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  • So today no new concepts, no new ideas,

  • you can release a little bit and I want to discuss with you

  • the connection between electric potential and electric fields.

  • Imagine you have an electric field here in space and that I

  • take a charge Q in my pocket, I start at position A and I

  • walk around and I return at that

  • point A. Since these forces are

  • conservative forces, if the electric field is a

  • static electric field, there are no moving charges,

  • but that becomes more difficult, then the forces are

  • conservative forces and so the work that I do when I march

  • around and coming back at point A must be zero.

  • It's clear when you uh look at the equation number three that

  • the potential difference between point A and point A is

  • obviously zero. I g- start at point A and I end

  • at point A and that is the integral in going from A back to

  • point A of E dot DL and that then has to be zero.

  • And we normally indicate such an integral with a circle which

  • means you end up where you started.

  • This is a line now this is not a closed surface

  • as we had in equation one. This is a closed line.

  • And so whenever we deal with static electric fields we can

  • add now another equation if we like that.

  • And that is if we have a closed line of E dot DL so we end up

  • where we started that then has to become zero.

  • Later in the course we will see that there are special

  • situations where we don't deal with static

  • fields when we don't have conservative E fields,

  • that that is not the case anymore.

  • But for now it is. So if we know the electric

  • field everywhere then we -- we can see equation number two then

  • we know the potential everywhere.

  • And so if we turn it the other way around if we knew the

  • potential everywhere you want to know what the electric field is

  • and that of course is possible. If you look at equation two and

  • three you see that the potential is the integral of the

  • electric field. So it is obvious that the field

  • must be the derivative of the potential.

  • Now when you have fields being derivative of potentials you

  • always have to worry about plus and minus signs,

  • whether um you have to pay MIT twenty-seven thousand dollars

  • tuition to be here or whether MIT pays you twenty-seven

  • thousand dollars tuition for coming here is

  • only a difference of a minus sign but it's a big difference

  • of course. And so let's work this out in

  • some detail. I have here a charge plus Q.

  • And at a distance R at that location P we know what the

  • electric field is, we've done that a zillion

  • times. This is the unit vector in the

  • direction from Q to that point and we know that the electric

  • field is pointing away from that charge

  • and we know that the electric field E we have seen that

  • already the first lecture is Q divided by four pi epsilon zero

  • R squared in the direction of R roof.

  • And last lecture we derived what the electric potential is

  • at that location. The electric potential is Q

  • divided by four pi epsilon zero R.

  • This is a vector. This is a scalar.

  • So the um potential is the integral of the electric field

  • along a line and now I want to try whether the electric field

  • can be written as the derivative of the potential.

  • So let us take DV DR and let's see what we get.

  • If I take this DV DR I get a minus Q divided by four pi

  • epsilon zero R squared.

  • Of course if I want to know what the electric field is I

  • need a vector so I will multiply both sides, which is completely

  • legal, there's nothing illegal about that, with unit vector in

  • the direction R so that turns them into vectors.

  • And now you see that I'm almost there, this is almost the same,

  • except for a minus sign. And so the derivative of the

  • potential is minus E, not plus E.

  • And so I will write that down here,

  • that E equals minus DV DR. So they are closely related if

  • you know what the -- oh I want -- I want this to be a vector so

  • I put here R roof. Vector on the left side,

  • you must have a vector on the right side.

  • And so if you know the potential everywhere in space,

  • then you can retrieve the electric field.

  • I mentioned last time that the electric field vectors -- e-

  • electric field lines, are always perpendicular to the

  • equipotential surfaces. And that's obvious why that has

  • to be the case. Imagine that you are in an --

  • in space and that you move with a charge in your pocket

  • perpendicular to electric field lines.

  • So you purposely move only perpendicular to electric field

  • lines. So that means that the force on

  • you and the direction in which you move are always at

  • ninety-degree angles. So you'll only move

  • perpendicular to the field lines.

  • These are the field lines, you move like this.

  • These are the field lines, you move like this.

  • So you never do any work. Because the dot product between

  • DL and E is zero and if you don't do any work the potential

  • remains the same, that's the definition.

