Today I'm going to work with you on a new concept and that is the concept of what we call electric field. We spent the whole lecture on electric fields. If I have a... A charge, I just choose Q, capital Q and a plus at a particular location, and at another location I have another charge, little q.
I think of that as my test charge. And there is a separation between the two, which is R, the unit vector from capital Q. To little q is this vector. And so now I know that the two charges, if they were positive, let's suppose that little q is positive, they would repel each other.
Little q is negative, they would attract each other. And let this force be F. And last time we introduced Coulomb's Law, that force equals little q.
times capital Q times Coulomb's constant divided by R squared in the direction of our roof. The two have the same sign, it's in this direction. If they have opposite sign, it's in the other direction.
And now I introduce the idea of electric field for which we write the symbol capital E. And capital E at that location P, where I have my test charge little q, at that location P is simply the force that a test charge experience divided by that test charge. So I eliminate the test charge. So I get something that looks quite similar, but it doesn't have the little q in it anymore. And it is also a vector.
And by convention, we choose the force such that if this is a positive test charge, then we say the E field is away from Q if Q is positive. If Q is negative, the force is in the other direction and therefore E is in the other direction. So we adopt the convention that the E field is always in the direction that the force is on a positive test charge. What you have gained now is that you have taken out the little q. In other words, the force here depends on little q.
Electric field does not. So the electric field is a representation for what happens around the charge plus q. This could be a very complicated charge configuration and electric field tells you something about that charge configuration. The unit for electric field, you can see, is newtons divided by coulomb. in SI units and normally we won't even indicate the unit, we just leave that as it is.
Now, we have graphical representations for the electric field. Electric field is a vector, so you expect arrows and I have here an example of a charge. Plus three, so by convention the arrows are pointing away from the charge, in the same direction that a positive test charge would experience the force, and you notice that very close to the charge the arrows are larger than farther away.
That it, that sort of represents, is trying to represent the inverse r square relationship. Of course it cannot be very quantitative. But the basic idea is, this is of course spherically symmetric, if this is a point charge, the basic idea is here you see the field vectors and the direction of the arrow tells you in which direction the force would be if it is a positive test charge and the length of the vector gives you an idea of the magnitude. And here I have another charge minus one. Doesn't matter whether it is minus one coulomb or minus microcoulomb, just it's a relative...
representation and you see now that the E field vectors are reversed in direction, they're pointing towards. The minus charge by convention and when you go further out they are smaller. And you have to go all the way to infinity of course for the field to become zero because the one over r square field falls off and you have to be infinitely far away for you to not experience, at least in principle, any effect from the from the charge. What do we do now when we have more than one charge?
Well, if we have several charges, here we have Q1 and here we have Q2 and here we have Q3 and let's say here we have Q of i, we have i charges. And now we want to know what is the electric field at point P. So it's independent of the test charge that I put here.
You can think of it, if you want to, as the The force per unit charge, you've divided out the charge. So now I can say what is the E field due to Q1 alone? Well that would be, if Q1 were positive, then this might be a representation for E1. If Q2 were negative, this might be a representation for E2. pointing towards the negative charge, and if this one were negative, then I would have here a contribution E3 and so on.
And now we use the superposition principle as we did last time with Coulomb's law, that the net electric field at point P is a vector, this E1, influence of charge Q1, plus the vector E2 plus E3 and so on. And if you have I charges, it is the sum over all I charges of the individual E vectors. Is it obvious that the superposition principle works? No. Does it work?
Yes. How do we know it works? Because it's consistent with all our experimental results. So we take the superposition principle for granted.
And that is acceptable. But it's not obvious. If you tell me what the electric field at this point is, which is the vectorial sum of the individual E field vectors, then I can always tell you what the force will be.
If I bring a charge at that location, I take any charge that I always carry in my pocket, I take it out of my pocket and I put it at that location, and the charge that I have in my pocket is little q, then the force on that charge is always q times E. Doesn't matter whether q is positive, then it will be in the same direction as E. If it is negative, it will be in the opposite direction as E.
