8.022 (E&M) – Lecture 7
Topics:
Electrical currents
Conductivity and resistivity
Ohm’s law in microscopic and macroscopic form
Electric current I
Consider a region in which there is a flow of charges:
E.g. cylindrical conductor
We define a current:
the charge/unit time flowing through a certain surface
I=
dQ
dt
Units:
cgs: esu/s
SI: C/s=ampere (A)
Conversion: 1 A = 2.998 x 109 esu/s
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Current density J
Number density: n = #charges / unit volume
Velocity of each charge: u
u∆t
θ
A
Current flowing through area A: I = ∆Q / ∆t
where ∆Q= q x number of charges in the prism
I=
∆Q q∆N qnV prism qnA cos θ u∆t
=
=
=
= qnu i A = J i A
∆t
∆t
∆t
∆t
Where we defined the current density J as: J ≡ qnu ≡
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8.022 – Lecture 7
ρu
3
ΝΒ: ρ=volume charge densi
ty
More realistic case…
We made a number of unrealistic assumptions:
only 1 kind of charge carriers: we could have several, e.g.: + and – ions
u assumed to be the same for all particles: unrealistic!
regular surface with J constant on it
Multiple charge carriers:
J ≡ ∑ qk nk uk ≡ ∑ ρ k uk
k
k
E.g.: solution with different kind of ions
NB: + ion with velocity uk is equivalent to – ion with velocity -uk
Velocity:
Not all charges have the same velocity
average velocity
J ≡ ∑ qk nk uk ≡ ∑ ρ k uk
k
uk =
1
Nk
∑ (u )
k i
i
k
Arbitrary surface S, arbitrary J: I =
∫ J idA
S
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Non standard currents
We usually think of currents as electrons moving inside a conductor
This is only one of the many examples!
Other kinds of currents
Ions in solution such as Salt (NaCl) in water (Demo F5)
+
+
+
+
-
Cl-
Na+
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The continuity equation
A current I flows through the closed surface S:
Some charge enters
Some charge exits
What happens to the charge after it enters?
Piles up inside
Leaves the surface
J
J
∫
J i dA = −
S
∂Qinside
∂t
J
J
J
NB: - because dA points outside the surface
S
J
Apply Gauss’s theorem and obtain continuity equation:
⎧ J i dA = ∇ i JdV
∫
⎪⎪ ∫S
V
⇒
⎨ ∂
∂
⎪ − Q inside = −
ρ
dV
∂ t V∫
⎪⎩ ∂ t
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⎛
∂
⎞
∫ ⎜⎝ ∇ i J + ∂ t ρ ⎟⎠ dV
V
8.022 – Lecture 7
= 0 ⇒ ∇iJ +
∂
ρ =0
∂t
6
3
Thoughts on continuity equation
Continuity equation:
∂
ρ =0
∂t
J
What does it teach us?
J
∇i J +
J
J
S
J
J
Conservation of electric charges in presence of currents
For steady currents:
no accumulation of charges inside the surface: dρ/dt=0
∇i J = 0
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Microscopic Ohm’s law
Electric fields cause charges to move
Experimentally, it was observed by Ohm that
J =σE
Microscopic version of Ohm’s law:
It reflects the proportionality between E and J in each point
Proportionality constant: conductivity σ
More stuff here please
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Macroscopic Ohm’s law
Current is flowing in a uniform material of length L in
uniform electric field E // L
J
A
E
L
Potential difference between two ends: V=EL
Ohm’s law J=σE holds in every point:
J =σE ⇒
I
V
=σ
A
L
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⇒
V = IR
where
R≡
L
σA
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Resistance R
Proportionality constant between V and R in Ohm’s law
ρL
L
R≡
≡
A
σA
Units: [V]=[R][I]
SI: Ohm (Ω) = 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
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Resistivity
Resistivity ρ = 1/σ
Describes how fast electrons can travel in the material
Units: in SI: Ω m; in cgs: s
Depends on chemistry of material, temperature,…
Demos F1 and F4
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Resistivity vs. Temperature
Does resistivity depend on T?
Demos F1 and F4
ρ
Why?
Room temperature:
ρ depends upon collisional processes
when T increases
more collisions
T
ρ increases
Very low temperature:
Mean free path dominated by impurities or defects in the
~ constant with temperature.
material
With sufficient purity, some metals become superconductors
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Application:
Resistance of a spherical shell
2 concentric spheres; material in between has resistivity ρ
Difference in potential V
current
φinner=V; φouter=0
Q: what is the resistance R?
ρ
Microscopic Ohm will hold: J=σE
spherical potential: φ ( r ) = A + B
Spherical symmetry
r
Boundary conditions: φ(a)=V and φ(b)=0
V
a
V=0
b
a ⎞
⎛ ab 1
−
⎟
−
−
b
a
r
b
a⎠
⎝
ab 1
ab 1
E=-grad(φ): E ( r ) = V
rˆ ⇒ J = σ V
b − a r2
b − a r2
b−a
ab
V
V
I = ∫ J i dA = J i A = 4πσ V
⇒R= =
=
ab
b
a
I
−
4
πσ ab
Sphere
4πσ V
b
a
−
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8.022 – Lecture 7
φ (r ) = V ⎜
13
What if σ is not constant?
Cylindrical wire made of 2 conductors with conductivity σ1 and σ2
I
σ1
σ2
What is the consequence?
Current flowing must be the same in the whole cylinder
I = Aσ 1 E1 = Aσ 2 E2
Electric fields are different in the 2 regions
E discontinuous
surface layer σq at the boundary
σq =
Esurface
4π
=
E2 − E1 I ( ρ 2 − ρ1 )
=
4π
4π A
When conductivity changes there is the possibility that some charge accumulates
somewhere. This is necessary to maintain steady flow.
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Thoughts on Ohm’s law
Ohm’s law in microscopic formulation: J = σ E
In plain English:
A constant electric field creates a steady current:
Does this make sense? F = ma ⇒ E ∝ a
E∝v
Charges are moving in an effectively viscous medium
As sky diver in free fall: first accelerate, then reach constant v
Why? Charges are accelerated by E but then bump into nuclei
and are scattered
the average behavior is a uniform drift
E
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Motion of electrons in conductor
N electrons are moving in a material immersed in E
Two components contribute to the momentum:
Random collision velocity u0:
Impulse due to electric field:
pRandom = mu0
pE = qEt
The average momentum is:
1 N
1 N
1 N
p = m u = ∑ (mui + qEti ) = m ∑ ui + qE ∑ ti
N i =1
N i =1
N i =1
N
For large N:
∑u
1
τ≡
N
i =1
i
→0
m u =
qE N
∑ ti ≡ qEτ
N i =1
N
ti is the average time between 2 collisions
∑
Where
i =1
Property of the material
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Conductivity
From this derivation we can read off the conductivity
⎧⎪ J = nq u
⎨
⎪⎩ m u = qEτ
⇒ J = nq
qEτ
=σ E
m
⇒
σ=
nq 2τ
m
For multiple carriers:
N
σ =∑
i =1
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nk qk 2τ k
mk
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