Prof: When I look at you
guys, I realize that I don't
know what most of you look like.
 
That's the problem in a big
class like this.
Usually it would be nice if you
know who your students are.
If they say hello,
you can say hello back,
but I don't know what I can do
about it, because if I focus on
you, I forget what I'm going to
talk about.
But you know what I look like,
and it's kind of an unnerving
asymmetry, because you might
become my doctor one day.
You'll be wearing a mask and
carrying a knife,
and you'll be thinking of
problem set number three,
right?
 
And I won't even know who did
it, so it's very asymmetric.
All right.
 
Okay, so anyway,
maybe I'll get to know you if
you ask a lot of questions,
or your favorite pastime,
finding something wrong on the
blackboard.
Look at that.
 
I didn't do that.
 
I cannot even reach that far.
 
So let me now go back a little
bit,
because I think I rushed some
of the topics near the end and I
thought about it some more,
so I think it will be helpful
to revisit this question about
image charges.
I want you to think about what
the point was.
So here's the main thing:
you all know how to think about
the potential or the field,
right?
I give you a bunch of charges
and you calculate the potential
and you take derivatives,
you get the field.
Sometimes that's not the only
kind of problem you have to
solve.
 
You may have to solve the
following problem.
There is some chunk,
maybe a potato,
and you bring some electric
charge, what does it do?
That's a meaningful question,
because it does something,
right?
 
And nature calculates right
away and does something.
But if you want to figure out
what happens,
it's quite complicated,
so a somewhat easier problem is
the case of the perfectly
conducting potato,
which is kind of solid metal,
shaped like whatever you like.
You bring a charge next to it,
what happens?
That question occurs a lot,
because we do electrical
experiments in the presence of
many conductors.
And the simple thing about a
conductor is,
the whole conductor is one
potential.
So I want to take a simple
problem and see if we can answer
it.
 
Now most problems,
you cannot answer.
By "cannot answer"
I don't mean in principle.
In principle,
we know the equations,
we can solve them,
including the real potato.
You can actually figure out
everything, but it's too hard.
You cannot write it in simple
form.
There are a few simple textbook
examples where you can ask the
question and you can answer.
 
And the question you want to
ask is, here is an infinite
conducting plane and it is
grounded.
Grounded means what?
 
You take a wire and you connect
it to the earth.
Now the point of connecting to
the earth is that the earth is
pretty close to being an object
at 0 potential.
In other words,
if you bring a charge from
infinity to the earth,
the amount of work you do you
can take to be 0.
 
And on a daily basis,
people are dumping charge,
people are taking charge out of
it, but it's so huge,
it doesn't matter.
 
It's like one of the reservoirs
you studied in thermodynamics.
You can take some heat,
you can put some heat.
It doesn't change its
temperature.
So the earth is a huge
repository where you can take a
few charges now and put a few
back.
It will always remain at 0.
 
It will force you also to go to
0 potential.
So that is that plane,
and you have another charge,
q, you slowly bring
towards that plane.
You may like to know what's the
answer to this problem.
What does the field look like
everywhere?
Now this plane,
for convenience,
I'm going to be taking to be
infinite.
If you want a side view of that
plane, it looks like this.
It divides the whole universe
into two parts,
where you are with your charge
and the rest on the other side.
So what will this plane do?
 
You can see that if it did
nothing, it's in trouble.
If it did nothing,
this charge at that distance
will produce a positive
potential.
At that distance,
it will produce an even bigger
positive potential,
because it's closer.
So different parts of the sheet
will be at different potentials.
That's not allowed.
 
It's an equipotential.
 
It's a metal.
 
It's got to arrange its
potential to be constant,
and because it's connected to
the ground, that constant has to
be 0.
 
So the way it will lower its
potential from your attempts to
raise it is to suck up some
negative charges from the
ground.
 
So negative charge will leave
the ground and somehow come and
stand in front of this positive
charge,
maybe something like this,
over some region roughly
proportional to the distance
between you and the plane.
Then what will happen is,
the electric field lines,
which normally go through the
metal if it wasn't there,
will now of course have to come
and terminate on that metal,
and they have to terminate
perpendicularly because the
electric field is always
perpendicular to the conductor.
You know why, right?
 
Because if you move along the
conductor, the potential cannot
change, so the line integral of
E must be 0,
but basically,
you should not feel any force
as you move along the plane.
 
So E is perpendicular.
 
So some bunch of charges will
be drawn from the ground,
as necessary,
and this is what will happen.
The question is,
can you say anything more in
detail?
 
Can you actually calculate the
charge distribution on this
infinite plane?
 
Can you calculate the force of
attraction between the charge
you have and the infinite plane?
 
You know there's going to be
attraction because your charge I
take to be positive,
and this has got negative.
They'll attract.
 
What is the attraction?
 
These are all well-defined
questions and they exist no
matter what this is made of.
 
But for this infinite plane,
we can actually answer it
So we answer by the following
device, which is what I was
trying to tell you last time.
 
If you take that infinite
plane, your conditions are that
it should be at 0 potential and
to the left of it should be the
charge q that I put in.
 
These are the requirements.
 
So you dream up another problem.
 
That's a charge -q,
same distance d from
this one.
 
Take out the plane.
 
Forget the plane now.
 
Think of a new problem,
and -q,
and you've done that many,
many times.
The lines go like this.
 
And you know that on the
perpendicular bisector,
the lines will be
perpendicular,
because the potential
everywhere is 0,
right?
 
Potential is a scaler and
whatever this guy does,
that will do - of that,
because it's at the same
distance, but has -q.
 
So V is 0,
so E will be
perpendicular.
 
Now if you take that
arrangement and look what's
happening to the left of the
plane, it looks like exactly
what we want in our problem.
 
What we demand in our problem
is in this part of the universe
where I live,
there should be charge
q, there should be
infinite plane at 0 potential.
That condition is satisfied by
this field configuration in the
following sense.
 
Start with this configuration
and stick in the infinite plane,
but give it the charge it needs
to terminate all these field
lines.
 
How much charge will it need?
 
You can tell right away.
 
What's the total charge on this
plane?
Because all the lines leaving
this have to terminate on this.
You can see it'll be -q.
 
It won't be point charge;
it'll be spread out,
but you need -q on this.
 
So what we say is,
if you brought the plane in and
either you give it -q in
advance,
or you connected it to the
ground, in which case it will
itself suck up -q.
 
It'll form this configuration
here, so the infinite plane is
able to remain where it is
without disturbing the solution
to the problem.
 
Therefore the answer to the
original question of what
happens when you put a q
in front of an infinite plane is
that as far as you're on the
left of this,
you can compute anything you
want by taking q and the
image charge,
-q.
The field the two of them will
produce will be exactly the
field this guy and all these
induced charges will produce.
So this is a way of taking a
solution for a simple problem
with an equipotential,
sticking a conductor into the
equipotential,
and giving it the right charge.
Then the problem of a charge in
front of a conductor is thereby
solved.
 
