Prof: Let's turn to the
subject proper. I will start by doing the usual
stuff of just refreshing your memory on what happened last
time. I mean, last time was pretty
heavy, a lot of mathematical machinery which I like to work
with. Now, you might say,
"I'm going to be a psychologist or art historian,
and I don't need all those details.
Just tell me what I need to get
through this course," so that's really a reasonable
question. I probably would have the same
attitude if I took a course in your field.
So here it is.
What I've told you so far is
that the electric field, which we understand now as
basically Coulomb's forces, allows you to define a
potential because it's a conservative force.
That means the integral of the
electric field on any closed loop is 0, and remember that's
what it takes to define a potential.
And the electric force,
which is Q times E will also satisfy the
condition. That means we can define for
this problem a potential V.
This is called a potential.
So that ½mv
_1^(2) (this is a little v this is big
V) V(r _1_ ) =
½ mv_2^(2 )
V(r _2). That is our goal.
Ah, missed one mistake already.
Anybody know what's missing
here? Student:
> Prof: Pardon me, yes?
What's missing?
Student: The charge.
Prof: Yes, the charge.
So this should really be
q times V and this should be q times
V. So let me remind you why it is.
The force on any charge is
really not just the electric field.
It's q times the
electric field, and therefore the potential
difference is associated with the work done by the electric
field times q. So you've got to remember q
times V is the real energy.
So V is like an
electrical height. For example,
if this is the ground there's a mountain like that.
You give me the height.
The potential energy is not
just the height. It's mg times h.
So let me say you give me
g times h. That's like the potential.
That tells you the whole story
because if you want to lift a 1 kilogram mass to that height you
do 1 times gh, 5 kilograms 5 times gh.
So depending on the mass you
want to lift the work done will be different and the potential
energy will be different, but this doesn't depend on the
mass you're planning to carry around.
So you take out the mass.
You can always put it back.
Similarly the potential
concerns itself with the unit charge.
If q were 1 nothing I
wrote is wrong, but in general q is not
1. So this is the actual energy.
So this is the old potential
energy you want. So the potential energy is
q times the potential, okay?
So think of V as the
electrical height at a point and q is like the mass,
and q times V is like mgh.
Okay, we've got a formula for
V. The general formula for
V is that V of r_2 minus
V of r_1 is the line integral of
E⋅dr from r_1_
to r_2. This minus sign has the
following meaning. The electric field exerts the
force E on the unit charge,
but if you want to drag the charge against the field from
r_1to r_2 you apply
force minus E to compensate that,
and the work done by you, this really work done by you to
move it from r_1 to
r_2. Let me give you the gravity
example. Gravitational force acts this
way, but if you want to lift something from here to here you
oppose it. Namely it's doing minus mg.
Your force mg you move
it to height h. That's the work you do,
and that you know is the potential at the top,
and that's potential at the bottom.
So that's the origin of the
minus sign. The electric field wants to go
one way, and if you want to take it
against that force and raise it electrically speaking the force
you apply is precisely -E.
That ⋅dr is
the work done by you. That's what you invest.
That's what you'll get back if
it rolls downhill again. Now if you want V at
some point r minus V at infinity that
integral will be minus starting from infinity to the point--
these should be all vectors--to the point r of
E⋅dr. We will now take the convention
that V of infinity is equal to 0.
Remember, potentials are
defined only by the differences so you can put the 0 anywhere
you like. We'll choose the 0 to be at
infinity. Then in this integral it's
going to be 0, and if you do this integral
from infinity to r you will find for 1 charge it is
q/4Π ε_0 times
r for a single charge, and r is the distance
from that charge. It looks like Coulomb's Law but
a couple of differences. There's no 1 over r^(2) and
there is no vector, okay?
The potential due to a charge
here at the point here is a number given by this,
and if you've got lots of charges that's just fine.
V due to many charges is
sum over all those charges, the value of each charge
(divided) by 4Πε_0
times the length from where that charge is to where you want the
potential. That's the principle of
superposition, okay?
The coulomb potential due to
one charge is 1 over r. Due to many charges is each
q times its own 1 over r times this 1 over
4Πε. And finally if you want the
potential difference due to the field you integrate it,
and if you want the field from the potential you take the
derivative called the gradient, which is identical to minus
ix partial derivative minus jy
partial derivative. You should expect two
derivatives because the force you are looking for now is a
vector. You're living in 2D so you
cannot say the force is a derivative of the potential,
in what direction? If you want the force in the
x direction take the x derivative.
If you want it in the y
direction, take the y derivative.
And you put them together into
a single vector and that's the electric field.
So you go from potential to the
field by taking the derivative or the gradient,
and you go from the field to the potential by doing the
integral. Again, the subtlety is the
integral is of a dot product of E with a displacement,
because everything is a vector now.
It used to be fx dx in
the old days, but f has become a
vector, dx has become a vector dr so you need the
dot product. Okay, so this is a summary,
but I want to make two course corrections from last time.
First one was when I tried to
prove to you that the electric field is conservative I did one
thing which I realized was a swindle.
I didn't realize it myself and
I wasn't waiting for anyone to correct me.
I did not think about it hard
enough. So here is the mistake I made.
Remember what I said.
I said that's a potential due
to point charge. I'm trying to look at it.
These are different lines of
the electric field. Let's say you want to go from
here to here. You do some work,
E⋅dr along this line.
The claim was that's the same
work you do if I take another path that looks like this,
go this way, go that way,
and go that way, same work, and that's still a
correct statement. It's the same work because here
dr is angular, E is radial,
the dot product is 0, likewise 0 here.
At every segment you have here
E is radial, dr is radial,
the dot product is the field E times that length.
Here it's the field here times
that length, but the field is constant at a given radius so
they're also equal; still no wrong statements.
Then I said,
"What if I do this, a little radial,
a little angular and so on," and that's also okay.
