usual by telling you what happened last time.

Not the whole thing, but just the highlights so you

can follow what's happening today.

The main things I did last time were the notion of an electric

field, which is going to be with you from now till the end of the

course.

The idea of the electric field is that if you've got lots of

charges, q_1,

q_2, q_3,

instead of worrying about the force they exert on each other,

you ask yourself, at a generic point where

there's nothing, if I put 1 coulomb here,

what will it experience?

What force will it experience?

You compute that.

So in your mind, imagine a coulomb,

and you find the force due to q_1,

it looks like that.

q_2 may exert a force that way.

q_3 could be of opposite sign,

so maybe it will exert a force that way.

You add all these vectors, they add up to something.

That something is called the electric field at that point.

There is nothing there except the electric field,

but it's very real, because if you put something,

something happens to it.

So the electric field is everywhere.

The charges are in a few places.

Electric field is defined everywhere, except right on top

of the charges, which is where it probably is

infinite.

Once you know the electric field anywhere,

if you put another charge q,

a real charge, it will experience a force

equal to qE, because electric field was the

force you would have had on the 1 coulomb,

and if you put q coulombs, it will be qE.

That's the electric field.

So you can imagine computing it for any given distribution of

charges, because you know what each one of them does.

Then I said there's one nice way to visualize the electric

field, which is to draw the field lines.

You go to each point and you ask, if I put a charge here,

a positive test charge, which way will it move?

Then you follow that thing as it moves and you get that line

and you get that line and you get that line.

You can draw these lines.

The lines give you one piece of information which is very

obvious, namely, if you are here,

the force is along that line.

But I also pointed out to you, you get more than just the

direction of the force.

You also can understand the strength of the force.

The strength of the force is contained in the density of

these lines.

Now density has to be defined carefully.

It's not like mass per unit volume.

It's an area density.

That means if you get yourself 1 square meter,

you know, meter by meter piece of wood,

like a frame, and you hold it there,

perpendicular to the lines and see how many lines go through,

that's called the area density of lines.

That you can see will fall like 1/r^(2),

because if you draw a sphere of radius r,

that area is 4Πr^(2), and the same number of lines

are going through that as any other sphere,

so it will be proportional to 1/r^(2).

But so is the electrical field proportional to 1/r^(2),

and I said, let us agree that we will draw

1/ε_0 lines per coulomb.

This is a necessity; it's just a convenience.

It's like saying I want to measure distances in inches and

centimeters for daily life.

You can measure them in parsecs and angstroms,

but you'll be dealing with nasty numbers.

So it's a convenience, and the convenience here is,

let's pick 1/ε_0

lines per coulomb.

You'll see the advantage of that, because if you ask,

what's the density of lines here on a sphere of radius

r, if there's charge q at

the center, I got that many lines,

and the area of the sphere is 4Πr^(2),

so the density of lines per unit area,

you see, is precisely equal in magnitude to the electric field

at that point.

If you had not drawn 1/ε_0,

but maybe 5/ε_0

lines, then the line density will be 5

times the electric field.

It will still represent the electric field,

but we don't want to simply represent it.

We want it to be the electric field.

It makes it easier.

Then I said, let's take a slightly more

complicated situation, two charges.

This is called a dipole, one pole and another pole,

and you can draw the field lines here.

You can see if you put a test charge, it'll run away from the

plus to the minus.

If you leave it somewhere here, it'll go like that and loop

around and come back, and you can calculate them.

This is no longer guesswork.

If you had enough time, I hope you all agree,

you can go to any point you like and find the force of

attraction due to this one, the force of repulsion due to

that one, add them up and you will get an

arrow that direction.

So one can draw these lines, and the lines tell you a story.

Then I said, let us find the field in an

analytic expression due to the dipole.

Yes?

Student: I have a question about the dipole.

If you put a test particle going in the positive direction

of the x axis, would it also _________?

Prof: No.

She's got a good point.

If you put somebody here, it will never loop around.

Can you see why?