Now there are some mathematical
theorems--I don't have time to
prove but they're not too hard
in fact.
I thought of lecturing on that,
but there isn't enough time--
that show that if you can find
one answer that has the right
behavior at the boundary of the
region,
that has got the charges in the
right place,
is the only answer.
 
Yes.
 
Student:  Is the -q
evenly distributed across
the plate?
 
Prof: Let us ask.
 
Where is the -q, right?
 
We know that it's got to be
-q because these lines
are all ending here.
 
But what's your intuition?
 
Where do you think it will be?
 
You think it'll be evenly
distributed?
Pardon me?
 
Student:  I don't think
so.
Prof: Where will it be?
 
What's your feeling?
 
Student:  Probably more
towards the middle.
Prof: Yeah,
more towards in front of this
guy, and less when you go up.
 
Now we can actually calculate
that quantity.
We can tell you exactly how
much it will be by the following
trick - we agree that to the
region to the left,
this charge is able to fake the
effect of all these charges.
That's why it's called a mirror
charge.
When you stand in front of a
mirror, you are here and there
is another person in the mirror.
 
And the light bouncing off and
coming here looks like it's
coming from this image.
 
So to the left of the mirror,
either you can talk about you
and the mirror and the reflected
light, or forget the reflected
light.
 
It's as if there's another
person behind the mirror.
Of course, you know there is no
other person behind the mirror,
right?
 
I don't want to be the one to
break that news.
There isn't.
 
But as far as this side is
concerned,
it's a lot easier to draw a
line from the mirror,
because that will simulate the
effect of what's happening
through the mirror.
 
Similarly, as long as you are
interested only in the left of
this region,
you cannot tell the difference
between the infinite plane with
its induced charge,
and this guy,
versus this guy and just one
other charge.
 
Another question was,
where is the charge and how is
it distributed?
 
You agree that if you knew the
electric field here,
you can find the charge
density, because if you draw a
little Gaussian surface like
that,
then we know the
σ/ε_0 is
the electric field.
 
So if you know the electric
field, you can find sigma,
and I find the electric field
very simply by saying that's the
repulsion from this q,
there's an attraction from that
q.
 
If you add the two,
you get the sum,
which I can calculate,
a very simple sum.
It's the contribution from two
point charges.
You can calculate it as a
function of this angle here or
as the function of the distance
here, and that will tell you
what σ is.
 
And it will have the property
you expect.
It'll be largely in front of
this guy and will vanish very
quickly when you move away from
him.
If you integrate that surface
charge density over the plane,
you will in fact get -q.
 
So you've answered some of the
questions, namely,
how will the charge distribute
itself on this infinite plane?
Next thing you can ask is,
what's the force of attraction
between this one and this one?
 
Now, is it really the same as
this one and this one?
The answer is yes,
and the reason is the following
- in this region,
for example,
the field is due to this guy,
and due to all these,
but that's the same as this guy
and this guy.
This one of course doesn't
respond to its own field.
It responds to the field of all
of these.
But all of these precisely have
the same field to the left as
this one charge,
therefore this will be pulled
towards the one charge.
 
It thinks it's falling towards
the other charge,
but it's not actually falling
towards the plane.
And the force will be q
times -q over
4Πε
_0 times
2d, right?
 
That's the force of attraction.
 
It will be
−q^(2)/4
Πε
_0,
the distance between them is
2d,
so you can calculate that too.
 
Normally it's a difficult
problem, because the image
charge reduces the problem of a
very complex conductor to that
of a point charge.
 
Yes?
 
Student:  Shouldn't it
be 2d squared?
Prof: Yes, thank you.
 
Then you can also ask,
what is the energy it takes to
assemble this distribution,
starting from this one at
infinity and this one here,
and if you integrate this
force, you can do the integral
and you'll find
−q^(2)/4
Πε
_0 divided by
4d.
That will be the energy.
 
Na�vely, you may have thought
it should be
(q^(2)/4Πε
_0)2d,
because 2d is the
distance between them,
but it's only half as much.
 
It's half as much because if in
real life you had a and a
-q and you brought them
together,
you will be exerting forces on
both of them and calculating the
work done.
 
But in this actual example of
an infinite plane,
you only do work on this guy,
therefore the answer will be
half.
 
So not everything is the same.
 
For example,
the energy in the
electromagnetic field,
which I'll tell you about later
today,
will be half as much as before,
because in the fake problem,
the energy is here and here.
In the real problem,
the field is non-zero only
here, but 0 here.
 
Yes?
 
Student:  For the
potential energy,
is it 1 over 4d or 1
over...?
Prof: You would think
it's 1 over 2d,
right?
 
Student:  Yes.
 
Prof: But I'm telling
you it's 1 over 4d.
Student:  Because when
you take the integral,
you have 2 on...
 
okay.
 
Prof: All right.
 
Now what if this plane was not
grounded, it's a standalone
plane?
 
So let's not take it to be
infinite in this problem.
It's 1 million miles in radius,
huge disc.
You bring the charge in.
 
What's it supposed to do?
 
It cannot suddenly acquire a
negative charge.
Charge conservation says you
only have the charge you have,
which was 0.
 
So what do you think you will
do if you have that infinite
plane?
 
You have to remain at some
potential.
Yes?
 
Student:  Push positive
charges out?
Prof: It will split into
positive and negative charges.
Let's see what to do with the
positive charges in a minute.
The negative charges,
it will arrange exactly the way
it wants to in this problem,
okay?
That takes care of this being
at a constant potential of 0.
But it's got an equal number of
positive charges,
which you've got to dump on
this infinite plane,
without ruining the constant
potential.
The way to do that is to put it
uniformly on this plane.
If you spread it uniformly on
the plane, you don't destroy the
equal potential nature.
 
But it turns out that field
will be very,
very, very small.
 
It will be very small,
because you're taking a charge
q and you're putting it
on a huge disc,
like the radius of the galaxy.
 
That's the charge density and
that over ε_0
will be the electric field.
 
You can see that that can be
made vanishingly small.
So the infinite plane,
basically you borrow from
somewhere a positive charge and
negative charge,
and you put the negative in
this particular arrangement.
Positive is smeared over the
whole thing uniformly.
It is so dilute,
because of the size of it,
it doesn't matter.
 
But it principle,
that amount of positive charge
will be a background constant.
 
That way the infinite plane
will be a little strange,
because you don't see the
richness of the phenomenon of
the infinite plane because it
can basically create negative
charge and not fully account for
the positive charge,
because it's smeared over a
huge distance.
So a problem that's more
interesting would be this one.
I take a sphere of radius
a, it's a conducting
sphere.
 
