Electrons in insulators are bound to the atoms and to the molecules, unlike conductors, where they can freely move, and when I apply an external field -- for instance, a field in this direction, then even though the molecules or the atoms may be completely spherical, they will become a little bit elongated in the sense that the electrons will spend a little bit more time there than they used to, and so this part becomes negatively charged and this part becomes positively charged, and that creates a dipole.
I discussed that with you, already, during the first lecture, because there's something quite remarkable about this, that if you have an insulator -- notice the pluses and the minuses indicate neutral atoms -- and if now, I apply an electric field, which comes down from the top, then, you see a slight shift of the electrons, they spend a little bit more time up than down, and what you see now is, you see a layer of negative charge being created at the top, and a layer of positive charge being created at the bottom.
That's the result of induction, we call that also, sometimes, polarization.
You are polarizing, in a way, the electric charge.
Uh, substances that do this, we call them dielectrics, and today, we will talk quite a bit about dielectrics.
The first part of my lecture is on the web, uh, if you go to 8.02 web, you will see there a document which describes, in great detail, what I'm going to tell you right now.
Suppose we have a plane capacitor -- two planes which I charge with a certain potential, and I have on here, say, a charge plus sigma and here I have a charge minus sigma.
I'm going to call this free -- you will see, very shortly why I call this free -- and this is minus free.
So there's a potential difference between the plate, charge flows on there, it has an area A, and sigma free is the charge density, how much charge per unit area.
So we're going to get an electric field, which runs in this direction, and I call that E free.
And the distance between the plates, say, is D.
So this is given.
I now remove the power supply that I used to give it a certain potential difference.
I completely take it away.
So that means that this charge here is trapped, can not change.
But now I move in a dielectric.
I move in one of those substances.
And what you're going to see here, now, at the top, you're going to see a negative-induced layer, and at the bottom, you're going to see a positive-induced layer.
I called it plus sigma induced, and I call this minus sigma induced.
And the only reason why I call the other free, is to distinguish them from the induced charge.
This induced charge, which I have in green, will produce an electric field which is in the opposite I- direction, and I call that E-induced.
And clearly, E free is, of course, the surface charge density divided by epsilon 0, and E induced is the induced surface charge density, divided by epsilon 0.
And so the net E field is the vectorial sum of the two, so E net -- I gave it a vector -- is E free plus E induced, vectorially added.
Since I'm interested -- I know the direction already -- since I'm interested in magnitudes, therefore the strength of the net E field is going to be the strength of the E fields created by the so-called free charge, minus the strength of the E fields created by the induced charge, minus -- because this E vector is down, and this one is in the up direction.
And so, if I now make the assumption that a certain fraction of the free charge is induced, so I make the assumption that sigma induced is some fraction B times sigma free, I just write, now, and I for induced and an F for free.
B is smaller than 1.
If B were .1, it means that sigma-induced would be 10% of sigma free, that's the meaning of B.
So clearly, if this is the case, then, also, E of I must also be B times E of F.
You can tell immediately, they are connected.
And so now I can write down, for E net, I can also write down E free times 1 minus B, and that 1 minus B, now, we call 1 over kappa.
I call it 1 over kappa, our book calls it 1 over K.
But I'm so used to kappa that I decided to still hold on to kappa.
And that K, or that kappa, whichever you want to call it, is called the dielectric constant.
It's a dimensionless number.
And so I can write down, now, in general, that E -- and I drop the word net, now, from now on, whenever I write E, throughout this lecture, it's always the net electric field, takes both into account.
So you can write down, now, that E equals the free electric fields, divided by kappa, because 1 minus B is 1 over kappa.
And so you see, in this experiment that I did in my head, first, bringing charge on the plate, certain potential difference, removing the power supply, shoving in the dielectric that an E field will go down by a factor kappa.
Kappa, for glass, is about 5.
That will be a major reduction, I will show you that later.
If the electric field goes down, in this particular experiment, it is clear that the potential difference between the plates will also go down, because the potential difference between the plates, V is always the electric field between the plates times D.
And so, if this one goes down, by a factor of kappa, if I just shove in the dielectric, not changing D, then, of course, the potential between the plates is also going down.
None of this is so intuitive, but I will demonstrate that later.
The question now arises, does Gauss's Law still hold?
And the answer is, yes, of course, Gauss's Law will still hold.
Gauss's Law tells me that the closed loop -- closed surface, I should say, not closed loop -- the closed surface integral of E dot dA is 1 over epsilon times the sum of all the charges inside my box.
All the charges! The net charges, that must take into account both the induced charge, as well as the free charge.
And so let me write down here, net, to remind you that.
But Q net is, of course, Q free plus Q induced.
And I want to remind you that this is minus, and this was plus.
