When positive charges move in
this direction,
then per definition,
we say the current goes in this
direction.
When negative charges go in
this direction,
we also say the current goes in
that direction,
that's just our convention.
If I apply a potential
difference over a conductor,
then I'm going to create an
electric field in that
conductor.
And the electrons -- there are
free electrons in a conductor --
they can move,
but the ions cannot move,
because they are frozen into
the solid, into the crystal.
And so when a current flows in
a conductor, it's always the
electrons that are responsible
for the current.
The electrons fuel the electric
fields, and then the electrons
try to make the electric field
zero, but they can't succeed,
because we keep the potential
difference over the conductor.
Often, there is a linear
relationship between current and
the potential,
in which case,
we talk about Ohm's Law.
Now, I will try to derive Ohm's
Law
in a very crude way,
a poor man's version,
and not really one hundred
percent kosher,
it requires quantum mechanics,
which is beyond the course --
beyond this course -- but I will
do a job that still gives us
some interesting insight into
Ohm's Law.
If I start off with a
conductor, for instance,
copper, at room temperature,
three hundred degrees Kelvin,
the free electrons in copper
have a speed,
an average speed of about a
million meters per second.
So this is the average speed of
those free electrons,
about a million meters per
second.
This in all directions.
It's a chaotic motion.
It's a thermal motion,
it's due to the temperature.
The time between collisions --
time  between the collisions --
and this is a collision of the
free electron with the atoms --
is approximately -- I call it
tau --
is about three times ten to the
minus fourteen seconds.
No surprise,
because the speed is enormously
high.
And the number of free
electrons in copper per cubic
meter, I call that number N,
is about ten to the
twenty-nine.
There's about one free electron
for every atom.
So we get twen- ten to the
twenty-nine free electrons per
cubic meter.
So now imagine that I apply a
potential difference  piece of
copper -- or any conductor,
for that matter -- then the
electrons will experience a
force which is the charge of the
electron, that's my little E
times the electric field that
I'm creating,
because I apply a potential
difference.
I realize that the force and
the electric field are in
opposite directions for
electrons, but that's a detail,
I'm interested
in the magnitudes only.
And so now these electrons will
experience an acceleration,
which is the force divided by
the mass of the electron,
and so they will pick up,
eh, speed, between these colli-
collisions, which we call the
drift velocity,
which is A times tau,
it's just eight oh one.
And so A equals F divided by M.
E F is in the A,
so we get E times E divided by
the mass of the electrons,
times tau.
And that is the
the drift velocity.
When the electric field goes
up, the drift velocity goes up,
so the electrons move faster in
the direction opposite to the
current.
If the time between collisions
gets larger, they -- the
acceleration lasts longer,
so also, they pick up a larger
speed, so that's intuitively
pleasing.
If we take a specific case,
and I take, for instance,
copper,
and I apply over the -- over a
wire -- let's say the wire has a
length of 10 meters -- I apply a
potential difference I call
delta V, but I could have said
just V -- I apply there a
potential difference of ten
volts, then the electric field
-- inside the conductor,
now -- is about one volt per
meter.
And so I can calculate,
now, for that specific case,
I can calculate what the drift
velocity would be.
So the drift velocity of those
free electrons would be the
charge of the electron,
which is one point six times
ten to the minus nineteen
Coulombs.
The E field is one,
so I can forget about that.
Tau is three times ten to the
minus fourteen,
as long as I'm room
temperature, and the mass of the
electron is about ten to the
minus thirty kilograms.
And so, if I didn't slip up,
I found that this is five times
ten to the minus three meters
per second, which is half a
centimeter per second.
So imagine, due to the thermal
motion, these free electrons
move with a million meters per
second.
But due to this electric field,
they only advance along the
wire slowly, like a snail,
with a speed on average of half
a centimeter per second.
And that goes very much against
your and my own intuition,
but this is the way it is.
I mean, a turtle would go
faster than these electrons.
To go along a ten-meter wire
would take half hour.
Something that you never
thought of.
That it would take a half hour
for these electrons to go along
the wire if you apply potential
difference of ten volts,
copper ten meters long.