And after distance b,
I put a charge q.
What will happen now?
 
Again, now let me say this is a
grounded sphere,
meaning it's connected to the
earth.
It's maintained at 0 potential.
 
So it's not very different from
the plane, except you've got a
huge sphere now instead of an
infinite plane,
but it's finite.
 
You come around from infinity,
you can see that some negative
charges will be induced on this,
and some lines will come here.
And some lines may go past like
this.
Because this is a finite
sphere, not every line has to
terminate here.
 
Now the question is,
what will be the charge
distribution?
 
How will it arrange itself,
and what will be the surface
electric field and so on?
 
That's what you want to
calculate?
So what kind of problem do you
think we should try to solve?
If it had an easy solution,
what form do you think it will
take?
 
Yes?
 
Student:  You could put
a charge on the other side of
the sphere?
 
Prof: Okay,
that's not a bad guess.
If you put a charge on the
other side of the sphere,
then in the universe in which
you're living,
namely outside the conductor,
things have changed,
right?
 
In the image charge problem,
remember, the fake charge was
never in your universe.
 
It was outside.
 
The problem you defined
yourself should remain the same
where you are,
but you can do stuff inside the
region that's excluded from you.
 
So you're almost right.
 
We want to put a charge inside
here.
And where do you think you
should put it,
roughly?
 
Can you make a guess where in
the sphere it will be?
Student:  Center?
 
Prof: Center is a good
guess, but it turns out,
if you put it in the center,
this doesn't turn out to be a
sphere.
 
But you agree that it should be
on this line,
just from symmetry.
 
You don't want to put it
anywhere else.
So I will give you the answer.
 
It's not very hard to calculate.
 
If you put a charge q'
at a distance s,
which is a^(2)/b,
and the size of q' is
-qa/b,
a/b is less than
1.
 
So q' is less than
q, and is at a distance
s from the center,
which is a^(2)/b.
This is something you can do
for fun.
You can go and calculate now
the potential of this negative
and smaller charge together with
this positive charge q
and you will find,
if you put q'/r
for that,
and q/r for this
guy,
it will be exactly 0 on this
sphere,
centered here.
So it's another fortunate thing
that you can get a spherical
equipotential by taking two
unequal charges--
not two equal,
two unequal charges--
and putting it slightly off
center by this amount.
Let's not do the algebra now.
 
You can all imagine doing
q'/r' and you'll
find it's equal to
q/r,
up to a sign,
and therefore it will be in
fact--
you can always make two things
cancel at one point,
but they cancel over an entire
sphere.
 
It's very amazing, but true.
 
Yes?
 
Student:  Are the field
lines always perpendicular to
the surface of the sphere?
 
Prof: They must end up
perpendicular.
Tell me why.
 
Can they end up anything but
perpendicular to a conductor?
Student:  Oh, right, no.
 
Prof: No,
because if it's got a
tangential part,
it will move the charges.
So charges, once they've come
to equilibrium,
the only force you can apply to
a conductor is normal to the
surface.
 
So the lines will terminate in
the normal way.
So the problem you want to
solve, the fake problem,
the image problem,
is a −q' and a
q so located you get a
sphere.
Now you go and say,
let me take another problem of
a conducting sphere.
 
There's nothing in it, grounded.
 
What it will do is suck from
the ground this charge and
spread it exactly the way you
want in this problem.
And then it will come to 0
potential.
So ground is here, 0 potential.
 
If you want to know,
what is the electrostatic
potential there in this problem,
we can do that now.
What's the electrostatic
potential here?
It is due to this guy at that
distance and the fake charge at
that distance.
 
You should add them with the
proper sign.
So the problem of the field due
to very complicated distribution
of charges on a conducting
sphere and a point charge is
reduced to the problem of 2
point charges.
Very easy to add their
potentials, take the derivative
and find the field.
 
If you want to know the surface
charge density,
you find the electric field at
the surface.
It will come out to be
perpendicular.
Then
σ/ε_0 =
E, therefore if you knew
E in magnitude,
you can find sigma and you can
find the charge distribution on
the whole sphere.
 
If you integrate that guy,
what will you get for the
charge distribution on the
surface?
What should it give for the
total charge?
Pardon me?
 
Use Gauss's law for the real
problem and the fake problem.
In the real problem,
surface integral of E is
the charge on the sphere.
 
But the same E is
produced in this combination
with this guy here.
 
The surface integral of
E is a charge enclosed,
which is q',
after some epsilons.
Therefore the charge on the
sphere will turn out to be
exactly the same as the
q'.
Just like in the plane,
there'll be some negative
charge, but it's spread out in a
particular way and you can
verify that.
 
Now a more interesting question
is, what if this is not
grounded?
 
In other words,
it's an isolated sphere and
you're bringing a charge near
it.
It wants a certain negative
charge to spread on the surface
to produce a zero potential
here, but you don't have
negative charge.
 
You have zero charge.
 
So what will you do if you are
that sphere?
You will say 0 = q'
-q'.
You would split into q'
and −q'.
The q' will arrange
itself on the sphere to exactly
terminate these field lines.
 
The only difference is,
now you've got a charge,
-q', left over.
 
You have to spread it around
and you don't want to screw up
the equal potential nature of
the sphere by doing that.
You can all guess what you
should do.
You should spread it uniformly
on that sphere.
So it will contain negative
charge distribution,
which is biased in this
direction, and a positive charge
distribution that's uniform in
it.
Then the three of them
together, this one,
the positive charge,
I put here of −q'
and this one,
three of them together will
keep the sphere at a constant
potential.
The potential it ends up with
will however not be 0,
because q' and
q made it 0,
but you have a −q'
at the center,
therefore you must find the
potential due to
−q' on the surface
of the sphere,
which is that.
 
That's the potential to which
the sphere will come.
So we have answered many
questions.
We have said if you bring a
charge q near the
sphere, what potential will it
acquire?
It will acquire exactly this
potential.
You can predict the charge on
the sphere.
You can predict the force of
attraction between this and that
sphere.
 
How?
 
Because the field that this one
feels generally,
the field at any point is due
to this one, this one and this
one.
 
But if you want the force on
this one, if you'd only find the
field due to the other two guys.
 
But the other two guys,
namely the sphere with all the
charges on it,
is simply equal to 2 point
charges.
 
So you must find the force
between these 2,
add to that the force between
these two and you'll get
something.
 
That will be the force.
 
It will be a force of
attraction, but you can actually
calculate it.
 
