surface of solid conductors but that it's not uniformly

distributed. Perhaps you remember that,

unless it happens to be a sphere.

And I want to pursue that today.

If I had a solid conductor which say had this shape and I'm

going to convince you today that right here the surface charge

density will be higher than there.

Because the curvature is stronger than it is here.

And the way I want to approach that is as follows.

Suppose I have here a solid conductor A which has radius R

of A and very very far away, maybe tens of meters away,

I have a solid conductor B with radius R of B and they are

connected through a conducting wire.

That's essential. If they are

connected through a conducting wire, then it's equipotential.

They all have the same potential.

I'm going to charge them up until I get a charge

distribution QA here and I get QB there.

The potential of A is about the same that it would be if B were

not there. Because B is so far away that

if I come with some charge from

infinity in my pocket that the work that I have to do to reach

A per unit charge is independent of whether B is there or not,

because B is far away, tens of meters,

if you can make it a mile if you want to.

And so the potential of A is then the charge on A divided by

four pi epsilon zero the radius of A.

But since it is an equipotential because it's all

conducting, this must be also the potential of

the sphere B, and that is the charge on B

divided by four pi epsilon zero R of B.

And so you see immediately that the Q, the charge on B,

divided by the radius of B, is the charge on A divided by

the radius on A. And if the radius of B were for

instance five times larger than the radius of A,

there would be five times more charge on B than there would be

on A. But if B has a five times

larger radius then its surface area is twenty-five times larger

and since surface charge density sigma is the charge on a sphere

divided by the surface area of the sphere, it is now clear that

if the radius of B is five times larger than A,

it's true that the charge on B is five

times the charge on A, but the surface charge density

on B is now only one-fifth of the surface charge density of A

because its area is twenty-five times larger and so you have

this -- the highest surface charge density at A than you

have at B. Five times higher surface

charge density here than there. And I hope that convinces you

that if we have a solid conductor like this,

even though it's not ideal as we have here with these two

spheres far apart, that the surface

charge density here will be larger than there because it has

a smaller radius. It's basically the same idea.

And so you expect the highest surface charge density where the

curvature is the highest, smallest radius,

and that means that also the electric field will be stronger

there. That follows immediately from

Gauss's law. If this is the surface of a

conductor, any conductor, a solid conductor,

where the E field is zero inside of the conductor,

and there is surface charge here, what I'm going to do is

I'm going to make a Gaussian pillbox, this surface is

parallel to the conductor, I go in the conductor,

and this now is my Gaussian surface, let this area be

capital A, and let's assume that it is positive charge so that

the electric field lines come out of the

surface like so, perpendicular to the surface.

Always perpendicular to equipotential,

so now if I apply Gauss's law which tells me that the surface

integral of the electric flux throughout this whole surface,

well, there's only flux coming out of this surface here,

I can bring that surface as close to the surface as I want

to. I can almost make it touch the

conductor. So everything comes out only

through this surface, and so what comes out is the

surface area A times the electric field E.

The A and E are in the same direction because remember E is

perpendicular to the surface of the equipotentials.

And so this is all there is for the surface integral,

and that is all the charge inside, well the charge inside

is of course the surface charge density times the area A,

divided by epsilon zero, this is Gauss's law.

And so you find immediately that the electric field is sigma

divided by epsilon zero. So whenever you have a

conductor if you know the local surface charge density you

always know the local electric field.

And since the surface charge density is going to be the

highest here, even though the whole thing is

an equipotential, the electric field will also be

higher here than it will be there.

I can demonstrate this to you in a

uh very simple way. I have here a cooking pan and

the cooking pan I used to boil lobsters in there,

it's a large pan. The cooking pan I'm going to

charge up and the cooking pan here has a radius whatever it

is, maybe twenty centimeters, but look here at the handle,

how very small this radius is, so you could put charge on

there and I'm going to convince you that I can scoop off more

charge here where the radius is small than I can scoop off here.

I have here a small flat spoon and I'm going to put the spoon

here on the surface here and on the surface there and we're

going to see from where we can scoop off the most charge.

Still charged from the previous lecture.

So here, we see the electroscope that we have seen

before. I'm going to charge this

cooking pan with my favorite technique which is the

electrophorus. So we have the cat fur and we

have the glass plate.

I'm going to rub this first with the cat fur,

put it on, put my finger on, get a little shock,

charge up the pan, put my finger on,

get another shock, charge up the pan,

and another one, charge up the pan,

make sure that I get enough charge on there,

rub the glass again, put it on top,

put my finger on, charge, once more,

and once more. Let's assume we have enough

charge on there now.

