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.
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.
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.