The free charge, positive there, is plus, and at that same plate, if you have your Gaussian surface at the top, you have the negative charged Q induced.
And so therefore, Gauss's Law simply means that you have to take both into account, and so, therefore, you can write down 1 over epsilon 0, times the sum of Q free, but now you have to make sure that you take the induced charge into account, and therefore, you divide the whole thing by kappa.
Then you have automatically taken the induced charge into account.
So you can amend Gauss's Law very easily by this factor of kappa.
Dielectric constant is dimensionless, as I mentioned already, it is 1, in vacuum, by definition.
1 atmosphere gases typically have dielectric constant just a hair larger than 1.
We will, most of the time, assume that it is 1.
Plastic has a dielectric constant of 3, and glass, which is an extremely good insulator, has a dielectric constant of 5.
If you have an external field, that can induce dipoles in molecules -- but there are substances, however, which themselves are already dipoles, even in the absence of an electric field.
If you apply, now, an external field, these dipoles will start to align along the electric field, we did an experiment once, with some grass seeds, perhaps you remember that.
And as they align in the direction of the electric field, they will strengthen the electric field.
On the other hand, because of the temperature of the substance, these dipoles, these molecules which are now dipoles by themselves, through chaotic motion, will try to disalign, temperature is trying to disalign them.
So it is going to be a competition, on the one hand, between the electric field which tries to align them and the temperature which tries to disalign them.
But if the electric field is strong, you can get a substantial amount of alignment.
Uh, permanent dipoles, as a rule, are way stronger than any dipole that you can induce by ordinary means in a laboratory, and so the substances which are natural dipoles, they have a much higher value for kappa, a much higher dielectric constant that the substances that I just discussed, which themselves, do not have dipoles.
Water is an example, extremely good example.
The electrons spend a little bit more time near the oxygen than near the hydrogen, and water has a dielectric constant of 80.
That's enormous.
Pengembangan Perkuliahan
1. Buatlah sebuah Esai mengenai materi perkuliahan ini
2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini
3. Lakukan Penelitian Sederhana dengan kelompok tersebut
4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat
5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya
Instructors: Dr. Peter Dourmashkin Prof. Bruce Knuteson Prof. Gunther Roland Prof. Bolek Wyslouch Dr. Brian Wecht Prof. Eric Katsavounidis Prof. Robert Simcoe Prof. Joseph Formaggio
Course Co-Administrators: Dr. Peter Dourmashkin Prof. Robert Redwine
Technical Instructors: Andy Neely Matthew Strafuss
Course Material: Dr. Peter Dourmashkin Prof. Eric Hudson Dr. Sen-Ben Liao
So far in these lectures we've talked about mass, about acceleration and about forces, but we never used the word "weight," and weight is a very nonintuitive and a very tricky thing which is the entire subject of today's lecture.
What is weight? Here you stand on a bathroom scale.
Gravity is acting upon you, the force is mg, your mass is m.
The bathroom scale is pushing on you with a force F scale and that F scale--
which in this case if the system is not being accelerated is the same as mg--
that force from the bathroom scale on you we define as weight.
When I stand on the bathroom scale I could see my weight is about 165 pounds.
Now, it may be calibrated in newtons but that's, of course, very unusual.
If I weigh myself on the moon where the gravitational acceleration is six times less then I would weigh six times less--
so far, so good.
Now I'm going to put you in an elevator and I'm going to accelerate you upwards and you're standing on your bathroom scale.
Acceleration is in this direction and I will call this "plus" and I will call this "minus." Gravity is acting upon you, mg and the bathroom scale is pushing on you with a force F.
That force, by definition, is weight.
Before I write down some equations, I want you to realize that whenever, whenever you see in any of my equations "g" g is always plus 9.8.
And my signs, my minus signs take care of the directions but g is always plus 9.8 or plus 10, if you prefer that.
Okay, it's clear that if this is accelerated upwards that F of s must be larger than mg; otherwise I cannot be accelerated.
And so we get Newton's Second Law: F of s is in plus direction...
minus mg--
it's in this direction--
equals m times a and so the bathroom scale indicates m times a plus g.
And I have gained weight.
If this acceleration is five meters per second squared in this direction I am one and a half times my normal weight.
If I look on the bathroom scale, that's what I see.
Last time we discussed that an acceleration is caused by a push or by a pull.
Today we will express this more qualitatively in three laws which are called Newton's Laws.
The first law really goes back to the first part of the 17th century.
It was Galileo who expressed what he called the law of inertia and I will read you his law.