This is the trick by which you
can solve a variety of problems
by finding equal potentials of
nice shapes.
Maybe one will be a nice
ellipsoid, because if it is an
ellipsoid, you can stick an
ellipsoid there.
And inside the ellipsoid will
be the image charge.
And if the ellipsoid is
grounded, you'll get whatever
charge is needed to maintain
that at 0.
If it's not grounded,
it will split into positive and
negative charges,
where the negative will
distribute this way,
and the positive will
distribute itself in some way,
so that the potential is a
constant.
 
For a sphere,
we know what that some way is,
which is uniform.
 
So the reason I took some time
to describe this is that
problems are not always finding
the potential due to a bunch of
charges.
 
That's the easiest problem,
but more generally you are
asked, if you're given a set of
conductors and a bunch of
charges, what happens?
 
If the conductors have a nice
shape,
like a sphere or a plane,
or certain solids of
revolution,
we can appeal to a different
problem with image charges and
solve it.
Now there's a wonderful theorem
that says if you get a solution
this way, it's the only
solution.
In other words,
there's a theorem,
which I'm not going to write
down,
which says if in all of space
that you're interested in,
or at least in a region bounded
by something,
if you know the potential of
the boundary and you've got a
whole bunch of conductors,
each one at a known potential,
and you've got a whole bunch of
charges,
there can be only one potential
function,
V(r),
in this whole region.
You cannot have two answers.
 
There's only one answer to that
question.
In other words,
the potential is completely
determined by knowing the charge
distribution,
and the potential on the
various conductors you stick
into that.
 
They don't all have to be to 0.
 
This can be at 5 volts,
this can be at 9 volts,
this can be at 13 volts.
 
This can be 1 coulomb,
2 coulombs, 3 coulombs,
-10 coulombs.
 
This is say this sphere at
infinity of 0 potential,
there's only one answer.
 
It's called the uniqueness
theorem,
which is why,
if you can find some way to
fudge the answer in a given
region,
which is completely specified,
that is the answer.
So that is the interlude on
metallic objects called
conductors.
 
Now I'm going to go to the
other problem that I did towards
the end, which is the notion of
a capacitor.
So if you take two blobs of
metal, they are both neutral.
Then you grab maybe a coulomb
from this and stick it there,
so that becomes positive and
this becomes negative.
Then you want to take more,
you take more stuff,
you can see that you're going
to run into resistance,
because these guys are getting
positively charged.
They don't want more positive
charges.
Meanwhile the negative charge
you leave behind wants it to
come back, so you're working
against that.
But you do some work and you
start pumping charge into this.
So in the end,
suppose you have a charge
Q there and charge
-Q there.
Then there will be a potential
energy difference between the
two, because there's a certain
amount of work needed to go from
here to there.
 
The potential difference is a
unique number,
because the whole solid has one
potential, other solid has
another potential.
 
No matter where you start and
where you end,
if you find the work done,
that potential difference we
like to call V.
 
Therefore V is always
going to be proportional to
Q and the constants of
proportionality we like to write
downstairs and call it
capacitance.
So that's the ability of the
system to hold charge.
If you've got any two metal
containers, you can store energy
by drawing the charges from one
to the other,
putting them there.
 
I did calculate the capacitance
for a very simple system and I'm
going to stick to the simple
system because we don't want to
get lost in the details.
 
It's the parallel plate
capacitor in which I put some
charge Q on the upper
plate and -Q on the lower
plate.
 
Someone says,
what's the capacitance of the
system?
 
For that, you must find the
voltage difference between the
two plates, and you want to take
Q over V.
Now there's going to be an
electric field here,
E, which is
σ/ε_0
pointing down.
 
So the voltages difference will
be the electric field times the
distance, because that's what
the line integral will be.
That's the distance between the
plates.
And σ is
Q/A.
Now in a real capacitor,
near the ends there'll be some
funny business.
 
It's not going to be simply
like this, but we take it to be
so large in this extent that the
edge effects are neglected.
Then this formula is valid.
 
Then you can compare it to
Q/C and you can
see C is
ε_0
a/d. So this is
what I have done towards the end
of class.
 
That's the capacitance of that.
 
Once again, you can talk about
capacitance, but you cannot
always calculate it.
 
You can take two irregular
metallic objects and in
principle for a transfer of
charge q.
There will be a voltage
v, and the ratio is the
capacitance, but you cannot
compute it.
But for a parallel plate
geometry, or another one I did
in class near the end,
you take two concentric spheres
and put some charge on this one,
maybe and - on that one,
you all know how to find the
field in the region in between.
You can integrate it and you
can find the potential
difference.
 
And then you can see it is
proportional to Q,
and you take the ratio,
you'll get the capacitance of
that.
 
Now the point,
one thing I want to calculate,
which also I think I started
doing is,
if I take a capacitor and I
charge it,
from 0 to some charge
Q_0,
what's the total amount of
energy stored in the capacitor?
So the way I think about it is,
I take some intermediate
situation, when there's a charge
Q on the capacitor,
not the final amount.
 
Then I want to take a little
amount of charge dQ and I
want to move it from the
negative to the positive.
Since at that stage the
potential difference is
Q/C that potential
difference times dQ is
the work you do.
 
That's the definition of
potential difference,
how much work it takes to move
a coulomb from one plate to the
other.
 
If you're moving dQ
coulombs, that's the work you
do.
 
You integrate that from 0 to
some maximum value
Q_0,
you see you get
Q_0
^(2)/2C.
So the energy in a capacitor is
q^(2)/2C.
Forget the 0, subscript 0.
 
Usually Q stands for the
charge of a capacitor.
But you can also write it in
another way.
Since Q = CV,
you can write it as
½CV^(2) .
 
There are two ways to write the
energy.
So the energy in the capacitor
is ½CV^(2),
but I'm going to play with that
expression and get a very
interesting result,
very profound.
Capacitance is
ε_0
A/d,
what about the voltage?
It is E/d.
 
Sorry, voltage E times
d.
E squared,
d squared.
So what do you get?
 
You get ε_0
E^(2)/2 times a
times d.
 
What is A times d?
 
It's a volume of the region
that contains the electric
field.
 
Before you charge a capacitor,
there is no energy and there is
no field.
 
Once you charge the capacitor,
you've got a field.
And if you wish,
you can say I've got to ascribe
that energy to the fact that
it's a non-zero field,
in which case,
this is the energy per unit
volume,
due to an electric field.
That's a very interesting
notion.
Little u is the energy
per unit volume and that happens
to be ε_0
E^(2)/2.
In other words,
it takes energy to establish
the electric field.
 
So in this room,
if you've go to a tiny region
where E is essentially
constant over the tiny region.
ε_0
E^(2)/2 times the tiny
volume is the energy in that
tiny region.
So once you've got an electric
field, it just cannot just
disappear.
 
