# 3. Gauss’s Law I

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Prof: All right,

class, I thought I’d start as 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, U is

½kx^(2), and F=-kx.

If you go to this one,

we tell you U(x_0) –

U(x) is equal to the integral of -kx dx

from x_0 to x.

So that gives you

kx_0^(2)/2 − kx^(2)/2.

And by comparison,

you can see U(x)=½kx^(2).

Actually, this is not the

unique answer. Do you know why?

Given this formula,

can I immediately say this guy corresponds to that,

this one corresponds to that one?

Is there some latitude here?

Yes?

Student: You can always

add in a constant. Prof: You can add a

constant to both, because if I said that,

that certainly works. If I add 92 to both,

it still works, because the 92 extra doesn’t

matter when you take the difference.

So it’s conventional to simply

pick the constant so that the formula looks simple.

Coming to other expression,

if you had a torque, which is

pEsinθ, you can ask–that’s the

torque–it’s minus, because it’s trying to reduce

the angle θ– you can ask what U leads

to that. And you can see it’s

-pEcosθ. See that, take the -U

and take the derivative, you’ll get the torque.

But if you got two vectors,

p and E and you see the cosθ,

I hope you guys know that you can write it as a dot product.

So that’s the end of what I did

last time, okay? The potential energy is

proportional to the dot product of p with E.

The torque is equal to the

cross product of p with E.

And what does

pEcosθ mean if you plot it as a function of

angle? It will look like this.

This is Π and this is 0.

That means it likes to sit here

and if you deviate a little bit and let it go there,

it’ll rattle back and forth, just like a mass spring system.

In fact, you can very easily

show, near the bottom of the well, the potential energy is

proportional to θ^(2).

That’s because

cosθ can be written as 1 −

θ^(2)/2 θ^(4)/4!,

etc. for small angles.

If you just keep that term,

you will find it looks like this.

Not very different from

U=kx^(2). So what x does with

forces, θ does with rotations.

All right, so this is what we

did last time. Now I’m going to do the new

stuff. So new stuff is going to give

you–I think it’s useful, because it tells you the level

at which you should be able to do calculations.

So here’s a typical problem.

You have an infinite line of

charge, of which I will show that part, and somebody has

sprinkled on it λ coulombs per meter.

So it is not a discrete set of

charges; it’s assumed to be continuous

and it’s everywhere. I’m just showing a few of them,

and if you cut out one meter, you’ll find there’s λ

coulombs there. And you want to compute the

electric field you will get due to this distribution,

everywhere. So you want to go somewhere

here and ask what’s the electric field.

That’s what we’re going to do.

Let’s go here.

You will see why.

Now first of all,

you’ve got to have an intuition on which way the electric field

will point. You have a feeling?

Yes.

It will point here, this way.

Why not like that?

Yeah?

Student:

>Prof: Okay.

She said the horizontal parts

will cancel. That’s correct.

Another argument from symmetry

is that if anybody can give you a reason why it should tilt to

the left, I can say, “Why don’t you

use the same argument to say it will tilt to the right?”

Because this is an infinitely

long wire and things look the same if you look to the left and

if you look to the right. And the field you get should

have the same property. If this was a finite wire,

I wouldn’t say that, because in a finite wire,

that can be a tilt, somewhere here.

But infinite wire,

it cannot tilt to the left or to the right,

because each point has the same symmetric situation to its right

and to its left. In a finite wire, it’s not true.

Life to the left is different

from the life to the right, but for infinite wire,

you know it cannot be biased one way or the other.

It’s got to go straight up.

Secondly, you can find the

field here, here, here or at anywhere at the same

distance, you’ve got to get the same answer.

Again, because if you move two

inches to the right, it doesn’t make any difference

with an infinite wire. You’ve still got infinite wire

on either side. So we’ll pick a typical point

and calculate the field and we know that answer is going to be

good throughout that line. So now I take this point here.

I want the field here.

I’m going to make that my

origin. Then I take a piece of wire of

length dx. dx is so small that I

can treat it as a point. Now the dx I’ve drawn is

not a point, but in the end, we’re going to make dx

arbitrarily small so that it’s good enough.

It’s like a point and it is at

a distance x from the origin, and let’s say that

distance is a. So let’s find the field due to

just this guy, the shaded region.

Think of a charge.

How much charge is sitting here?

I hope you all agree,

the charge sitting in this region is just λ

times dx. That’s just the definition.

If you’ve got that many

coulombs, imagine a test charge of 1 here.

Well, it will push it this way,

and the field due to that, I’m going to draw as

infinitesimal, so I’m going to call it

dE. You can call it E,

but dE is to remind you, it’s a tiny field,

due to a tiny section dx.

Now that electric field is

biased to the left, but for every such section you

find here, I’ll find a section on the

other side that’s precisely biased to the right by the same

amount. Therefore the only part that’s

going to survive due to this guy,

combined with this, will be the portion here,

which is the vertical part of that force.