"A body at rest remains at rest "and a body in motion continues to move "at constant velocity along a straight line unless acted upon by an external force." And now I will read to you Newton's own words in his famous book, Principia. "Every body perseveres in its state of rest "or of uniform motion in a right line "unless it is compelled to change that state by forces impressed upon it." Now, Newton's First Law is clearly against our daily experiences.
Things that move don't move along a straight line and don't continue to move, and the reason is, there's gravity.
And there is another reason.
Even if you remove gravity then there is friction, there's air drag.
And so things will always come to a halt.
But we believe, though, that in the absence of any forces indeed an object, if it had a certain velocity would continue along a straight line forever and ever and ever.
Now, this law, this very fundamental law does not hold in all reference frames.
For instance, it doesn't hold in a reference frame which itself is being accelerated.
Imagine that I accelerate myself right here.
Either I jump on my horse, or I take my bicycle or my motorcycle or my car and you see me being accelerated in this direction.
And you sit there and you say, "Aha, his velocity is changing.
"Therefore, according to the First Law, there must be a force on him." And you say, "Hey, there, do you feel that force?" And I said, "Yeah, I do! "I really feel that, I feel someone's pushing me." Consistent with the first law.
Perfect, the First Law works for you.
Now I'm here.
I'm being accelerated in this direction and you all come towards me being accelerated in this direction.
I say, "Aha, the First Law should work so these people should feel a push." I say, "Hey, there! Do you feel the push?" And you say, "I feel nothing.
There is no push, there is no pull." Therefore, the First Law doesn't work from my frame of reference if I'm being accelerated towards you.
So now comes the question, when does the First Law work? Well, the First Law works when the frame of reference is what we call an "inertial" frame of reference.
And an inertial frame of reference would then be a frame in which there are no accelerations of any kind.
Is that possible? Is 26.100...
is this lecture hall an inertial reference frame? For one, the earth rotates about its own axis and 26.100 goes with it.
That gives you a centripetal acceleration.
Number two, the earth goes around the sun.
That gives it a centripetal acceleration including the earth, including you, including 26.100.
The sun goes around the Milky Way, and you can go on and on.
So clearly 26.100 is not an inertial reference frame.
We can try to make an estimate on how large these accelerations are that we experience here in 26.100 and let's start with the one that is due to the earth's rotation.
...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 0.
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 completely 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 0.
And on this side of the plate, is that electric field, which is sigma divided by epsilon 0.
But inside the plate, which is a conductor, the electric field is 0.
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 0, so that's multiplied by 1.
And the reason why I do that is, because then I get another sigma divided by epsilon 0 here -- divided by epsilon 0, 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 0 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 0 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 0 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 electrostatic potential energy, is the integral of one-half epsilon 0 E-squared, over all space -- if necessary, you have to go all the way down to infinity -- and here, I have now, dV, 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 2 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 0 E-squared, over the entire volume of these two plates.
And since the electric field is outside 0, 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 0.
For E, if I want to, I can write sigma divided by epsilon 0, I can square that, and dV, 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 0 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 dL 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 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, 1 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 4 pi epsilon 0 R.
We've seen this many times, and so, by definition, the capacitance now is Q divided by the potential, and therefore, this becomes 4 pi epsilon 0 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 Van de Graaff, and I calculated it for the Earth.
Pengembangan Perkuliahan
1. Buatlah sebuah Esai mengenai materi perkuliahan ini
2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini
3. Lakukan Penelitian Sederhana dengan kelompok tersebut
4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat
5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya
Instructors: Dr. Peter Dourmashkin Prof. Bruce Knuteson Prof. Gunther Roland Prof. Bolek Wyslouch Dr. Brian Wecht Prof. Eric Katsavounidis Prof. Robert Simcoe Prof. Joseph Formaggio
Course Co-Administrators: Dr. Peter Dourmashkin Prof. Robert Redwine
Technical Instructors: Andy Neely Matthew Strafuss
Course Material: Dr. Peter Dourmashkin Prof. Eric Hudson Dr. Sen-Ben Liao
Last time I mentioned to you that charge resides at the surface of solid conductors but that it's not uniformly distributed.
Perhaps you remember that, unless it happens to be a sphere.
And I want to pursue that today.
If I had a solid conductor which say had this shape and I'm going to convince you today that right here, the surface charge density will be higher than there.
Because the curvature is stronger than it is here.
And the way I want to approach that is as follows.
Suppose I have here a solid conductor A which has radius R of A and very very far away, maybe tens of meters away, I have a solid conductor B with radius R of B and they are connected through a conducting wire.
That's essential.
If they are connected through a conducting wire, then it's equipotential.
They all have the same potential.
I'm going to charge them up until I get a charge distribution QA here and I get QB there.
The potential of A is about the same that it would be if B were not there.