If q is large, the force will be large. If q is small, the force will be small. So once you know the E field, which could be the result of very complicated charge configurations, the real secret behind the concept of an E field is that you bring any charge at that location and you know what force acts at that point on that charge.
If we try to be a little bit more quantitative, suppose I had here a charge plus three. And here I had a charge minus one. Here's minus one.
And I want to know what the field configuration is as a result of these two charges. So you can go to any particular point. You get an e vector which is going away from the plus three, you get one that goes two minus one and you have to vectorially add the two. If you're very close to minus one, it's very clear because of the inverse r square relationship that the minus one is probably going to win.
Let's, in our mind, take a plus test charge now. And we put a plus test charge very close to minus one, say we put it here. Even though plus three is trying to push it out, clearly minus one is most likely to win.
And so there will probably be a force on my test charge in this direction. The net result of the effects of the two. Suppose I take the same positive test charge and I put it here, very far away, much farther away than this separation. What do you think now is the direction of the force on my plus charge? Very far away.
Excuse me? Why do you think it's to the left? Do you think minus one wins?
So you really think the minus one is stronger than the plus three because the plus three will push it out and the minus one tries to lure it in, right, if the test charge is positive. Uh, imagine you're all the way on Mass Avenue. You think that this distance matters.
Who thinks the force is in this direction? Who thinks it's in this direction? Very good, you've helped him, really. The force is obviously in that direction because if you're very far away, the field will be the same as if you just had a plus three and a minus one somewhere here, which is plus two.
So if you're far away from a configuration like this, even if you're very far away, And if you're here or if you're there or if you weigh there, clearly the field is like a plus two charge and falls off as one over r squared. So therefore if you're far away, the force is in this direction. And now look what is very interesting. Here, if you're close to the minus one, the force is in this direction.
Here, when you're very far away, maybe I should be all the way here, it's in that direction. So that means there must be somewhere here a point where the E field is zero. Because if the force is zero, is here in this direction, it ultimately turns over in that direction, there must be somewhere a point where E is zero, and that is part of your assignment. I want you to find that point for a particular charge configuration. So let's now go to some graphical representations of a situation which is actually plus three minus one.
Try to improve on the light situation. And let's see how these electric vectors, how they show up in the vicinity of these two charges. So here you see the plus three and the minus one, relative units. And let's take a look at this in some detail.
First of all, the length of the arrows again indicates the strength. It gives you a feeling for the strength. It's not very quantitative, of course.
And so let's first look at the plus three, which is very powerful. You see that these arrows all go away from the plus three and when you're closer to the plus three they are stronger, which is a representation of the inverse R square field. If you're very close to the minus one, uh, the arrows are pointing in towards the minus one, because the one over r square, the minus one wins. And so you see they're clearly going into the direction of the minus one. Well, if you're in between the plus and the minus on this line, always the E field will be pointing from the plus to the minus, because the plus is pushing out and the minus is sucking in, so the two support each other.
But now if you go very far away from this charge configuration, anywhere but very far away, much farther than the distance between the two charges, so somewhere here or somewhere there or somewhere there or here, notice that always the arrows are pointing away. And the reason is that plus three and minus three are pointing away. is as good as a plus two if you're very, very far away. But of course when you're very close in, then the field configuration can be very, very complicated.
But you see very clearly that these arrows are all pointing outwards. None of them... come back to the minus one, none of them point to the minus one direction. And that's because the plus three is more powerful and then there is here this point and only one point whereby the electric field is zero.
If you put a positive test charge here, the minus will attract it, the plus will repel it and therefore there comes a point where the two cancel each other exactly. Now there is another way of electric field representation, which is more organized. And we call these field lines. So you see again the plus three and you see there the minus one.