Now, I want to massage this
further,
and see whether we can somehow
squeeze out Ohm's Law,
which is the linear relation
between the potential and the
current.
So let me start off with a wire
which has a cross-section A,
and it has a length L,
and I put a potential
difference
over the wire,
plus here, and minus there,
potential V,
so I would get a current in
this direction,
that's our definition of
current, going from plus to
minus.
The electrons,
of course, are moving in this
direction, with the drift
velocity.
And so the electric field in
here, which is in this
direction, that electric field
is approximately V divided by L,
potential difference divided by
distance.
In one second,
these free electrons will move
from left to right over a
distance V D meters.
So if I make any cross-section
through this wire,
anywhere, I can calculate how
many electrons pass through that
cross-section in one second.
In one second,
the volume that passes through
here, the volume is V D times A
but the number of free
electrons per cubic meter is
called N, so this is now the
number of free electrons that
passes, per second,
through any cross-section.
And each electron has a charge
E, and so this is the current
that will flow.
The current,
of course, is in this
direction, but that's a detail.
If I now substitute the drift
velocity, which we have here,
I substitute that in there,
but then I find that the
current -- I get a E squared,
the charge squared,
I get N, I get tau,
I get downstairs,
the mass of the electron,
and then I get A times the
electric field E.
Because I have here,
is electric field E.
When you look at this here,
that really depends only on the
properties of by substance,
for a given temperature.
And we give that a name.
We call this sigma,
which is called conductivity.
Conductivity.
If I calculate,
for copper, the conductivity,
at room temperature,
that's very easy,
because I've given you what N
is, on the blackboard there,
ten to the twenty-nine,
you know what tau is at room
temperature, three times ten to
the minus fourteen,
so for copper,
at room temperature,
you will find about ten to the
eighth.
You will see more values fro
sigma later on during this
course.
This is in SI units.
I can massage this a little
further, because E is V divided
by L,
and so I can write now that the
current is that sigma times A
times V divided by L.
I can write it down a little
bit differently,
I can say V,
therefore, equals L divided by
sigma A, times I.
And now, you're staring at
Ohm's Law, whether you like it
or not,
because this is what we call
the resistance,
capital R.
We often write down rho for one
over sigma, and rho is called
the resistivity.
So either one will do.
So you can also write down --
you can write down V equals I R,
and this R, then,
is either L divided by sigma A,
or L times rho -- let me make
it a nicer rho -- divided by A.
That's the same thing.
The units for resistance R is
volts per ampere,
but we call that ohm.
And so the unit for R is ohm.
And so if you want to know what
the
unit for rho and sigma is,
that follows immediately from
the equations.
The unit for rho is then
ohm-meters.
So we have derived the
resistance here in terms of the
dimensions -- namely,
the length and the
cross-section -- but also in
terms of the physics on an
atomic scale,
which, all by itself,
is interesting.
If you look at the resistance,
you see it is proportional with
the
length of your wire through
which you drive a current.
Think of this as water trying
to go through a pipe.
If you make the pipe longer,
the resistance goes up,
so that's very intuitively
pleasing.
Notice that you have A
downstairs.
That means if the pipe is
wider, larger cross-section,
it's also easier for the
current to flow,
it's easier for the water to
flow.
So that's also quite pleasing.
Ohm's Law, also,
often holds for insulators,
which are not conductors,
even though I have derived it
here for conductors,
which have these free
electrons.
And so now, I want to make a
comparison between very good
conductors, and very good
insulators.
So I'll start off with a -- a
chunk of material ,
cross-sectional area A -- let's
take it one millimeter by one
millimeter -- so A is ten to
the minus six square meters.
So here I have a chunk of
material, and the length of that
material L is one meter.
Put a potential difference over
there, plus here,
and minus here.
Current will start to flow in
this direction,
electrons will flow in this
direction.
The question now is,
what is the resistance of this
chunk of material?
Well, very easy.
You take these equations,
you know L and A,
so if I tell you what sigma is,
then you can immediately
calculate what the resistance
is.
So let's take,
first, a good conductor.
Silver and gold and copper are
very
good conductors.
They would have values of
sigma, ten to the eight,
we just calculated for copper,
you've seen in front of your
own eyes.
