Today in Physics 217: begin electrodynamics
Today in Physics 217: begin electrodynamics Constitutive relations we’ve seen in E&M Charges in motion, again: Ohm’s law Carrier collisions and drift velocity Carrier collisions and heating Resistance Examples: resistance and current 18 November 2002
Physics 217, Fall 2002
1
S
σ 2
1
Constitutive relations in E&M Constitutive relations are relationships between parameters of a theory that are derived empirically or that depend for their calculation on physics that is not part of the theory. We have met two constitutive relations so far and are about to meet a third: P = χeE M = χm H J =σE The last one is called Ohm’s Law. In general, the proportionality “constant” σ is a second rank tensor, as are the susceptibilities, but many common media are “linear” in the sense that the conductivity σ can be considered a scalar. 18 November 2002
Physics 217, Fall 2002
2
Ohm’s Law Suppose a chunk of material with cross-sectional area A and length L has equipotential ends, with potential difference V. Suppose furthermore that this gives rise to a uniform E in the chunk. Then the current that flows is I = JA = σ EA V σA = V ≡ R L 18 November 2002
A
E
L
. Physics 217, Fall 2002
3
Ohm’s Law (continued) Ohm’s law should strike you as strange, at first, because you also know that J = ρv = σ E Suppose J is constant. Then v and E should be constant. But if E is constant, charges should accelerate at a = qE/m, rendering it impossible to have a constant v. So which relation is wrong? Neither. What happens is that collisions between free charges in a current (like electrons) with fixed or slowermoving charges (like the ions the electrons leave behind), and other free current carriers, keep the acceleration from going on for very long. 18 November 2002
Physics 217, Fall 2002
4
Ohm’s Law (continued) In collisions with the ions, the kinetic energy gained by the free carriers from the field is largely transferred to the ions, and the electron starts over, like a car driving down Intercampus Drive, stopping at all the speed bumps, accelerating in between. We therefore reconcile Ohm’s Law with the definition of current density by supposing that the collision process results in a well-defined average velocity, and writing J = ρv . In fact, we don’t need to suppose; we can show that that’s the way it is, in a crude classical model of what is mostly a quantum phenomenon, as follows. 18 November 2002
Physics 217, Fall 2002
5
Collisions and drift velocity Consider a chunk of metal with N carriers in it and an electric field E present. If it has been a time ti since particle i last suffered a collision, and if it left that collision at speed vi , then the momentum of this particle is pi = mvi + qEti
.
So a snapshot of the metal would reveal an average value of carrier momentum given by 1 N 1 N p = mv = ∑ pi = ∑ ( mvi + qEti ) . N i =i N i =1 The “starting” speeds vi are endowed by collisions with the fixed ions; their energy, in turn, comes from the thermal energy (heat) of the medium. 18 November 2002
Physics 217, Fall 2002
6
Collisions and drift velocity (continued) Thermal motions in a solid or liquid are (essentially) random in magnitude and direction, so if N is large, N
∑ mvi = 0
; thus,
i =1
qE N qE v= ti = t ∑ mN i =1 m
,
Drift velocity
nq 2 t E ≡σE , and J = ρ v = nqv = m where n is the number density of carriers. Clearly J should be linear in E if the carrier velocities are mostly thermal, and if collisions take place. 18 November 2002
Physics 217, Fall 2002
7
Collisions and heating Energy is continuously added to such a medium, as the kinetic energy gained by carriers between collisions is passed on to the fixed ions during the collisions. Independent of the nature of the collisions, if a carrier passes through a potential difference V, it picks up kinetic energy qV. All this is dissipated in collisions. If a total current I flows, then the power dissipated in collisions is d P = NqV = IV = I 2 R = − ∫ J ⋅ da ∫ E ⋅ dA = σ ∫ E2 dτ . dt
(
)
The collisional transfer of energy from carriers accelerated by fields, to the medium in which the current flows, is called resistance. 18 November 2002
Physics 217, Fall 2002
8
Resistance For the wire with cross-sectional area A, length L, and conductivity σ, the resistance turned out to be L ρL R= = σA A
.
ρ = resistivity. Not to be confused with charge density.
Units: in cgs,
1 [σ ] = sec
[ ρ ] = sec
sec [ R] = cm
.
In MKS, volt ≡Ω [ R] = amp
18 November 2002
1 [σ ] = Ωm
Physics 217, Fall 2002
[ρ] = Ω m
.