If I release right here. Where I place here a positive test charge, all I know is that the force will be tangential to the field lines. That is the meaning of these lines. So if I'm here, the force will be in this direction. If I put a positive charge here, the force will be in this direction.
And of course, if it's a negative charge, the force flips over. So the meaning of the field lines are that it always tells you in which direction a charge experiences a force. A force, a positive charge always in the direction of the arrows, tangentially to the field lines, and a negative charge in the opposite direction. How many field lines are there in space?
Well, of course, they're an infinite number. Just like these little arrows that we had before, we only sprinkled in a few. But of course in every single point there is an electric field and so you can put in an infinite number of field lines and that would make this representation of course useless. So we always limit ourselves to a certain number.
If you look very close to the minus one, notice that all the field lines come in on the minus one. We understand that, of course, because the field lines want to go to the minus one. If you're very close to the plus, then they all go away from the plus, because they're being repelled.
You can sort of think as these field lines, if you want to imagine the configuration, that the plus charges blow out air, like a hair dryer, and that the minus suck in air. like a vacuum cleaner. And then you get a feeling for it that is on the left side here, this hair dryer which wants to blow out stuff and then there is that little sucker that wants to suck something in and it succeeds to some degree.
It's not as powerful as the plus three though. Have we lost all information about field strength? We had earlier with these arrows, we had the length of the arrow, the magnitude of the field was represented.
Yeah, you have lost that, but there is still some information on field strength. If the lines are closer together, if the density of the lines is high, the electric field is stronger than when the density becomes low. So if you look for instance here, look how many lines there are. per few millimeters.
And when you go further out, these lines spread out. That tells you the E field is going down. The strength of the E field is going down. It's the one over r square field, of course. If you want to make these drawings, what you could do to make them look good, you could make three times more field lines going out from the plus in this case, then return to the minus one.
So the field lines are very powerful and we will often think in terms of electric fields and the line configurations and you will have several homework problems that deal with electric fields and with the electric field lines. If an electric field line is straight, So I have electric fields, get some ram, chalk. Say we have fields that are like this, straight E field lines, and I release a charge there, for instance a positive charge, then the positive charge would experience a force exactly in the same direction as the field lines, because the tangential now is in the direction of the field line, it would become accelerated in this direction and would always stay on the field lines. If I release it with zero speed, it would start to accelerate and it would stay on the field lines. stay on the field lines.
In a similar way, if we think of the Earth as having a gravitational field, with eight oh one we may never have used that word gravitational field, but in physics we think of the, of gravity also being a field. If I have here a piece of chalk, the, um, the field lines, the gravitational field lines, here in twenty-six one hundred, nicely parallel and straight, and if I release this piece of chalk at zero speed, it will be to move in the direction of the field lines and it will stay on the field lines. So now you can ask yourself the question, if I release a charge, will it always follow the field lines? And the answer is no. Only in this very special case.
But suppose now that the field lines are curved. So here are field lines as you have seen in those configurations. It's very common. If now I release a...
A charge in here, say I have a point charge here, it will experience a force in this direction, so it will get an acceleration in this direction, so it will immediately abandon that field line. And so if now you ask me what is the trajectory of that charge, well, it could become very complicated, I really don't know. Maybe it's going like this, and by the time it reaches this point, what I do know, that then the force will be ten- to this field line, so we'll be in this direction. And so as it marches out and picks up speed, locally it will experience forces representative of those field lines, and so the trajectory can be rather complicated. So field lines are not trajectories and not even when you release a charge with, uh, with zero speed, only in case that the field lines are straight lines.
Let's now look at a field configuration which Maxwell himself, the great maestro in some of his publications, put there. There was a ratio one to four and whether It is plus four plus one or minus four minus one is not important because that's just a matter of the direction of the arrows. But uh, Maxwell didn't put arrows in.
So I leave it up to you. If it's plus four and plus one, you have to put arrows going outwards. And what you see now here is this air blower effect. Think of them as both being positive.