So that means rho would be ten
to the minus eight,
it's one over sigma.
And so in this particular case,
since A is ten to the minus
six, the resistance R is simply
ten to
the sixth times rho.
Because L is one meter.
So it's very easy -- resistance
here, R, is ten to the minus two
ohms.
One-hundredth of an ohm.
For this material if it were
copper.
Let's now take a very good
insulator.
Glass is an example.
Quartz, porcelain,
very good insulators.
Now, sigma, the conductivity,
is extremely low.
They vary somewhere from ten to
the minus twelve through ten to
the minus sixteen.
So rho, now,
the resistivity,
is something like ten to the
twelve to twelve to the plus
sixteens, and if I take ten to
the fourteen,
just I grab -- I have to grab a
number -- then you'll find that
R, now, is ten to the twenty
ohms.
A one with twenty zeros.
That's an enormous resistance.
So you see the difference --
twenty-two orders of magnitude
difference between a good
conductor and a good insulator.
And if I make this potential
difference over the wire,
if I make that one volt,
and if I apply Ohm's Law,
V equals I R,
then I can also calculate the
current that is going to flow.
If I R is one,
then the current here is
hundred amperes,
and the current here is ten
to the minus twenty amperes,
an insignificant  current,
ten to the minus twenty
amperes.
I first want to demonstrate to
you that Ohm's Law sometimes
holds, I will do a
demonstration,
whereby you have a voltage
supply  -- put a V in here --
and we change the voltage in a
matter of a few seconds from
zero to four volts.
This is the plus side,
this is the minus side,
I have connected it here to a
resistor which is fifty ohms --
we use this symbol for a
resistor -- and here is a
current meter.
And the current meter has
negligible resistance,
so you can ignore that.
And I'm going to show you on an
oscilloscope -- we've never
discussed an oscilloscope,
but maybe we will in the future
-- I'm going to show you,
they are projected -- the
voltage unintelligible go from
zero to four,
versus the current.
And so it will start here,
and by the time we reach four
volts, then we would have
reached a current of four
divided by fifty,
according to Ohm's Law,
I will write down just four
divided by fifty amperes,
which is point
oh eight amperes.
And if Ohm's Law holds,
then you would find a straight
line.
That's the whole idea about
Ohm's Law, that the potential
difference, linearly
proportional to the current.
You double the potential
difference, your current
doubles.
So let's do that,
let's take a look at that,
you're going to see that there
-- and I have to change my
lights so
that you get a good shot at it
-- oh, it's already going.
So you see, horizontally,
we have the current,
and vertically,
we have the voltage.
And so it takes about a second
to go from zero to four -- so
this goes from zero to four
volts -- and you'll see that the
current is beautifully linear.
Yes, I'm blocking it -- oh,
no, it's my reflection,
that's interesting.
Ohm's Law doesn't allow for
that.
So you see how beautifully
linear it is.
So now, you may have great
confidence in Ohm's Law.
Don't have any confidence in
Ohm's Law.
The conductivity sigma is a
strong function of the
temperature.
If you increase the
temperature, then the time tau
between collisions goes down,
because the speed of these free
electrons goes up.
It's a very strong function
of temperature.
And so if tau goes down,
then clearly,
what will happen is that the
conductivity will go down.
And that means rho will go up.
And so you get more resistance.
And so when you heat up a
substance, the resistance goes
up.
A higher temperature,
higher resistance.
So the moment that the
resistance R becomes a function
of the
temperature,
I call that a total breakdown
of V equals I R,
a total breakdown of Ohm's Law.
If you look in your book,
they say, "Oh,
no, no, no, that's not a
breakdown.
You just have to adjust the re-
the resistance for a different
temperature." Well,
yes, that's an incredible poor
man's way of saving a law that
is a very bad law.
Because the temperature itself
is a function of current,
the higher the current the
higher the temperature.
And so now, you get a ratio,
V divided by I,
which is no longer constant.
It becomes a function of the
current.
That's the end of Ohm's Law.
And so I want to show you that
if I do the same experiment that
I did here, but if I replace
this by a light bulb of fifty
ohms -- it's a very small light
bulb, resistance when it is hot
is fifty ohms,
when it is cold,
it is seven ohms.