  • And so you can see that therefore equipotential surfaces

  • must always be perpendicular to field lines and field lines must

  • always be perpendicular to the equipotentials.

  • And I will show you again the -- the viewgraph,

  • the overhead projection of uh the nice drawing by Maxwell.

  • With the plus four charge and the minus one charge.

  • The same one we saw last time. Only to point out again this

  • ninety-degree angle. I discussed this in great

  • detail last lecture so I will not do that.

  • R- the red lines are really surfaces.

  • This is three-dimensional, you have to rotate the whole

  • thing about the vertical. So these are surfaces.

  • And the red ones are positive potential surfaces and the blue

  • ones are negative potential surfaces.

  • That is not important. But the green lines are field

  • lines. And notice if I take for

  • instance this field line, it's perpendicular here to the

  • red, perpendicular there, perpendicular there,

  • perpendicular there. Perpendicular here.

  • Perpendicular here. Coming in here,

  • perpendicular, perpendicular,

  • perpendicular. Everywhere where you look on

  • this graph you will see that the field lines are perpendicular to

  • the equipotentials.

  • And that is something that we now fully understand.

  • The situation means then that if you release a charge at zero

  • speed that it would always start to move perpendicular to an

  • equipotential surface because it always starts to move in the

  • direction of a field line, a plus charge in the direction

  • of the field line, minus charge in the opposite

  • direction. So if you're in space and you

  • release a charge at zero speed it always takes off

  • perpendicular to equipotentials.

  • You have something similar with gravity.

  • If you look at maps of mountaineers,

  • contours of equal altitude, equal height,

  • if you started skiing and you started at that point,

  • and you started with zero speed, you would always take off

  • perpendicular to the equipotentials.

  • So this is the direction in which

  • you start to move. If you start off with zero

  • speed. I now want to give you some

  • deeper feeling of the connection between potential and electric

  • fields and I want you to follow me very closely.

  • Each step that I make I want you to follow me.

  • So imagine that I am somewhere in space at p- position P.

  • At that position P there is a potential, one unique potential,

  • V of P. That's a given.

  • And there is an electric field at that location where I am.

  • And now what I'm going to do, I'm going to make an extremely

  • small step only in the X direction.

  • Not in Y, not in Z. Only in the X direction.

  • If I measure no change in the potential over that little step

  • it means that the component of the electric field in the

  • direction X is zero. If I do measure a difference in

  • potential then the component, the X component of the electric

  • field, the magnitude of that, would be that little sidestep

  • that I have made, delta X, it would be the

  • potential difference that I measure divided by that little

  • sidestep. And I keep Y and Z constant.

  • And these are magnitudes. But that's why I put these

  • vertical bars here. Equally if I made a small

  • sidestep in the Y direction and I measured a potential

  • difference delta V keeping X and Z constant, that would then be

  • the component of the electric field in the Y direction.

  • Earlier we wrote down for E as a unit new- newtons per

  • coulombs. From now on we almost always

  • will write down for the unit of

  • electric field volts per meter. It is exactly the same thing as

  • newtons over coulombs, there is no difference,

  • but this gives you a little bit more insight.

  • You make a little sidestep in meters and you measure how much

  • the potential changes, it's volts per meters.

  • It is a potential change over a distance.

  • So now I can write down the connection between electric

  • field and potential in Cartesian coordinates.

  • It looks much more scary than the nice way that I could write

  • it down up there. When I had only a function of

  • distance R. And so in Cartesian coordinates

  • we now get E equals minus, that minus sign we discussed at

  • length, and now we get DV DX times X roof plus DV DY times Y

  • roof plus DV DZ times Z roof.

  • And what you see here, this first term here,

  • including of course the minus sign, that is E of X.

  • And this term including the minus sign, that is E of Y.

  • And so on. And the fact that you see these

  • curled Ds it means partial derivatives.

  • That means when you do this derivative you keep Z and Y

  • constant. When you do this derivative you

  • do X and Z constant and so on. And so this is the Cartesian

  • notation for which in eighteen oh two you will learn or maybe

  • you already have learned we would write this E equals minus

  • the gradient of G. This is a vector function.