Law of conservation of energy
will require that you account
for it, and this is the energy
per unit volume.
It turns out to work even if
you've got radio waves,
electromagnetic waves,
going anywhere,
to take that energy,
electric field squared times
epsilon over 2,
that's the energy density.
All right, so this is really
the end of what I wanted to
finish last time,
but I wanted to go back to the
study of equipotentials and
conductors.
But now I am setting the stage
for electrical circuits.
Now this is the kind of thing
some of you probably did
somewhere in high school.
 
So how many people have done
basic circuits in high school?
So I'm going to assume you've
done some of it,
so I won't do it in that
detail, but I will mention all
the essential facts.
 
For the few of you that didn't
do it, you have a chance to keep
up with the class.
 
The first thing in electrical
circuits is, you've got some
wire and you've got an electric
current flowing in it.
We need a description of the
current.
The current is defined as
follows.
Imagine this is the perfect
cylinder, cross section
A.
 
You cut it somewhere and you
watch all the charges go by,
and you see the number of
coulombs that go by per second.
That's called the electric
current and is measured in
amperes.
 
So 1 coulomb per second is 1
amp.
All right, now let us ask,
what's the connection between
the electric current and what's
going on microscopically?
So let n = number of
carriers per unit volume.
Let e be the charge of a
carrier.
Now here is one of the biggest
nuisances in life.
As you know,
in a wire, the current is
carried by electrons.
 
Because the charge is negative,
when you draw a picture like
this, the current to the right,
electrons are actually moving
to the left.
 
What we will instead do is to
just keep an eye on the
direction of the current.
 
You imagine there are
positively charged objects
carrying the current in the
direction of the current,
whereas in reality,
it's negatively charged objects
moving in the opposite direction
that produce the same current.
So you can ask yourself,
if you wait 1 second,
how many coulombs will go past
this checkpoint?
You can see that it's a
cylinder whose length is the
velocity, because in 1 second,
it will have gone v
times 1 second.
 
All those guys have crossed the
finish line.
Therefore the current will be
A times v (is the
volume of stuff that's gone),
that's the number of carriers
in the volume,
that's the charge of each
carrier.
 
That's the total current.
 
So we like to define a quantity
called the current density,
which is the current per unit
area.
That will be equal to
nev.
And actually,
current density is a vector so
if you want, because velocity is
a vector, you can make it a
vector like that.
 
In this problem,
the current density is uniform.
In fact, in any wire,
the current here and the
current there and the current
there do not change,
because if it changed,
charges will pile up in some
places like a traffic jam.
 
Then it will resist it till
this thing evens out.
Current is constant on a wire.
 
Need not be uniform across a
cross section of the wire,
but I'm going to take it to be
uniform.
But if it's not constant,
then you can have a J
that's varying with space,
so the current crossing that
surface will be the surface
integral of
J⋅dA.
 
In other words,
divide that area into tiny
little patches,
dA, then as we have seen
many times,
whenever something is flowing,
it's the dot product of the
flow rate with the area that
measures the actual flow.
 
You add it over the surface,
that's the current crossing a
surface.
 
We won't use this formula very
much.
I'm just mentioning it for
completeness.
All right, so the next question
is, I've told you there is no
electric field inside a
conductor.
But if you take a resistor,
even a nice thing like copper,
it's not a perfect conductor.
 
Actually, there's an electric
field inside a conductor.
That's what makes a current
flow.
If you want any wire to carry
current, except for ideal wires
that have no resistance,
any realistic conductor needs a
field to drive it.
 
The reason is that if you look
microscopically at these
carriers, they're all going like
crazy in all directions.
That's precisely why they don't
carry a current.
They don't have a common
direction.
They're going very fast but
they're going nowhere.
One of these guys is going to
the right.
For every one going to the
right, there's one to the left,
one going northeast and one
going southwest.
The velocity all averages out
to 0.
But if you apply an electric
field to the right,
you can imagine somehow in the
middle of all this chaos,
there'll be an overall tendency
to drift to the right.
So let's see how much there is.
 
I found that the current
density was nev.
Forget about the vector sign
now.
Just everything is along
x.
I want to find here an average
velocity.
What's going to be the average
velocity over all the particles
at a given time?
 
Now here's the picture of
conduction.
You should in fact find it very
paradoxical that when you have
an electric field,
the electric field produces a
force e times E,
has an acceleration a,
so the charges should
accelerate.
If they accelerate more and
more, the current should keep on
growing because velocity is
growing.
But you get a steady current,
in spite of a force acting to
the right.
 
Do you know why that happens?
 
Why doesn't everything
accelerate and pick up more and
more and more speed?
 
Student:  Resistance.
 
Prof: It is resistance,
but microscopically,
why don't these particles pick
up speed forever?
They collide.
 
They collide with basically the
impurities in the solid.
And every time they hit their
head on one of the impurities,
they don't know what happened,
and they bounce off the
collision in a totally random
direction.
So they may go in this
direction, hit an impurity.
There's no telling which way
they'll come out.
They can come out in any
direction with equal
probability.
 
So they'll lose all memory of
what they were doing after each
collision.
 
Now if I look at my clock,
and I look at the entire set of
electrons and I say,
what's the average velocity?
The average velocity,
which I denote with a bar,
is obtained by averaging
individual velocity.
Individual velocity is the
velocity at time 0 since the
last collision,
plus eE/m times
t_i.
 
In other words,
let me explain to you slowly
what I mean.
 
Take particle number i.
 
Let us say it has been
t_i seconds
since it had its last collision.
 
So just after the collision,
it has got initial velocity
which is completely random,
but since that thing,
because the electric field it
has been accelerating,
it has not collided with
anything yet,
so its velocity will be
eE/m times
t_i.
 
Therefore the average velocity,
which you find by the averaging
of everything,
has the following property -
the first one will give you 0.
 
If you average over all
possible initial velocities,
they'll be 0,
because they're pointing in
random directions.
 
So what you really need is
eE/m times some
τ,
where τ is the name for the
average time since the last
collision.
That time will have a range of
values.
Some guys will have just
collided;
some won't have collided for a
long time.
So you have to find that
average, and we don't know how
to compute it right now,
but that is some average
t_i for a
material.
The larger the
t_i is,
the better the conductor it is,
because it can go for a long
time on average without
colliding.
So the whole idea is,
the minute you collide,
you lose everything,
because you scatter off,
forgetting your memory.
 
So any coherent motion you have
in the direction of the field is
there only because you have not
yet collided.
So the net current is a
function of how many seconds
have elapsed since the last
collision for each guy.
On average, it gives you this.
 
So if you use the symbol tau
for that, therefore the current
density will be ne times
the average v.
The average v is
eE/mτ.
So if you write it as
ne^(2)τ/m
times the electric field.
 
So this is the very,
very important result.
So I'm going to go back and
study this result.
It tells you that the current
density in a wire is there
thanks to the electric field and
the number in front of it
happens to be the number of
carriers
e^(2)τ/m.
 