So R cold of the light bulb is
roughly seven ohms,
I believe, but I know that when
it is hot, it's very close to
the fifty ohms.
Think it's a little lower.
What do you expect now?
Well, you expect now,
that when the resistance is low
in the beginning,
you get this,
and then when the resistance
goes up, you're going to get
this.
I may end up a little higher
current, because I think the
resistance is a
little lower than fifty ohms.
And if you see a curve like
this, that's not linear anymore.
So that's the end of Ohm's Law.
And that's what I want to show
you now.
So, all I do is,
here I have this little light
bulb -- for those of you who sit
close, they can actually see
that light bulb  start glowing,
but that's not important,
I really want you to see that V
versus I is no longer linear,
there you go.
And you see,
every time you see this light
bulb go on, it heats up,
and during the heating up,
it, um, the resistance
increases.
And it's the end of Ohm's Law,
for this light bulb,
at least.
It was fine for the other
resistor, but it was not fine
for this light bulb.
There is another way that I can
That is the resistivity,
show you that Ohm's Law is not
always doing so well.
I have a hundred twenty-five
volt power supply,
so V is hundred and twenty-five
volts -- this is the potential
difference -- and I have a light
bulb, you see it here,
that's the light bulb -- the
resistance of the light bulb,
cold, I believe,
is twenty-five ohms,
and hot, is about two hundred
and fifty ohms.
A huge difference.
So if the  resistance -- if I
take the cold resistance,
then I would get five amperes,
but by the time that the bulb
is hot, I would only get half an
ampere.
It's a huge difference.
And what I want to show you,
again with the oscilloscope,
is the current as a function of
time.
When you switch on a light
bulb, you would expect,
if Ohm's Law holds,
that when you switch on the
current -- or switch on the
voltage, I should say -- that
you see this.
This is then your five amperes.
And that it would stay there.
That's the whole idea.
Namely, that the voltage
divided by the current remains a
constant.
However, what you're going to
see is like this.
Current goes up,
but then the resistance goes
down, then the resistance goes
up, when the current goes up,
the resistance goes up,
and then therefore the current
will go down,
and will level off at a level
which is substantially below
this.
So you're looking there --
you're staring at the breakdown
of Ohm's Law.
And so that's what I want to
show you now.
So, here we need a hundred and
twenty five volts -- and there
is the light bulb,
and when I throw this switch,
you will see the pattern of the
current versus time -- you will
only see it once,
and then we freeze it with the
oscilloscope -- turn this off --
so look closely,
now.
There it is.
Forget these little ripple that
you see on it,
it has to do with the way that
we produce the hundred and
twenty five volts.
And so you see here,
horizontally,
time, the time between two
adjacent vertical lines is
twenty milliseconds.
And so, indeed,
very early on,
the current surged toward -- to
a very high value,
and then the filament heats up,
and so the resistance goes up,
the light bulb,
and the current just goes back
again.
From the far left to the far
right on the screen is about two
hundred milliseconds.
That's about two tenths of a
second.
And here you get a current
level which is way lower than
what you get there.
That's a breakdown of Ohm's
Law.
It is actually very nice that
resistances go up with light
bulbs
when the temperature goes up.
Because, suppose it were the
other way around.
Suppose you turn on a light
bulb, and the resistance would
go down.
Light bulb got hot,
resistance goes down,
that means the current goes up.
Instead of down,
the current goes up.
That means it gets hotter.
That means the resistance goes
even further down.
That means the current goes
even further up.
And so what it would mean is
that every time you turn on a
light bulb, it would,
right in front of your eyes,
destruct itself.
That's not happening.
It's the other way around.
So, in a way,
it's fortunate that the
resistance goes up when the
light bulbs get hot.
All right.
Let's now be a little bit more
qualitative on some  networks of
resistors, and we'll have you do
a few problems like that,
whereby we just will assume,
naively, that Ohm's Law holds.
In other words,
we will always assume that the
values for the resistances that
we give you will not change.
So we will assume that the heat
that is produced will not  play
any important role.
So we will just use Ohm's Law,
for now, and if you can't use
it, we will be very specific
about that.
