Subtitles section Play video Print subtitles Today I'm going to work with you on a new concept and that is the concept of what we call electric field. We spend the whole lecture on electric fields. If I have a -- a charge, I just choose Q, capital Q and 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 li- 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 R 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 experienced 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. The electric field is a representation for what happens around the charge plus Q. This could be a very complicated charge configuration. An electric field tells you something about that charge configuration. The unit for electric field you can see is newtons divided by coulombs. In SI units and normally we won't even indicate the um 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 -- 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 give 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 Q one and here we have Q two and here we have Q three 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 Q one alone? Well, that would be if Q one were positive then this might be a representation for E one. If Q two were negative, this might be a representation for E two, pointing towards the negative charge. And if this one were negative, then I would have here a contribution E three, 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 as a vector is E one in reference of charge Q one plus the vector E two plus E three and so on and if you have I charges it is the sum of 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 would 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 it 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 to minus one, and you have to vectorially add the two. If you are 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 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? A: [inaudible]. Do 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. A: plus two. So if you're far away from a configuration like this, even if you were here, or if you were there, or if you're way 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 the point where the E field is zero. Because if the force is here in this direction but 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,