Does that formula make sense?
 
Take a minute to look at that
formula.
You apply a field.
 
That's why they even know which
way to flow.
If you don't have a field in a
wire, the current doesn't know
which way to flow.
 
It's just random motion,
not going anywhere,
like molecules in this room.
 
They're not going anywhere in
particular, even though they're
moving.
 
You are saying that the current
that you're going to get for
unit area is bigger if you have
a bigger density of carriers.
That makes sense,
because they're the ones
carrying the charge.
 
This is inversely proportional
to the mass.
You can understand why.
 
The electric field,
whatever force it produces,
the acceleration is inversely
proportional to the mass.
The bigger the τ,
the bigger the response,
because they can go for a
longer time on average before
colliding,
therefore they have more time
to pick up speed in the
direction of the field.
e^(2) is interesting.
 
One e comes because the
force on the carrier is little
e times big E.
 
The second e comes
because the current it carries
is itself proportional to
e.
You understand?
 
The charge of the carrier
affects it in two ways.
Any time it moves,
it carries a charge e.
How much it moves depends on
the force the electric field
exerts on it.
 
That's another e.
 
So this is our expression.
 
And we write it as σ
times E,
where sigma is called the
conductivity.
You notice that this argument
doesn't care what material it
is.
 
It could be copper,
it could be aluminum,
it could be some alloy.
 
For all of them,
the answer depends on
ne^(2)τ/m.
 
What depends on the actual
carrier, the mass is just the
mass of the electron.
 
τ is what varies from
problem to problem.
Some materials have very large
τ, some materials are very
small τ.
 
That's what decides how good or
bad a conductor it is
So this is the formula.
 
Now sometimes you can write it
in another way.
Let's see now.
 
Suppose you have a wire.
 
What is the total current in
the wire?
The total current in the wire
is the current density times
area, which is σAE.
 
Now let us say this E
was obtained by applying the
voltage difference V
between the two end points of a
wire of length L.
 
There is
σAV/L.
Do you understand?
 
The electric field times the
length of the wire is the
voltage difference between the
two end points.
So rather than saying the
current is driven by the
electric field,
let's say the current is driven
by the voltage difference
between the two ends of the
wire.
 
That's why I wrote E as
V/L.
But now you see it looks like
V/R that R
is by definition is called the
resistance of that wire.
It's measured in ohms,
as denoted by this symbol.
This is how you get ohm's law,
because just by going through
the microscopic equation,
and applying it for a wire of
length L,
I'm able to find that the
current is proportional to
V,
divided by some number.
 
That number looks like
L/Aσ,
also written as
Lρ/A,
where ρ = 1/σ
is called the resistivity.
It's just the inverse of the
conductivity.
So the bigger the resistivity,
the bigger the resistance,
but notice, if the wire is
twice as long,
the resistance of the wire will
be twice as big.
Does that make sense to you?
 
For a given material,
if you double the length of the
wire, resistance is double.
 
If you double the area,
resistance is turned into half
its value,
because if you've got a big
wire, you can think of it as two
wires that are carrying the
current together,
therefore the resistance is
half as much.
 
So this is the relation between
resistivity and the resistance.
As far as we are concerned,
the main thing for us is just
ohm's law,
which we are going to use -
V = IR,
and this is telling you roughly
where you get it.
 
So what's the summary of all
this talk?
In a wire, unlike in a perfect
conductor, there is an electric
field.
 
And it's the electric field
that keeps the charges moving.
But whereas an electron,
for example,
in the vacuum electric field
that will accelerate
indefinitely.
 
The carriers in the wire do not
accelerate indefinitely because
they keep bumping into stuff,
and every time they bump into
stuff,
they lose the gain they had.
They start all over again.
 
So at any given time,
the activity I have or the
motion I have depends on how
many guys are still around since
the last collision.
 
They're the ones who have been
picking up speed,
and that's how you get
conductivity proportional to the
applied field.
 
So it's hard to get a velocity
proportional to force.
You always get acceleration
proportional to force,
but when you've got random
motion, the velocity is
proportional to the applied
field.
By the way, I should tell you,
in a real solid,
if you ask what do they collide
into, can you imagine?
You have a mental picture of a
solid where the atoms form a
nice array.
 
The nuclei form a nice periodic
array.
The electrons in a metal are
free to travel the length and
width of the solid.
 
But who do you think I'm
talking about when I say
collisions?
 
Pardon me?
 
Student:  Other
electrons.
Prof: Other electrons is
possible.
How about with the nuclei?
 
If you're an electron,
you're going through,
you see a nuclei every
whatever, 10 to the -8
centimeters,
there's another nucleus,
but that's not what matters.
 
This is more advanced theory,
when you try to find the
conductivity of materials.
 
A perfectly periodic lattice of
nuclei, electrons have a way to
travel through them without ever
colliding.
It's like, if you cannot see
and the furniture is all in a
certain place,
you can navigate freely around
them.
 
But if somebody moves something
and you're not expecting it,
that's when you have a
collision.
And that motion comes about
when you heat the solid.
When you heat the solid,
the nuclei start vibrating,
so you don't know quite where
somebody will be.
That gives a small probability
for the electron to collide with
them.
 
Therefore the conductivity will
quite often depend on the
temperature, but it can also
depend on the force of
interactions between electrons.
 
But even if electrons don't
interact with each other in a
serious way, the collision with
the nucleus is controlled by
lattice vibrations.
 
But no matter how sophisticated
the calculation is,
you can go to my office and
ask, what are people doing for
conductivity?
 
Everyone is trying to compute
this quantity τ
and
ne^(2)τ/m.
The n,
the e^(2),
they're all what you think they
mean.
τ is more sophisticated,
and you have to calculate it in
a quantum theory.
 
But in the end,
after all the work,
you get a number τ,
you put it in the same formula,
ne^(2)τ/m to
get the conductivity of a
material.
 
So now I want to do a little
electric circuits for you.
So here's one simple circuit
I'm going to do.
I'm going to take a capacitor,
charge it up to some amount
Q,
then I'm going to put my
resistor here like this and ask
what happens when I close that
switch.
 
So when I close the switch,
we can all imagine what will
happen.
 
These positive guys are dying
to get over to the negative
side, but they cannot jump the
gap here, because it's a vacuum.
But if you give them a path,
they will go through that and
come back to the other side.
 
So charges will start going
round if only I close the
switch.
 
So when I close the switch,
let me write down an equation
for what will happen.
 
There's a resistance here.
 
This is the current I flowing
here.
The fundamental equation you
write down in any circuit is if
you start at any point,
you take any closed path and
find all the changes in
potential,
the change has to be 0.
 