So suppose I have here,
between point A and point B,
suppose I have two resistors,
R one and R two.
And suppose I apply a potential
difference between A and B,
that this be plus,
and this be minus,
and the potential difference is
V.
And you know V,
this is known,
I give you V,
I gave you this resistance,
and I gave you that one.
So I could ask you now,
what is the current that is
going to flow?
I could also ask you,
then, what is the potential
difference over this resistor
alone --
which I will call V one -- and
what is the potential difference
over the second resistor,
which I call V two?
Very straightforward question.
Well, you apply,
now, Ohm's Law,
and so between A and B,
there are two resistors,
in series.
So the current has to go
through both,
and so the potential difference
V, in Ohm's Law,
is now the total current times
R one plus R two.
Suppose these two resistors
were the same,
they had the same length,
same cross-sectional area.
If you put two in series,
you have twice the length.
Well, so, twice the length,
remember, resistance is
linearly proportional with the
length of a wire,
and so you add them up.
So now you know R one and you
know R two, you know V,
so you already know the
current, very simple.
You can also apply Ohm's Law,
as long as it holds,
for this resistor alone.
So then you get that V one
equals I times R one,
so now you have the voltage
over this resistor,
and of course,
V two must be the current I
times R two.
And so you have solved your
problem.
All the questions that I asked
you, you have  the answers to.
We could now have a slightly
different problem,
whereby point A is here,
but now we have a resistor
here,
which is R one,
and we have here,
R two.
This is point B,
and this is R two.
And the potential difference is
V, that is, again,
given, and now I could ask you,
what, now, is the current that
will flow here?
And then I can also ask you,
what is the current that would
go through one -- resistor one,
and what is the current that
could go through
resistor two?
And I would allow you to use
Ohm's Law.
So now you say,
"Aha!
The potential difference from A
to B going this route,
that potential difference,
is V, that's a given." So V
must now be I one times R one.
That's Ohm's Law,
for this upper branch.
But, of course,
you can also go the lower
branch.
So the same V is also I two
times R two.
But whatever current comes in
here must split up between these
two, think of it as water.
You cannot get rid of charges.
The number of charges per
second that flow into this
juncture continue on,
and so I, the total current,
is I one plus I two.
And so now,
you see, you have all the
ingredients that you need to
solve for the current I -- for
the current I one,
and for the current I two.
And you can turn this into an
industry, you can make extremely
complicated networks of
resistors -- and if you were in
course six, you should love it
-- I don't like it at all,
so you don't have to worry
about it, you're not going to
get very complicated resistor
net- networks
from me -- but in course six,
you're going to see a lot of
them.
They're going to throw them --
stuff them down your throat.
The conductivity of a substan-
substance goes up if I can
increase the number of charge
carriers.
If we have  dry air,
and it is cold,
then the resistivity of cold,
dry air at
one atmosphere -- so rho for
air, cold,  dry,
one atmosphere -- cold means
temperature that we have outside
-- it's about four times ten to
the thirteen.
That is the resistivity of air.
It is  about what it is in
this room, maybe a little
lower, because the temperature
is a little higher.
If I heat it up -- the air --
then the conductivity will go
up.
Resistivity will go down,
because now,
I create oxygen and nitrogen
ions by heating up the air.
Remember when we had this
lightning, the unintelligible
came down, and we created a
channel full of ions and
electrons, that had a very low
resistivity,
a very high conductivity.
And so what I want to
demonstrate to you,
that when I create ions in this
room, that I can actually make
the conductivity of air go up
tremendously.
Not only will the electrons
move, but also the ions,
now, will start to move.
And the way I'm going to do
that is, I'm going to put charge
on the electroscope -- oh,
that is not so good --
no harm done.
I'm going to put charge on the
electroscope,
and you will see that the
conductivity of air is so poor
that it will stay there for
hours.
And then what I will do,
I will create ions in the
vicinity of the electroscope.
But let's first put some charge
on the electroscope.
I have here a glass rod  and
I'll put some charge on it.
OK, that's a lot of charge.
And, uh, the r- the air is
quite dry, conductivity is very,
very small, and so the charge
cannot go off through the air to
the surroundings,
to the earth.