Then I finally said,
"What if I have some nice smooth path,
which is not made up of radial and angular,"
and my claim was, well, you take for this
segment, for example, you take a radial segment,
then you take an angular segment,
then you may take another radial segment and another
angular segment. So you approximate the smooth
pass by a bunch of radial, angular, radial,
angular, and from a long distance it all looks the same,
so it should be the same amount of work.
That's wrong.
I'll give you an example of why
such arguments can be wrong. Let's take a triangle.
This side is 1,
and this side is 1, and I want to walk from here to
there. So you know I go a distance
√2, but then you say I'm going to take a different path.
I'm going to go horizontal,
vertical, horizontal, vertical, equal amounts like
that. And these are very tiny
differences. You can say it's the same
thing, but this path is of length two.
Understand why?
All the horizontal parts add up
to 1. All the vertical parts add up
to 1, so going zigzag is longer by a factor √2.
So just because things are
approximately the same it doesn't mean the answer will be
the same because if you make an infinite number of infinitesimal
mistakes you can add up to something finite.
Likewise here,
you've got to make sure that when you deform your path from
what was really given to you to what you did,
which is radial and angular, the work done should be the
same. The path length is clearly not
the same in this example, and it can all add up,
but the work will be the same. That is what I did not show to
you. So let me take a segment here
where that's the radial part and an angular part,
but the person really wanted it on that segment along some
straight line joining the two. What I want to show you is that
not only does that approximate this,
from a long distance, the work done on that plus that
is the work done here, okay?
So for your convenience I'm
going to blow up that picture here so you can all see what I'm
trying to do. This should be the radial
direction. This is the radial direction.
This is E.
This is some perpendicular
direction, and this is the direction in which the move
dr was made. That's the actual path.
It's going like that,
and you manage to just deform it.
So what's the line integral on
the radial path? Well, it's the magnitude of the
electric field at that distance times the difference in radius
between this and then this end. Let me call it delta r,
delta r is the spacing here between that curve and that
curve. We agree the dot product is
trivial. E is this way.
Displacement is this way,
no cause thetas to worry about. That's the work done.
Work done here is 0 because
perpendicular to the field. So that's the total work done
going like that. How about going along this line?
It's going to be
E(r)⋅ dr.
When I say E at
r, r is, of course, not 1 point it's
kind of spread out, but this is an infinite decimal
region, so you can talk about the
E at one point and not worry about the variation.
So don't forget it's an
infinite decimal triangle. E⋅dr
will be E at the distance r,
the length of dr and the cosine of the angle between them
which is this angle. The length of dr times
cosine is precisely that distance, Δr and
therefore you do the same amount of work.
So basically I am saying I
should wander around. You may not choose to go radial
angle or radial angular. If you go at a slant to the
radial the work you do really depends on how much radial
distance you cover. That's what
E⋅dr does for you.
That's why you can deform the
fact anyway and get the same answer.
That's the first thing.
Second thing is I forgot to
tell you anything about units. So the electric field is force
on the unit charge, so the unit for that is
Newton's per coulomb. The electric potential looks
like the work done on a unit charge,
so the units for that is joules per coulomb,
and the name for joules per coulomb is a volt.
It means if there's a voltage
difference between two points of 12 volts it's going to take you
12 joules of work to lug a coulomb from the lower point to
the higher point. So you guys have to get your
units right, okay? Now, you might say,
"Why don't you use units," right?
You know why?
Because I have tenure;
when you have tenure you don't have to use units.
I park next to the fire hydrant.
I have tenure.
I tear up the tickets.
I don't do jury duty.
I don't pay taxes,
you don't have to do anything when you reach--and Alan is
going to tell you the joys of being a graduate student.
The joys of being tenured prof
are hard to describe, okay, priceless.
In fact, one of my sons,
when he saw my lifestyle, said it reminded him of a
jellyfish deep in the ocean. As soon as the little guy can
crawl he attaches himself to a rock,
and he never moves, and eats his own brain for food
because he has no further use for it.
So apparently that's what I
reminded him of, anyway.
So you can, if you work hard,
you can be like me, okay?
You can do that.
Anyway, look,
so you guys have to get the units,
because seriously speaking if you don't write any units we
don't know if you're right or wrong,
right? For any problem the answer is
19 in some units, right?
So if you don't give the units
the number is not worth anything.
Another advantage of units is
you can keep track of your calculation.
You multiply one quantity by
another quantity. Everything has units.
You cross out the lengths and
the meters. You've got to make sure the
stuff you get has the right units on the two sides.
That helps you also keep track
of calculations. Okay, so let's go back now.
This is all stuff left over
from before. So now for the new stuff.
I'm going to tell you the first
part of this lecture is going to be what are the advantages of
V. I will just mention them.
Then I will elaborate on them.
The first advantage is the
conservation of energy. The second advantage is the
computation of E, and the third advantage is a
lot of good visual pictures which are very helpful.
So you won't know what these
mean, but this is what we're going to do so you know where we
are going. First thing is conservation of
energy, there's no need to belabor this
point, I mean, that's why we did the
whole thing, the whole conservative forces,
gradient, this, that, all that was to
make sure in the end you can get a potential energy out of it.
And as you know from the roller
coaster problem if you can do kinetic plus potential = kinetic
plus potential it saves you a lot of trouble.
In other words,
in this room right now there's a lot of electric fields,
a lot of charges everywhere applying a field here,
and I want to carry a charge from here and I want to drag it
over here, or maybe the charge was shot by
a cyclotron or something. It came flying out at some
speed here, and it comes here a little later,
and I want to know how fast it's moving.
One way to do that is to follow
the charge as it moves. At every instant find the force
on it, the acceleration from that,
change in velocity, add up all the changes you'll
get the final velocity, but as you know with the law of
the conservation of energy you can skip all the intermediate
stuff. You just have to know the
initial kinetic energy, initial potential energy,
final kinetic energy, final potential energy.
They are equal,
if you sum them- so you can find one number from it.
One of the numbers you can find.