So let me find the contribution

first, only from this one, then we will add the

contribution for this one. For that one,

you just find the vertical projection.

So how much is that?

Remember, the Coulomb’s law for

the electric field is q/4Πε

_0r^(2). So this is the q.

That’s the 4Πε

_0. r^(2) is that distance

squared, which is x^(2) a^(2).

That’s really like the field of

a point charge at that distance. But now this is the magnitude

of the electric field vector at this angle, but I want the part

along the y direction. So I’ve got to take cosine of

that θ. You guys follow that?

If you took this vector,

this part is dEcosθ.

That part is

dEsinθ. But that angle and that angle

are equal, and cosθ for this

triangle, you can see,

is a divided by (a^(2)

x^(2))^(½). So this is the electric field

in the y direction. I’m going to call it

dE, in the y direction,

due to the segment dx. The total electric field is

obtained by adding all the dEs or adding all

the contributions from all the segments on this line.

And that goes from – infinity

to infinity. All right, so now it’s a matter

of just doing this integral. So this gives me

(λa/4 Πε

_0) dx/(x^(2) a^(2))^(3/2),

integrated from – to infinity. Now I can make life a little

easier by saying that this function is an even function of

x. That means when you change

x to -x, it doesn’t care.

Therefore the contribution from

a positive x region is the same as the contribution

from a negative x region. That also makes sense in this

picture here, because if you look at the

field I’m computing, this section and this section

give equal contributions in the y direction.

But even if you did not know

any of that background, as a mathematician,

if you see this integral, you would say,

“Hey, put a 0 there and put a 2 here”

namely,integrate over half the region,

because the second half is giving you the same answer.

So you double the integral,

but cut the region of integration in half.

So at this point,

you are free to look up a book, if it was an exam,

but maybe not even if it was an exam at your level.

You should be able to do this

integral. So integrals have been around

from the time of Newton and the question of an integral is,

find the area of some graph with this particular functional

form. And the answer to any integral

is that function whose derivative is the integrand.

So what you have to do is guess

many answers until you get the right one.

But people have been guessing

for hundreds of years, and there’s big tables of

integrals with all the integrals you want.

But you should still be able to

do some integrals from scratch and I’m going to tell you how to

do this one. But before you do the integral,

you’ve got to have some idea what the answer is going to look

like. I want you to get some feeling

about this. Answer depends on what,

is the first question. What’s the answer going to

depend on? Student: a.

Prof: a,

you understand? Whatever this is depends only

on a because 0 and infinity are not going to be

present as part of the answer. If the lower limit was 5,

it can depend on 5, but it doesn’t depend on any

other thing, other than a.

Then from dimension analysis,

I got a length squared to the 3/2,

that’s length cubed, and a length on the top,

so whole thing should look like something over length squared.

The only length I have is

a, so it’s going to look like 1 over a^(2) times a

number. Once you got the number,

you’re done. So I’m going to do all the work

now to show you that the number is actually just 1.

This number will turn out to be

1, in which case, you will find it’s

λ/2Πε _0a.

Well, let’s see how we get the

number to be 1. So does anybody know what trick

you use to do this integral? This is whatever,

math 120 or– yes? Student: Use

substitution? Prof: Yes.

What substitution?

Student: x^(2)

a^(2)=U. Prof: That won’t help

you. Yes?

Student: Can you use

trigonometric substitution? Prof: Yes.

Trigonometric substitution.

Which one?

Okay, look–no,

no, I don’t blame you. I know the answer because I’ve

seen it, but if I have to work on it, I’ll try for a while

before I got it right. The whole idea is,

we don’t like all these 3 _______ here.

We want to turn that into

something nice. So I’ll tell you what the

answer is. You can all marvel at how

wonderfully it works. So what we are going to do is

to introduce an angle theta–nothing to do with the

angle in the problem–so that x=tanθ.

That means instead of going

over all values of x, I’ll go with the suitable

values of θ– I’m sorry, this would not even

be correct dimensionally. x=

atanθ. You can see that every x

that I want, I can get by some choice of

theta, because tanθ goes from 0 to infinity when

θ goes from 0 to Π/2.

You cannot say x=

acosθ, for example.

You are doomed.

If x is

acosθ, the biggest x you can

get is a, whereas I want this x to

go from 0 to infinity. So when you make the change of

variables, you’ve got to make sure that

for every x you want, there is some θ

that will do it. Then the next thing you do,

you say dx/dθ=

a times derivative of tanθ which is

sec^(2)θ. Then you write that as

dx=that. What that means is an integral

dx is related to an integral dθ

by this factor. Therefore going to the

integration here, I’m just doing that part,

which is going to be a sec^(2)θ

dθ. θ goes from 0 to

Π/2. Now let’s look downstairs.