Because B is so far away that if I come with some charge from infinity in my pocket that the work that I have to do to reach A per unit charge is independent of whether B is there or not, because B is far away, tens of meters, if you can make it a mile if you want to.
And so the potential of A is then the charge on A divided by 4 pi epsilon 0 the radius of A.
But since it is an equipotential because it's all conducting, this must be also the potential of the sphere B, and that is the charge on B divided by 4 pi epsilon 0 R of B.
And so you see immediately that the Q, the charge on B, divided by the radius of B, is the charge on A divided by the radius on A.
And if the radius of B were for instance 5 times larger than the radius of A, there would be 5 times more charge on B than there would be on A.
But if B has a 5 times larger radius, then its surface area is 25 times larger and since surface charge density, sigma, is the charge on a sphere divided by the surface area of the sphere, it is now clear that if the radius of B is 5 times larger than A, it's true that the charge on B is 5 times the charge on A, but the surface charge density on B is now only one-fifth of the surface charge density of A because its area is 25 times larger and so you have this -- the highest surface charge density at A than you have at B.
5 times higher surface charge density here than there.
And I hope that convinces you that if we have a solid conductor like this, even though it's not ideal as we have here with these two spheres far apart, that the surface charge density here will be larger than there because it has a smaller radius.
It's basically the same idea.
And so you expect the highest surface charge density where the curvature is the highest, smallest radius, and that means that also the electric field will be stronger there.
That follows immediately from Gauss's law.
If this is the surface of a conductor, any conductor, a solid conductor, where the E field is 0 inside of the conductor, and there is surface charge here, what I'm going to do is I'm going to make a Gaussian pillbox, this surface is parallel to the conductor, I go in the conductor, and this now is my Gaussian surface, let this area be capital A, and let's assume that it is positive charge so that the electric field lines come out of the surface like so, perpendicular to the surface.
Always perpendicular to equipotential, so now if I apply Gauss's law which tells me that the surface integral of the electric flux throughout this whole surface, well, there's only flux coming out of this surface here, I can bring that surface as close to the surface as I want to.
I can almost make it touch the conductor.
So everything comes out only through this surface, and so what comes out is the surface area A times the electric field E.
The A and E are in the same direction, because remember E is perpendicular to the surface of the equipotentials.
And so this is all there is for the surface integral, and that is all the charge inside, well the charge inside is of course the surface charge density times the area A, divided by epsilon 0, this is Gauss's law.
And so you find immediately that the electric field is sigma divided by epsilon 0.
So whenever you have a conductor if you know the local surface charge density you always know the local electric field.
And since the surface charge density is going to be the highest here, even though the whole thing is an equipotential, the electric field will also be higher here than it will be there.
I can demonstrate this to you in a very simple way.
I have here a cooking pan and the cooking pan, I used to boil lobsters in there, it's a large pan.
The cooking pan I'm going to charge up and the cooking pan here has a radius, whatever it is, maybe 20 centimeters, but look here at the handle, how very small this radius is, so you could put charge on there and I'm going to convince you that I can scoop off more charge here where the radius is small than I can scoop off here.
I have here a small flat spoon and I'm going to put the spoon here on the surface here and on the surface there and we're going to see from where we can scoop off the most charge.
Still charged from the previous lecture.
So here, we see the electroscope that we have seen before.
I'm going to charge this cooking pan with my favorite technique which is the electrophorus.
So we have the cat fur and we have the glass plate.
I'm going to rub this first with the cat fur, put it on, put my finger on, get a little shock, charge up the pan, put my finger on, get another shock, charge up the pan, and another one, charge up the pan, make sure that I get enough charge on there, rub the glass again, put it on top, put my finger on, charge, once more, and once more.
Let's assume we have enough charge on there now.
Here is my little spoon.
I touch here the outside here of the can -- of the pan.
And go to the electroscope and you see a little charge.
It's very clear.
Pengembangan Perkuliahan
1. Buatlah sebuah Esai mengenai materi perkuliahan ini
2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini
3. Lakukan Penelitian Sederhana dengan kelompok tersebut
4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat
5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya
Instructors: Dr. Peter Dourmashkin Prof. Bruce Knuteson Prof. Gunther Roland Prof. Bolek Wyslouch Dr. Brian Wecht Prof. Eric Katsavounidis Prof. Robert Simcoe Prof. Joseph Formaggio
Course Co-Administrators: Dr. Peter Dourmashkin Prof. Robert Redwine
Technical Instructors: Andy Neely Matthew Strafuss
Course Material: Dr. Peter Dourmashkin Prof. Eric Hudson Dr. Sen-Ben Liao
You can relax 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 look at the equation number three that the potential difference between point A and point A is obviously zero.
I 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 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 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 -- 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 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.
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, 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 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 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 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.
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