  • This is a scalar function. And this is just a s- a

  • different notation, just a matter of words,

  • for this mathematical recipe. And you'll get that with

  • eighteen in eighteen oh two if you haven't seen that yet.

  • So I now want a straightforward example whereby we assume a

  • certain dependence on X and I give you it is a given that V,

  • the potential, is ten to the fifth

  • times X. So that is a given.

  • And this holds between X equals zero and say ten to the minus

  • two meters. So it holds over a space of one

  • centimeter. So the potential changes

  • linearly with distance. What now is the electric field?

  • In that space? Well, the electric field I go

  • back to my description there. There's only a component in the

  • X direction. So the first derivative becomes

  • minus ten to the fifth times X roof and the others are zero.

  • So EY is zero and EZ is zero. So you may say well yeah whoa

  • nice mathematics but we don't see any physics.

  • This is more physical than you think.

  • Imagine that I have here a plate

  • which is charged, it's positive charge.

  • And the plate is at location X and I have another plate here,

  • it's say at location zero. I call this plate A and I call

  • this plate B and this plate is charged negatively.

  • So X goes into this direction. So I can put the electric field

  • inside here according to the recipe minus ten to the fifth

  • and it is in the direction of minus X roof so e- X roof is in

  • this direction, the electric field is in the

  • opposite direction, and it's the same everywhere

  • and that is very physical. We discussed that.

  • When we discussed the electric field near very large planes,

  • that the electric field inside was a

  • constant, remember, and the electric field inside

  • was sigma divided by epsilon zero if sigma is the surface

  • charge density on each of these plates.

  • And we argued that the electric field outside was about zero and

  • that the electric field outside here was about zero.

  • So it's extremely physical. This is exactly what you see

  • here. The electric field minus ten to

  • the fifth times e- so the magnitude of this electric field

  • here, the magnitude, is ten to the fifth volts per

  • meter. What now is the potential

  • difference? Well, VA minus VB minus VB is

  • the integral in going from A to B of E dot DL.

  • Well I go from here to here so I write down for DL I wrote down

  • DX of course. Because I called that the X

  • direction now. So I will write down here dot

  • DX. And so this is minus ten to the

  • fifth times the integral in going from A to B of X roof dot

  • DX. It looks scary but it is

  • trivial, the X roof is the unit vector in this direction.

  • And DX is a little vector DX in this direction.

  • So they're both in the same direction.

  • So the cosine, the angle between the two is

  • one. So I can forget about vectors,

  • I can forget about the dot. And so this becomes minus ten

  • to the fifth times the integral in going from A to B of DX and

  • that is trivial. That is minus ten to the fifth

  • times the location. I have to do the integral

  • between A and B. So I get here X of B minus X of

  • A. And if this is ten to the minus

  • two meters, to go from here to here is one centimeter,

  • I must multiply this by ten to the minus two,

  • so I get that this is minus one thousand volts.

  • So A is a thousand volts lower than B.

  • That's what it means. And that's something that's

  • very physical. Notice that if you go from left

  • to right that the potential grows linearly,

  • this is lower than that, and if you in your head use

  • planes like this parallel to the other planes,

  • each one of those planes would be equipotentials,

  • they everywhere have the same potential.

  • And gradually when you move it up your potential increases,

  • but notice the electric field goes

  • from plus to minus in the opposite direction.

  • That's always the reason that's behind that -- that minus sign.

  • Well clearly I'm always free to choose where I choose my zero

  • potential. We discussed that last time.

  • You don't always have to choose infinitely far zero.

  • So I could choose this arbitrarily to be zero

  • potential. This would then be plus a

  • thousand. And so you then find that the

  • potential V is then simply ten to the fifth times X,

  • when X is zero you find the potential to be zero,

  • and when X is one centimeter you find the potential to be a

  • thousand volts and that then goes together with the electric

  • field equals minus ten to the fifth in this direction.

  • And so this is extremely physical.

  • This is something that you would have whenever we deal with

  • parallel plates. As long as there's no charge

  • moving, and we're dealing with solid

  • conductors, so we have static electric fields,

  • the charges are not heavily moving, then the field inside

  • the conductor is always zero. Not the case in a nonconductor.