Downstairs I’ve got

x^(2) a^(2). x itself is

atanθ, therefore a^(2) times 1

tan^(2)θ. 1 tan^(2)θ happens

to be sec^(2)θ. That to the power 3/2,

which is what I want, will give me an

a^(3)sec^ (3)θ.

So what do we get?

You can see as promised I get a

1/a^(2) and I get integral of 1/secθ,

which is cosθdθ from 0

to Π/2. Yes.

And integral of

cosθ is sinθ from Π/2 to

0. That just happens to be 1.

So the final answer is what I

gave you here, E.

Well, E is the vector.

I’ve just shown you the

magnitude, but we’ve all agreed what the direction is.

The direction is away from the

wire. So if you like,

if you look at this wire from the end, the lines will look

like this. If the infinite wire is coming

out of the blackboard towards you and you look at it this way,

if you go too close, you’ll poke your eyes out.

Look from here,

you’ll see the lines are going out radial everywhere.

The question is,

how do the fields get weak? How does it weaken with

distance? It weakens like 1/a.

That’s a big of a surprise,

right? The field away from the wire

doesn’t fall like 1 over distance squared,

but falls like 1 over distance. The reason is that every

individual portion of the wire has a contribution that does

fall like 1 over distance squared, but it is an infinite

wire. When you add it all up,

the net answer goes like 1 over the distance.

The field away from a wire

falls like the distance from the wire, on the perpendicular from

the wire and there’s pointing away from the wire.

That’s it.

Okay, so that’s one calculation.

Then I’m going to do one more

and that’s going to be the end of the tough calculations.

Second calculation is going to

be an infinite sheet. On the infinite sheet,

the appropriate quantity is called the charge density,

which is coulombs per meter squared.

That means if you cut out a

tiny piece, the charge on it will be sigma times the area of

that piece. So there is positive charge

everywhere here, and the number of coulombs per

unit area is called sigma. These are standard.

λ is coulombs per meter,

σ is used for coulombs per unit area.

The question is,

what’s going to be the electric field at some point away from

that plane? Once again, I think we can all

agree that the electric field at some point from the plane will

not depend on where in front of the plane you are standing.

Are you standing here or are

you standing there? It doesn’t matter,

because it’s an infinite plane. If I moved 1 inch–I’ll tell

you why it won’t matter. If I moved 1 inch and the

answer changed, I should get the same change if

I didn’t move and somebody moved the sheet 1 inch the other way.

But when I move an infinite

sheet the other way by 1 inch, it looks exactly the same.

It’s got to produce exactly the

same field. So you can always ask,

what will happen if I move to the left, the same as what will

happen if the sheet moves to the right?

The sheet moving to the right

looks exactly like the sheet before.

The answer won’t change,

therefore the answer won’t change for you if you move to

the left. I’ve got infinite plane below

you. As long as you don’t change the

distance from the plane, you navigate perpendicular to

it, no matter where you are, you will get the same answer.

Same answer,

meaning same direction of the field, same magnitude.

And that direction has to be

perpendicular to the plane, again for symmetry reasons.

If you tilt it in any one

direction, you have no reason to do it.

For example,

if you tilt it this way, I can take the infinite plane

and rotate it, then the tilt will be in some

other direction, maybe like that,

but the rotated infinite plane looks the same.

In other words,

if the cause does not change, the effect should not change.

If I can do certain things to

the infinite plane that leave it invariant,

then I can do the same transformation to the location

of the point, and that shouldn’t have a

different answer. So the plane has the property

that when you slide it up and down parallel to itself,

or twist it and turn it, it looks the same,

therefore the field pattern should have that property.

Therefore the field has to be

the same at all distances from the plane anywhere on top of the

plane, and it’s going to point this way.

But you can also find out in a

minute–by the way, you don’t need any of the

symmetry arguments. You just do the calculation by

brute force, it will have these properties.

But it’s good to know what to

anticipate, because maybe you made a mistake somewhere.

It’s good to know some broad

features. So none of this is needed to

calculate, even in that problem. Go ahead and find the electric

field not where I found it, but 2 inches to the right.

You’ll find the answer looks

the same. So those symmetry properties

will come out of the wash, but it’s good for you to

anticipate that, and that’s where you should

look at the symmetry of the source.

For example,

the source was a ball of charge.

You know if you rotate the

ball, when I’m not looking to rotate the ball,

it’s going to look the same. That means the field pattern

should have the property, when you rotate it,

it looks the same, because the same cause should

produce the same effect. Anyway, going to this problem

now, let’s find the electric field here.

Okay, now this is going to be a

stretch for me to draw, so I’m going to try,

but you’ll have to go look at some textbook if you want a

really nice looking picture, but this is the best I can do.

I take a ring of radius

r and thickness dr.

I take an annulus,

and I ask, what will that ring do to this point?

So let’s take a tiny part of

that ring, this guy. Well, for that,

you just did what you did, you draw the line here.

You’ll produce a dE

that looks like this. What is its magnitude?

