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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,