A set of answers to the
questions on Homework Assignment #1 were handed out today and
Homework Assignment #2 was collected. I'm not sure whether
I'll be able to get that back to you by Thursday.
Homework #3 was handed out
today. I originally planned on having only 2 questions but
added a third at the last minute. The new question (Question
#2 on the assignment) was worked out in class today. That
seemed like a good compromise.
Currents in the atmosphere and small ions
Today we'll mostly leave electric field, electrostatic potential
and those sorts of things behind for a few lectures and
concentrate more on air conductivity and currents in the
atmosphere. The first figure below is a recap of what we
covered earlier in the course (the first day of class).
Currents in the atmosphere and in a wire are compared at the
top of the figure. A battery drives a current through a wire
and resistor in the picture at right. Free electrons alone
transport the current.
In the atmosphere, at left, a large voltage difference between the
ground and the ionosphere drive the global electric circuit.
We normally speak of a current density (Amps/square meter) rather
than current. Positive and negatively charged "small ions"
move in opposite directions in response to an electric field and
carry the weak currents that flow through air. We'll
refer to this as a conduction current.
The small ions are created when ionizing radiation of some kind
strips an electron from an N2 or O2 molecule
leaving behind positively charged N2 or O2.
The electrons then rapidly attach to O2
creating a negatively charged O2 molecule
(they don't attach to N2). The
charged nitrogen and oxygen molecules are then surrounded by water
vapor molecules. Note that because of the structure of the
water molecule, the positively charged small ions end up with more
H2O molecules surrounding them and are slightly less
mobile. We'll look at the creation of small ions in more detail in
a later lecture.
Electrical mobility, Be
The electrical mobility is a
measure of how readily or quickly charge carriers will move in
an electric field. The drift velocity is simply the
electrical mobility multiplied by the strength of the electric
field.
Typical values of Be and drift velocity are shown above.
These drift velocities are much smaller than the random thermal
motions of atoms and molecules in air which are typically 100s of
meters per second.
Wind motions (typically a few or a few 10s of meters per second)
can also potentially transport charge, but for this to occur there
must be a gradient is space charge density (so that the amount of
charge being carried in by the wind is different from the charge
carried away by the wind). These would be referred to as
convection currents. Current density J
Next we'll see what the current density J is comprised of.
Here we consider N charge carriers per unit volume (each with
charge q) moving in the same direction and derive an expression
for J, the current density. ΔQ is the charge in the cylinder
above that will pass through dA in time Δt.
We will only concern ourselves with conduction currents
that are linearly proportional to the electric field.
This is known as Ohm's law and
the constant of proportionality is the conductivity.
Conductivity
We can use our newly derived expression for current
density together with the definition of electrical mobility to
obtain an expression for conductivity.
Let's go back to the earlier diagram where we determined
the expression for J. In air we will
generally need to sum contributions from charge carriers of
two polarities: +q moving in one direction and oppositely
charged carriers (-q) moving in the opposite direction.
In this case we've assumed that the number density of
positive and negative charge carriers is the same and that the
charge carriers carry the same charge and move at the same
speed.
The expression for conductivity would also need to include
both positive and negative charge carriers. The earlier
expression would have two terms:
We've made allowance for the fact that negatively charged
small ions generally have slightly higher electrical mobility
than positively charged small ions. Often we will assume
that the mobilities and concentrations of positive and
negative charge carriers are equal in which case the
conductivity would just be 2 N q Be. A note about conductivity units
Conductivity has units of 1/(ohms m). In some older
literature you may find this written as mhos/m (mhos is ohms
spelled backwards) though Siemens/m are now the officially
adopted units. Resistivity and resistance
Resistivity is the reciprocal of conductivity.
Resistivity is not the same as resistance, but it is easy to
relate the two.
I've just left λ (conductivity) in the work above to avoid any
confusion between the symbols for resistance and resistivity.
Here's the relationship between resistance and resistivity.
Homework Assignment #3 Question #2
This where we worked out the Question #2 on the new homework
assignment. Here's the question:
2.
(10 pts) What resistance would you measure across
the ends of a 1 cm x 1 cm x 10 cm
long piece of graphite assuming a resistivity of 7 x 10-6
ohm m.
Would you measure the same resistance if you connected
your ohmmeter across opposite sides of the piece of
graphite?
We worked out three possibilities in class (R is
resistance in ohms, Re is resistivity in ohm m)
I just left Re in the results and didn't substitute in the actual
value of Re. Even though we're dealing with the same piece
of graphite it does depend on how you connect the ohmmeter.
I included the cube because it gives you a clue about why the
resistance changes. Imaging that the cube is just a resistor
of 100 ohms. Figure (a) is just ten cubes connected in
series. The total resistance of ten 100 ohms resistors
connected in series would be 1000 ohms. Figure (b) is just
10 cubes connected in parallel. Ten 100 ohm resistors
connected in parallel would have a total resistance of 10
ohms.
Resistivities of some common materials are listed below
(conductivity is shown in the case of air)
Note how the conductivity of air is a function of air
temperature. A lightning return stroke will heat air to a
peak temperature of about 30,000 K for a short time.
Later in the semester we will look at some fast time resolved
measurements of lightning electric fields that were being used to
try to determine characteristics of the fast time varying currents
in lightning strokes. The measurements were made in a
location where propagation between the lightning source and E
field antenna was over salt water to preserve as much of the high
frequency content of the signals as possible. Continuity equation
The last item in class today was to derive a continuity
equation that relates changes in the space charge density in a
volume to net flow of charge into or out of the volume.
Starting at upper left, the charge, Q, contained in the volume is
just the volume integral of the volume space charge density.
A positive value for the surface integral of J at upper right
would mean net transport of charge out of the volume. This
would reduce the total charge in the volume and dQ/dt would be
negative.
We use the divergence theorem to write the surface integral of J
as a volume integral of the divergence of J. Then the two
volume integrals are set equal to each other. The only way
the two integrals can be equal for an arbitrary volume is if the
integrands themselves are equal. We end up with the
continuity equation in the middle of the figure.
In this class we'll often assume steady state conditions
and that J has just a z component. In that
case the divergence of J is zero and Jz
is constant with altitude.
Time needed to neutralize the charge on surface of the
earth
We didn't have
time for this last item in class today.
It returns to something we covered on the first
day of the course
A small portion of the earth's negatively charged surface is shown
above. The area shown contains a total charge Qo.
How quickly will the flow of current shown above to neutralize
all the charge on the earth's surface. J depends on
E. We consider the earth's surface to be a perfect conductor
so we can write E in terms of the surface charge density. J
is proportional to the charge on the earth's surface. So as
Q on the surface is neutralized Jz
will weaken
The first equation above shows how Q will change as current
flows to the ground. The product J x A is just current I
(Coulombs/second); multiplying by t gives the charge transported
to the surface (Coulombs) during time t.
We find that the Q on the surface will decay exponentially from
its initial value Qo
If we insert reasonable values for εo and λ we obtain
This is the time for the charge on the surface to decrease to
1/e of its original value, not the time needed to neutralize it
completely and is comparable to what we determined early in the
semester.