Magnitude is just given by

Coulomb’s law. The q there is sigma

times the tiny area, dA.

Let’s call this dA.

dA is the name for a

small area. σ times dA is the

name for a small charge. That charge will produce a

force, 1/4Πε_0,

square of the distance, r^(2) a^(2).

Finally, here is where the

symmetry comes in, can you see that for every

section here, I can find an opposite section

that will cancel everything but the part perpendicular to the

plane? So I should only keep this

portion of it. Namely, I should take the

cosθ. The cosine of that

θ is the distance a,

just like in the other problem, a/(r^(2)

a^(2))^(½). This is now dE.

If you want,

you can put this following symbol,

dE_perp, meaning perpendicular to the

plane. Yes?

Student: Do you need to

multiply by 2 again, because you’re __________?

Prof: Let’s be careful.

Her question was,

should I multiply by 2, because of this guy here,

right? In fact, I should multiply by

all kinds of other numbers, because so far,

I’ve found the field only due to this segment here.

I’ve got to add the field due

to that and that and that and so on, right?

What will that contribution be?

For every one of them,

this factor, (r^(2)

a^(2))^(3/2) is the same. They all contribute to the same

factor, so when I added the shaded region,

I’ll just get the area of the shaded region.

All these dAs,

if you add them up, what will I get?

It will be sigma over

4Πε _0.

Now you’ve got to ask yourself,

what’s the area of an annulus of radius r and thickness

dr? So take that annulus,

take a pair of scissors and you cut it, and you stretch it out

like that, it’s going to look like this.

This is dr and this is

2Πr. So the area of an annulus is

just 2Πr dr. So the sum of all these areas

is 2Πr dr and then I’ve got here (r^(2)

a^(2))^(3/2). But now this is the dE,

due to annulus of thickness of dr.

Then I’ve got to integrate over

all values of r, but r goes from 0 to

infinity. So I have here

σ/2ε_0 times rdr,

divided by (r^(2) a^(2))^(3/2),

0 to infinity. Student:

>Prof: Did I miss a pi?

Student: Shouldn’t it

be sigma over ____ pi? Prof: There is a 2 pi

here. Student: Oh, okay, yeah.

Prof:

2ε_0. So do you understand what I did?

I broke the plane into

concentric rings and I took one ring and looking head on at that

ring, I took a portion of that ring

and see what field is produces. And I know that even though the

field due to that is at an angle,

the only part that’s going to remain is the part perpendicular

to the plane, because the counterpart to this

one on the other side will produce a similar field with the

opposite angle here that will cancel,

so only the part perpendicular will survive.

Then I found out that the

contribution from every dA had exactly these

factors. They all had the same r

and they all had the same a,

so some of all the dAs, all I have to add is

2Πr dr. And that’s the contribution

from this annulus, then I still have to look at

annulus of every radius, so that’s the integral over dr.

Yes?

Student: What happened

to the a _______? I thought it was a

over– Prof: Oh, I’m sorry.

It’s there.

Thank you.

There is an a still here.

Yeah, I would have caught that

guy in a while, but I’m always happy when you

do that. That’s correct.

Okay, so now about how this

integral. Do you have any idea what you

might do now? Yes?

Student: Use

substitution. Prof: Right.

What substitution?

“Use substitution”

is a pretty safe answer, but you’ve got to go a little

beyond that. Student: Substitute

r^(2) a^(2) for the ________.

Prof: Yes.

You can do that in this

problem, because there’s an r on the top.

If you didn’t have the

r, you couldn’t do that, but now you can.

I’ll tell you how it works.

First of all,

you can always do that tanθ substitution

even here. It will always work.

The tanθ

substitution, if you put it here,

it will still work. You can go home and verify

that, but I will do it a different way now.

I will say, let w=

r^(2), then dw is equal to

2r dr. So if I come here,

I can write it as aσ/

4ε_0. I borrow a 2 top and bottom to

make it dw. w also goes from 0 to

infinity, but now I get (w a^(2))^(3/2).

Now this is simple integral,

dx/x some number to some power is x^(n

1)/(n 1), but n is now -3 over 2.

So you get aσ/

4ε_0, divided by (w

a^(2))^(-½), divided by -½,

which is -2 on the top, and that goes from infinity to

0. So I’m not going to do this

much slower than this. This is the kind of integral

that you can see right away, or you can go and work out the

details. This is something you should do.

If you have trouble with such

integrals, then you should work harder

than people who don’t have trouble with such integrals,

because you should be able to do x^(n 1)/(n 1),

and know that n 1 is -½, and when it comes upstairs,

it becomes -2. Now if you look at this

integral, in the upper limit omega’s infinity,

you get 1 over infinity, which is 0.

The lower limit when omega is

0, you get 1/a, and that will cancel the

a here, and you will get

σ/2ε_0. So that’s the final answer.

