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  • 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.

When positive charges move in this direction,

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B1 中級

Lec 09: 電流,電阻率和歐姆定律 | 8.02 電和磁 (Walter Lewin) (Lec 09: Currents, Resistivity and Ohm's Law | 8.02 Electricity and Magnetism (Walter Lewin))

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    Cheng-Hong Liu 發佈於 2021 年 01 月 14 日
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