In particular if you didn't
know the final kinetic energy you can find it.
So that's the first advantage,
conservation of energy. Second thing is that it makes
life easy when you want to compute the electric field.
So this is an example I did,
again, near the end of the class,
and I think it was needlessly complicated,
so I will try to repeat the derivation a little more quickly
because it's a lot easier than I gave the impression.
So we are trying to find,
let us say, the electric field you'll do a dipole -q and
q separated by a distance 2a.
If you did it with forces,
with forces I didn't even try to do it everywhere.
I remember I did it here and I
did it there. The force here was like this.
The force there was like that.
Then I said if you do a lot of
work you'll get all these lines. What makes life difficult is
that if you go to this point here,
this I explain to you last time, this guy will repel it
with some force, this guy will attract it,
and you've got to add the two arrows and get a new arrow.
That's the net electric field
at that point. So you've got to add arrows and
that's more difficult than adding the potential which is
just a number form this guy and that guy.
Then once you've got the
potential you find the field by taking derivatives.
You want E_x
take the x derivative, the minus sign.
If you want E_y
you take minus dV/dy.
So what's the E here?
You remember it's
q/4Πε _0 times
[1/r^( ) (r^( ) is that length) −
1/r^(-)]. That's it.
No arrows, nothing,
that's the total V. This is r^(-).
Let's define r to be if
you want the distance from the center.
And I want you to imagine this
is long way off. This picture is not drawn to
scale. You should imagine these two
points are like this and I'm somewhere there.
That's when the dipole
approximation's working. This is not an approximation.
This is exact.
So let's do the following then.
We come by the denominators.
You get r^( ) times
r^(-). Numerators r^(-) −
r^( ). Now, you can see that r^(
) and r^(-) differ from r because of this
little guy a. If you use the law of cosines
you know it's that squared plus that squared minus 2ar
cosine theta or something. In the bottom we ignore that
difference and we write everything as 4Πε
_0r^(2). In other words r^( ) I
approximate with r, r^(-) I approximate with
r. So this is not the exact answer.
There are corrections here that
look like 1 plus maybe a number a divided by r and also
a^(2) divided by r^(2),
but these are all negligible. In the numerator you can again
say, "Hey, why don't you replace this by
r and this by r then you get 0?"
It is true.
If you replace everything by
r then you do get 0 and that's because in the difference
the big guy cancels and what's left over in the difference is
what's going to keep this from vanishing.
So we have to find the
difference in that distance and that distance.
So for that what you do you
draw a line here. Now, you cannot draw a real
perpendicular here, but if these things are very
far and these lines are almost parallel you can draw a
perpendicular there. Then if that angle is theta,
theta is the angle then it's very easy to show that this
distance is simply 2a cosine theta because if theta is
this angle then you can say cosine of this angle here will
be by whatever complimentary angle is 2a cosine theta.
Therefore in the top I have
2a times cosine theta, and that I will write as dipole
moment p cosine theta over 4Πε
_0r^(2). That's what I was trying to get.
Yep?
Student: Sorry,
which angle is theta? Prof: Theta is the angle
measured from here. Let me draw this.
One can worry about is it this
angle or that angle, but I'm saying draw the angle
from the middle, okay?
That angle is theta,
and roughly speaking that angle is also theta.
Now drop a perpendicular here
and this side is 2a, therefore this difference is
2a cosine theta. So now I have V in terms
of r and theta, but I can write it in terms of
x and y very easily,
it is p/4Πε
_0r^(2), and cosine theta is
x/r. So the whole thing is
x/r^(3) which is the same as
x/(x^(2) y^(2))^(3/2),
right? r is the square root and
r^(3) is three powers of that.
Now, you can easily take the
derivative of this guy with respect to x,
or with respect to y. Derivatives are very easy to
take. Integrals are hard to do.
And that'll give you the
electric field E_x and
E_y everywhere, okay?
So E_x is
equal to -dV/dx, and E_y is
-dV/dy. It's one of the homework
problems is to really calculate this.
And if you calculate it and you
plot it you will see this thing emerging.
Okay, in other problems most of
the time it's easy to find V and then to find E.
For example,
if I have a ring of charge and I want to find the V here
then let's say the ring has charge of q and some
radius a. Then a little guy here produces
a little potential dV which is the charge dq
(dq is not any distance it's like the differential)
divided by 4Πε _0.
And what's the distance?
If that is z then simply
a root of z squared a^(2). That's due to this little
charge, but every unit of charge on this ring is equal distance
from this point. You understand?
It's a ring and I'm
perpendicular to the ring. So they all contribute the same
number and sum of all the dq's is just the
q. So that's the potential on the
axis. Remember, you can take a
z derivative to get the field, okay?
Generally it is easier to do
the V. Either it's a sum or integral
over the charges producing the V then take derivatives.
But I want to give you one
example in which the opposite is true,
where in fact it's easier to get E than to get
V, and that is--there are many
such examples. I'm just giving you one.
Let's say there's a hollow
conducting sphere. It's got charge Q.
So this guy's hollow and
conducting. I want to find the potential
everywhere due to this, and the charge Q is
uniformly spread on the surface. So what's going to be the
potential? First of all,
by symmetry you can see that if you pick any point anywhere the
answer's going to depend only on the radial distance from the
center because it has spherical symmetry.
But you still have a lot of
hard work to do because you can divide the sphere into little
rings like this, slice it at different
latitudes, and find the potential due to each ring using
this formula I just did somewhere,
then integrate it over all the rings and you will get an
answer, but that's a decent amount of
calculus. But you don't have to do that
because in this problem because it's spherically symmetric and
there's a charge Q here we know from Gauss's Law the
electric field is going to be purely radial and will look like
Q/4Πε _0r^(2) as long
as r is outside this sphere.
So here's one problem where
it's easier to get E because E is
pretty trivial, it's radial.
