we discussed that earlier. And we call that electrostatic

potential energy. Today, I will look at this

energy concept in a different way, and I will evaluate the

energy in terms of the electric field.

Suppose I have two parallel plates, and I charge this one

with positive charge, which is the surface charge

density times the area of the plate, and this one,

negative charge, which

is the surface charge density negative times the area of the

plate. And let's assume that the

separation between these two is H, and so we have an electric

field, which is approximately constant, and the electric field

here is sigma divided by epsilon zero.

And now, I'm going to take the upper plate, and I'm going to

move it up. And so as I do that,

I have to apply a force, because these two plates

attract each other, so I have to do work.

And as I move this up, and I will move it up over

distance X, I am creating here, electric field that wasn't

there before. And the electric field that I'm

creating has exactly the same strength as this,

because the charge on the plates is not changing when I am

moving, the surface charge density is not changing,

all I do is, I increase the distance.

And so I am creating electric field in here.

And for that, I have to do work,

that's another way of looking at it.

How much work do I have to do? What is the work that Walter

Lewin has to do in moving this plate over the distance X?

Well, that is the force that I have to apply over the distance

X. The force is constant,

and so I can simply multiply the force times the distance,

that will give me work. And so the question now is,

what is the force that I have to apply to move this plate up?

And your first guess would be that the force would be the

charge on the plate times the electric field strength,

a complete reasonable guess, because, you would argue,

"Well, if we have an electric field E, and we bring a charge Q

in there, then the electric force is Q times E,

I have to overcome that force, so my force is Q times E." Yes,

that holds most of the time. But not in this case.

It's a little bit more subtle. Let me take this plate here,

and enlarge that plate. So here is the plate.

So you see the thickness of the plate, now, this is one plate.

We all agree that the plus charge is at the surface,

well, but, of course, it has to be in the plate.

And so there is here this layer of charge Q, which is at the

bottom of the plate. And the thickness of that layer

may only be one atomic thickness.

But it's not zero. And on this side of the plate,

is there electric field, which is sigma divided by

epsilon zero. But inside the plate,

which is a conductor, the electric field is zero.

And therefore, the electric field is,

in this charge Q, is the average between the two.

And so the force on this charge, in this layer,

is not Q times E, but is one-half Q times E.

So I take the average between these E fields,

and this E field is then this value.

And so now I can calculate the work that I have to do,

the work that I have to do is now my force,

which is one-half Q times E, and I move that over a distance

X. And so what I can do now is

replace Q by sigma A, so I get one-half sigma A times

E times X, and I multiply upstairs and downstairs by

epsilon zero, so that multiply by one.

And the reason why I do that is,

because then I get another sigma divided by epsilon zero

here -- divided by epsilon zero, and that is E,

and therefore, I now have that the total work

that I, Walter Lewin have to do -- has to do is one-half epsilon

zero, E-squared times A times X. And look at this.

A X is the new volume that I have created,

it is the new volume in which I have created electric field.

And this, now, calls for a work done by

Walter Lewin. Per unit volume,

and that, now, equals one-half epsilon zero

times E squared. This is the work that I have

done per unit volume. And since this work created

electric field, we called it "field energy

density". And it is in joules per cubic

meter. And it can be shown that,

in general, the electric field energy density is one-half

epsilon zero E squared, not only for this particular

charge configuration, but for any charge

configuration. And so, now,

we have a new way of looking at the energy that it takes to

assemble charges. Earlier, we calculated the work

that we have to do to put the charges in place,

now, if it is more convenient,

we could calculate that the energy electrostatic potential

energy, is the integral of one-half epsilon zero E-squared,

over all space -- if necessary, you have to go all the way down

to infinity -- and here, I have now, D V,

this is volume. This has nothing to do with

potential, this V, in physics, we often run out of

symbols, V is sometimes potential, in this case,

it is volume. And the only reason why I chose

H there is I already have a D

here, so I didn't want two Ds. Normally, we take D as the

separation between plates. And so this,

now, is another way of looking at electrostatic potential

energy. We look at it now only from the

point of view of all the energy being in the electric field,

and we no longer think of it, perhaps, as the work that you

have done to assemble these charges.

I will demonstrate later today that as I separate the two

plates from these charged planes,

that indeed, I have to do work.

I will convince you that by creating electric fields that,

indeed, I will be doing work. So, from now on,

uh, we have the choice. If you want to calculate what

the electrostatic potential energy is, you either calculate

the work that you have to do to bring all these charges in

place, or, if it is easier, you can take the electric field

everywhere in space, if you know that,

and do an integration over all space.