  • It's only in a conductor because conductors have free

  • electrons to move and if these free electrons see electric

  • fields inside which they may they start to move until they

  • experience no longer a force, thereby they kill the electric

  • field inside.

  • So the charge in a conductor always rearranges itself so that

  • the electric field becomes zero. If the field is a static field,

  • not rapidly changing. And so now I want to evaluate

  • with you the situation that I'm going to charge a solid

  • conductor and ask myself the question where does the charge

  • go. In honor of um Valentine's Day

  • let's take a solid heart, steel heart,

  • it's solid all the way throughout.

  • So this is a solid conductor and I bring on this conductor

  • charge from the outside. Plus or minus,

  • let's just take plus for now. And so the question that I'm

  • asking you now, this is a conductor,

  • this is not an insulator, the story for insulators is

  • totally different, this has free moving electrons

  • inside, I'm asking you now if I touch this conducting heart --

  • by the way, your heart is a very good conductor -- um if you

  • touch this conducting heart where would this charge end up?

  • Where would it go to? And I leave you with three

  • choices. And we'll have a vote on that.

  • The first choice is that the plus charges would uniformly

  • distribute for throughout. A possibility.

  • The second possibility, less likely,

  • I think, that all the charge will go to one place there.

  • I don't know which place that would be, but maybe.

  • And then the third possibility is that maybe the charge will

  • uniformly distribute itself only on the outer surface and then

  • the fourth possibility is none of the above.

  • All these suggestions I made were wrong.

  • Who think that the charge might uniformly distribute it

  • throughout the conductor? I see one or two hands.

  • That's good. Don't feel ashamed of raising

  • your hands, in the worst case you're wrong,

  • I've been so many times wrong when it comes to this.

  • Don't feel bad about that. Who thinks that the charge will

  • all go to one point in the heart?

  • You have the courage? You think it will go to one

  • point? Charge repels each other,

  • right, so that doesn't seem likely.

  • Who thinks that it will uniformly distribute itself on

  • the outer surface? Who thinks none of the above?

  • Very good. Well, those who suggested that

  • it might be uniform on the outside I would still give them

  • a B but it's not uniform, as you will see.

  • But it will go exclusively to the outside.

  • And I will prove that now to you.

  • Let us first look for that ridiculous possibility that the

  • charge would somehow end up in the conductor itself.

  • I take here a Gaussian surface which is a closed surface.

  • I know inside the conductor if we have electrostatic fields,

  • not fastly moving charges, but it's a static field,

  • I know that the E field everywhere must be zero on the

  • surface, this is a closed surface, so the integral of E

  • dot DA equation one is zero. That means the charge inside my

  • sphere is zero and so there cannot be any charge.

  • So Gauss's law immediately kills the possibility that there

  • would be any charge inside this conductor.

  • So that's out of the question. So that leaves you only with

  • one choice, that is on the -- at the surface.

  • So the charge must be at the surface.

  • And later in a later lecture I will discuss with you the

  • details why that charge is not uniformly distributed.

  • It would be uniformly distributed at the surface if

  • this were a sphere. But not if it has this funny

  • shape. But it will be at the surface.

  • Now I'm going to make this heart a very special heart,

  • more like a real heart, it's open here,

  • but it is solid here, so this is a conducting,

  • the heart muscle, and here it's open,

  • there's nothing here. And again I'm going to charge

  • it. Bring charge on the outside.

  • So now it's obvious that we don't expect that there is any

  • charge that will be inside the conductor.

  • That's clear, the same argument

  • holds with the Gauss's law argument.

  • But now is it perhaps possible that some of the positive charge

  • will go on the inside of this surface and some on the outside?

  • Who thinks that maybe some will now go on the inside,

  • because now the situation is different, right,

  • there is now, it's now a hollow conductor.

  • Anyone in favor of some of that charge maybe going on the

  • inside? I see one hand,

  • two hands, who says no it's not possible,

  • it will not go to the inside? It will still go to the

  • outside. Well, most of you are very

  • careful now, you don't want to vote anymore.

  • It cannot go to the inside. Why can it not go to the

  • inside? Let this be my Gauss surface.