So the electric field of this

infinite plane, if you look at it from the

side, looks like this. The

σ/2ε_0. So what do you notice about

this one that’s interesting? Student: It doesn’t

depend on the distance. Prof: It doesn’t fall

with distance. No matter how far you go from

this infinite plane, the field is the same.

Again, each part of it makes a

contribution that falls like 1 over distance squared.

As you go further and further

out, you might think the field should get weaker,

right? How could it not get weaker?

They’re moving away from

everything. At least with the line charge,

it didn’t go weaker like 1/r^(2),

but it did get weaker. How can you go further and

further from a plane? You are going further away from

everybody? How could it not matter?

Yes, any ideas?

For example,

if you go very close to the plane, what happens?

If you go really close to the

plane what happens is, the field in each section here

looks like this. Therefore the part that’s

useful is very small. If you go further away,

you get things like that. Maybe it’s a little weaker,

but the part that’s useful, this one, is getting bigger.

So by magic,

these tendencies cancel in the end.

It doesn’t depend on distance.

Now unless you do the integral,

you will not know it doesn’t depend on the distance,

because you can give arguments for why it’ll get weaker,

arguments for why it’ll get stronger.

The fact that it’ll precisely

be independent of distance, you have to do the calculation.

Yes.

Student: What’s the

negative sign? Prof: Negative sign

where, here? Student: Yes.

Prof: -2 is there,

but the upper limit is infinity.

Student: Oh, okay.

Prof: All right.

Now here’s the third problem,

and the good news is, I’m not going to solve it for

you, but I’ll tell you what it is.

Here is a solid ball of charge.

It’s got some charge density

ρ coulombs per meter cubed. So ρ is the standard name.

You use density for mass over

volume and you use the same symbol rho for charge per unit

volume. So somebody’s assembled a blob

of electrical charge, and 1 cubic meter of that has

ρ coulombs. You want to find the field due

to this one. Now when you do a similar

problem in gravitation, it is generally assumed that

when you’re outside the sphere, the whole sphere acts like a

point charge with the entire charge sitting at the center.

But you actually have to prove

that. That’s what took Newton a long

time to prove. He knew it was true but he

couldn’t prove it, because for that,

you’ve got to be able to do integral calculus.

And even today,

to find the field due to a sphere using integration is

quite difficult. Think about what you have to do.

You want to sit somewhere here.

First of all,

for a sphere, we know the field is going to

be radial. It doesn’t matter where you

pick, everything looks the same. You can decide to be

horizontally here at that point. Then you’ve got to divide the

sphere into tiny pieces, tiny little cubes,

each with some charge rho times the volume of the cube,

and that will exert a force like this.

And you’ve got to integrate

over the volume of the sphere, but each portion is at a

different distance and a different angle.

You’ve got to add it all up.

That’s why it’s a tough problem.

So to solve that tough problem,

we’re now going to use a very powerful trick and that trick is

called Gauss’s law. So we’re going to learn today

about the Gauss’s law. Now a prelude to that,

you need a little more mathematical definitions,

but they’re not bad. I just have to tell you what

the definition is. Suppose I have in three

dimensions a tiny little area, like a snowflake,

but it’s flat and it’s rectangular, let’s say.

I want to tell you everything

about it. I want you to be able to

visualize the area. What can I do to specify this

little thing? First I have to tell you how

big it is. If it’s a tiny area,

let this area be dA meters squared,

but that doesn’t tell me the orientation of this area,

because that area could be like this,

it could be like this, it could be tilted in many

ways. So I want to tell you it’s an

area in a certain plane, what should I do?

How do I nail down the plane in

which the area is located? Yeah?

Student: Define the

vector that’s perpendicular to that surface.

Prof: Define a vector

normal to that surface, because if you draw that

vector, then there’s only one plane perpendicular to it.

Then we can follow,

we can then form a vector, dA.

It’s a tiny vector whose

direction is perpendicular to its area and whose magnitude is

the value of the area itself. So areas can be associated with

vectors. You may not have thought about

it that way, but you can by this process.

I’ve told you,

there’s only one ambiguity even now.

Do you know what that one– yes?

Student: Which

direction. Prof: There are two

normal’s you can draw to an area, right?

We’ve got an area like this,

it can come out towards you or go away from you.

Therefore simply drawing that

rectangular patch is not enough to nail that.

That’s like saying,

“Here is a vector.” That’s not enough.

Where is the head and where is

the tail? That’s not a vector.

That is a vector.

Similarly, this area has to be

specified some more and here is what you’re supposed to do.

You take that area here,

I’m just drawing it another place, draw some arrows,

then circulate around it in one sense or another.

I picked a particular sense in

which they’re going around. Then use the famous right hand

rule, where your fingers curl along

the arrow and your thumb points in some direction,

that is the direction, the area of the vector.

If the arrows are running round

the opposite way, then your thumb will point into

the blackboard. So an area like this is like a

vector without a head. Area like this is a signed area.