Here is my little spoon. I touch here the outside here

of the can -- of the pan. And go to the electroscope and

you see a little charge. It's very clear.

What I want to show you now it's very qualitative is that

when I touch here the handle, it's a very small radius,

that I can take off more charge.

There we go. Substantially more.

That's all I wanted to show you.

So you've seen now in front of your own eyes for the first time

that even though this is a conductor that means that it is

an equipotential, that the surface charge density

right -- right here is higher than the surface charge density

here. Only if it is a sphere of

course for s- circle symmetry reasons will the charge be

uniformly distributed. If the electric field becomes

too high we get what we call electric breakdown.

We get a discharge into the air.

And the reason for that is actually quite simple.

If I have an electron here and this is an electric field,

the electron will start to accelerate in this direction.

The electron will collide with nitrogen and oxygen molecules in

the air and if the electron has enough kinetic energy to ionize

that molecule then one electron will become two electrons.

The original electron plus the electron from the ion.

And if these now start to accelerate in this electric

field, and if they collide with the molecules,

and if they make an ion, then each one will become two

electrons, and so you get an avalanche.

And this avalanche is an electric breakdown and you get a

spark. When the ions that are formed

become neutral again they produce light and that's what

you see. That's the light that you see

in the spark.

And so sparks will occur typically at the -- at sharp

points -- at areas where the curvature is strong,

whereby the radius is very small, that's where the electric

fields are the highest. How strong should the electric

field be? Well, we can make a back of the

envelope calculation. If you take air of one

atmosphere, dry air, at room temperature,

then the -- the electron on average, on average,

will have to travel about one micron, which is ten to the

minus six meters, between the collisions with the

molecules, it's just a given. On average.

Sometimes a little more, sometimes a little less.

Because it's a random process of course.

To ionize nitrogen, to ionize oxygen,

takes energy. To ionize an oxygen molecule

takes twelve-and-a-half electron volts.

And to ionize nitrogen takes about fifteen electron volts.

What is an electron volt? Well, an electron volt is a

teeny weeny little amount of energy.

It's one point six times ten to the minus nineteen joules.

Electron volt is actually a very nice unit of energy.

Because once you have an electron and it moves over a

potential difference of one volt,

it gains in kinetic energy, that's the definition of an

electron volt, it gains one electron volt.

It's the charge of the electron, which is one point six

times ten to the minus nineteen coulombs, multiplied by one

volt. And that gives you then the

energy, one electron volt. And so what it means then --

let's assume that this number is ten electron volts.

Do we -- we only want a back of the envelope calculation.

So we want the electron to move over a potential difference

delta V which is roughly ten volts and we want it to do that

over a distance delta X which is ten to the minus six meters,

that's your one micron. And if that happens you'll get

this enough kinetic energy in the electron to cause an ion.

So what electric field is required for that,

that is delta V, the potential difference,

divided by the delta X, so that is ten divided by ten

to the minus six, so that's about ten to the

seven volts per meter. That's a very strong electric

field. In reality when we measure the

electric fields near breakdown, it is more like three million

volts per meter. But it's still very close.

This was only a back of the envelope calculation.

So very roughly at one atmosphere air room temperature

when the air is dry we get electric breakdown at about

three million volts per meter. When the ions neutralize you

see light, that's why sparks can be

seen. They heat the air,

they produce a little pressure wave, so you can also hear

noise. If you had two parallel plates

and you would bring those plates closely together and suppose

they had a potential difference of three hundred volts,

then you would reach an electric field of three million

volts per meter when the distance D is about one tenth of

a millimeter. So that's when you expect

spontaneous discharge between these two plates.

In practice however it probably will happen when the plates are

further apart than one tenth of a millimeter.

And the reason for that is that there is no such thing as

perfect plates. The plates have imperfections.

That means there are always areas on the plate which are not

flat, which are a little bit like what you see there,

small radius, and that's of course where the

electric field then will be larger and that's where the

discharge will occur first.

However, if you touch the doorknob and you get a spark,

you feel a spark, and you look at the spark and

you see that when you're three millimeters away from the

doorknob that the spark develops, you can s- pretty sure

that the potential difference between you and the door was of

the order of ten thousand volts, several thousand volts,

at least. Because over three millimeters

it requires ten thousand volts to get the three million volts