Prof: Let's start with a
brief recall of Gauss's Law. It is so important I wanted to
make sure you guys understand not only how it is used but how
it's derived. I derived it for you last time,
but I thought the presentation could be improved a little bit,
so I'm going to do it one more time,
just this time only hitting the highlights.
The content of Gauss's Law is
very easy to visualize, so I'm going to draw you a
picture that'll tell you the whole story,
then there are all the equations.
The picture is the following.
You take one charge q
and you draw the lines of force coming from the one charge.
It's a point charge.
Therefore, you have no freedom
but to draw lines that look like this.
Now, how many lines you draw is
a matter of convention. You can draw eight or twelve,
whatever it is, but you've got to agree that
the number per coulomb should be fixed.
So we'll fix some number.
So I'm going to pick for
whatever charge is here these lines.
You cannot have a distribution
of lines in which it's suddenly more dense in just this
direction, for example. That's not allowed,
because if I take this point charge and I rotate it,
it looks the same, but if I took the line
distribution with more dense here than other places if I
rotate it, it looks different.
So if the cause of the field
lines looks the same after rotation, the effect,
which is the field lines, should not change.
You can only have a
distribution of lines with respect to symmetry of the
source of the cause of the field lines.
You understand that?
It cannot be biased one way or
the other, whereas this distribution,
if you turn it around, looks exactly the same.
So we draw those lines,
and the content of Gauss's Law is really the following.
We are going to count the lines
crossing a sphere, S, and we're going to count the
lines crossing some generic surface S'.
You can see at a glance without
doing any calculations that the following is true.
Lines crossing S = lines
crossing S', right? That's clear to everybody?
You don't need to do any
equation for that. That's because the lines have
nowhere to go but out, and you can count them on this
circle, that circle as long as you
enclose the whole thing, the solid surface.
Enclosing this charge you're
going to get the same lines. That's the heart of Gauss's
Law, but that's not the final form, but that's the part that
starts everything. Now I'm going to calculate the
left hand side, and I'm going to calculate the
right hand side. I'm going to equate them.
From that I will get almost
everything I want. Now, how many lines are coming
out of S? We know--first the notion of
line density. I've got to remind you what a
line density is. Line density is the lines per
area, but it's not enough to say area.
I'm going to put a
perpendicular symbol there. That means the lines are going,
say, from left to right like this.
You want to put an area that
intercepts it perpendicularly, or if you like,
the area vector which is always normal to the area should be
parallel to the lines. So it's the number of lines
crossing unit area perpendicular to the lines.
That's called the line density.
If you tilted this area;
the same area, but at a bad angle,
it'll intersect fewer lines. That you know.
Okay, so this is the meaning of
line density. What's the line density going
to look like in this problem? Line density is going to be
proportional to the charge you have because we agreed we'll put
some number of lines per coulomb,
and it's inversely proportional to the radius because the area
of a sphere of radius r is 4Πr^(2).
So if you cut it with a bigger
sphere of bigger radius the density, lines per area,
will go like 1/r^(2). So it's proportional to
q. It's proportional to
1/r^(2), okay?
So let me write not equal to
but proportional to. You have to stop me if you
don't follow anything I'm saying.
There's no reason why you
shouldn't follow this argument. Then I say how about the
electric field? The strength of the electric
field, not the direction,
the electric field E at radius r is also
proportional to q and also proportional to
1/r^(2). Because from Coulomb's Law the
field is the force on a unit test charge, it is
q_1q _2.
q_2 is 1
because the test charge is 1 and there's
q/4Πε _0r^(2).
Forget the 4Π.
It's proportional to
q/r^(2). From this we may conclude that
the line density is proportional to the electric field because
they are both proportional to q/r^(2).
Therefore, we can say line
density is equal to some constant c times the electric
field in magnitude; direction of the lines is along
E, but the magnitude. Do you understand that?
This guy is proportional to
this. This may have some
proportionality constant, this may have another one,
but nothing changes the fact that this expression's
proportional to that one. The constant will depend on
details. Let me just call it some number
c. c is up to you.
It really depends on how many
lines you want to draw per coulomb.
I made a particular choice last
time, 1/ε_0
lines per coulomb, but you don't have to make that
to prove the law. So this time I'm just assuming
the lines is proportional to charge, but you pick your own
proportionality constant. So I've got this line density
that's proportional to electric field.
Now I'm going to go back and
I'm going to calculate the left hand side.
Now, let's give this equation a
number, equation number 1. I'm going to calculate the left
hand side of 1. Left hand side of 1 is the
number of lines crossing this sphere, S.
So that's going to be equal
to--let me call it Φ_S is equal to
number crossing S, crossing this sphere S.
Φ is the standard symbol
for flux. So what is it?
Well, Φ_S = line
density times area of sphere, because on a sphere the lines
are uniform in density so the total number crossing the sphere
is the line density times area of the sphere.
But the line density is a
constant times the electric field at radius r (if you
like) times area of the sphere which is 4Πr^(2).
That's the lines crossing the
inner surface which is this sphere of radius r.
Now, Φ_S' is
lines crossing S'. Now you've got to be a little
careful when you find the lines crossing S'.
You understand why that's more
difficult? What are the two complications
in finding the lines crossing S' compared to this one which I did
in one second? Any idea?
Yes?
Student: The normals
are all parallel to the field lines?
Prof: Which is not
parallel? Student: The normal to
the surface isn't parallel to the ________.
Prof: Okay,
the surface is not--surface vector at each point is not
parallel to the electric field. For example,
if you've got a crazy surface like this--
I'll try to replicate that surface,
but I don't remember what I drew, some weird object like
this. Then the electric field point
this way, but the normal to the surface will point that way.
See, E points that way.
Tiny area there,
which I'm calling vector dA, points in a different
direction. So each region has a different
direction. Secondly, the strength of the
field itself is varying on the surface.
Unlike on a sphere where
everything is at the same distance r some are
further, some are closer. But it is still true that if
you want the number of lines cross a patch,
if you pick a little area here, if you want to know how many
lines cross that you take the normal to the area,
call it a vector dA, and you multiply by line
density. Therefore, this is equal to the
integral of the electric field at some point r dot
product with a tiny area which is there.