It’s an area that’s got a

magnitude and unique direction. So get used to the notion that

a tiny planar area can be represented as a vector.

Another way to see that is,

if you took any area, a rectangular area like

this–square is a special case– if you took two vectors

A and B that form the two edges,

then A x B is just double the area.

It’s the fact that given two

vectors, I can find the third vector

perpendicular to them, up to a sign,

is what makes a cross product possible,

only in three dimensions. You cannot have a cross product

of two vectors and four dimensions because in four

dimensions, if I pick two vectors,

they’ll be two other directions perpendicular to that plane.

Only in 3D, there’s only one

direction left. The question is,

is it in or out? That you pick a sign in the

cross product, A x B is

something that goes from A to B. Or for an area,

you draw arrows around the edge in a certain direction.

So area is a vector.

You have to get used to that

notion, along with all the other vectors you know.

Now I’ll tell you why that

becomes useful. So we’re going to take–let’s

see– there is a rectangular tube

which has got a height h and a width w,

and some fluid is flowing along the tube with the velocity

v along the length of the tube.

You got that?

It’s like an air duct.

Stuff is going through that

tube. It’s got a rectangular cross

section. The cross section area is

hw. If the fluid is going velocity

V along the length of the tube,

what is the flow rate, which is equal to meter cubed

of stuff flowing per second. I’m going to denote it by the

symbol Φ. If I wait 1 second and I watch

all the fluid go by me, past any cross section,

how much stuff goes by? I think you can all see,

if I wait 1 second, the fluid whose front was here

would have advanced to here, and the volume here will be

v times 1, because in 1 second,

it goes a distance v. So the flow rate will be

hwv or area times v.

That makes sense?

The faster it’s going,

the more stuff you get. The bigger the cross section,

the more stuff you get. This doesn’t depend on a

rectangular cross section. It can be cylindrical pipe

carrying oil. Again, the flow rate is area

times velocity. But I’m going to write this in

terms of vectors, because I know the velocity is

a vector. But now I have also learned

area is a vector, because area vector here,

you can draw a vector perpendicular to this area,

and I’ll draw up this convention that it’s area

vectors along this way, rather than the other way.

Then I can write this

v⋅A. Let’s check that

v⋅A is correct.

v⋅A

is the length of v, the size of v,

the size of A and cosine of the angle between them.

Here we’ve got to be very

careful. The velocity vector is like

this. It’s perpendicular to the plane

but don’t say it’s cosine of 90 degrees.

The area vector is

perpendicular to the area itself.

Do you understand that?

When I take the dot product and

I will ask for the angle between the area vector and the velocity

vector, that angle is 0. For the area,

there’s a little confusion. The vector representing it

happens to be perpendicular to the plane of the area itself.

So if you remember that,

then v⋅Ais

vA times cos 0, so this indeed is one way to

write the flow. This flow is also called a flux.

But now let us do the following.

Let’s take the same problem,

and I have this area here. Let me now take a tilted area

like this. It also goes from the ceiling

to the floor but still turned at an angle θ.

So it’s a bigger area than the

original one. How much bigger?

That area prime,

I claim, is equal to the base w, times this side.

This side is

h/cosθ. θ is the angle

between these two planes. It’s a bigger area,

but you all know that just because it’s a bigger area,

it doesn’t intercept more fluid per second.

Any stuff crossing this guy

also crosses this at the same rate.

So how am I going to get the

same rate? The flux is not going to be

v times A’. It’s going to be

vA’cosθ. But θ is the angle

between the area vector and the velocity vector,

which is the same angle here. So the moral of the story is,

v⋅A’, or v⋅A in

general is the flux or the flow, off of any vector across an

area. If it’s a fluid that’s flowing,

then v⋅A is the fluid flow past that area.

If you need the dot product,

you need the cosine of the angle, because the area,

if it’s not perpendicular to the flow, it’s not useful.

In fact, you can take a huge

area parallel to the flow and nothing goes through it.

So area is most effective if

the plane of the area is normal to the fluid,

or the area vector is parallel to the velocity of the fluid.

So that’s that lesson.

Okay, this had nothing to do

with the electric field, but we’re going to come to the

electric field now. This is just a warm up.

Let’s come to the electric

field and see what’s going on. So I take a charge q and

I draw the lines coming out of it.

How many lines do I get

crossing a sphere S? Well, we know that we have

agreed to draw 1/ε_0 lines

per coulomb, so this q here,

that many lines cross the sphere.

I’m now going to relate it to

something I can do with the electric field as follows.

I’m going to say that if I go

to that sphere, I look at the electric field.

Electric field is in that

direction, E. And any portion of the

sphere–and I want this is where you got to __________

imagine–take a tiny part of the sphere.

There’s a tiny area that’s got

a size dA. What is its direction?

Direction of the area vector is

radial. You understand?