You just need to know the
magnitude. Gauss's Law tells you the
magnitude. You put the unit vector back
and you get this. So as long as you're outside
this sphere it's as if Q were sitting at the origin and
the potential V(r),
which is -Q Q/4
Πε_0r outside that sphere,
because remember V(r) minus
V of infinity is equal to integral of
E⋅dr from infinity to r with a
minus sign, and if you just take the field
of a point charge and do the integral you will get the
potential of a point charge which is this.
So if I draw the potential as a
function of radius it will look exactly like that of a point
charge falling like 1/r. Now, what about for r
less than r? How do I continue this graph?
Student:
> Prof: Pardon me?
Student: Use Gauss's
Law. Prof: You can use
Gauss's Law. If you use Gauss's Law inside
what will you get for E? Student: 0.
Prof: E is 0.
Integral of
E⋅dr is 0, so potential does not
change inside this sphere, but don't think potential is 0.
It does not change.
It's stuck at this value.
So throughout the hollow sphere
there's only one potential, okay?
Electric field is a different
story. Electric field is,
in fact, 0 inside then falls like 1/r^(2).
The electric potential is a
constant. So there are certain problems
like a cylinder of charge surrounded by another cylinder
of charge where, again, it's easier to get
E from symmetry arguments then integrate that E to
get the potential. Okay, so this is my second
point. Namely, barring a few
exceptions like this one, with high symmetry the
potential has the advantage that it lets us compute the electric
field by first doing a sum of scalar quantities rather than
vectors, then taking derivatives.
The third advantage has to do
with being able to visualize things.
It gives you a landscape and
that's intuitively very helpful. So here's one example.
Let's take two parallel plates
and put a lot of plus charge here and equal amount of minus
charge here. What will the electric field
look like in this region? Let's say both have the same
density σ for unit area and they're
infinite. I'm just showing you a finite
portion. This guy will produce sigma
over 2ε_0
here and here. This one, because of negative
charge, will do this; σ/2ε
_0 on this side. In the region between them you
can see the two arrows are additive, so the field will look
like this. In the region outside the two
arrows are opposite. This guy's coming in.
That guy's going out.
And remember,
the field doesn't diminish with distance due to either plate so
they can completely cancel each other.
So the electric field is
trapped between the two plates, and has the strength which is
double what you have on either side if you had only one sheet.
That's the electric field
E. That's my electric field.
You can ask,
"What's the potential difference,"
or "What's the potential at various points?"
This whole thing being metallic
is at one potential. I say V = 0.
Throughout this line let's say
V is 0. In fact it's very clear
V is 0 because if I move horizontally on this line I
don't do any work because the electric field is perpendicular
to me. So if V was 0 at one
point it is 0 throughout this plane.
Yes, I should qualify this
statement. You cannot pick V equal
to 0 over an extended region. You have freedom to pick it to
be 0 at one point, but I'm telling you if you pick
the one point in this plane you can navigate throughout the
plane without changing V because you move perpendicular
to E. So V is 0 in the lower
plate, but it won't be 0 anywhere
else, but you can see that may be V = 1,
that may be V = 2, V = 3,
upper plate maybe V = 4 volts.
So lines of E are
vertical and lines of constant V are horizontal.
This is really like the
gravitational problem. An electric field raining down
on you is just like the force of gravity.
This might as well be gh.
This might as well be g.
You multiply a mass you get
mg. You multiply by the q
you get the force on the charge. It's straight down and the
potential just increases with height.
Yep?
Student: Can you just
go over real quick, please, how you do the calculus
to get the constant voltage _________?
Prof: Oh, okay.
Think of it as a plate coming
outside the blackboard. Student: For the hollow
sphere. Prof: Oh,
for the hollow sphere. The potential is always the
work done to go from infinity to wherever you are,
right? So when I come from here what
do I think is at the origin? I think there is a point charge
at the origin, so I pushed up against that
until I come here. So I do that amount of work.
Once I'm in here there is no
electric field inside, so
E⋅dr is 0 from then on,
but don't forget the V that you got up to that point.
You don't drop it.
You keep on adding 0 to it.
So it is 0 everywhere.
You add 0 to all points here
because you can go from here to here, no electric field.
So the whole sphere has the
same potential as the surface. So now, here I've got this
potentials and it gives you a picture that you're running
uphill and these are the steps you've got to climb to go
uphill. And if you've got 4 volts here,
if you take a coulomb and you let it drop it'll pick up 4
joules by the time it hits this lower plate.
That's the meaning.
That's the work done by the
electric field just like gravity is the potential difference
which is the work done on unit charge times the charge which is
1 coulomb here. If it was 10 coulombs you do 40
joules of work. If an electron is released
you've got to be a little careful.
By definition the charge of an
electron is negative. See, if an electron is released
here in fact it'll go like this. You don't have any analogs in
gravity. Gravity everything falls down.
Suppose some things had
negative mass and some things had positive mass.
All guys with negative mass
you've got to nail them to the ground otherwise they'll just
float up. That happens in electricity,
so any electron you leave here will go up.
So the force on the electron is
-q times E, so everything has got a minus
sign. So an electron going from here
to here will in fact lose energy.
We will lose potential energy
and it will gain and turn in kinetic energy.
So an electron falls through 1
volt. The work done is charge of an
electron times 1 volt which is 1.6 times 10^(-19) joules.
That happens to be called an
electron volt, just a convenient thing to use.
When the charges that are
pushing around are not coulombs but this tiny electron it's good
to know in electron volts. For example,
in the hydrogen atom the coulomb potential is negative
because the charge at the origin is a proton q (electron
has got some -q) divided by 4Πε
_0r. That's the potential.
And in quantum mechanics only
at certain radius you can have electrons.
So the lowest electron is
somewhere here, and they say it is minus
whatever, some number of electrons, 13 point something
electron volts. That means if you want to drag
an electron from this hole and liberate it from the atom you've
got to do 13 times 1.6 times 10^(-19) joules of work.