Because it's coming from a
conservative field,
the integral of the electric
field on a loop is 0.
That means the total change in
potential from anywhere back to
the same place is 0,
because it's like a height.
So when I start here and I go
through the capacitor,
I go up in electrical height by
an amount Q/C.
This conductor is assumed to be
a perfect conductor,
so there is no electric field
inside here,
there's no change in potential
until I come to this resistor.
A resistor will not carry
current unless there's a voltage
applied to it.
 
You've seen there,
and this is the higher end of
the voltage, this is the lower
end of the voltage,
because it's flowing downhill.
 
So you have a drop in voltage
by an amount RI,
then you come back to the
starting point.
All the changes you had better
add up to 0.
That's the fundamental
statement that in any electrical
circuit,
if you start anywhere and go
through any path in the circuit
and come back to a starting
point,
the net change in voltage has
to be 0,
because every point has an
electrical height.
 
So this is our equation.
 
Now we can write it as RI =
Q/C.
But what's the relation between
I and Q?
Can you find a relation between
I and Q?
Suppose a current flows for a
short time.
Who's paying for it?
 
Where is the charge coming from?
 
Yes?
 
Student:  Isn't I
Q divided by time?
Prof: It's not
necessarily divided by time,
but it's a derivative.
 
I is dQ /dt,
because in a small time
dt,
when the current I
flows, Idt coulombs
flow through here,
but they've got to come from
there.
But the only thing missing is
the - sign.
Because if I is defined this
way and is positive,
a positive I depletes the
charge, so I is
−dQ/dt.
 
So I have the equation here
that looks like--so go back to
RI = Q/C and write
it as RdQ/dt.
This is my equation.
 
That can be written as
dQ/Q =
-dt/RC.
 
Now if you integrate that from
start to finish and start to
finish,
this will give you log of
Q over the initial Q
will be - the time divided
by RC.
 
Or Q of time t
will be Q_0e^(-t)
^(/RC).
 
That means after you close the
switch,
if you measure the charge and
the capacitor as a function of
time,
it starts with some
Q_0 and it
decays exponentially.
How long does it take for it to
completely discharge?
The answer is infinite amount
of time.
Why is the capacitor not able
to discharge it completely?
Can you think about that?
 
Why doesn't it just get it over
with, right?
Just dump all the charge to the
other plate?
Why is it taking forever?
 
Student:  There's always
resistance.
Prof: There is always
resistance.
There has always been
resistance, so what's happening
as time goes by?
 
Student: 
>
Prof: The voltage on the
capacitor plate is decreasing
with time, right?
 
That's the voltage driving the
current through the resistor.
So as it drives current through
the resistor and it begins to
become empty,
it's able to drive less current
through the resistor.
 
So it's trying to work against
itself, but it doesn't have
enough Q on it to drive
more Q down.
That's why.
 
That's the meaning of the
equation.
The rate of flow of Q is
proportional to Q itself.
So as long as there is some
Q left,
it will be decaying,
but it will never come to 0,
because the driving force for
the decay is Q itself.
And if you say how long should
I wait?
There's a certain time called
1/RC, which is called the
time constant.
 
But any exponential function,
e^(-t)/t_0,
if t is much bigger than
t_0,
then you've got e to the
minus huge number and that's a
negligible number.
 
So whenever you say something's
falling exponentially,
it doesn't mean that it may
take forever,
or It may happen very quickly.
 
It depends what's up in the
exponent.
All exponential functions will
look like e to the minus
time over some other number with
dimension of time.
That's the unit in which you
measure time.
If little t is many
times big t_0,
then it's e to the minus
big number, which is negligible.
So you've got to wait many time
constants before a capacitor
will discharge.
 
So capacitors are pretty
dangerous.
In fact, your computer,
if you open it,
there can be capacitors inside
which are charged,
even though it's not plugged
into the mains.
So don't think it's safe to
open an electrical device,
just because it's not plugged
into the mains.
Because people tell you,
"Hey, pull the plug,
then you can do what you
want."
Not quite, because if you do
that, that's a big fat
capacitor, that R stands
for you.
That current is going to go
right through you.
That's why they always tell
you, "Do not take this
computer to your bathtub"
for example.
So these warnings,
even though they sound
ridiculous, part of it is true.
 
Capacitors are very dangerous.
 
What you want to do is
discharge all your capacitors
first.
 
Even in flash bulbs,
that's what happens.
You charge up a lot of charge
in a capacitor,
and the resistance there is the
bulb itself, the little coil in
the bulb.
 
When we close the circuit,
the capacitor dumps all its
charge and then in the brief
moment,
the coil heats up,
namely R heats up and
glows and you have a flash.
 
But there you want time
constant to be very small,
because how long are you going
to tell people to keep smiling?
So you've got to get on with
it, so you put in a very quick,
very rapid time constant,
so R is going to be very
small.
 
Sometimes you want the decay to
be very, very slow.
People are even thinking of
getting rid of batteries and
just buying capacitors.
 
With a capacitor,
you can use it to drive
something, but slowly the
voltage will go down.
So if your device can operate
over a range of voltage,
the voltage is just
Q(t)/C.
That's the voltage at a given
time.
If it can operate up to that
voltage, between that and that
voltage, you can run it for that
time.
Okay, so this is the fate of
the current.
Now what is
Q_0?
At t = 0,
what is the charge in the
capacitor?
 
Well, that's the initial charge
in the capacitor.
What is the current in this
problem, I?
It's just dQ/dt.
 
It's −dQ/dt,
and if you do
−dQ/dt,
you'll find there's
Q_0/RC.
 
Just take the derivative of
this function times
e^(−t/RC).
 
At t = 0,
Q_0/C is the
voltage in the capacitor,
divided by R as a
current.
 
That's a current it starts
with, but the current also
decays exponentially,
so with another prefactor.
But there's one calculation I
wanted to do,
which is the energy
calculation.
If I had a capacitor that was
charged, I had an energy
½CV^(2),
or if you want,
Q^(2)/2C.
 
At the end of the day,
my capacitor's completely
discharged.
 
You want to ask yourself,
what happened to the energy I
had in the capacitor?
 
And we all know the answer is
that it went through the
resistor and heated it up,
and the heat energy or light
energy is where you got your
rewards.
But we've got to make sure that
the energy deposited in the
resistor over all time is equal
to the energy you had in the
capacitor.
 
That's the last one thing I
wanted you to check.
So when current goes through a
resistor--here's a current going
through a resistor--what is the
rate at which energy is being
consumed?
 
So you must think,
let me draw the resistor this
way.
 
Its' very suggestive.
 
You're flowing downhill
electrically through voltage
V.
 
Every coulomb that falls down
loses an energy Q times
V as it falls.
 
