Download Chapter 5 Steady Electric Currents

Chapter 5 Steady Electric Currents
5-1 Introduction
5-2 Current density and Ohm’s law
5-3 Electromotive force and Kirchhoff’s voltage law
5-4 Continuity of equation and Kirchhoff’s current law
5-5 Power dissipation and Joule’s law
5-6 Boundary conditions for current density
5-7 Resistance calculation
§5-1 Introduction
Types of electric currents:
n  Conduction current: Drift motion of conduction electrons and/or holes in
conductors and semiconductors; very low for the average drift velocity of
the electrons (1/1000 m/s);
n  Electrolytic current: Migration of positive and negative ions;
n  Convection current: Motion of electrons and/or ions (positively or
negatively charged particles) in vacuum or in rarefied gas; not governed
by Ohm’s law.
§5-2 Current density and Ohm’s law
Consider a steady motion of one kind of charges q over a differential surface Δs
with velocity u and the total charges ΔQ passing this surface in a time interval Δt is
Total current over the differential surface Δs:
Current density:
The total current I flowing through an arbitrary surface S:
Noting that the charge density ρ = Nq, we have
Which is the relation between the convection current density and
the velocity of the charge carrier
p Drift motion of charge carriers under electric field
p Atoms remain neutral (ρ=0)
For most of metallic conductors in which the average electron drift is
proportional to electric field, we write
Electron mobility µe:
Cu:
Al:
Then we have,
Where ρe=-Ne is the charge density of the drifting electrons. We
rewrite it by
Conductivity, σ: σ = -ρeμe; a macroscopic parameter of a medium;
Unit: amper per volt-meter (A/Vm) or siemens per meter (S/m)
Drude model: conductivity for metals
(Microscopic Ohm’s law)
Metal and Free electron gas
Harmonic oscillation of electrons
under the excitation of electric fields



m e a el = F E _ local + F damping

 −iωt

du
me
= −eE L e − meγ u
dt
Collision frequency

⇒ u=
 −iωt
e
EL e
imeω − meγ


⇒ J = −Neu = (
Nuclei fixed; electron are free moving
 −iωt
−Ne 2
)E L e
imeω + meγ
Ne 2
Ne 2
1
⇒ σ (ω ) =
=
−imeω + meγ meγ (1− iω / γ )
Conductivity (static)
Ne 2
σ (0) =
meγ
Resistivity: the reciprocal of conductivity (ٟm)
Count Alessandro Giuseppe
Antonio Anastasio Volta,
1745--1827, Italy
André-Marie Ampère,
1775--1836, France
Georg Simon Ohm,
1789—1854, German
Resistance and conductance
Voltage between terminals 1 and 2:
The total current:
Point for microscopic
form of Ohm’s law
Fig. 5-3 Homogeneous conductor
with a constant cross section
Resistance
Macroscopic Ohm’s law
Resistance and conductance
§5-3 Electromotive force and Kirchhoff’s voltage law
Ohmic material under
static electric field
Static electric field
is conservative
A steady current cannot be maintained in the same direction in a closed
circuit by an electrostatic field. (D. K. Cheng, p. 206)
(steady: motion and constant velocity；static: no motion)
n  Electric fields inside an electric battery?
Impressed electric field, Ei
p  non-conservative electric field caused by chemical action;
p  Electromotive force (emf) is the integral from the
negative to the positive electrode inside the battery.
The conservative electrostatic field satisfies
emf:
or
Two different electric fields:
(1)  Conservative fields from the
charges, both inside and outside
the battery;
(2)  Nonconservative fieldsimpressed fields, from chemical
reaction, only inside the battery
R
Connect the battery electrodes by resistance R
Emf: (integral over the closed loop)
J = I/S
n  For one source of electromagnetic force,
n  For more than one source of electromagnetic force and more
than one resistor,
Kirchhoff’s
voltage law
Around a closed path in an electric circuit,
the algebraic sum of the emf’s (voltage rises)
is equal to the algebraic sum of the voltage
drops across the resistances.
§5-4 Equation of continuity and Kirchhoff’s current law
From the principle of conservative of charge, the current leaving one region
is the total outward flux of the current density through the enclosing surface S,
Apply divergence theorem, we obtain for a stationary volume
The above equation holds regardless of V, and the integrands must be equal.
Equation of
continuity
The charge conservation law can also be given as follows
For steady currents,
Steady electric currents are divergenceless or solenoidal. Over any
enclosed surface, we have
Which can be written as
Kirchhollf’s
current law
The algebraic sum of all currents flowing out of a junction in
an electric circuit is zero.
Relaxation time of charges in a metal?
Combining the Ohm’s law with the continuity equation and assuming σ,
we have
Relaxation time, τ
The solution is
Eg: Copper
τ = 1.52×10-19s
The charge density at a given location will decrease with time exponentially.
§5-5 Power dissipation and Joule’s law
Power p provided by an electric field E in moving a charge q:
Total power P delivered to all charge carriers in a volume dv:
= J (current density)
or
= power density
Total power converted into heat in a volume V
Joule’s
law
In a conductor of a constant cross section, dv=dsŸdl, with dl measured in the
current direction
§5-6 Boundary conditions for current density
In the absence of non-conservative energy source, we shall have
We can obtain the boundary conditions for J
(as in Fig. 3-23 and in Sec. 3-9):
(1) Normal component of boundary current, Jn
The normal components of J at two sides of an interface is continuous.
(2) Tangent component of boundary current, Jt
The ratio of the tangential components of J at two sides of an
interface is equal to the ratio of the electric conductivities.
Q5-3 Two conducting media with conductivities σ1 and σ2 are separated by
an interface. The steady current density in medium 1 at point P1 has a
magnitude J1 and makes an angle α1 with the normal. Determine the
magnitude and direction of the current density at point P2 in medium 2.
J1 cos α1 = J 2 cos α2
σ 2 J1 sin α1 = σ1 J 2 sin α2
tan α2 σ 2
=
tan α1 σ1
If medium 1 is a much better conductor than medium 2 σ 1 >> σ 2 or
σ2
→0
σ1
α2 →zero, and J2 emerges almost perpendicularly to the interface (normal to the
surface of the good conductor). The magnitude of J2 is
J 2 = J 2 2t + J 2 2 n = ( J 2 sin α 2 )2 + ( J 2 cos α 2 )2
= [(
1
1
σ2
σ
J1 sin α1 )2 + ( J1 cos α1 )2 ] 2 =J1[( 2 sin α1 )2 + cos 2 α1 ] 2
σ1
σ1
Boundary between two lossy dielectrics
Q5-4