So there is the plus four trying to blow air out like a hairdryer and the plus one is trying to do its own thing and so you get a field configuration, field lines, which are sort of not perhaps easy but you can sort of imagine why it has this peculiar shape. If you, um, put a plus test charge in between the one and the four, then the four will repel it, but the one will also repel it. And so there's going to be a point somewhere, probably close to one, whereby the two forces exactly cancel out. Therefore E will be zero there.
In a similar way, between the moon and the earth, there is a point not too far away from the moon, where the gravitational attraction from the earth and the gravitational attraction from the moon exactly cancel each other out. That's not too dissimilar from this situation. So when you have charges of the same polarity, you always find in between somewhere a point where the electric field is zero. Let's now go to a very special case whereby I make the two charges equal in magnitude but opposite in sign and we have a name for that. We call that a dipole.
The plus charge is here and the minus charge is there. Situation is extremely symmetric, as you would expect, because they have equal power. There's one air blower upstairs and one vacuum cleaner downstairs.
If you close to the plus charge, notice that all the field lines go away from the plus and if you close to the minus, notice that all the field lines come in on the minus. You expect that. If you are far away from this... Dipole, now you have a problem.
Before we had a plus three and a minus one, and when you're far away, the plus three wins. So it's like having a plus two charge. If you're far away, you always expect the electric field then to be pointing away from the equivalent charge of plus two.
But if you add up plus and minus and they have an equal magnitude, let's say plus one and minus one, you get zero. So neither one wins if you're far away. And notice carefully, if you're very far away, indeed you do not see arrows either pointing out.
nor pointing in. Nature cannot decide. There isn't one that is stronger than the other. And that makes dipole fields very, very special. In the case of the plus three and the minus one, if you're very far away, it's like a having a plus two charge and the E field, when you go further and further out, will fall off as one over R squared.
With a dipole, your intuition sort of tells you that it will probably fall off faster than one over R squared. And that is part of a homework assignment that you have this week. In fact, I can already give you the answer.
You have to prove it. If you're far away from an electric dipole, the electric field falls off as one over R cubed. It goes faster than one over R squared.
r squared. There is not a single point in space where the electric field is zero. You can think about that, why that is the case. So these field configurations can be rather complicated and can be very interesting and each one has its own applications. Are dipoles rare in physics?
Not at all. In fact, they are extremely common. You cannot avoid them. Remember last time I told you, if you have a spherical atom or you have a spherical molecule and you bring that close to a charge, let's now think of it, you bring it in an electric field, it's another way of saying the same thing. So we have a nice spherical atom or a nice spherical molecule and we bring it in an electric field.
The electrons want to go upstream the electric field vectors. They go against the direction of the electric field. And the positive charge wants to go in the direction, wants to go downstream. And so what are you going to do? The electrons will spend a little bit more time on one side of the nucleus than they would in the absence of that electric field.
And therefore you are, through induction, turning that atom, turning that... molecule in becoming a dipole. If you have a little bit more charge on this side, averaged over time, you have the same amount of extra charge plus on that side, averaged over time.
So you make dipoles very often, whether you like it or not. And later in this course, we will learn more about the polarization of atoms and molecules creating dipoles when we talk about dielectrics. And you will see that it will have an eno- can have an enormous impact on the properties of the material. Could I make you a dipole here in class? Oh yeah, that's very easy.
To make one of non-conductors is not so easy in class. To make one of conductors is very easy. And I'm going to do that with these two spheres that you have.
Look at these two metal spheres. Conductors, free electrons, very easy for them to move. And I'm going to bring. This rubber rod, which I will rub and becomes, I think, negatively charged, if I remember correctly, and I will bring that, say, close to these two, which are touching each other.
So here is this one metal sphere and here's the other metal sphere and here. Comes the rubber, negatively charged. Ah, what's going to happen?
The electrons want to go away, so this becomes negatively charged and therefore this remains a little bit positively charged. For every one electron that has is excess here, When I start it's neutral. There will be a positive excess there because charge is conserved. You can't create charge out of nothing.