At each region you've got to
draw a different dA. So you take the surface,
you chop it up into tiny little tiles.
Each tile is small enough to be
considered a plane. Then the area vector is
perpendicular to the plane, outward from the surface,
and of magnitude equal to the numerical area of the plane.
So the lines crossing every
tiny little square on the big surface is the number of lines
per unit area times the area times the cosine of the angle
which you need, because the area vector may not
be parallel to the flow. I have explained many times why
the flow there, the flow vector and the area
vector are not parallel you've got to put in a
cosθ. That's what comes into this dot
product. And we write the symbol like
this to tell you not only is it a double integral,
meaning integral over a two-dimensional region,
but it's over a closed two-dimensional region.
That symbol is to remind you.
So now, Φ_S equal
to Φ_S' is what I'm going to write.
So I'm going to write--oh,
I made one mistake. Can anybody see one thing
missing in this equation here? This equation's not correct.
To find the lines crossing S' I
took E times dA. It's a constant c
because the line density is not simply the electric field it's
the constant. So I forgot the constant,
but I knew that I would need it because when I now equate
Φ_S to Φ_S' I get
constant times the electric field at r times
4Πr^(2) is equal to constant times the surface
integral of E⋅dA on some
surface S'. But E(r) times
4Πr^(2) is c times
q/ε_0 because E is
q/4Πε _0r^(2).
You see that?
This much you should know.
I'm not going to write the
details. So that is equal to c
times surface integral of E⋅dA.
Now you see the constant does
not matter, so whatever the constant is I
can make the following statement and that's going to be the
beginning of Gauss's Law. So Gauss's Law tells me that if
you took a charge and you sat under it with any surface
whatsoever--I'm not going to put a prime on this anymore.
S is going to be the surface.
Then the surface integral of
the electric field due to that charge over a surface is equal
to the charge that is sitting inside divided by
ε_0, and it does not depend on the
details of the surface.
Okay, everybody should know in
principle how to do the surface integral.
If I give you a surface,
and at the surface at every point I give you a vector
E, which is electric field,
you should know how to numerically calculate it.
You may not know how to do the
integral analytically, but operationally you should
know the meaning of this, okay?
If E were actually the
velocity of some stuff running out of the volume you're
counting in each region how many guys are escaping per second and
you're adding it all up. And to find how many guys are
leaving a tiny area, you multiply the area by the
velocity of flow times the cosine of the angle between the
velocity in the area. That's it.
Okay, now we do the next thing.
Let's call this charge
q_1 because I'm now going to put a second guy
q_2 here. If you now think in terms of
lines of force you're going to get into a big mess because
lines of force, due to two charges is a
complicated thing, you remember drawing some
pictures. But now here's the beauty.
Forget the lines of force
completely, not interested. I'm interested in the electric
field. Focus on the electric field for
which there's a very simple rule.
The electric field due to two
charges 1 and 2 is the electric field due to 1 plus the electric
field due to 2. That's the great principle of
superposition. Therefore, on the left hand
side you might say if both charges are inside the surface
integral of E_1,2_
meaning when both are present is equal to surface
integral of E_1 ⋅dA surface
integral of E_2 ⋅dA,
because E_1,2 is E_1
E_2. But the first guy is
q_1 /ε
_0, second is q_2
/ε _0.
Once you can do two charges you
can do any number of charges. And now we write Gauss's Law
almost in its final form. The final form is that if you
took the total electric field, I'm not going to put total,
E will stand for the total,
on a surface S that is equal to the sum of all the
charges, i from 1 to whatever
many charges you have, divided by
ε_0. The condition is these are
charges inside the volume bounded by the surface.
You do not count the charge
here. You only count charges which
are inside not charges which are outside.
So let's make a notation that
tells you that i is inside the volume V.
And this surface is the
boundary of the volume V. This partial derivative symbol
is the universal convention for the boundary of any region.
Okay, if you draw a real
circle, this is the circle. The boundary of the circle is
this. If you a have a cube,
or if you have an orange the boundary of the volume orange is
the skin of the orange. That's the definition of
boundary. You can see this is like the
skin of the solid region, and the surface integral on the
skin of the electric field is a charge inside the volume
whatever be the shape of the volume.
And it works for all kinds of
surfaces. I mean, here's one example.
Suppose you have a charge here.
You take a surface that kind of
intersects itself and so on. Then what can happen is if I
put a charge here the lines may go like this.
It doesn't matter because over
this when it switches back on itself some lines leave,
some lines enter, and some lines leave again.
So it might as well be that
surface because double entering and leaving doesn't matter.
So the rule is if q is
inside it contributes to the surface integral,
if q is outside it doesn't.
The last variation is what if
q is not made up of a discrete set of charges,
which you can say one is here one is there,
but there's some continuous blob, some lump of charge where
the charge density-- so then we say suppose q
is continuous. You've got to realize in real
life it's never continuous. Real charges are made up of
electrons and protons. They are discrete,
but if your eye is not looking at it in great detail the charge
can appear to be continuous. For example,
if you take a glass of water the water is really made up of
point molecules, water molecules,
but to your eye it looks like a continuous medium.
So for certain purposes we can
replace a discrete distribution by a continuous one.
For a continuous distribution
what you have to tell me is the following.
This is space x,
y, and z. Here's the point.
I take a tiny cube whose sides
are dx, dy and dz,
and ρ of x, y, and z times
dxdydz is equal to charge in that little cube.
In other words,
like the usual density with this mass per unit volume the
ρ is the charge per unit volume.
So you multiply by volume you
get the charge, but it need not be uniform.
So if you have a solid region
then the charge density could be varying from point to point,
so. Oh, let's keep it here.
In that case you will write the
surface integral of E⋅dA over a
surface that binds a volume V would be
1/ε_0 times the volume integral of ρ
of (x, y, z) dxdydz.
This is the final form of
Gauss's Law. Now, how many people have ever
done integrals in more than one dimension?