The area is on the surface and

normal to that is the radial vector, which I always denote by

e_r. The electric field is equal to

q/4Πε _0(1/

r^(2))e _r.

Therefore

E⋅dA=q/4Πε

_0(1/ r^(2))dAe

_r ⋅e_r.

This is a dot product of the

area vector with the electric field vector.

The two unit vectors are

parallel. The dot product of that guy

with itself is just 1. So this is the number of lines

crossing the tiny area, because electric field

numerically is equal to the line density.

So those lines crossing this

area, this is the number of lines crossing that patch,

which is given by E⋅dA.

If you add up all the lines,

you must add up all the dAs.

Sum of all the dAs on a

sphere of radius r is just q/4Πε

_0r^(2) times sum of all the dAs on a

sphere, that is just 4Πr^(2) .

So you get

q/ε _0.

In other words,

either you can draw the picture and it’s immediately obvious to

you the lines crossing is q/ε

_0 by construction,

or you can remember, “Hey, the electric field is a direct

measure of the number of lines per unit area,”

and the electric field times area times the cosine of the

angle between the area and the electric field will count the

lines going through a tiny area. If I add them all up,

I’d better get q/ε

_0, and indeed you do.

So the moral of this little

exercise is that the surface integral,

let’s call it the surface integral of the electric field,

on a surface is equal to q/ε

_0 where this was a sphere.

In other words,

even if you’ve never heard of field lines, just take the

electric field and do the surface integral,

you get this. So surface integral is a new

concept. You probably have not done that

before and I’ve got to remind you how it is done at least

operationally. If you’ve got a computer,

you can find the surface integral of anything as follows.

Take the surface over which

you’re doing an integration. Divide it into tiny pieces,

each is a little area dA.

Take the dot product of the

dA with the electric field there, and sum over all

the patches covering the sphere. Then take the limit in which

every patch gets vanishingly small.

That is called the surface

integral of the electric field and we see that if you do that,

you get the charge inside divided by

ε_0. But that was on a sphere,

and the interesting thing was, the answer was independent of

the radius of the sphere, because the 1/r^(2) in

the field canceled the r^(2) in the area.

But it is even better than

that, because it’s clear to you know,

without any calculation, that if I took some crazy

surface like this, the lines crossing that is also

the same. Because the lines leave the

origin, they go radially outwards.

They don’t terminate on

anything. They’re only supposed to

terminate on other charges, therefore you can count them

anywhere you like. You can take a census here,

you can take a census there or there.

You’re always going to get the

same number of lines. Therefore

q/ε_0 is also going to be

equal to the line count on some weird surface enclosing the

charge. So how am I going to count the

lines on a weird surface? I take the surface,

divide it into little pieces, but now the area vector and the

electric field vector may not be parallel,

because it’s not a sphere. But it is still true that if

you’re trying to count the lines going through,

just like in the velocity, you must take the dot product

of these. Therefore you will again find

E⋅dA for any surface,

any closed surface, is equal to

q/ε_0 .

If you got lost in the

mathematics, the physical picture is very clear.

If you want to count how many

lines leave the charge, you can surround it with any

surface you like. If it’s a sphere,

it’s very easy for you to do the check.

If it’s a crazy surface,

it’s harder but they all come from the fact that

E⋅dA, count the number of lines and

that’s independent of the surface at stake.

So now that it what is called

Gauss’s law, so I’m going to write it down here.

What I’ve shown you is the

following. Here is some strange closed

surface that’s a charge q inside.

Then E⋅dA

on that surface is equal to q_inside

/ε _0.

This is not yet the theorem,

but this is the case for 1 charge.

If the charge were outside,

suppose the charge were here, then the lines would go like

this, and the total number leaving the surface would

actually be 0. You might say,

“How is that? I see all these lines

penetrating the surface,” but you’ve got to remember,

if you take an area vector here, with the definition for a

closed surface, the area vector is the outward

pointing normal. For a closed surface,

every area is a vector pointing out from the closed surface.

Then you can see that in this

surface, lines are coming out; on this surface,

lines are going in. It’s very clear what’s going in

is coming out, because nothing is terminating.

Therefore if you took a surface

that did not enclose the charge, this answer would be 0,

but if it enclosed the charge, it will be the charge inside,

but it won’t matter where it is.

You can also see that you can

move this charge around anywhere you like, you don’t change the

number of lines coming out. Then the most important

generalization is, if instead of 1 charge,

I have 2 charges, what do you do?

If you had 2 charges,

let us take the total E⋅dA

on a surface. The total electric field is the

electric field due to the first charge on that surface,

the electric field due to the second charge on the surface,

thanks to superposition. It’s the fact that the electric

field is additive over charges. But this one is

q_1 /ε

_0. This is q_2

/ε _0.

Therefore, we can now write,

generalizing to any number, whether one draws a double

integral with this thing, meaning it’s a closed surface.

You can also have an open

surface, take the skin off an orange and cut it in half.