And now what I do, while this rubber is still here, while that rubber rod is there, I separate them. So here they are in contact with each other first. They have to be in contact. Wow, we get some visitors.
Don't be late. That means you, pretty boy. I'm impressed. Thank you.
Um... So what I do now is, while this rubber rod is still in place, I take them apart. And when I take them apart, this negative charge is trapped and this positive charge is trapped. And so I have thereby created negative charge on this one, positive on this one, and it's equal in magnitude, so I have a dipole.
What I want to demonstrate to you is that indeed I have positive charge here and negative here, that there is a difference in polarity between these two. And that's the way that I will do the experiment. I will not show you that the amount of charge is exactly the same on each, which of course it has to be. So let me give you some better light or we have to get the, uh, view graph off, the overhead.
You see there for the first time an electroscope. We discussed it last time. It is a... Piece of aluminum foil, very thin, with a metal rod next to it, and when I put charge on the rod it will also go onto the aluminum foil and they will repel each other and so the aluminum tinsel will go to the right.
And the more charge there is on it, the farther it goes to the right. So let me first put these two together, make sure they are completely discharged, and now I'm going to bring these two into an electric field, which is produced by this rubber rod. I have to rub with the cat fur and I believe it was negative. But if you've, you never have to remember whether it's negative or positive, of course.
That is not so important. What is in a name, after all? But it is, happens to be negative.
Okay, so now we go here. I bring it here, I hope that no sparks will fly over because that ruins the demonstration. And now, notice what I do, while the rod is here, I separate them.
So as I was holding it there, things were going on in there that you and I couldn't see, but the electrons, the rubber rod is negative, electrons were shifting in this direction and this is now positive and that is now negative. If I take this one and I touch it with the electroscope, you clearly see that there is charge on this. How can I show you now that there is charge of different polarity on the other one? Well, the way I will do that is I will approach this electroscope by bringing this sphere very close to it. And if this charge is different than the charge that is on it, the electroscope will, the reading will become smaller.
And why is that? Why will the reading become smaller? Well... Here is the situation of the electroscope now.
And here is that ball that you see on top. This is upside down there. If this is all negative, That's why there's a part. If now I approach this here with an object which is positively charged, and I claim that this one now is positively charged because this one was negatively charged, the electrons love the positive charge, so the electrons want to charge.
to come to the positive charge, so these electrons drift down again. And so if they come down, fuel will be here and so you will see this. If however I put here a negative rod, then the electron will drift down again.
which are here, want to go further away, they will stream up, and therefore the reading will become larger. So you can always, through induction, test what the polarity is of your charge. Let's hope that this one is still holding its charge while I was talking. So I claim now that if this polarity is different, and if it's still there when I approach the electroscope, come very close, that the reading should become a little smaller without even touching it. Let's see whether that works.
You see? You see, it goes down. It goes down. So through induction I have demonstrated that this has indeed a different polarity from this one. If I approach it with this one, it would go further out, unless it already is at a maximum.
Let's try that. You see, it goes further out. So not only have I demonstrated that I created the dipole, but you've also seen that by means of induction, that you can demonstrate, um, that the, there's a difference in polarity between the two spheres. If I create a dipole and I put that dipole in a, in an electric field, The dipole will start to rotate.
Let's first talk about it, why it rotates, and then I will try to demonstrate that by making a dipole. A big one, this big, right in front of you, almost as big as the one there. So let's have a, an electric field, like so.
And I bring in this electric field a dipole. A biggie. Here, this is the one I'm going to use for this demonstration.
Ping-pong balls on either side, they are conducting and they are connected with a rod which is not conducting. And so here is this dipole. So this rod is not conducting, and this is a conducting, and this is a conductor.
And let's suppose this is positive and this is negative for now. And I'll show you how we get the charge on it. Well, the positive charge will experience a force in this direction, always in the direction of the electric field, and the negative charge will experience a force always upstream. And now there is a torque on this.