Great, very good,
made my day because I didn't have Plan B.
But I did--no actually I had a
Plan Tiny B which is to give you a quick refresher.
If there is time at the end of
the class I will remind you of how the multiple integrals are
done. But right now I'm going to show
you the efficacy of this law and that doesn't require doing
complicated integrals. All the integrals will be so
trivial you can just read them off.
So when we have time we'll do
questions of how to do these integrals, but I'm assuming this
symbol means something to you. In words it means all the
charge inside. So, again, by taking the
volume, dividing into tiny regions,
multiplying each volume, tiny volume,
by the density to get the charge there,
and summing over all the volumes inside the big volume
V. That's a charge.
So Gauss's Law still says the
surface integral of E is the charge inside over epsilon
0. All right, so this is the great
law. Now we're going to get some
mileage out of it to solve certain problems which normally
would be very difficult. The first problem,
I think I probably did that last time but it's worth going
over, is what is the electric field of a uniform density of
charge. Suppose there's a charge
Q and this is sphere of radius r,
and it is a ball of radius r.
You take all these charges and
you pile them on top of each other.
They don't like it,
but you glue them or do whatever you have to do.
You've got a ball of charge
Q. You want the field due to that.
It's the same problem Newton
had with gravity. The earth is a solid ball.
It's got mass everywhere and
you would like to find the force on an apple.
And it's not so evident that
the force of the earth on the apple is as if the entire mass
of the earth but at the center. You have to prove that.
You have to prove that by
dividing the earth into little pieces and finding the
gravitational force of every little piece on the apple and
adding all those vectors up. And when you do all the hard
work you will find, in fact, that it's true,
but it took Newton several years to prove that even though
he was pretty certain that was true.
That's very important because
when you do the force law for apple versus the earth the
distance you use for m_1m
_2/r^(2) is a center-to-center distance as
if the entire earth were at the center.
A similar law is true for
electric forces. In other words,
if you've got a spherical distribution of charge the field
due to that outside that ball is as if all the charge were at the
center. That's what we want to prove,
but we want to prove that without doing any difficult
integrals, and the trick is the following.
You are going to use Gauss's
Law, and you can use Gauss's Law on any surface you like.
For example,
you can take that surface and you can say surface integral of
E on the surface is Q/ε_0,
but that's not very helpful because what you're trying to
find out is not the electric field integrated over a whole
region, but the electric field at each
point. You want to know what's going
on right there. So even though it's a true
statement that whatever E is doing its integral on any
surface is Q/ε_0 that
doesn't tell you what is E at every point,
and we're going to find that by the following trick.
The trick is to first decide.
This surface S,
by the way, it's called a Gaussian surface.
It's not a real surface.
In real life there are no
Gaussian surfaces. You draw them for a
calculation, and you can move it around in your mind anyway you
like, for each surface you'll get a different answer.
We'll pick a Gaussian surface
equal to a sphere of radius r.
So I'm going to pick a Gaussian
surface of radius r. I'm going to argue two things
and you have to follow the reasoning because all the work
is in the argument. You don't do any integrals,
but you do some serious arguing.
The arguing says it's a ball,
positive charge. It's going to repel a test
charge. Which way is the repulsion
going to be at this point? It has to be radially away from
the center. That's your argument.
At each point the repulsion has
to be away from the center. In other words the electric
field has to have this distribution.
We're not saying how it varies
with distance, but it has to be radial.
Again, the argument is the
radially outgoing thing like a hedgehog.
It's the only distribution of
lines which will look the same if you rotate them because then
one line will go to another line.
The hedgehog will look exactly
the same. And that is a necessary
requirement because if you took this ball of charge you all
understand that if you rotate the ball the field lines have to
rotate due to any calculation you did,
but if you rotate the ball it looks the same.
So the field lines have to look
the same. So the distribution you draw
must be invariant under that rotation.
This is a very powerful
principle of symmetry and invariance.
The principle says if something
is the cause of something, and if I do something to the
cause that leaves it alone, namely it looks the same,
the effect has to look the same.
It's a very reasonable argument.
And the cause here is a ball of
charge which is completely isotropic and spherical.
You turn it,
it looks the same, so the field distribution looks
the same. That's all you need to proceed
because that distribution then is going to have radial lines of
unknown strength, but the strength can vary only
with r, but not at a given r.
At a given r the
direction varies, but the strength is some E
of r, and we're asking what is
E of r? So we know that E looks
like r, which is the vector of length
in the radial direction times E of r.
And sometimes I also write this
as e_r times E of r,
and we just want to know what E is a function of r.
But now we can use Gauss's Law.
So the left hand side is a
surface integral of this radial E on a sphere of radius
r. That's going to be E
times 4Πr^(2), because E is a constant
on the surface. The integral of a constant on
the surface is the constant times the area of the surface.
It's like saying if you want to
integrate a function which is constant from here to here it's
the constant height multiplied by the interval over which
you're integrating. There's no integral to do.
It's very trivial.
It's a rectangle.
Similarly a constant function
integrated on a sphere is the area times this,
but E is the function of, of course,
r. That's going to be charge
enclosed. The charge enclosed is Q
and you've got to put the ε_0,
and then you find E(r) is
Q/4Πε _0r^(2).
And if you want to put all the
vectors back you can write here times e_r.
So by this trick one can show
that the field of a sphere is as if the charges were at the
center, provided you are talking about
a point outside the sphere. As long as this field is
outside the real ball no matter what its radius is,
the charge enclosed is always Q.
The right hand side doesn't
vary with Q. The left hand side looks like r
squared times electric field so E goes like
1/r^(2). But now I'm going to do some
variations on this. The first variation is what if
I want the electric field at some point here,
in here? You're allowed to ask that.
You can take an instrument and
put it there and ask what field you have.
You have to calculate that too,
so let's calculate it for a sphere.
The formula is the same,
same formula. Everything is the same,
but Q is now the charge enclosed.
So I'm going to write,
again, 4Πε _0,
I'm sorry, 4Πr^(2) times E(r) is the
charge enclosed over ε_0.