But now I'm going to create
ions there by heating it up,
and I decided to do that with a
candle, because a candle is
very romantic,
as we all know.
So here I have this candle  --
look how well the charge is
holding, eh?
-- and here's my candle.
And I will bring the candle --
oh, maybe twenty centimeters
from the electroscope.
Look at it, look at it,
already going.
It's about fifteen centimeters
away.
I'll take my candle away,
and it stops again.
So it's all due to the fact
that I'm ionizing the air there,
creating free electrons as well
as ions, and they both
participate now in the current,
and the charge can flow away
from the electroscope through
the earth, because the
conductivity now is so much
higher.
I stop again,
and it stops.
You see in front of your eyes
how important the temperature
is, in this case,
the presence of the ions in the
air.
If I have clean,
distilled water -- I mean,
clean water.
I don't mean the stuff that you
get in Cambridge,
let alone did I mean the stuff
that is in the Charles River,
I mean clean water,
that has a pH of seven.
That means one out of ten to
the seven of the water molecules
is ionized, H plus and O H
minus.
The conductivity,
by the way, is not the result
of the
free electrons,
but is really the result of
these H plus and O H minus ions.
It's one of the cases whereby
not the -- the electrons are
maj- the major responsibility
for the current.
If I have add three percent of
salt, in terms of weight,
then all that salt will ionize,
so you get sodium plus and C L
minus ions, you increase the
number of ions by an enormous
factor.
And so the conductivity will
soar up
by a factor of three hundred
thousand, or up to a million,
because you increase the ions
by that amount.
And so it's no surprise then,
for you, that the conductivity
of seawater is a million times
higher -- think about it,
a million times higher -- than
the conductivity of distilled
water.
And I would like to give you
the number for water --
so this is distilled water  --
that is about two times ten to
the fifth ohm-meters.
There is another way that I can
That is the resistivity,
two times ten to the five
oh-meters.
I have here,
a bucket of distilled water.
I'll make a drawing for you on
the blackboard there.
So here is a bucket of
distilled water,
and in there,
is a copper plate,
and another copper plate,
and here is a light bulb,
and this will go straight  to
the outlet [wssshhht],
stick it in,
hundred ten volts.
This light bulb has eight
hundred ohm resistance when it
is hot.
You see the light bulb here.
You can calculate what this
resistance is between the two
plates, that's easy,
you have all the tools now.
If you know the distance,
it's about twenty centimeters,
and you know the surface area
of the plates,
because remember,
the resistance is inversely
proportional with A,
so you have to take that into
account -- and you take the
resistivity of water into
account, it's a trivial
calculation, you can calculate
what the resistance is of this
portion here.
And I found that this
resistance here is about two
megaohms.
Two million ohms.
So, when I plug this into a
wall, the current that will flow
is extremely low,
because it has to go through
the eight hundred ohms,
and through the two megaohms.
So you won't see anything,
the light bulb will not show
any
light.
But now, if I -- put salt in
here, if I really manage to put
three percent in weight salt in
here, then this two megaohm will
go down to two ohms,
a million times less.
So now, the light bulb will be
happy like a clam at high tide,
because two ohms here,
plus the eight hundred,
the two is insignificant.
And this is what I want to --
to demonstrate to you now,
the
enormous importance of
increasing ions.
I increased ions here by
heating the air,
now I'm going to increase the
ions by adding salt.
And so the first thing that I
will do is, I will stick this in
here.
There's the light bulb.
And I make a daring prediction
that you will see nothing.
There we go.
Nothing.
Isn't that amazing?
You didn't expect that,
right?
Physics works.
You see nothing.
If I take the plates out,
and touch them with each other,
what will happen?
There you go.
But this water has such a huge
resistance that the current is
too low.
Well, let's add some -- not
pepper -- add some salt.
Yes, there's salt in there.
It's about as much as I would
put on my eggs in the morning --
stir a little -- ah,
hey, look at that.
Isn't that amazing?
And when I bring them closer
together, it will become even
brighter, because L is now
smaller, the distance is
smaller.
I bring them farther apart,
it's amazing.
