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• 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,

• relative units, and let's take a look at this

• in some detail. First of all the length of the