That hemispherical skin is also

a surface, but to signify it’s a closed surface,

we do it like this. On a closed surface of

E⋅dA=q_enclosed

/ε _0,

which means sum of all the q’s.

And sometimes we write this as

follows. Suppose inside you don’t have a

discrete set of charges that you can count,

but a continuous blob of charge, and the charge has a

certain density, ρ is the charge density.

What is the

q_enclosed? I claim the answer is due to a

volume integral of ρ or any point r inside,

times dxdydz inside that volume.

You know, if you want to say

how many coulombs are there, well divide the volume into

tiny cubes of size dxdydz.

That times the ρ

of r, meaning ρ

at x, y and z,

is the charge inside the cube. You add it all up,

you get the charge enclosed in that funny shaped object.

So this is the final form of

Gauss’s law that we’re going to use.

I’m going to use the symbol,

which is very useful.

This is the theorem of Gauss.

And S=DV.

DV is the boundary of V.

In other words,

V is like a potato, the skin of the potato is DV.

Therefore if the potato is full

of charge, it will emit some electric flux

and the surface integral electric flux over the skin of

the potato is the charge enclosed inside it.

The charge enclosed inside it

is not simply some constant density times volume,

if the charge density varies from point to point.

In each neighborhood inside

their volume, you take a tiny section,

a tiny little cube of size dxdydz,

see how much is in there, as the function of x,

y and z, and you do the integral over

the volume. Now these integrals may be hard

to do, but you should know at least what I’m saying.

This is just a way of counting

the total charge inside a volume when the charge is continuously

distributed. So I’m going to show you one

application of this and we’ll come back to more next time.

And the one application I’m

going to show you is to find the electric field due to a solid

ball, solid ball of charge. So I’m going to use Gauss’s

theorem to do that. This is true for any surface

you pick. It doesn’t matter what surface

you pick. So if I want to find the

electric field here, I’m going to pick a surface of

radius little r. Let the sphere have a radius

big R. I want to find the field here

and I’m going to use Gauss’s law.

So on the left hand side and

the right hand side are two different things.

On the right hand side,

what is the charge enclosed? That region.

Well, the charge enclosed is

just some number q. This whole thing is q,

q spread over a sphere of radius r,

over ε_0.

That’s going to be the surface

integral of the electric field on that sphere.

E⋅dA

on that sphere. Now normally,

if you knew the value of an integral over a region,

you cannot deduce anything but the integrand,

unless what? There’s one exception where if

you know the integral of a function, you can find the

integrand. You know when that might be?

Here’s a function.

I tell you its integral from

here to here, but I don’t show you the

picture. What’s the integrand?

You don’t know.

But if I also tell you the

function is a constant, and I tell you the integral,

and you know the width of this region, you can find the

integrand. So you’re going to cook this up

so that this whole integral is a constant times the area of

integration. And we argue that if you’ve got

a spherical charge density, the electric field must be

radial everywhere. You’ve not proven this,

but you argue that because it’s the only distribution with lines

coming out everywhere, invariant under every possible

rotation of the sphere. Because if you rotate the

sphere, I won’t know you rotated it, so the field pattern cannot

look different. If the field pattern looked

different when you rotated it, then you have a problem because

the cause looks the same, but the effect looks different.

That’s not allowed.

The only allowed this is the

radially outgoing electric field.

Therefore the electric field is

radial and also constant throughout the sphere,

because all points on the sphere are equivalent.

There’s no reason why this is

any better than that. The sphere looks the same for

all directions. Therefore this whole integral

is going to be 4Πr^(2) times the electric field at that

radius r, because the area vector and the

electric field vector are both parallel,

so in this dot product, forget the dot product.

It’s just E times

dA. There’s no cosθ,

just 1. You can pull the E out

of the integral, because E is constant in

magnitude on the sphere. And the integral of the

dA is just 4Πr^(2).

Therefore you deduce

E(r) is q/4Πε

_0r^(2). It’s a very profound result.

It looks very simple,

but it is true for not a point charge, but for the spherical

distribution of charge. That the field goes like that

of a point charge sitting at the origin, is a consequence of

Gauss’s law. If the charge inside was not

uniform, suppose it’s a charge q,

but there’s more stuff here, less stuff here,

this theorem would still be true up to this point,

but you can never deduce that E is a constant on a

sphere, because even though a sphere is

nice and symmetrical, the charge distribution is not.

It could be big here,

it could be small here. You know something about the

integral over a surface of a varying function,

then Gauss’s law is no good, not useful.

True, but not useful.

Gauss’s law is useful only when

in a given problem, there’s only one number you

don’t know. That number here happens to be,

what’s the strength of E at a radius r?

I know it’s direction is

radial, I know it’s magnitude is constant on the sphere by

symmetry, but what is the number?

You can trade that one number

for this one number on the left hand side,

q/ε_0, you can calculate it.

I’ll come back and do more

examples for you guys next time.