Because as it goes further away, this is trying to pull it

back.

It's always going to be closer to this guy, so it's never going

to come back.

This line will go like that and this line will go like this.

But anything else at any other angle will loop around and come

back.

All right, now the field strength, if you want to

calculate it, you can use the formula for

E due to this one and E due to that one and add

them, and I did that for you.

I don't want to go into the details,

but I remind you in all cases, the electric field fell like

1/r^(3), because the field of each

charges goes like 1/r^(2).

And if these were on top of each other, they will completely

cancel each other.

So the reason you have a non-zero field is thanks to

a.

Therefore the answer has to contain an a in front of

it, at least n the first approximation.

But that a, from dimensional analysis,

has to come with a 1/r so that the whole thing has the

same dimension as before.

That's where you get a/r^(3).

That a times q times 2 and so on,

that became the dipole moment.

That was true here.

We verified that's true here.

Later on, we'll verify it everywhere, because there are

easier ways to calculate than what I'm using.

Then I said, forget about the field due to

charges.

Let's look at what charges do when you put them in a field.

So I took two examples.

One was a very simple example.

These are two parallel plates.

They are not two lines; they are plates coming out of

the blackboard.

They're filled with charge and this has got charge on it.

This has got - charge on it.

And therefore, the electric field will look

like this, right?

Because if you leave a test charge, it will go away from the

positive, towards the negative.

Then I said, suppose there really is a

particle here with some velocity, v_0,

what will it do?

You can see that the force on this guy is going to be q

times E.

E is pointing down.

If you divide by the mass, that's the acceleration,

also pointing down, and it's constant.

So that's like a particle in a gravitational field,

except g is replaced by this number.

So it will just curve like that, and you can calculate the

trajectory.

The final thing I did was, what happens when you put a

dipole in a uniform field.

Here as well, I think I was rushing near the

end, and even I couldn't read my stuff in the corner.

So I will go over that.

If there's something that you didn't follow,

then I will be happy to repeat that part for you.

But you should understand what the question is.

There is an electric field which is pointing like this,

as if you have two plates here, charge is here,

- charge is there.

They're producing a constant electric field in the horizontal

direction.

In that environment, I take an electric dipole whose

- charge and charge, q and -q,

happen to be oriented like that.

Question is, what will happen to this guy?

If you want, you can imagine that it's a

little massless stick, and one end you glue q

coulombs, other end you glue -q

coulombs, and your let it sit there.

What will it do?

First of all, it won't feel any net force,

because the force in this direction is q times

E and the force that direction is also q times

E, if you want,

it's -q times E and they cancel.

But that doesn't mean it won't react.

It will react, because you can all see

intuitively, it's trying to straighten this guy out and

applying a torque like that.

You follow that?

That's what it will do.

And the way to find the torque, the torque is the product of

the force and the distance between the point of rotation

and the force, and the sine of the angle

between them, that is to say,

sine of this angle.

What the sine of the angle does is to take the component of the

force perpendicular to this axis,

because if you resolve the force into that part and that

part, this part is no good for

rotation.

That's trying to stretch the dipole along its own length.

It's the perpendicular part that's going to rotate

something, so you get that times sine theta.

That you can write now as the vector equation

p x E.

Because p is equal to--I'm sorry.

I need a 2 here.

I forgot the 2, because this charge will have a

torque and that charge will have a torque and the two torques are

additive.

They are both going the same way.

Then 2q times a is p, and this is

E and the sinθ comes in the

cross product.

I'm assuming all of you know about the cross product.

Okay, final thing I did, which is, if you have a force,

you can associate with that force a potential energy.

Again, this is something you must have seen last time,

but I will remind you.

As long as it's not a frictional force,

you can say the force is connected to potential energy in

this following fashion.

Or the potential energy at x minus potential energy

at some starting point x_0 --

I'm sorry, x_0 - x is the integral of

the force from x_0 to

x.

This is the relation between the force, as delivered to the

potential, and the potential is the integral of the force.

For example, for a spring,