More commonly, Ω cm. 9
Examples: resistance and current Griffiths problem 7.3: (a) Two metal objects are embedded in weakly conducting material of conductivity σ. What is the resistance between them, in terms of their capacitance? (b) Suppose you connect a battery between them and charge them up to a potential difference V0. If you then disconnect the battery, the charge will leak away. Find the time dependence of the potential difference between the metal objects. 18 November 2002
Physics 217, Fall 2002
1
σ 2
10
Examples: resistance and current (continued) (a) Enclose the positively charged conductor with a surface S; then I=
v∫
J ⋅ da = σ
S
V = 4πσ CV R 1 ⇒R= 4πσ C
18 November 2002
v∫
E ⋅ da = 4πσ Q
1
S
S
ε0 = σ C in MKS .
Physics 217, Fall 2002
σ 2
11
Examples: resistance and current (continued) (b) Q = CV = CIR dQ Q = −I = − dt RC Q
∫
Q0
t
1
S
1 dQ′ =− dt ′ ∫ Q′ RC 0
Q ( t ) = Q0 e −t Q0 = V (t ) C
RC
−t e
= V0 e −4πσ t 18 November 2002
σ RC
= V0 e −t
RC
2
= V0 e −σ t ε 0 in MKS . Physics 217, Fall 2002
12
Examples: resistance and current (continued) Griffiths problem 7.2: A capacitor C is charged up to potential difference V0; at time t = 0 it is connected to a resistor R, and begins to discharge. (a) Determine the charge on the capacitor, and the current through the resistor, as a function of time. (b) What was the original energy stored in the capacitor? Show that the heat delivered to the resistor is equal to the energy lost by the capacitor. 18 November 2002
Physics 217, Fall 2002
C
R
13
Examples: resistance and current (continued) Q (a) V = = IR ; C Q ( t ) = Q0 e −t
RC
dQ Q = −I = − dt RC = CV0 e −t
RC
, so, as before,
, and
1 −t RC V0 −t RC dQ = CV0 = . I=− e e dt RC R 1 In capacitor, at start. = W CV02 . (b) 2 ∞ ∞ 2 ∞ V0 2 −2 t RC ∆E = Pdt = I Rdt = e dt R Dissipated 0 0 0 In resistor. ∞ V02 RC −2t RC 1 2 = − = CV0 . e R 2 0 2
∫
18 November 2002
∫
∫
Physics 217, Fall 2002
14
Physics 217, Fall 2002
1
S
σ 2
1
Constitutive relations in E&M Constitutive relations are relationships between parameters of a theory that are derived empirically or that depend for their calculation on physics that is not part of the theory. We have met two constitutive relations so far and are about to meet a third: P = χeE M = χm H J =σE The last one is called Ohm’s Law. In general, the proportionality “constant” σ is a second rank tensor, as are the susceptibilities, but many common media are “linear” in the sense that the conductivity σ can be considered a scalar. 18 November 2002
Physics 217, Fall 2002
2
Ohm’s Law Suppose a chunk of material with cross-sectional area A and length L has equipotential ends, with potential difference V. Suppose furthermore that this gives rise to a uniform E in the chunk. Then the current that flows is I = JA = σ EA V σA = V ≡ R L 18 November 2002
A
E
L
. Physics 217, Fall 2002
3
Ohm’s Law (continued) Ohm’s law should strike you as strange, at first, because you also know that J = ρv = σ E Suppose J is constant. Then v and E should be constant. But if E is constant, charges should accelerate at a = qE/m, rendering it impossible to have a constant v. So which relation is wrong? Neither. What happens is that collisions between free charges in a current (like electrons) with fixed or slowermoving charges (like the ions the electrons leave behind), and other free current carriers, keep the acceleration from going on for very long. 18 November 2002
Physics 217, Fall 2002
4
Ohm’s Law (continued) In collisions with the ions, the kinetic energy gained by the free carriers from the field is largely transferred to the ions, and the electron starts over, like a car driving down Intercampus Drive, stopping at all the speed bumps, accelerating in between. We therefore reconcile Ohm’s Law with the definition of current density by supposing that the collision process results in a well-defined average velocity, and writing J = ρv . In fact, we don’t need to suppose; we can show that that’s the way it is, in a crude classical model of what is mostly a quantum phenomenon, as follows. 18 November 2002
Physics 217, Fall 2002
5
Collisions and drift velocity Consider a chunk of metal with N carriers in it and an electric field E present. If it has been a time ti since particle i last suffered a collision, and if it left that collision at speed vi , then the momentum of this particle is pi = mvi + qEti
.