  • Closed surface. Think of this as

  • three-dimensional. Everywhere on that line the

  • electric field is zero because you're inside the conductor.

  • So the surface integral is also zero.

  • So Gauss's law says there cannot be any charge inside that

  • box. And so again the charge has to

  • go to the outer surface and

  • nothing will go to the inner surface.

  • And so the conclusion then is that the electric field is zero

  • in the conductor but the electric field is also zero in

  • this opening. There's never any charge there.

  • And so the whole heart including the cavity is an

  • equipotential. There is never any electric

  • field anywhere. The only electric field's

  • outside the heart. And there are field lines and

  • these field lines everywhere are perpendicular to the surface of

  • the heart because the heart is an

  • equipotential. So here you get very funny

  • field lines that go like this. They have to be perpendicular

  • locally where they reach the heart wall.

  • Earlier in my lectures I showed that a uniformly solid sphere

  • has electric field zero inside and I even showed to you that a

  • hollow conducting sphere also has zero

  • electric field inside. Today I have demonstrated that

  • it doesn't have to be a sphere. You don't need spherical

  • symmetry. That any shape provided that

  • it's a hollow conductor, it has to be a conductor,

  • any shape will give you an electric field of zero inside.

  • And I first want to demonstrate that.

  • I have here something that is not a sphere.

  • It's a paint can. It has some aluminum on top.

  • It has an opening there. It's not

  • perfect. It's not really closed like

  • this is. So the electric field inside

  • will not be exactly zero. But it will be very close.

  • I must have an opening because I want to get in.

  • I want to get charge, see whether there is any charge

  • on the inside. So I must be able to get

  • through. So I'm going to charge this one

  • and then I will take some charge off the outside and take some

  • charge off the inside and use the electroscope and see whether

  • we can demonstrate that indeed there is charge on the outside

  • but there is nothing on the inside.

  • I will use the same method that I used last time when I

  • challenged you to figure out how this works.

  • This is this crazy method which we call electrophorus elec-

  • electrophorus, it's difficult to pronounce.

  • Electrophorus. We have here a glass plate.

  • I rub it with cat fur. Think about it again.

  • It's a little problem inside the

  • problem. Metal plate.

  • I put it on top. I touch it.

  • I get a shock. I touch it here.

  • I touch it again. I get again a shock.

  • And I charge this up. I touch it again.

  • I get another shock. And I touch it again.

  • Let's get a little bit more on it.

  • The charge on this plate is positive by the way.

  • That I create on the glass. I touch it.

  • The charge on here is negative. Not positive.

  • Put it on again. Touch it.

  • OK. So I should have negative

  • charge on there now. Here is little test sphere.

  • It's a conductor. I'll take some charge off from

  • this side. Touch it.

  • Boy. There's charge.

  • There's no question. We agree, right?

  • There's charge. OK.

  • Now I touch the inside, let's hope that no sparks fly

  • over. I touch it.

  • Nothing. See that?

  • Absolutely nothing. So there's no charge inside,

  • the charge is on the outside. Which is what I've just

  • demonstrated. You see it in front of your own

  • eyes. All the charge goes to the

  • outside. Not so intuitive but an

  • immediate consequence of the fact that it's a conductor that

  • the electrons will move freely so that the electric

  • field in the conductor itself is zero and we have argued that

  • no charge can ever go on the inside of the surface.

  • It all stays on the outside. So when I touch the inside

  • there was no charge. So if you are inside that

  • conductor, if your house is a conducting house,

  • and someone in the outside world charges your house up when

  • you're inside, you have no knowledge of that.

  • It's quite amazing, isn't it?

  • You are electrically shielded from the outside world.

  • Now I'm going to make the situation even more complicated.

  • I now take a conducting object, doesn't have to be a sphere.

  • And I bring that conducting object hollow in an external

  • electric field. So someone outside your house

  • is turning on a VandeGraaff creating an electric field.

  • What now is going to happen? Well, due to induction,

  • you're going to get some charge polarization in the conductor.

  • One side may end up negative and the other side may end up

  • positive. But what happens on the inside?

  • Nothing. The electric field in the

  • conductor must stay everywhere zero if it is a static electric

  • field. And so no charge will -- can

  • accumulate here and no charge can accumulate on the inside.