(a) J normal component
Kirchhoff’s voltage law


J1 = J 2
V =（R1 + R2）I =（
Hence,
J=
d1
d
+ 2 ）I
σ1S σ 2 S
σ1σ 2V
I
V
=
=
S ( d1 ) + ( d 2 ) σ 2 d1 + σ1d 2
σ1
σ2
( A / m2 )
(b) To determine E1 and E2: 
V = E1d1 + E2 d 2
σE =σ E
J1 = J 2
1
1
E1 =
σ 2V
σ 2 d1 + σ1d 2
E2 =
σ1V
σ 2 d1 + σ1d 2
2
2
(V/m)
(V/m)
(c) The surface charge densities on the upper and lower plates : 
ρs1 = ε1E1 =
ε1σ 2V
σ 2 d1 + σ1d 2
ρs 2 = -ε2 E 2 = -
(C/m 2 )
ε2 σ1V
σ 2 d1 + σ1d 2
(C/m 2 )
The surface charge densities at the interface: 
ρ si = D2 n − D1n = ε 2 E2 n -ε1E1n =
(ε 2σ 1 -ε1σ 2 )V
σ 2 d1 + σ 1d 2
(C/m2 )
From these results we see that ρ ≠ - ρ but that ρ + ρ + ρ = 0
s2
s1
s2
s1
si
§5-7 Resistance calculations
The basic formula for capacitance can be written as
Fig. 5-7 Two conductors in a lossy
dielectric medium
Since the metallic conductors are equipotential media, you can choose anyone of
the integral paths for calculating the
electric potential difference.
Note: The dimension of RC and ε/σ is [time].
Calculate resistance
Conductor resistance
 
 
V − ∫ L E ⋅ dl − ∫ L E ⋅ dl
  =
R= =
 
I
∫ J ⋅ dS ∫ σ E ⋅ dS
S
S
The procedure for computing the resistance of a piece of conducting material
between specified equipotential surfaces (or terminals):
1. Choose an appropriate coordinate system.
2. Assume a potential difference V0 between conductor terminals.
3. Find electric field intensity E within the conductor.

homogeneous material: Laplace’s equation ∇ V = 0
E = -∇V
2
4. Find the total current
5.
I=
∫
s

J ⋅ ds = ⋅
s
 
σ E ⋅ ds
Find resistance R by taking the ratio V0/I.
Q5-6 A conducting material of uniform thickness h and conductivity σ has
the shape of a quarter of a flat circular washer, with inner radius a and
outer radius b. Determine the resistance between the end faces.
appropriate coordinate: cylindrical coordinate
assume a potential difference V0 between the
end faces: V=0 on the end face at y=0
V=V0 on the end face at x=0.
solve Laplace’s equation subject to the
following boundary conditions
d 2V
=0
2
dφ
V=0 at Φ=0
V=V0 at Φ =π/2
V = c1φ1 + c2 ,
V=
2V0
π
φ
1 ∂ ∂V
1 ∂ 2V ∂ 2V
∇V =
(r
)+ 2 2 + 2
r ∂r ∂r
r ∂φ
∂z
2
The current density is 


 ∂V
 2σ V0
J = σ E = -σ∇V = -aφσ
= -aφ
r∂φ
πr

The total current I can be found by integrating J over the φ = π surface at
2
 
which-ds = -aφ hdr
I=
∫
s
  2σ V b dr 2σ hV
b
0
0
J ⋅ ds =
h∫
=
ln
π
r
π
a
a
V0
π
R= =
I 2σ h ln(b / a)
Note that, for this problem, it is not convenient
to begin by assuming a total

I because it is not obvious how J varies with r for a given I. Without
current

J , E and V cannot be determined.
0