And when there's a torque on this dipole, it will start to rotate clockwise. And of course, if it overshoots the field lines, when it is in this direction, the torque will reverse. It's very easy to see. And so what you will see, it's going to oscillate and if there is enough damping, it will come to a halt, more or less, in the direction of the field lines.
And this is something that I can demonstrate. First, I have to make a dipole of this kind. And the way I will do that is the following.
This is a metal bar. This is insulator. And here is this, there's these two ping pong balls, the one on this side has a yellow marker, the one on that side has an orange marker.
And I'm going to attach them, holding them up against this metal bar. In other words, here is... This dipole, it's not a dipole yet, metal, metal, and here is a metal bar, this is a conductor, which connects them.
I'm going to turn on the VandeGraaff here, and the VandeGraaff creates an electric field, so we have the VandeGraaff here. And let's suppose that this Van der Graaf creates positive charge. Sometimes a Van der Graaf creates positive charge on the dome, others can be designed to create negative charge on the dome.
And remember, for now I assume that's positive. What will happen now? Electrons want to go in this direction, so this becomes negative. Protons, positive charge stays behind, so that becomes, through induction, a dipole. Because I have them connected.
I have them connected with this metal bar. So these electrons can flow through this bar and end up here. Now I remove the bar.
And so when I remove the bar, I have created now a dipole. I have here... An insulating thread and I have a fishing rod and at the end of my fishing rod I have now a permanent dipole.
With that permanent dipole I'm now going to probe the electric field around this VandeGraaff. I could have chosen the same VandeGraaff but there's a reason why I picked this one. And as I walk around this VandeGraaff you will see that this at the end is this dipole, that the dipole always wants to go radially inwards or outwards, depending on how you look at it, of this field. So I can probe this field and make you see for the first time that there is indeed somewhere here a strong radial field going in or out of the VandeGraaff.
And now comes something very interesting, which I found out this morning for the first time when I did this experiment. If the other Van der Graaf there is also positive when I run it, how do you think this dipole is going to align then if I walk into it? Will the negative ball?
Be closer to the VandeGraaff or will the positive one go closer to the VandeGraaff? So I give you thirty seconds to think about it. So I make the dipole as it is here. Let's assume this one is positive, this VandeGraaff. So this side becomes minus, I call that A, and this side become positive, that's B.
I now walk with this. Dipole, I bring it in this field, and let's assume that one is also positive. We don't know that yet. How will the dipole align now? Will A go inwards or will A go outwards?
Who thinks A goes inward? Very good. Who says A goes outwards?
OK, A will go inwards if the two VandeGraaffs have the same polarity. So if that doesn't happen, that doesn't mean that physics doesn't work, it means the two VandeGraaffs have different polarities. And we'll see what happens.
So let me first then create a dipole. So here is the the dipole. It's shorted out now. I turn on the, this VandeGraaff, so induction takes place. Remember that the yellow is pointing towards the VandeGraaff and that the orange is away from the VandeGraaff.
OK? So I induce a dipole. Oh, I really should redo that.
I don't know what happens when it... Have to remove the field first. OK.
The yellow was inside, right? Was that the way it was? OK.
Yellow inside. There we go. So now it's creating a dipole through this metal boy.
And I break contact. And this should now be a dipole. Now I turn on the field of the...
So if the polarity is the same, yellow will go in. I will try to swing it a little. Notice two things.
It's going to line up beautifully radially, but the yellow is not in, the yellow is out. So the two Vandegrounds have different polarities. But you will see, they rotate nicely. And they end up beautifully radial. And when I go all the way around here, again, they may swing a little, they may oscillate a little, but through damping they will come to a halt.
And look, the field is indeed beautifully radial and the yellow is pointing outwards. The two Van der Graafs have different polarities. So you've seen how we can create a dipole, and you've also seen how we often can make statements about, uh, the specific polarity.