  • And so as you bring this electric

  • field on the outside you may get negative and positive charge

  • on the outside, maybe negative here and

  • positive there, but inside nothing.

  • You are inside electrically shielded from the outside world

  • in the same way that you were when someone was trying to put

  • charge onto your house, now someone is trying to zap

  • you with electric fields, nothing will happen inside.

  • You will never see an electric field inside.

  • I will show you a interesting drawing, interesting figure,

  • which is a conducting box. It's closed.

  • The cup that you see open is just to allow you to look inside

  • but it is closed from all sides. And there are some negative

  • charges here and there are positive charges in the

  • foreground which you don't see. The red field lines come from

  • positive charges, end up on the box,

  • and the negative field lines go from the box to the negative

  • charges. There is clear polarization.

  • The box itself is neutral. I started with a neutral box.

  • But because of this electric field I get polarization.

  • I end up with negative charge on the box here,

  • only on the outside, positive charge on the box

  • here, only on the outside. Inside electric field is zero.

  • No charge anywhere inside. Due to this crazy electric

  • field the free moving charges in the conductor will rearrange

  • themselves in such a way that the electric field is zero

  • everywhere in the conductor, is zero inside the cavity,

  • and that the closed loop integral of E dot DL is zero

  • everywhere if these are static fields.

  • And it is clearly impossible for us to ever calculate how

  • that charge configuration at the surface will

  • have to be in order to meet all those conditions.

  • But nature can do this effortlessly.

  • And it can do it extremely fast, obeying all the laws of

  • physics. It puts very quickly plus

  • charge here and minus charge there.

  • Make sure that there is no charge on the inside of the

  • surface. It makes sure that the electric

  • field is everywhere zero inside and in the box and it also makes

  • sure that the integral E dot DL is zero everywhere in space.

  • And therefore the box and everything inside becomes an

  • equipotential so it also arranges matters so that the

  • field lines wherever they intersect with the box are

  • always perpendicular to the box. And all of that is done in

  • almost no time at all by nature. It is an amazing thing that

  • this happens and something that as I said would be impossible

  • for us to calculate because the field configurations are

  • extraordinarily difficult. So if you were inside this

  • metal box, no matter what happens

  • on the outside, you would be electrically

  • isolated from the outside world. You would not notice that there

  • is a strong electric field outside, nor would you notice

  • that people are trying to charge up your house.

  • We call that electrostatic shielding and we give that a

  • name, that house of yours would be called a Faraday cage.

  • It's called after the great physicist Faraday.

  • You will learn a lot more about him during this course.

  • Before I demonstrate this I want to address an issue which

  • is related to problem two-one. Which is your next assignment.

  • And I want to urge you, I make myself no illusion,

  • but I want to urge you to start working on that assignment this

  • weekend, not next week. These assignments are not just

  • baby assignments. These are MIT assignments and

  • you got to put in a lot of work to do them, so please start this

  • weekend not to do me a favor but to do yourself that favor.

  • But let's talk about problem two-one.

  • In other words I will help you with that problem two-one.

  • I said several times that it is not possible to get an electric

  • field inside a hollow conductor. Well, suppose I go inside the

  • conductor. I go inside there.

  • And I put sneakily a charge in my pocket.

  • And I sit inside there and you close it.

  • Then there is a charge inside, there's nothing you can do

  • about it. And if there is a charge

  • inside, there is an electric field.

  • So now we have a situation and since it is post Valentine's Day

  • my heart has evolved into a sphere again.

  • So now we take a spherical conductor, solid,

  • this is solid material, and somehow I'm sitting inside

  • here with a charge plus Q. Can make it minus if you want

  • to. That's exactly the problem

  • two-one is about. Now clearly there is positive

  • charge inside. So clearly there has to be an

  • electric field. But the electric field inside

  • the conductor, that means the electric field

  • anywhere here, must be zero.

  • If it's not zero the electrons will keep moving until it is

  • zero. So the conducting material

  • itself has no electric field. What does that mean now with

  • respect to any charge on the inside surface?

  • Now there must be charge on the inside surface.