So a snapshot of the metal would reveal an average value of carrier momentum given by 1 N 1 N p = mv = ∑ pi = ∑ ( mvi + qEti ) . N i =i N i =1 The “starting” speeds vi are endowed by collisions with the fixed ions; their energy, in turn, comes from the thermal energy (heat) of the medium. 18 November 2002
Physics 217, Fall 2002
6
Collisions and drift velocity (continued) Thermal motions in a solid or liquid are (essentially) random in magnitude and direction, so if N is large, N
∑ mvi = 0
; thus,
i =1
qE N qE v= ti = t ∑ mN i =1 m
,
Drift velocity
nq 2 t E ≡σE , and J = ρ v = nqv = m where n is the number density of carriers. Clearly J should be linear in E if the carrier velocities are mostly thermal, and if collisions take place. 18 November 2002
Physics 217, Fall 2002
7
Collisions and heating Energy is continuously added to such a medium, as the kinetic energy gained by carriers between collisions is passed on to the fixed ions during the collisions. Independent of the nature of the collisions, if a carrier passes through a potential difference V, it picks up kinetic energy qV. All this is dissipated in collisions. If a total current I flows, then the power dissipated in collisions is d P = NqV = IV = I 2 R = − ∫ J ⋅ da ∫ E ⋅ dA = σ ∫ E2 dτ . dt
(
)
The collisional transfer of energy from carriers accelerated by fields, to the medium in which the current flows, is called resistance. 18 November 2002
Physics 217, Fall 2002
8
Resistance For the wire with cross-sectional area A, length L, and conductivity σ, the resistance turned out to be L ρL R= = σA A
.
ρ = resistivity. Not to be confused with charge density.
Units: in cgs,
1 [σ ] = sec
[ ρ ] = sec
sec [ R] = cm
.
In MKS, volt ≡Ω [ R] = amp
18 November 2002
1 [σ ] = Ωm
Physics 217, Fall 2002
[ρ] = Ω m
.
More commonly, Ω cm. 9
Examples: resistance and current Griffiths problem 7.3: (a) Two metal objects are embedded in weakly conducting material of conductivity σ. What is the resistance between them, in terms of their capacitance? (b) Suppose you connect a battery between them and charge them up to a potential difference V0. If you then disconnect the battery, the charge will leak away. Find the time dependence of the potential difference between the metal objects. 18 November 2002
Physics 217, Fall 2002
1
σ 2
10
Examples: resistance and current (continued) (a) Enclose the positively charged conductor with a surface S; then I=
v∫
J ⋅ da = σ
S
V = 4πσ CV R 1 ⇒R= 4πσ C
18 November 2002
v∫
E ⋅ da = 4πσ Q
1
S
S
ε0 = σ C in MKS .
Physics 217, Fall 2002
σ 2
11
Examples: resistance and current (continued) (b) Q = CV = CIR dQ Q = −I = − dt RC Q
∫
Q0
t
1
S
1 dQ′ =− dt ′ ∫ Q′ RC 0
Q ( t ) = Q0 e −t Q0 = V (t ) C
RC
−t e
= V0 e −4πσ t 18 November 2002
σ RC
= V0 e −t
RC
2
= V0 e −σ t ε 0 in MKS . Physics 217, Fall 2002
12
Examples: resistance and current (continued) Griffiths problem 7.2: A capacitor C is charged up to potential difference V0; at time t = 0 it is connected to a resistor R, and begins to discharge. (a) Determine the charge on the capacitor, and the current through the resistor, as a function of time. (b) What was the original energy stored in the capacitor? Show that the heat delivered to the resistor is equal to the energy lost by the capacitor. 18 November 2002
Physics 217, Fall 2002
C
R
13
Examples: resistance and current (continued) Q (a) V = = IR ; C Q ( t ) = Q0 e −t
RC
dQ Q = −I = − dt RC = CV0 e −t
RC
, so, as before,
, and
1 −t RC V0 −t RC dQ = CV0 = . I=− e e dt RC R 1 In capacitor, at start. = W CV02 . (b) 2 ∞ ∞ 2 ∞ V0 2 −2 t RC ∆E = Pdt = I Rdt = e dt R Dissipated 0 0 0 In resistor. ∞ V02 RC −2t RC 1 2 = − = CV0 . e R 2 0 2
∫
18 November 2002
∫
∫
Physics 217, Fall 2002
14