What is the charge enclosed in
this sphere, now, which is a mathematical sphere
with radius as little r, but it's smaller than big
R? Can you make a guess?
How much charge do you think
that is in a smaller sphere? Yeah?
Student: The charge
density times volume. Prof: Right.
So he said it's the charge
density times the volume. I'm going to write down the
answer without going through the intermediate step,
but it'll coincide with what he said.
The answer I'm going to write
down is the total charge is Q on the whole ball.
The charge is proportional to
the volume. Then the volume of this little
guy is r^(3) and the volume of the big guy big
R^(3). That's how much charge will be
enclosed. Can you see that?
If you take a square,
and you take a part of the square that's inside that's half
as big, the area of it will be
¼, because areas go like 1 over
the distance squared. Volumes go like 1 over distance
cubed. But we can also do what he just
said. You can take the Q
divide it by the volume 4/3Πr^(3).
That's the density.
Then you can multiply the
density by the volume of the little sphere 4/3Π
little r cubed, and all I'm saying is forget
these 4 thirds pis. It's just little r^(3)
over big R^(3). Therefore, the electric field
now, E(r), looks like Q over
4Πε _0r divided by
r^(3). This is for r <
R, is equal to Q/4Π
ε_0 (1/r^(2)) for
r > R.
Student: Isn't it
supposed to be r cubed ________? Prof: Sorry, here?
Student: Yeah,
down on the second one. Prof: Here you mean?
This line?
Student: No,
I mean like on the line above. You have like r over
r cubed. Prof: I have little r
cubed over big R cubed.
Student: Right,
but no. I mean the next line.
Prof: Here?
Student: Yeah.
Prof: This is r
less than R, r bigger than R.
Student:
> Oh, gotcha.
Prof: Oh,
you mean this 4Πr squared came downstairs?
Yeah, that's right.
Student: Gotcha.
Prof: Okay,
eternally vigilant, that's good.
Yes?
Student: What if little
r equals big R? Prof: Let's ask.
What if little r grows
up and become big R, right?
We'll find out.
What do you expect?
You've got to get the same
answer at the field at a certain point no matter how we
approached it, right?
You'll find that's true in this
formula, because if you put little
r equals big R and big R over big
R^(3) you'll get 1 over big R^(2) and this will
also give you 1 over big R^(2). These formulas
will match on the surface of this sphere.
Yes?
Student: So can you say
again why we can assume that we don't have to do the integral
________________________? Prof: Why did I not do
the integral? Student: Why we
________ 4 pi squared times E of r.
Prof: Why do I do that
you mean? Student: Yeah.
Prof: Right.
So the correct way to do it is
to really do E⋅dA, right?
If I did that it'll be fine
with you, correct? This is what you're saying I
should do, right? You asked me why I didn't do
this integral, why I just wrote
4Πr^(2), is that your question?
Right.
The assumed form for E
is the function that depends on r times unit vector in
the radial direction. And since I surrounded it with
this sphere the tiny area vector is also e_r
times the value of the tiny area.
In other words,
dA and E are both parallel if the surface is a
sphere, right? So when you take the dot
product this will just become E(r) times
dA times cosine of 0 which you don't care about.
That's the first thing.
You still have to integrate
over this sphere, but E of r is a
constant on this sphere because it depends only on r.
So throughout the whole sphere
if it's a constant, it's like the number 19,
you just pull it and then you get E of r times
integral dA and that is my 4Πr^(2).
So in general you have to do an
integral, but if you're lucky and the integrand is a constant
then the integral is trivial. So for a charge at the center
of the spherical ball, and if you went to a sphere
then the integrant is a constant.
Okay, so this is what you get.
If you draw a picture of this
it looks like this. Electric field as the function
of r it will grow with r then it will decline as
1/r^(2). And here is little r
equals big R. Yep?
Student: What if it
were hollow and all the charges were on the outside ________?
Prof: We'll come to that.
Yes?
Student: Why can we
ignore the charge that's outside the small sphere?
Prof: Why don't we count
it? Student: Uh-huh,
did you explain that already? Prof: Yeah,
well it's a very interesting point, right?
Suppose you make a hole in the
earth, of if you go deep into the earth.
We said if you're outside the
earth the entire ball is pulling you in,
but the claim I'm making is if you're inside the earth,
if you're here, only the ball underneath your
feet is pulling you, but not these guys, right?
That's the question,
why don't they exert a force, right?
Well, Gauss's Law says,
no, but we can try to verify that explicitly.
So we are going to do the
hollow sphere, but you followed the solid
sphere now? So why is the field getting
stronger as you leave the center, why is it falling like
1/r^(2) once you leave the sphere?
You've got to think about that.
Why is it growing when you
leave the origin? Yep?
Student: Because there
is more charge _________. Prof: Right.
The charge you are enclosing
grows like r cubed, and the distance from the
center of that charge is r squared which comes downstairs.
So as you go further out inside
the ball you are gaining, but once you reach the surface
of the ball and you go further out you're getting more area for
Gauss's Law, but not more charge which is
stuck at capital Q. That's why once you leave the
sphere it starts declining, but if you're inside the sphere
it increases. This is also true for the
gravitational force. The gravitational force if you
go to the center of the earth and you move away by 1 inch
you'll be pulled back with an extremely weak force that's
proportional to the distance from the center of the earth.
Okay, then we come to another
question which came up, which is let's take a hollow
sphere. I take a sphere and I scoop a
hole in it and I want the field here.
Okay, so if you want the field
here you take a Gauss's Law for a circle like that,
or a sphere like that and you'll get E times
4Πr^(2) is equal to 1 over ε_0 at
times what? Student: Nothing.
Prof: Nothing.
So the electric field inside a
hollow sphere is 0. In other words,
if you took the earth and you scooped a big hole in the middle
you can float. It won't pull you at all.
So inside a hollow sphere there
is no electric field. Even though there's charge
everywhere there's no electric field.
We can understand that in some
cases, and other cases require a little more thought.
Let me first show you that for
a very, very thin shell why that happens.