And that's the energy that's
given to the wire by colliding
with the stuff in it and heating
it up.
Therefore in 1 second,
the number of coulombs falling
down is VI and that's the
power.
So power in the resistor is
VI.
That's the rate of energy
consumption.
So in this problem,
V is Q/C and I,
I got somewhere here--I'm too
close to the board to see where
anything is.
 
I want to first find the
current.
Here is my current.
 
I'm sorry, there's an easier
way to do this.
Instead of writing VI,
let me write it as
I^(2)R,
because 1V is IR.
So I want to integrate from 0
to infinity the quantity
I^(2)R.
 
Now I'm not going to waste your
time, guys.
First of all,
you know how to do this
integral.
 
Square this,
multiply by R and do the
integral, and believe me,
you will just get
Q_0
^(2)/2C.
I don't want to spend time
doing that.
So that's where the energy goes.
 
 
 
You follow that?
 
The capacitor discharges
through the resistor and the
energy you pumped into it comes
out in the form of heat,
or light if it's glowing.
 
And we have seen the balance of
energy.
But now the trouble with this
circuit is that it doesn't last
very long.
 
You've done it once,
you're finished.
The capacitor discharges.
 
If you want an experiment where
the current can keep on running,
then you need what's called a
cell or a battery.
I've got to explain to you a
little bit about the cell.
It's something I did not fully
understand the first time
around, so I want to share with
you whatever understanding I
have, how the cell works.
 
People generally tell you a
cell provides a certain voltage
difference between the end
points, maybe 1.2 volts between
this and the terminal.
 
That's certainly true,
but in light of what we have
learned, here is what is the
correct way to think about it.
The analogy is with the ski
resort.
So here is the ski resort.
 
All these guys are coming down.
 
Let's follow one person coming
down the ski.
Gravity's acting down here,
then you sort of loop around
and you come here for free.
 
Then there is a lift that takes
you to the top.
So let's call it the force of
the lift.
The force of the lift has
nothing to do with gravity,
driven by some other engines
and so on.
And we define force of the lift
on the closed loop to be
something called curly E.
 
This is the mechanical analog
of what's called electromotive
force.
 
Now this line integral of this
force is not 0 on a closed loop,
because on the way up,
it is doing some work,
but it doesn't do anything the
rest of the circuit.
It's always acting up,
moving a distance h.
Maybe that force times h is the
electromotor force.
So you go to the top,
you come down,
and the electromotive force
comes and pushes you back to the
top, and again,
gravity brings you down.
So in a battery,
what you have is the following
- that's where you have to
follow this very closely.
Here is an electrical cell.
 
In an electrical cell,
there are a lot of positive
charges and there are a lot of
negative charges,
and the field in fact looks
like this.
So when a current flows,
it goes like that,
goes through resistor,
comes down.
But the current,
you realize,
has to flow up here.
 
Inside the cell,
if the current flows this way,
the current has to go like
this.
The electric field is actually
pointing down between the
plates.
 
That's not what makes it move,
because there's an extra force.
This is your electric field.
 
There's an extra force called
E',
which is of chemical origin,
which pushes your charges
against their will from lower
potential to higher potential.
That is like the ski lift.
 
Electric field everywhere is
like gravity.
Its line integral that way and
line integral that way are all
the same.
 
But this E' has a line
integral,
E'⋅dr
on a closed loop,
which is not equal to 0 and in
fact is called the emf.
You can see it's not 0,
because it's non-zero here.
Everywhere else it is 0.
 
And in the region where it's
non-zero, there's no
cancellation.
 
The displacement and the force
are all in the same direction.
So it's a non-conservative
force sitting inside a battery.
It's a chemical force.
 
The usual electric field that
we all like,
E⋅dr,
in fact has line integral equal
to 0.
 
But when you're inside this
battery,
if you ask how big is this
E' and how big is this
E,
what happens inside the battery
is,
once you start piling up
charges there,
this electric field opposed to
the charge is coming,
the chemical force will exactly
balance the electrical force
inside.
So inside this region,
the electric field will be -
that chemical force.
 
E' is the chemical force
per charge.
Inside the battery,
there is something that takes
the charge against its natural
instinct, against gravity if you
like, and takes it up.
 
You need something like that to
make this thing run,
otherwise charges will just sit
in the bottom.
They won't go up.
 
So you need an external agency.
 
Sometimes you can have a belt.
 
In a Van der Graaf generator,
you have a belt that runs up
and down and carries positive
charge to the top,
even though the top is already
positively charged.
Left to itself,
a positive charge would never
like to go to the dome which is
positively charged,
but you drag it on that belt,
kicking and screaming.
You've got to do that to charge
it up against its will.
That's what provides the force
there.
You can connect that guy to the
resistor and the current will
keep flowing.
 
Now here is the interesting
thing.
Since inside this loop,
the electric field and the
chemical force are exactly
balanced.
Now let's look at
E⋅dr
between here and here,
between the positive and
negative terminal.
 
That's the same as
E⋅dr
inside.
 
So
E⋅dr
from the positive to the
negative terminal is
−E'
⋅dr from the
positive to the negative
terminal is V_-
- V_ .
 
Therefore V_ 
- V_- is the
emf.
 
So if the chemical force has an
emf,
which is the line integral of
the chemical force around a
loop,
which is basically inside a
battery,
that will be also equal to the
voltage difference for a person
living outside the battery.
If a person outside the battery
went on a loop like this from
here to here,
the line integral done by the
electric field,
which is the voltage dropped,
if you want,
the voltage gain between this
one and this one,
will be exactly equal to the
emf.
 
So V_  and
this is V_-.
Let me just make sure this part
is correct.
The line integral is 0,
or if you like,
the integral of the electric
field from here to here,
doesn't matter which path you
take,
but the electric field is
conservative.
That's equal to the integral of
-E from here to here,
but that's the same as being
the polarities reversed in the
integral,
and that will be the emf.
So that = integral
E⋅dr
from - to .
 
And that will = V_
 -
V_-,
if you write it that way.
Okay, so the point is,
when you have a battery,
if you don't want to get into
any of the details,
this is what you guys have to
know.
If you don't look under the
hood, just look at those two
terminals.
 
There'll be a voltage
difference between them equal to
what's called the emf of the
battery.
The subtle thing to remember
is, there are non conservative
forces at work inside the
battery,
this is what you may not
realize, whose line integral is
in fact not 0,
but you don't have to know all
that.
 
You just have to know the
integral, you have to know what
happens if I go from here to
here in an electrical circuit.
I go up in voltage equal to
V .
That's what I want you to
remember.
For those of you who want to
know the details,
this is the reason that
happens.
From now on,
we won't look inside the
terminal.
 
We'll take every battery to
have an emf E,
meaning if you jump from a
negative to positive,
you gain a voltage.
 
 
 