So as if this were 13 volts
down compared to infinity. It's a hole in the ground 13
point 4 or 6 volts deep. So if you want to liberate that
electron you have to furnish that amount of energy.
Okay, now going on with the
visualization let me say--let's take another case we can all do,
which is a single point charge. Here's a point charge.
The electric field due to that
looks like this; we've done it a million times.
Now, what about the electric
potential? For a point charge the
potential is 1/r. That means it's a constant on a
circle of radius r. That means if you pick any
circle, or a sphere, I'm sorry, V is one
number here, V is a different number
there, a different number there.
So that's the whole surface
surrounding the charge on which V is some fixed number.
It depends on that radius.
You go further out it's a
different number depending on that radius and it falls like
1/r. You'll notice once again that
the lines of constant potential are perpendicular to the lines
of the field. That was true there.
That is true here.
In fact, let me do one more
example. If you take a dipole negative
and positive, and the field lines go like
this, if you compute the potential,
if you go very, very close to this guy,
I think you all seem to understand if you're very near
one charge forget about the rest of the world.
It dominates everything.
Potential would look like
circle centered here, but as you go further out the
shape will change and it will look like this.
A similar thing on the other
side then there will be this huge infinite sphere that
becomes plainer that goes all the way to infinity.
These are contours of constant
V. If you really plot them you'll
again find contours of constant V or equipotentials are
perpendicular to the electric field lines,
perpendicular to E. So I want to explain to you
that this is a very general result.
This is not something true in
these isolated cases of nice symmetry.
Why is that?
So let's understand why that is
true. You remember that the
V_2 - V_1 is -
integral of E⋅dr
from 1 to 2. So let's say
V_2 and V_1,
2 and 1 are very close. In that case let's call it
delta V = -E ⋅dr.
There's no integral to do
because I've picked two points which are near by an infinite
decimal amount dr, so
E⋅dr is equal to delta V which
is a minus of V_2 -
V_1. What does that mean?
That means if you start at some
point, and you move a distance
dr, you take a step dr,
and you find the change in potential,
the answer depends not only on the length of your step,
but on which direction you move. Because of this dot product the
change in potential will be the magnitude of the electric field
at that point, the magnitude of the step
length and the cosine of the angle between them.
So if the electric potential is
replaced by height in two dimensions,
you know, top of Sleeping Giant and the hill is going up and
down, up and down,
you're at some point. There's a height function
h of x and y instead of V of
x and y, and I want to take one-step,
and I can ask, "What are the consequences
to my height?" The answer is if the electric
field is pointing in some direction and you take a step
dr in a generally different direction this is the
change you get. So you can ask,
"How do I get the biggest drop in height,
or the biggest drop in potential?"
And cosine theta is 1, right?
Because then there's this,
that times 1 with a minus sign. So the electric field points in
the direction of the greatest rate of change of potential.
It drops in that direction.
If you release a marble in the
gravitational field it'll roll along the gravitational field
which is obtained the same way from the gravitational
potential. If you want to climb up the
mountain, suppose there's a tsunami
coming and you want to get to the top as quickly as possible,
don't panic, compute the gradient and move
along the gradient, okay?
Then stop, think again,
recompute the gradient and in that manner you will get to the
top in quickest time if you can compute quickly enough,
okay? But that's the process.
At every point there's a vector
pointing in the direction of greatest increase.
The electric field happens to
be the direction of greatest decrease because of this minus
sign in our convention. Now, let's us say you're on
this mountain and you don't want to go down and you don't want to
go up, but you still want to take a walk.
Can you do that?
Yes?
Or are you stuck at one place?
What do you have to do guys?
Yes?
Student: You move
perpendicular to the gradient. Prof: You move
perpendicular to the field. To move perpendicular to the
field the cause theta is 0. That's why, now going back to
this, this is called Mt. Coulomb, it's coming out of the
blackboard really tall and you are here.
If you want to go downhill you
go along the field. You run away from the hill all
the way to the valley, to the plains.
If you want to go to the top
you go opposite of the field, but if you go along this
equipotential at every point you do not do any extra work so your
potential does not change. So it's clear from this that
the equal potential is always-- in fact in general the change
in any potential is gradient of the potential dot dr.
Except for the minus sign it's
the same story that if dr is parallel to grad
V or anti-parallel to E you get the biggest
change in voltage. All right, so this is what you
should understand in general. What is good about the
potential is you can take a whole bunch of charges put them
here, there, there,
there, and you can draw little contours,
maybe of constant V. It's just like the graph you
have of the weather, of the temperature,
for example, or if you have a topographic
map of the mountain there are some valleys,
and there are plains, and there are some
mountaintops, and these lines of constant
height are like constant V,
so it helps you visualize. And then if you're told you
release something here and it rolled down and came here you
can find the speed here given the difference of these two
numbers. You don't have to go through
the intermediate step. So these are the advantages.
Okay, now I'm going to return
to the conductors briefly to tell you something.
One other thing that is new
compared to what we did before. Remember what I told you about
a conductor? A conductor is one in which the
electrons do not belong to any one parent.
They run around to the whole
sample, but they cannot leave the metal.
At the boundary of the metal
there are forces bringing them back.
Inside they are free to move.
Therefore, if you take a
conductor and you stick it into a field initially the electric
field will penetrate the conductor.
The charges,
because they are free to move, will move.
The negatives charges will move
one way, positive charges will move the
other way until they set up a back reaction to this that
blocks the field inside the metal until the field inside is
0. So here's a generic conductor.
If you put it in an external
field it will have some charges on its surface,
will produce a field this way that will oppose the external
field in which you put it until at the end there's no field
inside the metal. Or, if you did not put it in
the field and just took this material and dumped some charge
on it, threw some charge on it and
asked where does it go, well, the charges are all plus
and they don't like each other. They want to move away from
each other, and they move as far as they can until they come to
the boundary and they cannot go any further.
That's where they stop.
That we saw before,
so we know all the charges in the metal are in the boundary.