Once it's true for a very,
very thin shell and it doesn't do anything you can put one
shell inside another and make it as thick as you like because 0
is going to be the contribution from each guy.
So let me take a very thin
shell. Here's a very thin shell.
So remember, this is not a ball.
This is a shell whose thickness
is the thickness of the chalk here.
I think you will agree that at
the center of this sphere the force has to be 0.
You agree with that?
Because which way should it go?
For every guy pushing,
say from here in that direction, there's another one
from there pushing in this direction.
They all cancel.
It's very clear that at the
center you don't feel a force. But what is truly amazing is
that even when you're off center you don't feel a force.
So I'm going to give you the
argument for that one, argument for that namely I'm
once again going to pair off canceling charges,
okay? I'm going to cancel one thing
against another thing. When you were at the center you
canceled a little charge here with a little charge here,
right? When you are here here's what
you do. Take a cone like that.
It is not two lengths.
It is a three dimensional cone,
but I'm able to show you only a slice right down the middle,
you understand? This is really a cone that'll
cut the sphere there; it'll cut the sphere there.
Take all these guys.
They have a certain charge,
which is some charge density which is uniform,
times the area of this thing divided by the square of the
distance r_1. Here, these charges here will
push you to the right with a force which is the area of
density, the charge density per unit
area times the area of the smaller circle divided by
r_2 squared. But now I claim that I can
equate these two. I can equate these two if this
area of vector and this area of vector were radiating outwards,
and this is actually a true statement.
This area is proportional to
r_1 squared. This area is proportional to
r_1 squared, therefore they will cancel.
But this is a very subtle
argument. Sometimes even textbooks get it
wrong. And generally if you take two
cones like this you can see that the area vector points that way,
but the electric field vector points this way.
They are not the same.
The area vector,
area on the sphere, right, will point in one
direction, but the r vector is different.
So you have to take the cosine
θ on both sides, but it turns out cosine
θ is also the same because if you took a line and
you cut it this way that angle will equal that angle.
So that's a subtly we don't
need, but that's how you really prove for every shell.
There's a cancellation.
If you know what a solid angle
is it's very easy to say. This is called,
with the cosine thetas in it, it's called a solid angle
enclosed by this point, and the two things have the
same solid angle and they cancel.
But if you don't want the most
complicated proof you can take points like this where it's
clear that this area vector and this area vector are both
perpendicular, and to the radial direction,
and therefore the electric field and this vector are
parallel. Then you don't need the cosine
thetas. They are clearly equal at
possibly even 1. Then A/r^(2) being the
same is the reason it works. In other words,
if the force of gravity or the electric force did not fall
exactly like 1/r^(2) the field inside a hollow sphere
will not be 0. So this is how people tested
the 1/r^(2) force law in the old days.
They went inside a hollow
sphere and tried to see if there is any field inside by putting
test charges and finding they don't respond.
That's the most reliable way to
prove 0 field. Yes?
Student: Can you
explain again why the A over r squareds are equal?
Prof: Well,
you can see that if you took a cone and if you just did similar
triangles you can see that this area is growing like that radius
square. It's a matter of how areas grow
when you scale anything. Take a cone,
okay, and you slice it at some point, and slice it even higher.
Here's a cone.
You slice it there and you
slice it there, and my claim is this area is to
that area that distance squared is to that distance squared.
Why is that, right?
That's your question.
The area of this cone is
something, something radius times something,
something height, right?
So you can calculate,
if you like. The radius of this circle and
the radius of that circle are growing linearly as you go out.
If you go distance r
here, and you get a radius here, if you go distance r
there you get another radius. Is it clear to you that the
radii are proportional to the distance from the center in this
cone? Let me draw it again.
This is a bad picture.
Here's my cone.
Take two circles.
Well again, I screwed up.
This is supposed to be circles.
This is the center.
That radius is to that height
what this radius is to this height.
They're just similar triangles.
Therefore, this height is what
I'm calling r_1 and this height is what I'm
calling r_2. Therefore, if this radius is
proportional to r_2 the area of that circle is
Πr_2^(2). This is
Πr_1^(2). So that'll go like that
distance squared is to that distance squared.
Okay, whenever you scale things
for an area by factor of 2 it'll go like the square of that
factor. You can already see that if you
draw these cones this'll intersect the circle,
the sphere on the small circle. This will intercept it on a
bigger circle. That much is clear,
but the extra is not linearly proportional.
it is quadratic because the
circle has two dimensions both of which are growing linearly
with the distance from the apex of the cone.
That's why it's r
squared. All right, another thing one
can calculate is the electric field due to an
infinitely long wire. Here's an infinitely long wire
with lambda coulombs per meter. So you want to find the
electric field. By the way, I should tell you
that Gauss's Law--let me do this one example.
I'll tell you what the
restriction of Gauss's Law is. So we come to this problem and
we can again argue by symmetry the field lines have to be
radially away from the axis of their line,
and if you look at it from the edge,
from one edge it'll look like this.
If you look at the wire from
the edge the lines should be going out like that.
It's like a hedgehog but it's
cylindrical. It's not in all three
dimensions. It's radial in this direction.
And the question is and the
field has to be constant along the length of the infinite wire
at each point at the same distance because if you move the
wire horizontally it looks the same so the field distribution
cannot vary. If it is a finite wire you
cannot make the argument. In a finite wire as you come
near the edges lines will start tilting.
But a finite wire doesn't look
the same when you move it; an infinite wire does.
Therefore, for an infinite wire
if you don't stop here the lines will always look the same,
so that if you shift them over they should look the same.
So the only unknown question is
- I know the field is in this direction radially away from the
wire in all directions whose magnitude is fixed at a distance
r, but I want to know what the
magnitude is. Again, I'm going to take
Gauss's Law and I'm going to apply to the following surface.
Surface is a cylinder and the
cylinder has some length l,
and it's got some radius r and the flat faces of
the cylinder are parallel, so if you really want to give
the nice picture here they look like this.
So Gauss's Law can be applied
only to a closed surface. You understand that?
