7. Resistance

Uploaded by YaleCourses on 23.03.2011

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,
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
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,
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
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
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
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,
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
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
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
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.
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
It's as if there's another person behind the mirror.
Of course, you know there is no other person behind the mirror,
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
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.
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
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.
Student: For the potential energy,
is it 1 over 4d or 1 over...?
Prof: You would think it's 1 over 2d,
Student: Yes.
Prof: But I'm telling you it's 1 over 4d.
Student: Because when you take the integral,
you have 2 on...
Prof: All right.
Now what if this plane was not grounded, it's a standalone
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
You have to remain at some potential.
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
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
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
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,
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,
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
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
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
It's very amazing, but true.
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
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
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
Because the field that this one feels generally,
the field at any point is due to this one, this one and this
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
So you must find the force between these 2,
add to that the force between these two and you'll get
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
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
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
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
Someone says, what's the capacitance of the
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
And then you can see it is proportional to Q,
and you take the ratio, you'll get the capacitance of
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
If you're moving dQ coulombs, that's the work you
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
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
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
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
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
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
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
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
So they'll lose all memory of what they were doing after each
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
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
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
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
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
It could be copper, it could be aluminum,
it could be some alloy.
For all of them, the answer depends on
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
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
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
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
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
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
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
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
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
So when I close the switch, we can all imagine what will
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
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?
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
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 =
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)
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
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
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
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
Just take the derivative of this function times
At t = 0, Q_0/C is the
voltage in the capacitor, divided by R as a
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,
At the end of the day, my capacitor's completely
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
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
So you must think, let me draw the resistor this
Its' very suggestive.
You're flowing downhill electrically through voltage
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
Now I'm not going to waste your time, guys.
First of all, you know how to do this
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
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
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
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
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
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
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
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
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
We'll take every battery to have an emf E,
meaning if you jump from a negative to positive,
you gain a voltage.