3. Gauss's Law I - VoiceTube: Learn English through videos!

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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.