Uploaded by MIT on 29.12.2010

Transcript:

Now we go to problem number seven, which is another

classic, and that has plane geometry.

We have now done circle geometry, we have done

cylindrical geometry, and now we're doing plane geometry.

Here is a plane which is infinitely large in size--

x equals 0 here, x equals plus d, and equals minus d.

We have a charge density here-- rho

coulombs per cubic meters--

so it's uniformly distributed throughout.

The density is rho coulombs per cubic meter, and you're

being asked: what is the electric field

here, here, and there?

Let me do the electric field outside, and let me leave you

largely with the electric field inside.

I choose here a cylinder, but it doesn't

have to be a cylinder--

it could be a box like this.

I don't want mislead you on the cylinder,

but I choose a cylinder.

The cylinder sticks all the way through this slab--

and here's the bottom of the cylinder--

and it is crucial, essential, and an absolute must that this

distance here above the slab must be the same as the

distance below the slab.

You will see shortly why that is.

The charge inside this cylinder is the charge which

is only here.

For reasons of symmetry, the electric field could only be

pointing radially outward--

I shouldn't say radially, because it could only be

perpendicular to the surface, and nature cannot decide

between left and right--

so the E vector must be right here.

The E vector here must also be perpendicular to this flat

surface, but since I have chosen this h the same, this E

must be the same as that E. If you choose this h different

from this h, you have no knowledge yet that those two E

vectors are the same.

It turns out that they are the same, but you don't know that

yet, so you must use this h the same at this h, and only

then can you justify saying that the magnitude of the E

vector here is the same as the magnitude of

the E vector there.

The ds's are pointing outwards, so they are parallel

to E, and the ds's here are also parallel to E. Both here

and here are the cosine of the angle between ds and E plus 1.

What, now, is the electric flux that is escaping from

this cylinder?

You could have taken a box.

The electric flux escaping here is the same as the

electric flux escaping here, because ds and E are both in

the same direction, so I get A times E, if A is

this surface area.

I have two surfaces, so I have a 2, and that now equals the

charge inside.

Maybe I am forgetting something--

how about the flux escaping from the sides?

How about this flux escaping from the cylindrical surface?

That flux is 0, for the reason that ds everywhere on the

surface of this cylinder--

and even if you had chosen this box--

everywhere on these vertical services, ds and E would be

perpendicular to each other.

The dot product would therefore be 0, so there is no

contribution and there is no electric flux coming out from

the cylindrical side.

There is no contribution from the vertical parts off the

box, but only from the two end pieces

I can indeed now say this must be the total charge inside.

The total charge inside is the volume, which is A times d--

times 2d--

A times 2d, divided by epsilon 0, and the volume must be

multiplied by rho, which is the number of coulombs per

cubic meter.

A cancels--

of course, A cancels, because nature couldn't care less how

large you choose this A--

and so you find the famous result that the electric field

outside the plane equals rho times d divided by epsilon 0.

What the big surprise is is that it is independent of h--

it doesn't matter how far you are from the plane, it's

always the same.

That's not so intuitive.

In fact, if I have a surface which was modest in size--

this is that plane--

of course, if I go very, very far away the electric field

here must be smaller than the electric field here.

Why, then, do we not find that?

Why do we find that the electric field is independent

of the distance to this plane geometry?

I want you to think about that a little bit, but I want to

help you, though.

I think it has to do with this absurd geometry--

namely, that we have assumed that our plane is infinitely

large, and so give that some thought.

Now you have to do the electric field inside.

I will only start you off on that, but no more.

Here is the slab, and here equals x equals 0.

You can either choose a box, or you can choose a cylinder,

but the cylinder now has to be

completely inside the material.

This distance now is x, and this distance now must be x.

You're going to calculate now the flux that escapes from

this surface, and the flux that

escapes from this surface.

There is no flux escaping from the vertical surface, and you

are now going to put that equal to all the charge inside

divided by epsilon 0.

You will find an electric field which is

now linear with x.

It turns out that it is 0 at x equals 0, and it grows to a

maximum value at this point.

That maximum value at this point is E--

no surprise--

will be rho time d divided by epsilon 0.

I would like to leave you with that.

It's an interesting exercise, and as I said, I need a lot of

time for the problem.