  • Because now if I made this my Gaussian surface,

  • which is now a spherical surface, a closed surface,

  • Mr. Gauss says that the closed

  • surface integral of E dot DA over this surface must be zero

  • because the electric field is zero anywhere.

  • That's the same as all the charge inside divided by epsilon

  • zero. So this -- the charge inside

  • must be zero. Since there can be no charge in

  • the conductor itself, negative charge must now

  • accumulate on the inside of that surface.

  • So that the net charge inside this surface is zero.

  • So now we do get charge on the inside, and how much charge do

  • you get on the inside? Exactly minus Q.

  • So that the sum of the two is zero.

  • Now this conductor originally was neutral.

  • It had no net charge. So therefore on the surface of

  • the conductor we must now see charge plus Q.

  • Because the minus charge on the inside came from the

  • induct- conductor itself, and so the sum must be zero.

  • So now you get a peculiar situation that the plus Q charge

  • inside which creates an E field inside creates negative charge

  • on the inside, the same in magnitude,

  • opposite in sign, and plus Q charge on the

  • outside. And the electric fields,

  • they're very complicated. The electric fields,

  • let me try to put them in, I would imagine that if this

  • charge Q is closer to this wall than to this wall that the

  • negative charge here will be larger in density than there.

  • It's really an induction effect.

  • The negative charge wants to go to this plus.

  • That's really what's happening. And so since this charge is

  • closer to this wall than to that wall it will be able to attract

  • more electrons and so it's clear that the density of charge here

  • should be higher than there and so the field lines always

  • perpendicular to the equipotential,

  • so they must be always perpendicular to the wall,

  • sort of like this. So I put in a few field lines.

  • But here the field will be stronger than there.

  • So there is a field inside. What now is the charge

  • distribution on the outside? That is the hardest of all.

  • And by no means so obvious. It turns out that the charge on

  • the outside on this sphere, because it is a sphere,

  • will be uniformly distributed. And it is not intuitive and it

  • is not obvious. Nature must obey all laws of

  • physics. The conductor must become an

  • equipotential. There can be no electric field

  • inside the conductor. Electric field lines have to be

  • everywhere perpendicular to the surface.

  • The closed loop integral of E dot DL must be zero everywhere.

  • And the only way that nature can do that is by making the

  • charge distribution on the surface uniform.

  • And that is amazing when you think

  • of that. It's independent of the

  • position of that charge plus Q inside.

  • So if you start to move around with that charge plus Q inside,

  • the outside world will not know.

  • The outside world only knows that there is a charge plus Q

  • uniformly distributed on the outside because it is a sphere.

  • That would not be the case if it were a heart.

  • But the outside world has no way of knowing that you are

  • moving that charge inside around.

  • So I'm sitting inside there and suppose I crawl inside there

  • with a rubber rod and with a cat.

  • And I use the rubber rod on the cat creating positive and

  • negative charge, same amount.

  • The outside world will not know because I don't change the

  • charge inside. Only if there's plus Q in my

  • pocket. The fact that I create plus on

  • the cat and maybe minus on myself, the outside world will

  • never know because the sum of the charges is still Q.

  • They may hear the cat scream, that's all they can hear.

  • But they have no way of knowing that I'm fooling around there

  • with charges. And so the outside world has no

  • way of knowing what happens on the inside.

  • And we call that electrostatic shielding.

  • That's the effect of a Faraday cage.

  • I want to demonstrate when I bring this can in the electric

  • field, it's a conducting hollow object, I bring it in an

  • electric field of the VandeGraaff, the thing is a

  • conductor, I will show you that because of

  • the induction you're going to get like you see on that figure

  • you're going to get negative charge on one side,

  • positive charge on the other side, and zero charge on the

  • inside. That's quite amazing,

  • isn't it? If this were positive,

  • let's assume it is, then this side becomes

  • negative, this side becomes positive, the whole thing is an

  • equipotential, no charge inside.

  • Quite amazing. So let's turn on this,

  • the VandeGraaff. So we create that electric

  • field. We turn on the electroscope.

  • Here is my little Ping Pong ball conducting and I'm going to

  • touch first the can on your side, on your left side,

  • there we go, and I bring this charge on the

  • electroscope. Boy, nice charge.

