Physics 122B
Electricity and Magnetism
Lecture 8 (Knight: 28.1 to 28.5)
Review of Material to be examined in
Midterm 1
Current and Resistance
April 11, 2007
Martin Savage
Lecture 8 Announcements
The Midterm examination is this Friday during lecture time.
You should bring with you to the Exam a good calculator, a Scantron
sheet, and one sheet of 8.5”x11” paper on which you have written
anything you wish on both sides.
No cell phones or wireless communication allowed.
 Lecture HW Assignment #3 will be been posted on Tycho and is due
at 10 PM on Wednesday, April 18
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Summaries
Knight,
Chapters 25, 26 and 27
Summary: Chapter 25 (1)
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Summary: Chapter 25 (2)
Charges
Fields
Not really true…
semi-conductors
Not always true…
Only if object
is polarizable
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Chapter 26 Summary (1)
The E field exists at each point in space
Non-zero values of the E-field are induced by
electric charges
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Chapter 26 Summary (2)
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Chapter 27 Summary (1)
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Chapter 27 Summary (2)
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Chapter 27 Summary (3)
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The Electron Current
We start with some “thought
experiments” on a simple system. We
have a parallel plate capacitor that
has been charged, e.g. with glass and
plastic rods. Now we connect the
plates with a wire. What happens?
The plates quickly become neutral,
and we say that the capacitor has
been discharged.
Further study shows that while the
discharge is taking place, the wire
gets warm, a light bulb can be made
to glow, and a compass needle can be
deflected. These are indicators of
current flow in the wire.
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Charge Carriers and Inertia
Does the discharge occur because positive charges are moving to the
negative plate, or because negative charges are moving to the positive plate?
We have asserted that the current in metals is
caused by the flow of negative electrons. The first
direct evidence that this was the case was provided in
1916 by the Tolman-Stewart experiment, which showed
that negative charges “go to the bottom” of an
accelerated conductor.
We model a metallic conductor as a rigid lattice of
positive charges pervaded by a “sea” of conduction
electrons, 1 per atom, that move freely in the material.
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The Electron Current
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In a metallic conductor in electrostatic
equilibrium, the conduction electrons move
around quite rapidly, but there is no net
movement of charge.
This can be changed by “pushing” on the
sea of electrons with an electric field,
thereby causing the entire sea of
electrons to move in a particular direction,
like a gas or liquid flowing through a pipe.
The net motion, the “drift speed” vd, is
superposed on the random thermal motions
of the individual electrons, and it is very
slow, typically around 10-4 m/s.
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We define the electron
current i as the number of
electrons Ne that pass through
a cross section of wire or other
conductor in a time interval Dt.
In other words: Ne = i Dt.
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Current and Drift Velocity
If the electrons have an average drift
speed vd, then on the average in a time
interval Dt they would travel a distance Dx in the wire,
where Dx = vd Dt. If the wire has cross sectional area
A and there are n electrons per unit volume in the wire,
then the number of electrons moving through the cross
sectional area in time Dt is Ne = n A Dx = n A vd Dt = i Dt .
Therefore,
This table gives n
for various metals.
i  nAvd
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Question 1
Which wire has the largest electron current?
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Conservation of Current
Question: An electron current iA flows
to the light bulb, passing point A, where
it delivers some energy and makes the
bulb glow. How much electron current iB
then passes point B?
Answer: All of it! iA=iB. Reason: the
electrons don’t have anywhere else to go.
What goes to the bulb must return from
the bulb. The bulb cannot “use up” the
electrons.
Plumber’s Analogy 1: If water flows
into a constant diameter pipe at 2.0 m/s,
it must flow out of the pipe at the same
speed. It cannot “pile up” in the pipe.
This principle is called “Conservation of Current”.
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A Puzzle
We discharge a capacitor that
has been given a charge of Q = 16 nC,
using a copper wire that is 2 mm in
diameter and has a length of L = 20
cm. Assume that the electron drift
speed is vd = 10-4 m/s.
How long does it take to discharge
the capacitor? (Note that L/vd =
0.2m/10-4 m/s = 2000 s = 33.3 min.)
Points to consider:
1. The wire is already full of electrons.
2. The wire contains about 5x1022
conduction electrons.
3. Q = 16 nC requires about 1011 electrons.
4. A length L’ of wire that holds 16 nC
of conduction electrons is 4x10-13 m.
5. L’/vd = 4x10-9 s = 4 ns. That is roughly the discharge time.
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Creating a Current…
Non-Static Situation
Suppose you want to slide a book
across a table. If you give it a quick push,
it moves but slows due to friction as soon
as you remove your hand, and its kinetic
energy becomes heat. The only way to
make the book move at a constant speed
is to continue pushing it.
The sea of conduction electrons is
similar to the book. If you push the
electrons, you create a current, but they
are not moving in vacuum, and collisions
with other electrons and atoms soon
dissipates their kinetic energy as heat.
The only way to maintain the current of
electrons is to push them, using an
electric field.
An electron current is a non-equilibrium motion in an E field.
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Establishing the
Electric Field in a Wire (1)
The figure shows two metal wires
attached to the plates of a parallel plate
capacitor, with their ends close together but
not touching. The wires are conductors, so
some of the charge from the capacitor
plates spreads out along the wires as surface
charge. E=0 inside all conductors..plates and
wires.
Now we connect the wires. What
happens? The surface electrons can move,
and do so. In ~10-9 s the sea of electrons
shifts slightly, and the surface charges are
rearranged into a non-uniform distribution of
charges, as shown. Surface charges near the
+ and – plates reflect these charges, but
surface charges become near-neutral halfway along the wire
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Establishing the
Electric Field in a Wire (2)
The figure shows the region of the wire near the neutral midpoint. The
surface charge rings become more positive to the left and more negative
to the right.
In Chapter 26, we found that a ring of charge makes an on-axis E field that:
1. Points away from a positive ring and toward a negative ring;
2. Is proportional to the net charge of the ring;
3. Decreases with distance from the ring.
The non-uniform surface charge distribution creates an E field
inside the wire. This pushes the electron current through the
wire
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Example: The Surface Charge
on a Current-Carrying Wire
Two 2.0 mm diameter charged
rings, with charges ±Q, are 2.0 mm
apart.
What value of Q causes the
electric field at the midpoint to be
E=0.010 N/C?
E 
1
zQ
4 0  z 2  R 2 3/ 2
4 0  z  R
2
Q
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z