You cannot do it for an open
surface because only if you trap the charge completely in all
directions will you count every line.
If you've got holes in your
surface then stuff can escape and you cannot promise anything.
So I need a closed surface and
my closed surface is the cylinder.
So I'm going to write,
once again, the surface integral of the electric field
on that cylinder is the charge enclosed.
The charge enclosed is the
easiest part. I'm going to give you 10
seconds to think in your head. What is a charge enclosed by
the cylinder? Okay?
It's charge per unit length
times the length of wire trapped inside the cylinder and then,
of course, I have this 1/ε_0.
It still is a statement about
the integral of E on the surface, and it could be any
surface, but the beauty of this surface is the following.
There are contributions to
E⋅dA from the curvy side of the cylinder
and from the flat side. In the flat side the surface
vector, area vector dA for any small portion,
or in fact for the entire face, is like that.
The electric field runs along
that face so the dot product is 0.
So I got 0 from the left side,
0 from the right side, then I got non-zero from the
curvy cylinder. On the curved cylinder you can,
I hope, see that every area vector is actually parallel to
the field lines. So just like on a sphere this
integral will be E(r) times the surface of the curvy part of the
cylinder which is 2Πr times L.
That's the area of the curved
part of the cylinder. So you see L cancels
out, and it better cancel out because L is an
artificial construction. We made up L.
The answer should not depend on
the Gaussian surface you picked. I told you it's a figment of
our imagination, so it doesn't depend on it,
and we find E(r) = λ/2Πε
_0r, which is the result we got
earlier on by doing all the brute force integration.
Remember we took a point here.
We took some segment.
We drew the arrows,
did the integrals, sine thetas and whatnot,
but you can get that in one shot.
So it looks like Gauss's Law is
the easy way to do stuff, and you may wonder why we
bother to do any difficult integrals.
The reason is that if I change
the symmetry of the problem in the slightest way I cannot
calculate the field. For example,
if on the spherical charge distribution instead of a sphere
I made a little blip here, maybe did a little surgery and
put that guy here. We're dead.
I mean, there's a formula for
the field, but no one can calculate it in any simple way.
You can, again,
take a Gaussian surface and it'll still be true that the
integral of the electric field on that surface will be the same
Q/ε _0,
but the problem is the E on the surface is no longer
constant. You understand?
Not every point is symmetric
any more. E may be stronger where
there's a bulge. E may be weaker where
there is hole, so and also its direction is
changing in a crazy way. So you can make one true
statement about the integral of that crazy function over the
whole region. That cannot be used to deduce
the value of E at every point.
So you still need the integral.
You may have to do the integral
maybe on a computer, but that's the answer to all
problems, but for simple problems with a
great deal of symmetry we can use Gauss's Law to get these
things very easily, okay?
That's the Gauss.
Now, I'm going to introduce you
to a second notion which is pretty important to study
electricity, and that's the notion of conductors.
So we're going to divide the
world into two things, conductors and insulators.
As you know,
matter is made up of positive and negative charges and the
negative charges circle the positive charges,
and they pretty much stay near their parent atoms,
near the parent nuclei, and you cannot separate them.
But in a metal at least some of
the electrons from each atom become communal.
In other words,
they can run around the whole solid.
They don't belong to any one
nucleus. So that's a conductor.
In a conductor the negative
charges, if you like, are free to move.
So here's the first result.
E is equal to 0 inside a
conductor. That really follows from the
meaning of the word conductor. If you took a chunk of perfect
conductor, maybe copper is good enough;
they'll be no electric field inside copper.
Why?
Because if this is the chunk of
material, this electric field,
then the charges that I said are free to move will respond to
the electric field and they'll be moving.
So I should say E equal
to 0 inside a conductor in a static situation.
In other words,
once the charges have stopped moving the electric field will
be 0 in a conductor. Let me explain to you a little
more what's going on so it's not a big mystery.
Suppose there is a uniform
electric field going from left to right?
In that uniform field I take a
chunk of copper like a nice rectangular chunk and I stick it
in there. What will happen?
In the beginning the electric
field will penetrate the copper and the field lines say to the
positive charges, "You go to the
right," and to the negative charges,
"You go to the left." Negative charges will race to
the left until they cannot go anymore without leaving the
solid, and that they're not allowed to
do, leaving behind some deficit on
the other side. But look what's happening now.
These guys produce their own
electric field which goes from here to here.
Therefore, inside the
conductor, the electric field by the superposition principle is
the field due to whatever outside agency produced this
field, plus the field due to these
guys, and they will not stop until the field they produced
exactly cancels the external field.
Then the migration will stop.
So how do they know when to
stop? They are not that smart, right?
But the point is once they've
produced the potential, I mean, once they produce the
field that cancels the external field there'll be no field
inside the bulk to encourage charges to move any more and
that's when they stop moving. Okay, so a conductor,
at least in this case, not so hard to understand what
they have to do. Negative charges go to one side.
Positive go to the other side
until the field they produce is an arrow going to the left of
the same magnitude as this one then E is 0 inside.
But what's amazing about these
metals is that if you take a potato shaped metal it's not so
easy to see what charge arrangement will exactly
neutralize the field everywhere in sight.
Here it's easy to see.
I want a right moving one
canceled with a left moving one. I draw a line,
a plane of charges here and here, and you can see they will
cancel. But even this oddball object,
I claim, will eventually acquire some
density of charges, they're not uniform or
anything, in some complicated fashion until the field inside
the solid becomes 0. So these particles will always
figure out a way to make the field 0 inside,
because it's the definition. If it's not 0 they've got more
work to do, and the charges have to
separate even more until there is no hunger for the separation
because they managed to produce a field inside that is 0.
Then nobody else will join this
flow and it'll stop. So that's a very short period,
10 to the minus something, when charges rearrange when you
put a conductor in the field, then quickly it'll stop.
If you put an alternating field
going back and forth then, of course, it depends on how
rapidly it's oscillating, and charges may not be able to
keep up with that. That's called a plasma
frequency, and beyond that a field would start penetrating
because charges cannot keep in step.