  • I now touch the other side, and I will approach the -- I

  • heard a spark, sparks are always bad.

  • I approach that electroscope, and if the reading of the

  • electroscope, if the deflection becomes less,

  • as we discussed earlier, it means that the polarity that

  • I have on here is different from the polarity on the electroscope

  • and you clearly see that. The deflection becomes less.

  • So the charge that I took off from this side has a different

  • polarity than the charge that I took on from that side.

  • But yet, it's an equipotential. All that strange polarization

  • of charges takes place at the

  • surface. And now I will try to get

  • inside. To see whether I can get some

  • charge from the inside, and there shouldn't be any,

  • ooh, I have to take this charge off of course,

  • and I touch this and you see no charge.

  • So you've seen three things, which is quite amazing.

  • That the charge on this side has a different polarity from

  • the charge on that side, and that everything happens on

  • the surface, nothing happens on the inside.

  • I could not get any charge from the inside.

  • Now we're going to experiment with more dangerous stuff and

  • that is with the VandeGraaff. Here you see a Faraday cage

  • with hat -- has some openings, it's not solid conductor,

  • but it has small openings, which doesn't change the

  • situation too much, maybe only a little,

  • and I'm going to go inside that cage, this would also be a nice

  • Faraday cage but it's very hard for me

  • to crawl in there. And if I go in there with a

  • radio just like the radio in your car, then you may not be

  • able to hear the radio, even though radio waves is a

  • difficult story because the shielding that we discussed is

  • only electrostatic shielding and radio waves are electromagnetic

  • radiation which strongly changing fields.

  • So it may not be as perfect a shielding as you may think.

  • But we all know if someone breaks off the antenna of your

  • car which happens in Cambridge all the time,

  • you have no reception inside. Because your car is a Faraday

  • cage. And so what I will do,

  • I will go into the cage, I will first show you that when

  • we charge that cage, that we bring it up to a few

  • hundred thousand volts, I'll just hold some tinsel in

  • my hand, to convince you that yes indeed this cage will be

  • charged by the VandeGraaff provided there

  • is contact here. And we'll see that yes Marcos

  • give me the full blast, let's just look at the tinsel,

  • you see this tinsel clearly indicates like an electroscope

  • that I'm being charged now. So I'll jump off,

  • if you can discharge. Then I will go inside.

  • I will have the tinsel with me. So I will show you that inside

  • there when they charge that cage that the tinsels will not spread

  • out, and I will bring with me this

  • wonderful radio and -- Radio: Said a woman who opposes

  • embalming is a suspect in the murders -- [laughter] I didn't

  • plan that, believe me. Radio: [unintelligible] --

  • self-proclaimed prophet, Catherine Padilla,

  • grandmother of ten, denies the charges and tells a

  • reporter she's not quote -- I'll first go in without any charge.

  • But don't do anything. Radio: [unintelligible] And

  • she's one that -- when she calls her own --

  • [unintelligible] Nothing. I'm shielded.

  • However, there is a problem. You can still hear me and I'm

  • wearing a transmitter and the receiver is somewhere outside

  • this box. So why can you still hear me?

  • That means that the kind of radio waves that I am

  • transmitting are very high frequency, it's not a static

  • field, so somehow they can get through.

  • So the shielding is not perfect for fast changing electric

  • fields. But it's good enough for AM

  • radios. So now I'll go in and I'll try

  • to be brave and he's going to try to zap me now,

  • to electrocute me. But since I've taken eight oh

  • two, I'm not afraid. I burned my hand,

  • that's a different story. OK.

  • Marcos. Do the best you can.

  • Here are the tinsels. Run it up a hundred thousand

  • volts. Two hundred thousand volts.

  • I feel as happy like a clam at high tide inside here.

  • Nothing is happening. I'm not worried at all.

  • If lightning were to strike me who cares?

  • I'm in a Faraday cage. Not going to spoil my weekend.

  • I can touch the inside. There's no charge anywhere

  • here. My weekend won't be spoiled.

  • And I hope that the new assignment is not going to spoil

  • yours either. See you next Tuesday.

So today no new concepts, no new ideas,

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