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• assemble charges, I have to do work,

• 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,