But a DC field,
where you've got all the time in the world to settle down,
they will very quickly come to this arrangement.
They will stop.
So remember field in a
conductor is 0 by definition. Okay, now the next thing I will
show you is that charge inside a conductor equal to 0.
By that I mean the following.
If you took a conductor and you
threw some charge on it where will it go?
Okay, here's a chunk of copper.
Throw 10 million electrons.
Well, the 10 million electrons
don't like each other. They will try to run away from
each other. In the end they will all sit
somewhere here, but they will sit in such a way
that the field inside is 0, because you cannot have an
electric field. If the electric field is 0
everywhere then the charge is also 0 everywhere,
because you can take any surface, any volume you like
with the integral of E⋅dA,
you're going to get 0. If you can get
E⋅dA is 0 for every possible surface then the
charge enclosed by every possible surface is 0,
therefore there is no charge in sight.
So where does the charge go?
It goes to the boundary of the
metal. Once they're at the boundary of
the metal the metal starts calling you back because there
is something called a work function.
It's like all the electrons in
a swimming pool. If they don't like each other
they can go to the edge of the pool,
but to leave the pool they've got to climb up over the
vertical walls, and they cannot scale the walls.
If you rip them hard enough
they will. If you put enough charges,
charges will start flying from here and maybe land there;
that we call lightening. For that they'll have to break
the air and produce a conducting path, but if you don't put such
strong fields the charges will remain on the surface.
Now, it turns out we can
actually relate the electric field at the surface to the
charge density of the surface by a following trick.
Let's go to the surface here
and ask, "What's the electric field?
Which way can it point?"
I claim the electric field can
only point perpendicular to the surface,
because if you had parallel components then the charges can
move along the surface. No one says you cannot run it
on the edge of the swimming pool, you just cannot leave it.
So the electric field lines
must point radial, I mean, normal to the surface.
And I claim that we can
actually calculate the electric field here if you knew the
charge density here. So let's do that.
So here's some surface,
and I have--take a tiny region here and I'm going to take a
Gaussian surface that looks like this.
It's a cylinder,
very tiny cylinder, and the field lines are like
that there. There is no field lines here
because there's no field inside. And the field here all of these
things are 0, and the field here is parallel
to the cylinder so it doesn't contribute.
You see that?
I got a cylinder.
I rammed it into this solid.
It's a mathematical Gaussian
surface. It's got following faces,
flat face, no E because there's
nothing inside the metal, this part of the curvy-face,
no E, nothing inside the metal.
Here E is parallel to
the sides of the cylinder, so there's no flux.
The top end,
if the area's A or dA,
the top end will have a flux which will be the E that
I'm trying to calculate times dA.
That's going to be equal to the
charge enclosed is σdA/
ε_0, because σ
is the charge density which I assume I'm given.
Then you can see the dA
cancels. The electric field is
σ/ε_0. Do you remember ever seeing a
formula like this sigma or anything like this?
Yes?
Okay.
Let me remind you where you saw
it. I showed you that if you took
an infinite plane of charge density σ
the electric field was σ/2ε_0,
and I did that with a long calculation.
I took a point there.
I took rings and so on.
Now, you don't have to do that.
Let's use Gauss's Law again to
calculate it. Once again, you argue that if
you're in front of an infinite plane the only distribution of
field lines that makes sense is if the field line's always
perpendicular to the surface. They can get weaker as you go
away, but at a given distance from
the plane they should have the same value,
because if you slide the plane up and down it shouldn't matter.
So let's take the side view of
the plane. There are your charges.
And I'm going to take a
Gaussian cylinder that looks like this.
The field lines are going like
that. They're going like that.
There is some charge trapped
here. And the area of that guy is
some dA. So again, I apply Gauss's Law
here. Of the cylinder I got the curvy
sides with no contribution, because the field is parallel
to the curvy side. I got the flat faces.
So I get E times
dA for one of them, and E times dA
for the other one, because both are outgoing.
That is equal to charge
enclosed with the σ times dA divided by
ε_0. If you cancel the dA you
find E is σ/2ε_0.
So the electric field on either
side of the plane is σ/2ε_0.
You just get it from Gauss's
Law. Again, there's no reason to do
the complicated integral. That's because we've reduced
everything to one unknown, namely what was the magnitude
of the electric field at a certain distance from the plane?
Direction is known to be
perpendicular and to be constant on this line.
Then the one number came out to
be, in fact, independent of how far you are.
But now go back to the
conductor. If you go back to the conductor
you find the field at the surface of a conductor is
σ/ε_0 where σ
is the charge density there, whereas for an infinite plane
it's σ/2ε_0
on either side. And you can ask,
what's going on? Why is it
σ/ε_0? Can anybody guess what may be
happening? In other words,
if you take a tiny area on a conductor and you go very,
very, very close it should be as if you're next to an infinite
plane, because if you're very,
very close to the surface you don't know if it's finite or
infinite. So you should get the same
answer as for the infinite plane.
Therefore, if you took only the
charge sitting there it should give you
σ/2ε_0 going out and
σ/2ε_0 going in.
In other words,
let me draw a picture here. That's my tiny area.
I expect the field due to just
the charge there to be like this,
σ/2ε_0, and
σ/2ε_0. But that is not the whole story
because I've got the rest of the conductor that's got its own
charge. It is going to produce a field.
That field will have some
value, and I claim I know the value of that field.
It'll be precisely enough to
cancel it on the inside, but if you cancel it on the
inside you'll double it on the outside because it's got
σ/2ε_0 pointing outwards to cancel this
guy, but then it'll aid the one due
to the little area by doubling it.
That's why it's
σ/ε_0 outside and 0 inside.
In other words,
here is the full story. If you go to a conductor,
if you pick a region here, the field at this point is due
to what's here and what is everywhere else.
What is here does that as if it
was an infinite plane. The rest of the guys do that
and that cancels this here, but adds there,
so you can see more ε_0 outside.
Okay, then there are other
variations to this theme. So here's a conductor,
and I make a hole in it, and I put some charge.