Just a teeny,
weeny little bit of salt,
about as much as I use on my
egg, let alone -- what the hell,
let's put everything in there
-- that's a unintelligible I put
everything, then,
of course, you go almost down
to the two ohms,
and the light bulb will be just
burning normally.
But even with that little bit
of salt, you saw the huge
difference.
My body is a fairly good
conductor -- yours too,
we all came out of the sea --
so we are almost
all water -- and therefore,
when we do experiments with
little charge,
like the van der Graf,
being a student,
then we have to insulate
ourselves very carefully,
putting glass plates under us,
or plastic stools,
to prevent that the charge runs
down to the earth.
In fact, the resistance,
my resistance between my body
and the earth is largely
dictated by the soles of my
shoe,
not by my body,
not by my skin.
But if you look at my soles,
then you get something like
this, and it has a certain
thickness, and this,
maybe one centimeter.
This, now, is L in my
calculation for the resistance,
because current may flow in
this direction,
so that's L.
Well, how large is my foot?
Let's say it's one foot long --
no pun implied -- and let's say
it's about ten centimeters wide.
So you can calculate what the
surface area A is,
you know what L is,
and if you know,
now what the resistivity is for
my sole, I can make a rough
guess, I looked up the material,
and I found that the
resistivity is about ten to the
tenth.
So I can now calculate what the
resistance is in this
direction.
And I found that that
resistance then,
putting in the numbers,
is about ten billion ohm.
And you will say,
"Wow!" Oh, it's four,
actually.
Well, big deal.
Four billion ohm.
So you will say,
"That's enormous resistance!"
Well, first of all,
I'm walking on two feet,
not on one, so if I would be
standing one the whole lecture,
it would probably be four
billion, but if I have two feet
on the ground,
it's really two billion,
you will say,
"Well, that's still extremely
large!" Well,
it may look large,
but it really isn't,
because all the experiments
that we are doing here in twenty
six one hundred,
you're dealing with very small
amounts of charge.
Even if you take the van der
Graff -- the van der Graff,
say, has two hundred thousand
volts -- and let's assume that
my resistance is two times ten
to the nine ohms,
two feet on the ground.
So when I touch the van der
Graff,
the current that would flow,
according to Ohm's Law,
would be hundred microamperes.
That means, in one second,
I can take hundred
microCoulombs of the van der
Graff, but the van der Graff has
only ten microCoulombs on in.
So the resistance of four
billion or two billion ohms is
way too low for these
experiments that we have been
doing in twenty six one hundred,
and that's why we use these
plastic stools,
and we use these glass plates
in order to make sure that the
current
is not draining off the  the
charge that we need for the
experiments.
I want to demonstrate that to
you, that, indeed,
even with my shoes on -- that
means, even with my two billion
ohm resistance to the ground --
that it will be very difficult
for me, for instance,
to keep charge on an
electroscope.
I'm going to put charge on this
electroscope by scuffing my
feet.
But, since I keep my -- I have
my shoes on, I'm not standing on
the glass plate,
the charge will flow through
me.
You can apply Ohm's Law.
And you will see that as I do
this -- I'm scuffing my feet now
-- that I can only keep that
electroscope charged as long as
I keep scuffing.
But the moment that I stop
scuffing, it's gone.
Start scuffing again,
that's fine,
but the moment that I
stop scuffing,
it goes off again.
Even though this resistance is
something like two billion ohms.
Let alone if I take my shoes
off.
I apologize for that.
If now I scuff,
I can't even get any charge on
the electroscope,
because now,
the resistance is so
ridiculously low,
I don't even have the two
billion ohms,
I can't even put any charge on
the electroscope.
It's always very difficult for
us to
do these experiments unless we
insulate ourselves very well.
And if, somehow,
the weather is a little damp,
we can very thin films of water
onto our tools,
and then the current can flow
off just through these very thin
layers of water.
That's why we always like to do
these experiments in winter,
so that the conductivity of the
air is very low,
no water anywhere.
Here you see a slide of a
robbery.
I have scuffed my feet across
the rug, and I am armed with a
static charge.
Hand over all your money,
or I'll touch your nose.
This person either never took
eight oh two,
or he is wearing very,
very special shoes.
See you on Wednesday.