There is no electric field
inside because of Gauss's Law. If I want to find the charge
here I do a tiny sphere and do integral
E⋅da. I get 0, and that's the charge
inside. And I can move this little
microscope all over the inside. I keep getting zero.
That's why I know there's no
charge inside. There's no charge inside.
There is no field inside.
Q = 0 inside metal.
This is a solid metal,
and E = 0 inside solid. The other variation you can
have is suppose they make a hole in this.
There is no electric field
anywhere here because it's a metal.
The whole logic is if you've
got a field you're not in a static situation.
Let the charges move.
They'll keep moving until they
have no reason to move. That reason will stop when the
field vanishes. It's quite amazing.
They can always find an
arrangement in which they kill the external field.
That's the perfect metal.
So there's no field here.
So if you do a Gaussian surface
like this the charge inside that is 0,
but that does not preclude the possibility that there are maybe
some plus charges there and some minus charges there.
Last time I told you that if
that happens then the pluses and minuses will run around and meet
and destroy each other and neutralize,
but that's a little subtle because maybe if they were the
only two things in the world I can understand that,
but there's also these charges outside maybe that'll impede
their motion. But here's one way to prove
that there can be really no charge even on the inner
surface. If there were some plus charges
here and minus charges there--remember Q is 0
inside so you can only have equal number.
Then electric field lines will
have to leave the plus charge and terminate on the minus
charge, so there'll be an electric
field inside the cavity, and I'm saying that's not
allowed because of the following.
If there is a field inside the
cavity let's take that path and another path which are two ways
to go from here to here. On one path I get a line
integral of E to be not 0.
Second one I get line integral
equal to 0 because there is no E, but the line integral
cannot vary with the path in an electrostatic situation.
Therefore, there cannot be any
electric field inside the cavity due to anything you put outside.
You can have an electric field
inside the cavity if you go into the cavity and put by hand some
extra charge there. Then what will happen is that
guy will produce a field and the field lines will come to the
boundary and terminate on some negative charges.
Again, there'll be no field
here. But a hollow cavity with no
charges inside the hole will not have charges on the edge either.
So now, what else can we say
about conductors? Today I used the fact that the
line integral of E is 0 to tell you that in a hollow
cavity you cannot have any charges in the wall.
Second thing is every conductor
is an equipotential. It's an equipotential.
Why is that?
Can you tell me why that is
true if it is not equipotential, the whole lump?
Yep?
Student: If it wasn't
equal potential then there would be an electric field inside
there. Prof: And how do we know
that? Student: Because
there's a potential difference ________________.
Prof: That's right.
In other words,
if you take the gradient, namely taking derivatives,
you'll get a non-zero derivative because I'm telling
you V is not constant. In some direction it's got to
change. There's an electric field in
that direction, at least, but that's not
allowed. So V is a constant,
so take any metal shaped like anything.
You can associate a single
potential to all of it, okay?
If it's got charge on it that's
the same, and what it means if you start
at infinity the work done to come there,
the work done to come there, the work done to come there are
all the same because they're all measuring the potential.
The whole metal is at one
potential. That allows you to do certain
tricks. So here's one trick you do.
There are more tricks,
but this is the only one we can discuss in this class.
So here's an infinite
conducting plane, a sort of ground,
attached to the ground, and over this guy I put a
charge q. And I ask you,
"Find the electric field everywhere now."
That's your challenge.
Without this guy fields are
just radial, so radial, radial, but the fields come
near this metal they cannot penetrate.
They've got to stop at the
surface of the metal. So the field lines will do that.
They will all terminate on the
metal, and to terminate you need some negative charge which the
metal will suck out of the ground.
That's the reason for grounding
it. So the question is what kind of
charge distribution do you get on the metal?
How much charge is there on the
metal? And because there are positive
charge and negative charge here what's the force of attraction
between this charge and the plane?
In general that's a very
difficult problem. If I gave you a potato,
aluminum potato, put a charge next to it,
it will have a force, but it's not easy to calculate
it. But this infinite sheet turns
out very easy to calculate. The reason is that we can solve
this problem by borrowing from another problem.
The problem I borrow from is
the following. Forget this plane.
Continue these lines,
and think of the problem of a dipole with an equal negative
charge here. So the lines from them go like
this, right? And if you slice it in the
perpendicular bisector of this distance between them you will
hit every line perpendicularly. You guys remember that part
about the dipole field at somewhere there?
You cut it right down the
middle. So this plane was an
equipotential at V = 0. If it's an equipotential there
is nothing there. It was just a mathematical
surface on which V was constant and equal to 0.
If V is constant you can
slide in a conductor there and nobody will know,
because the field is 0 along the surface of the conductor,
nothing happens. So you can reconcile the fact
that conductors are equipotentials,
and solve problems in which a charge is in front of a
conductor if you're lucky enough to find the situation where
that's an equipotential surface of the same shape as your
conductor then you can slip your conductor there.
So in this example what happens
is that the electric field here throughout the plane is exactly
what you would get due to a dipole minus and plus.
That you can easily calculate
at all points on the plane. And once you've got the
electric field you remember σ/ε
_0 is the electric field so you can find
σ, the charge density.
And if you integrate the charge
density you will get exactly -q,
because the number of lines coming from here have to
terminate on the plane rather than on this and that'll be
-q. And finally,
what's the force of attraction between the plane and this
charge? Can you make a guess?
What's the force with which the
plane attracts the charge? I'm asking you to guess,
so it cannot be too difficult. Students:
Same as that of the other charge?
Prof: Same as the force
that this negative charge would exert on this positive charge,
and I'll tell you why. This guy is looking around,
okay, seeing some forces. It's due to these negative
charges on the plane, but they're identical to the
field created by this guy. So it cannot tell the
difference and therefore if you're releasing it it'll move
the same way in this problem as in the other problem.
So it'll be drawn to the plate
with an attractive force which is q times -q
divided by square of that distance.
