# 8.02x – Lect 9 – Electric Currents, Resistivity, Conductivity, Ohm’s Law

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 this 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 feel

the electric field, 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 a speed,

between these collisions, which we call the drift velocity,

which is a times tau. It’s just 801. And so a equals F divided by m e. F is e E [misspoken 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 this 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,

this electric field E. When you look at this here, that really depends

only on the properties of my 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 for 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 for 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 to ten to the minus sixteen. So rho, now, the resistivity, is something like

ten to the twelve to ten to the plus sixteens and if I take ten to the fourteen,

just I gra– 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,

of 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–

I 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, which goes 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 0.08 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 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 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 voltage–

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. 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. Ehm– 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 towards– 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 gets 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 give you this resistance

and I give 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. The number of charges 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 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 step leader 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– [cracking sound] 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 a 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. 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 VanderGraaf,

beating 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 an

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 VanderGraaff– the VanderGraaff, 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 VanderGraaff,

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 VanderGraaff, but the van der Graff has only

ten microCoulombs on it. 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, then the conductivity of many materials

can increase 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 802, or he is wearing very,

very, special shoes. See you on wednesday.