I can probe an electric field using grass seeds in oil. Grass seeds... I elongate it and when I put a grass seed in an electric field, it will become polarized.
There's nothing you can do about it. Here's a grass seed and the electric field is like so. And so the electrons want to go as far away in this direction as they can, through induction, and so this side remains positive.
And so what is this grass seed going to do? It's going to rotate. It's going to line up with the electric field. And this is the way that I'm going to show you now field configurations in the vicinity of a dipole. And I will also show you then field configurations in the vicinity of two charges which have equal polarity.
You may have seen this in high school with magnetic fields, with iron files, that's kid stuff, that's the easiest thing to do. This is the real thing, this is the electric field. I bet you've never seen electric fields which are traced by these mysterious seeds.
So I'll give you some light that may optimize the demonstration. These seeds first have to be oriented in a way so that it is chaos. The first thing you see is I'm going to make this, I believe it's going to be a dipole first, almost certain. So I'm going to charge one positive and charge the other negative and then we'll see how these grass seeds will form each other.
Watch closely. There you go. My goodness, that is a wonderful dipole field. Of course, we don't know which one is plus or minus because the grass seeds have no arrows on them. But you clearly see these incredible lines, radially inwards or outwards on each one of the charges and then you see these nice arcs in between.
Who could see it easily? Okay, you got something worse for your tuition. Put a little bit more charge on maybe. Very clear. And now, which is perhaps more interesting, I'd like to show you the field surrounding two charges, but now the charges are both the same polarity.
So we have to undo the memory of the grass seeds. Okay, now we'll try to make them both the same polarity. And then watch this hair blower effect that I told you about.
Ah, look at that. Great. Now you really cli- clearly see these, these field lines, you see in between how the two air blowers are competing with each other.
Very impressive. All right, so that's the way you see field lines now, electric field lines. And some of you may have seen with iron files, uh, magnetic field lines.
If I have the VandeGraaff, And I have the Van der Graaf here. And let's suppose the Van der Graaf is positive. I don't know whether it's positive or negative, let's suppose it's going to use the one over there.
And I'm going to stand here on the ground, Walter Lewin, what is going to happen with me? Through induction, the electrons being sucked out of the earth and coming up because they want to go close to the positive charge. So...
I will become negatively charged. What will the field lines do? Oh, they will be extremely complicated, very complicated.
But something like this maybe, maybe something like this. Uh, some may come out here, some may end up on my neck here, some may go here, like so. Very complicated field configurations. But I want to probe that field, somehow, a little, get a feeling for how that, what that field is like. And the way I'm going to do that is I'm going to put a charged balloon, there you see the balloon, it's a conductor, I'm going to put a charged balloon and put it here, say.
Well, if it is a positively charged balloon, it will take off in that direction, right? The force is always tangential to the field lines. It will abandon the field lines, it won't stay on the field lines.
There's a lot of damping on the balloon, that's why I chose a balloon, so it will move relatively slowly and it will ultimately maybe add up on my head right here. Once it adds up on my head there, so it comes in maybe like this, now it will get the negative charge from my head and so it will become immediately negatively charged and so the force now will reverse and will be in this direction, tangential to this field line. And so it will go back.
When it hits the Van der Graaf again, it will get positive charge. Reverses polarity and it will go back. And so it will bounce back in force between me and the VandeGraaff and that gives you some rough feeling of what this field conflagration is about, although I want to remind you that the charge does not follow exactly field lines. So I'm going to sit here and I will be part of this. That's probably going to be positive.
I will automatically become negative, there's nothing I have to do. All I turn is the VandeGraaff and we have to put a little bit of charge on that balloon. It will probably do that by itself but I can always give it a little kick. So that it goes through the Vantagraph, there it goes. Oh, I had to...
Because my, my glasses are a good insulator, so I better take my glasses off, so that every time it hits me, changes polarity. So this is a way you can do physics and have fun at the same time. See you Monday.