 5 103 N/C
2 3/ 2
E  1.6 1018 C  10e
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Turning the Corner
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Question 2
Which surface charge distribution will
produce the largest electron current?
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A Model of Conduction (1)
Suppose E  0 : K  12 mv 2  32 kT
gives v  105 m/s. However, v  0.
Now turn on an E field. The straight-line
trajectories become parabolic, and because of the
curvature, the electrons begin to drift in the
direction opposite E, i.e., “downhill”.
ax=F/m=eE/m so
vx=vix+ axDt = vix+ Dt eE/m
This acceleration increases an electrons
kinetic energy until the next collision, a “friction”
that heats the wire….energy is imparted to the
atoms of the lattice.
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A Model of Conduction (2)
One Collision
Time-History over many Collisions
eE ___ eE ___ e
vd  vx  vix 
Dt 
Dt 
E
m
m
m
___
___
i  nAvd 
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ne A
E
m
Physics 122B - Lecture 8
Collision is independent of E
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Example: The Electron Current
in a Copper Wire
The mean time between collisions for electrons in room-temperature
copper is  = 2.5x10-14 s.
What is the electron current in 2.0 mm diameter copper wire where
the internal field strength is E = 0.010 N/C?
e
(1.60 10-19 C)(2.5 1014 s)(0.010 N/C)
5
vd  E 

4.4

10
m/s
-31
m
(9.1110 kg)
n  8.5 1028 m-3 ; A   r 2   (1.0 103 m)2  3.14 10-6 m2
i  nAvd  1.2 1019 electrons/s
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Batteries and Current
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Electrical Current
 dQ

I 
, in the direction of E 
 dt

1 ampere  1 A
 1 coulomb per second
= 1 C/s
DQ  I Dt
DQ eN e
I

 ei
Dt
Dt
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Example: The Current in a
Copper Wire
From the previous example,
i  nAvd  1.2 1019 electrons/s
What is the current I?
I  ei  (1.60 10-19 C)(1.2 1019 s-1 )  1.9 A
How much charge passes through the wire in an hour?
D Q  I Dt  (1.9 A)(3600 s)  6840 C
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Current and Electrons
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The Current Density in a Wire
I  ei  nevd A
I
J  current density   nevd
A
Example: A current of 1.0 A passes through a 1.0 mm diameter
aluminum wire. What is the drift speed of the electrons in the wire?
I
I
(1.0 A)
6
2
J  2 

1.3

10
A/m
A r
 (5.0 104 m)2
J
(1.3 106 A/m2 )
4
vd 


1.3

10
m/s
28
-3
19
ne (6.0 10 m )(1.60 10 C)
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End of Lecture 8
 Before the next lecture, read Knight: 29.1 to 29.4.
 Exam Friday….study everything in chaps 25,26,27
 Will NOT cover material on currents and non-static
situations.
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