So the summary of what I did
now is that if you solve any problem, let's say the dipole
problem, and you get some shape. It's an equipotential.
You're allowed to,
then, replace that with a chunk of aluminum and keep the field
lines the way they are because that aluminum will be an
equipotential and it'll happily sit on equipotentials because
there are no forces on the surfaces of the conductor moving
the charges. So the moral of the story is
you can sneak in a solid object bounded by the surface of an
equal potential for any problem. So if you can solve certain
problems and the equal potentials have nice shapes you
can put conductors of that shape in that precise location.
That'll answer a different
problem in which the second charge is absent.
You only have the first charge.
It's like a phantom charge,
an image charge. So it's like this charge looks
into the mirror and sees the image.
The image is minus of itself
and it's drawn to it, because the field produced by
this guy and this guy is the same as the field produced by
all of these and this. All right, so now we've got a
final problem with energetics. The question is I want to bring
a whole bunch of charges, q_1,
q_2, q_3 from
infinity and bring them to this arrangement.
So when they're infinitely far
they don't even know about each other.
They don't feel any force.
So here is my goal.
I've got to do some work,
bring everybody together. They may all be positive.
They may not want to be in the
same region, but I'm going to force them.
The question is,
"What work do I have to do?"
We're going to calculate that.
We do that by saying first
let's take charge 1. Let it be wherever it has to be.
Then I bring charge 2 from
infinity and put it here. The work done for that is by
definition q_2 times the potential here,
potential there is q_1
/4Πε _0r_12,
r_12 is the distance between 1 and 2.
Now I nail the two charges.
Nobody is allowed to move so I
don't do further work. Then I bring a third charge
from infinity to its location here.
How much work should I do?
The work that I have to do is
the potential energy from infinity to here due to these
two charges. So that means the work I'll
have to do is equal to q_1q
_3/4 Πε
_0r_13 q_2q
_3/4 Πε
_0r_23. You follow that?
That's the work that I have to
do. The first guy,
q_2, had to fight only this one.
The third guy has to fight
these two, and the fourth guy will have to fight these three
and so on. I'm just going up to three
charges. From this it should not be too
hard to guess in general if you've got many,
many charges you do q_iq
_j/4Πε _0r_ij
where i goes from 1 to-- let's see, from 1 to n,
where j goes from 1 to n,
but i cannot be equal to j and you divide by 2.
So if you're not able to read
this I'll tell you the summation is 1 over 2,
i from 1 to n, j from 1 to n,
but with the condition that i should not be equal to
j times all these. You can see that's valid here.
Yep?
Student: V or
U? Prof: This is the--oh,
I'm sorry. You're absolutely right.
This is not the potential.
It's the potential energy.
It should be called U.
Did I say--yeah, that's right.
Because
q/4Πε _0 is the
potential, but I multiply by q_2.
That is actually an energy,
so these are all energies denoted by U.
Now, if you have a continuous
set of charges you must replace a sum by the integral.
Do you see what's going on here
why there's a half? Because this one counts
q_1q _3,
then again it counts q_3q
_1, but there's only one such thing
so I divided it by 2 and I have nothing like q_1q
_1, so you don't allow that.
That's the energy to assemble a
whole bunch of charges together. So that's the work you do.
If you do that work and you
take your hands off it'll fly apart and give you back that
energy in the form of kinetic energy.
Now, let me do another simple
example. I want to take a sphere of
radius r, and I want to put q
coulombs on it, and I want to know how much
work I have to do for that one. What work do I have to do to
charge this sphere with q coulombs?
Well, first couple of coulombs
come in they don't run into anything, so I put some charge
here. That makes it harder to bring
the next set of coulombs and the next set of coulombs.
At some intermediate stage,
when the charge on this sphere is q and I want to bring
in a charge dq, how much work do I have to do?
When the charge is q the
potential of the surface of the sphere is
q/4Πε _0 over R.
The whole sphere is at that
potential and I'm trying to bring in a tiny more dq
from all sides and smear it on this.
The work for that will be that
times dq. Then the total work I do when I
go all the way from 0 to capital Q will be a
Q^(2)/2 times 1/4Πε
_0R. Sorry, capital R.
Again, you see why you get the
half because the last coulomb is much harder to bring than the
first one. The first one is free.
It's like a spring.
When you pull it the first
inches are easier. The other ones are harder
because the force is kx, so the work done is the
integral of x that becomes x^(2)/2.
Similarly to bring in an extra
charge you've got to worry about how much charge is already there
on this sphere that's opposing you with this potential.
You work against that by
bringing dq from infinity and you integrate it you get
q^(2)/2. So this is a good problem to
explain the difference between potential and energy.
This is the potential and this
is the charge I brought from infinity,
and that together is the du and integral of the
dU is my total u. So one can ask,
"What's the energy to produce any arrangement of
charges," and you should be able to
either do the sum or do the integral.
One homework problem will be
how to assemble a solid ball of charge, not a sphere,
solid ball. It's got no radius and slowly
it grows up to some size and you can do similar integrals where
you slap on more and more spherical shells on this onion
to build it up. Again, if you knew the
potential at any stage you can find the work to add a thin
layer to it and integrate it. Okay, so final topic is one of
capacitors. So a capacitor is like a way of
storing charge and storing energy.
In other words,
if I want to store gravitational energy I can do
the following. I build this tank,
and maybe there's some water in the lake here at the foot of the
tank, and I go to the top,
and go to the top and empty little buckets until I've got
some water at the big height. Then I've done some work and
it'll pay me back when I open the tap and let it all rush
down, and it may turn a turbine blade
and give me back thermo electric power.
You want to do similar things
electrically. You want to invest with some
work and it'll give you back. So one thing I can do is I can
take a plus and minus charge which love to be next to each
other and separate them. Now, I've done some work,
but I cannot walk around like this.
I've got other things to do.
I've got to find a better way
to store energy. How do you shake hands with
people? So you have to find another way.
So here's what you do.