We could do that, for instance,

for these two parallel plates now, and we can ask what is now

the total energy in these plates -- uh, in the field.

And at home, I would advise you,

to do that the way that it's done in your book,

whereby you actually assemble the charges minus Q at the

bottom and plus Q at the top, and you calculate how much work

you have to do. That's one approach.

I will now choose the other approach, and that is,

by simply saying that the total energy in the field of these

plane-parallel plates, is the integral of one-half

epsilon zero E-squared, over the entire volume of these

two plates. And since the electric field is

outside, zero, everywhere, it's a very easy

integral, because I know the volume.

The volume that I have, if the separation is H -- so we

still have them H apart -- this volume that I have is simply A

times H, and the electric field is constant, and so I get here

that this is one-half epsilon zero.

For E, if I want to, I can write sigma divided by

epsilon zero, I can square that,

and D V, in- doing the integral over all space,

means simply I get A times H, it is the volume of that box.

So I get A times H. And so this is now the total

energy that I have, I lose one epsilon here,

I have an epsilon zero squared and I have an epsilon.

I also remember that the charge Q on the plate is A times sigma,

and that the potential difference V,

this now is not volume, it's the potential difference

between the plates, is the electric field times H.

The electric field is constant, it can go from one plate to the

other, the integral E dot D L in going from one plate to the

other, gives me the potential difference.

And so I can substitute that now in here, I can take for A,

sigma, I can put in the Q, and you can also show that this

is one-half Q V.

V being, now, the potential difference

between the plates. And so this is a rather fast

way that you can calculate what the total energy is in the

field, or, say, the same thing,

the total work you have to do to assemble these charges.

Or, to say it differently, the total work you have to do

to create electric fields. You have crela- created

electric fields that were not there before.

I now will introduce something that

we haven't had before, that is the word "capacitance".

I will define the capacitance of an object to be the charge of

that object divided by the potential of that object.

And so the unit is coulombs per volt, this V is volt,

now, it's potential. Uh, but we never say that it is

coulombs per volt in physics, we write for that a capital F,

which is Farad, we call that,

one farad is the unit of capacitance, undoubtedly called

after the great maestro Faraday, we will learn more about

Faraday later in this course. So let us go to,

um, a sphere which has a radius R, and let us calculate what the

capacitance is of this sphere. Think of it as being a

conductor, and we bring a certain charge Q on

this conductor, it will then get a potential V,

which we know is Q divided by four pi, epsilon zero R.

We've seen this many times, and so, by definition,

the capacitance now is Q divided by the potential,

and therefore, this becomes four pi epsilon

zero R. So that is the capacitance of a

single sphere. And so we can now look at the

values as a function of R.

I have here some numbers, I calculated it for the

VandeGraaff, and I calculated it for the Earth.

If you want one Farad capacitance, that's a real

biggie, you need a radius of 9 times ten to the 9 meters,

that's the four pi epsilon zero that comes in there.

That's huge, that's twenty-five times the

distance from the Earth to the moon, that's a big sphere to

have a capacitance of one Farad. The Earth itself,

with a radius of sixty-four hundred kilometers,

would have seven hundred microfarad, the VandeGraaff

thirty centimeters radius would be 30 picofarad,

the pico is ten to the minus twelve.

And if you take a sphere with a radius of one centimeter,

then you have, uh, roughly,

one picofarad, ten to the minus twelve Farad.

So this gives you a rough idea about the size of objects,

and how they connect to their capacitance.

So if I bring all these spheres,

uh, at the same potential, so I charge them all up to the

same potential, then the one with the largest

capacitance, uh, will have the most charge.

And that, of course, is where the word "capacitance"

comes from, it is the capability of holding charge for a given,

uh, electric potential. Don't confuse that with

electric fields, because if you bring all these

spheres at the same potential, then the one with the strongest

electric field, that's the one which has the

short -- smallest radius,

we discussed that last time. Now, I will look at the

situation a little bit differently.

I have, here, a sphere, B,

positively charged, and I place it close to another

sphere, A, which is negatively charged.

And so, by my definition, I can say that the capacitance

of B is the charge that I have on B

divided by the potential of B. That will be my definition.

But, there is here, this object which charged

negative. And how did we define

potential? Potential was work per unit

charge. I go to infinity,

I put plus Q in my pocket, I approach B,

and the work I have to do per unit charge is the potential of

B, that's the definition of potential.

But B is repelling me. So I have to do positive work.

But A is now attracting me. And so the work I have to do is