You can ask if I throw some
charge on it where will it sit. Well, some charges will sit
here if you put some coulombs, but what will happen on the
inner wall? This is a hole.
The claim is that there'll be
no charges here. They'll all be outside even
though you've got a hole in the middle.
Again, Gauss's Law tells you
why. If I take a surface like this
and do Gauss's Law, since the electric field is 0
inside the metal, Q enclosed is 0.
So the Q that's
enclosed, which is 0, means that maybe some positive
guys and some negative guys, but there's nothing from
stopping the positive guys from rushing to meet the negative
guys and canceling. They will, for there is no
reason they just stay on opposite sides of the island.
They'll just come together.
Therefore, there is no stable
equilibrium in which the charges will be ever inside a conductor.
So if you have a conductor and
you put charge on it, it goes to the surface even if
there are holes inside the conductor.
Now, that's a very interesting
paradox. I showed you that the field
inside a conductor even if there's a hole is 0,
but let's ask the following question.
I take a conductor with the
hole in it and I put a charge q here.
Will you know about it from the
outside? You're not allowed to enter it.
Will you know from the outside
there's a charge inside or will it get also shielded?
Will there be any flux lines
coming out here? Yep?
Student: Yes.
Prof: Because?
Student: Because
there's a charge. Prof: Right.
If there are no field lines
here, and I do Gauss's Law on this surface,
if the surface integral of E is 0
q_enclosed is 0,
but I know there is some q inside.
So what's going to happen?
I want the field lines to be
present here and absent here. So what'll happen is if you put
a q here the material conductor will separate into
-q's where these lines can terminate,
and then out here will be some compensating plus charges
that'll produce the field that'll produce the lines going
out. In other words,
this is really like a chunk of copper I showed you where it
screens the field by polarizing into a negative part in this
wall and positive contribution in the outer wall so that inside
there is no field. Okay, so these are different
variations of this theme and you should be able to do a whole
bunch of problems connected with that.
Okay, I've used up my board,
so I think I'll give you guys a five minute tour on how to do
these integrals in case it ever comes up.
I'm just going to do two very
trivial integrals and stop, and I'm going to do them in two
dimensions and you can worry about higher dimensions.
It's just generalization.
Suppose you're in the xy
plane and there's a function in the xy plane,
f (x, y), and someone says find f (x,y) dx dy.
What's the procedure and what
are you supposed to do is the question.
What you do in Cartesian
coordinates is you take a line y equal to something,
y equal to something delta y,
x equal to something, x equal to something
delta x, and that region (bounded) by
these contours of constant x at different values and
constant y at different values,
has an area dxdy. You multiply that area by this
function. This could be the number of
people living per square mile and this could be the number of
square miles, and you add up all the little
squares over the area that was given to you.
So whoever tells you to do
integrals better tell you over what range you're going to do
that integral. Okay?
So let's find out the area for
triangle by this process. Here's a triangle.
Let's say there's one here and
one here, 0,1; 1,0.
This is y.
This is x.
I want to find the area of the
triangle. Yep?
Student: Isn't that
point 1,1? Prof: Yes, thank you.
That point is 1,1.
You're supposed to take
dxdy you write it like this.
There's no function or integral.
I just want 1 times
dxdy, so the 1 is implicit.
dy, you pick a certain
x and y should go from here to here.
The range of y is
clearly 1 - x because this is a curve x
y = 1. So y goes from 0 to 1 -
x, and for every choice of x, x goes from
0 to 1. Yep?
Student: Isn't the line
defined as y = x ________. Prof: Pardon me?
Student: Is the line
________? Prof: Oh, let me see.
No, no.
This is not the equation of the
line. This height here is equal to 1
- x. Oh, you think that's wrong?
Student:
> Prof: The height of
y, isn't there 1 - x?
Student:
> Prof: Or it's just
x. I'm sorry.
Okay, caught on tape.
Very good.
All right, so here is x,
goes up to x and the y integral you do it as
lower limit, upper limit gives you an
x and x goes from 0 to 1.
That gives you x^(2)/2
from 0 to 1. That gives you 1 over 2.
If you had a function of
x and y you wanted to integrate,
in other words instead of just the number one,
if you had a function of x and y the rule
is the following. You put the function here,
and put x equal to a fixed value,
don't take it as a variable, and treat y as a
variable and do dy from this limit to this limit.
When you are done you will get
some mass that depends only on x that you integrate from
0 to 1. But the final thing I'm going
to do is to do the same integral in polar coordinates,
area of a triangle in polar coordinates looks like this.
In polar coordinates,
as you know, you draw circles of r
and r dr, and lines at θ and
θ dθ, and this shaded region here has
got one side equal to rdθ and the other
side equal to dr. So area integral will look like
rdrdθ times some function of r and
θ over the allowed region,
but I want to do now a triangle of height 1 and base 1.
Here's what I do.
I'm going to pick theta first
and hold it fixed, and theta will go from 0 to
Π/4. Can you see that?
This is supposed to be a square.
Theta goes from 0 to
Π/4, and I have rdr.
What is the range of r?
How far is this thing?
You can see that if this angle
is theta then rcosθ is 1,
and for r is 1/cosθ.
So r dr is really
d of r squared. d of r squared is
equal to d of 1/cos^(2)θ,
right? I mean, let me do it this way.
d of r squared is
r^(2)/2 from 1/cosθ to 0.
So that gives me 1/2 integral
dθ 0 to Π/4 of 1/cos^(2)θ.
Well, you may not know this,
but sec^(2)θ is the derivative of
tanθ. So derivative of integral of
tanθ will turn out to be simply tanθ
calculated between Π/4 and 0.
That'll be just 1.
So you're not supposed to
follow at lightening speed this integral, okay?
Either you've seen it before or
if you need some help you can look at my math book I mentioned
to you. All I'm trying to tell you here
is if when you have multiple integrals you pick one
coordinate and you freeze it, for example,
x when you integrate to the other coordinate over the
allowed region that keeps you in the boundary then integrate over
the second variable. You may need that,
but for most of the problems I've given you won't have to do
that. They're pretty simple.