You take one conductor and you
take another conductor, any shape you like,
it doesn't matter. They are both electrically
neutral, but they're conductors. So you take a tiny charge from
this guy and put it there. Then take more and more charge
and start putting it there. At any stage,
when you stop, you will have a lot of positive
charges on this conductor and a lot of negative charges on this
conductor. This conductor will be at one
potential because it's a conductor.
That conductor will be at
another potential because it's a conductor, but both have unique
potentials. They are not the same,
but this whole conductor has one potential,
this has another potential, V'.
And because it's harder and
harder to pump the coulombs you can imagine that there is a
potential difference between them,
because you've got to go uphill to pump even more charge into
this already charged object. So we define the capacitance as
the charge that you transferred divided by the voltage
difference you got. So let me call this 0 by
convention. That is called the capacitance.
That's the ability of that
system to hold charge. If they're very,
very tiny metal spheres there's some amount of work.
If they're huge metal spheres
you can transfer a lot of charge, okay?
That's the definition.
It's measured in farads.
Again, it's got some units
coulomb for volt, but again it's got a different
name, named after Mr. Faraday, measured in farads.
Usually you will have
microfarads in a lab. Farad is a huge unit.
So let's find the capacitance
of certain simple arrangements. Simplest arrangement is the
parallel plate capacitor. Here it is.
There's one plate.
Here's a second plate.
The area of the plate is
A. The distance between them is
d. And I put plus charges on this
one and I got them from this one.
Do you understand that?
You don't have to bring your
own charge. You remove charge from one
conductor you therefore leave it negatively charged and you dump
it on the other conductor thereby leaving it positively
charged. So you don't need to bring
charge from the outside world. You just redistribute the
neutrality so that one is now negative and one is positive.
So Q stands for the
magnitude of charge that you transferred.
So suppose I transfer some
charge Q. My job is to find Q
divided by V and that's my capacitance.
Okay, so how am I going to find
the voltage difference between these two?
I think you guys should know by
now voltage difference is the electric field times the
distance between the plates. There's no line integral.
E⋅dr
is just E times d. E, if you remember,
is σ/ε_0
between the plates, and sigma is Q/A.
Charge density is the total
charge over that. And that is by definition
Q over C, therefore the capacitance of
this parallel plate capacitor is ε_0
A/d. So that's one of the easier
calculations to do. So anytime someone gives you
two objects and says find for me the capacitance you will take
charge from one, you will put it on the other
one and then you will-- hey, where is all these
blackboards? Ah, bad choice.
So you guys took all this down?
Yes?
Student: Should that
A be on the bottom for the A Q over
________________? Prof: Oh, yes.
σ is Q/A.
Yeah, this is what comes from
knowing the answer. This is what you guys do in the
exam. When you turn that page then
all kinds of stuff happens. Signs change,
numbers change and the right answer appears on the next page.
That's really what I did
because I knew this formula. We're getting there no matter
what, but you're quite right. I didn't put the A in
the right place. So any way, it's not good to
know the answer every time because you're not that alert
doing this. Anyway, this is the correct
answer. So here's the capacitance of
this parallel plate. Here's another problem you
could do. These are also common.
Spherical capacitors are one
sphere of radius a surround by second sphere of
radius b and you shove charges,
maybe plus on this one and equal number of minus on that
one. So what's my goal?
Goal is to find the electric
field in the region between them,
right, and integrate it from here to here,
and to give the potential difference,
and divide by the charge. So if I put a charge Q
here you can see that the electric field in this region is
just Q over 4Πε
_0r^(2), because by Gauss's Law if I
draw a sphere here that encloses charge Q,
and it's isotropic and uniform, so it's like that of a point
charge. Therefore, the potential
difference between these two points will be like the
potential difference between these two points with a point
charge of the origin, so V_2 -
V_1 will be Q/4Πε
_0 times [1/a - 1/b].
Well, we are done because this
is what I call V. The capacitance is
Q/V and that's going to be 4Πε
_0 divided by [1/a −
1/b] which is 4Πε
_0ab/(b − a).
So in every geometry if they
want capacitance move charge from one guy to the other guy,
and find the field between them, then integrate it to find
the potential. The potential will always be
proportional to the charge. Take the ratio and that's
capacitance. I'm going to put this guy to a
test. Let me write down clearly the
answer here. 4Πε
_0ab/(b − a).
What test can I do if I want to
compare it to a parallel plate capacitor?
Can you see how I can relate it
to parallel plate capacitor? Yep?
Student: Make the a
close. Prof: a what?
Student: b -
a is close. Prof: Right,
you can take b - a very close,
but it still may not look like a parallel plate capacitor.
So here's what I say we do.
Make b and a
astronomical, size of the galaxy,
okay? And b - a is just
1 inch. Then in that case it will
really look like two infinite planes.
The fact that it gets wrapped
around over the length of the galaxy is just like saying the
earth is round which is not obvious from short excursions.
In that case we can write
4Πε _0.
a and b are
really equal to some common number R and b -
a is the distance between the inner shell and the outer
shell. That you can see.
Wait.
I'm sorry, 4ΠR^(2) is
the area times ε_0
over d, right?
So that's one test.
It doesn't mean it's right,
but if this test failed it means you're definitely wrong.
Okay, now how much energy does
it take to charge up a capacitor?
It's what I did earlier,
but let me do it again. If you take two plates or any
two surfaces, and there's a certain amount of
charge q on it and I transfer delta q the work
I do is, the delta q times the
voltage at that stage. Voltage at that stage is equal
to q/C, which is the q you have
up to that point. Therefore, the total work you
do is ½q^(2)/C.
So when you want to charge a
capacitor with a charge q on it either you can write it
that way or you can also do the following.
You can write it as
½CV^(2) because Q/C is a V.
Multiply top and bottom by
C. You can write the energy in a
capacitor either in terms of the charge stored or in terms of the
voltage between the plates. So we'll resume this next time.
