outside the surface

Apply Gausss theorem and obtain continuity equation:

inside J dA

inside

J dA JdV

dV dV

3

7

Thoughts on continuity equation

For steady currents:

S

J

J

J J

J

J

0J t

+ = G i

0J = G i G. Sciolla MIT 8.022 Lecture 7

Continuity equation:

What does it teach us? Conservation of electric charges in presence of currents

no accumulation of charges inside the surface: d /dt=0

8

Microscopic Ohms law Electric fields cause charges to move

conductivity

J E= G G

e


Experimentally, it was observed by Ohm that

Microscopic version of Ohms law: It reflects the proportionality between E and J in each point

Proportionality constant:

More stuff here pleas

4

9

in E // L

Macroscopic Ohms law

RI V LV IJ E A AL

R = = =

A

L

J E


Current is flowing in a uniform material of length L uniform electric field

Potential difference between two ends: V=EL Ohms law J= E holds in every point:

where

Resistance R Proportionality constant between V and R in Ohms law

R L A

L A

Units: [V]=[R][I]

SI: Ohm () = V/A cgs: s/cm

Dependence on the geometry: Inversely proportional to A and proportional to L

Dependence on the property of the material: Inversely proportional to conductivity

G. Sciolla MIT 8.022 Lecture 7 10

5

Resistivity = 1/ Units: in SI:

Depends on chemistry of material, , Demos F1 and F4

Resistivity


Describes how fast electrons can travel in the material m; in cgs: s

temperature

Demos F1 and F4

Why? Room temperature:

when T i increases

Very low temperature:

material

Resistivity vs. Temperature

T


Does resistivity depend on T?

depends upon collisional processes ncreases more collisions

Mean free path dominated by impurities or defects in the ~ constant with temperature.

With sufficient purity, some metals become superconductors

6

E Spherical symmetry

):

Application:

Resistance of a spherical shell Difference in potential V current

=0 Q: what is the resistance R?

b

a V

( ) B r A r

= + 1( ) ab a r V =

2 2

1 1( ) ab abE r V r J V= = G

4 44 ab V VI V abb a I V

b a

b aR ab

= = = = = =

G GG G i i G. Sciolla MIT 8.022 Lecture 7 13

Microscopic Ohm will hold: J=spherical potential:

Boundary conditions: (a)=V and (b)=0

E=-grad(

2 concentric spheres; material in between has resistivity

inner=V; outer V=0

b a r b a

b a r b a r

Sphere

J dA J A

What if is not constant? ivity 1 and 2

El i i q

l

1 1 2 2I A E = =

1 2 I

2 1 2 1( ) 4 4 4q

E E E I A

= = =


Cylindrical wire made of 2 conductors with conduct

What is the consequence? Current flowing must be the same in the whole cylinder

ectric fields are d fferent in the 2 regions E d scontinuous surface layer at the boundary

When conductivity changes there is the possibility that some charge accumulates somewhere. This is necessary to maintain steady f ow.

A E

surface

7

Thoughts on Ohms law

In plain Engli

A constant el

ll i

and are scattered the average behavior is a uniform drift

J E= G G

E v G G F ma E a= G GG G

E G. Sciolla MIT 8.022 Lecture 7 15

Ohms law in microscopic formulation: sh:

ectric field creates a steady current: Does this make sense?

Charges are moving in an effectively viscous medium As sky diver in free fa : first accelerate, then reach constant v Why? Charges are accelerated by E but then bump into nucle

Motion of electrons in conductor E

i 0: Impul l

The average momentum is:

i i ial

0p = G G Ep =

GG

1 1 1

1 1 1( ) N N N

i i i i i i i

p m u qE tN N N= = =

= = + = + G GG G

1 0

N

i i

u =

G

1

1 N i

i t

N

=

1

N

i i

qEm u tN

=

= G G


N electrons are moving in a material immersed in Two components contribute to the momentum:

Random collision veloc ty use due to e ectric field:

For large N:

Where is the average t me between 2 coll sions Property of the mater

Random mu qE t

m u mu qEt

qE

8

Conductivity

For multiple carriers:

2

= u qE nqE

m mm u

= = = =

G GG G G GG

2

1

N k k k

i k

n q m

=

=


From this derivation we can read off the conductivity

J nq J nq

qE

9

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Current- Continuity Equation