you can release a little bit and I want to discuss with you

the connection between electric potential and electric fields.

Imagine you have an electric field here in space and that I

take a charge Q in my pocket, I start at position A and I

walk around and I return at that

point A. Since these forces are

conservative forces, if the electric field is a

static electric field, there are no moving charges,

but that becomes more difficult, then the forces are

conservative forces and so the work that I do when I march

around and coming back at point A must be zero.

It's clear when you uh look at the equation number three that

the potential difference between point A and point A is

obviously zero. I g- start at point A and I end

at point A and that is the integral in going from A back to

point A of E dot DL and that then has to be zero.

And we normally indicate such an integral with a circle which

means you end up where you started.

This is a line now this is not a closed surface

as we had in equation one. This is a closed line.

And so whenever we deal with static electric fields we can

add now another equation if we like that.

And that is if we have a closed line of E dot DL so we end up

where we started that then has to become zero.

Later in the course we will see that there are special

situations where we don't deal with static

fields when we don't have conservative E fields,

that that is not the case anymore.

But for now it is. So if we know the electric

field everywhere then we -- we can see equation number two then

we know the potential everywhere.

And so if we turn it the other way around if we knew the

potential everywhere you want to know what the electric field is

and that of course is possible. If you look at equation two and

three you see that the potential is the integral of the

electric field. So it is obvious that the field

must be the derivative of the potential.

Now when you have fields being derivative of potentials you

always have to worry about plus and minus signs,

whether um you have to pay MIT twenty-seven thousand dollars

tuition to be here or whether MIT pays you twenty-seven

thousand dollars tuition for coming here is

only a difference of a minus sign but it's a big difference

of course. And so let's work this out in

some detail. I have here a charge plus Q.

And at a distance R at that location P we know what the

electric field is, we've done that a zillion

times. This is the unit vector in the

direction from Q to that point and we know that the electric

field is pointing away from that charge

and we know that the electric field E we have seen that

already the first lecture is Q divided by four pi epsilon zero

R squared in the direction of R roof.

And last lecture we derived what the electric potential is

at that location. The electric potential is Q

divided by four pi epsilon zero R.

This is a vector. This is a scalar.

So the um potential is the integral of the electric field

along a line and now I want to try whether the electric field

can be written as the derivative of the potential.

So let us take DV DR and let's see what we get.

If I take this DV DR I get a minus Q divided by four pi

epsilon zero R squared.

Of course if I want to know what the electric field is I

need a vector so I will multiply both sides, which is completely

legal, there's nothing illegal about that, with unit vector in

the direction R so that turns them into vectors.

And now you see that I'm almost there, this is almost the same,

except for a minus sign. And so the derivative of the

potential is minus E, not plus E.

And so I will write that down here,

that E equals minus DV DR. So they are closely related if

you know what the -- oh I want -- I want this to be a vector so

I put here R roof. Vector on the left side,

you must have a vector on the right side.

And so if you know the potential everywhere in space,

then you can retrieve the electric field.

I mentioned last time that the electric field vectors -- e-

electric field lines, are always perpendicular to the

equipotential surfaces. And that's obvious why that has

to be the case. Imagine that you are in an --

in space and that you move with a charge in your pocket

perpendicular to electric field lines.

So you purposely move only perpendicular to electric field

lines. So that means that the force on

you and the direction in which you move are always at

ninety-degree angles. So you'll only move

perpendicular to the field lines.

These are the field lines, you move like this.

These are the field lines, you move like this.

So you never do any work. Because the dot product between

DL and E is zero and if you don't do any work the potential

remains the same, that's the definition.

And so you can see that therefore equipotential surfaces

must always be perpendicular to field lines and field lines must

always be perpendicular to the equipotentials.

And I will show you again the -- the viewgraph,

the overhead projection of uh the nice drawing by Maxwell.

With the plus four charge and the minus one charge.

The same one we saw last time. Only to point out again this

ninety-degree angle. I discussed this in great

detail last lecture so I will not do that.

R- the red lines are really surfaces.

This is three-dimensional, you have to rotate the whole

thing about the vertical. So these are surfaces.

And the red ones are positive potential surfaces and the blue

ones are negative potential surfaces.

That is not important. But the green lines are field

lines. And notice if I take for

instance this field line, it's perpendicular here to the

red, perpendicular there, perpendicular there,

perpendicular there. Perpendicular here.

Perpendicular here. Coming in here,

perpendicular, perpendicular,

perpendicular. Everywhere where you look on

this graph you will see that the field lines are perpendicular to

the equipotentials.

And that is something that we now fully understand.

The situation means then that if you release a charge at zero

speed that it would always start to move perpendicular to an

equipotential surface because it always starts to move in the

direction of a field line, a plus charge in the direction

of the field line, minus charge in the opposite

direction. So if you're in space and you

release a charge at zero speed it always takes off

perpendicular to equipotentials.

You have something similar with gravity.

If you look at maps of mountaineers,

contours of equal altitude, equal height,

if you started skiing and you started at that point,

and you started with zero speed, you would always take off

perpendicular to the equipotentials.

So this is the direction in which

you start to move. If you start off with zero

speed. I now want to give you some

deeper feeling of the connection between potential and electric

fields and I want you to follow me very closely.

Each step that I make I want you to follow me.

So imagine that I am somewhere in space at p- position P.

At that position P there is a potential, one unique potential,

V of P. That's a given.

And there is an electric field at that location where I am.

And now what I'm going to do, I'm going to make an extremely

small step only in the X direction.

Not in Y, not in Z. Only in the X direction.

If I measure no change in the potential over that little step

it means that the component of the electric field in the

direction X is zero. If I do measure a difference in

potential then the component, the X component of the electric

field, the magnitude of that, would be that little sidestep

that I have made, delta X, it would be the

potential difference that I measure divided by that little

sidestep. And I keep Y and Z constant.

And these are magnitudes. But that's why I put these

vertical bars here. Equally if I made a small

sidestep in the Y direction and I measured a potential

difference delta V keeping X and Z constant, that would then be

the component of the electric field in the Y direction.

Earlier we wrote down for E as a unit new- newtons per

coulombs. From now on we almost always

will write down for the unit of

electric field volts per meter. It is exactly the same thing as

newtons over coulombs, there is no difference,

but this gives you a little bit more insight.

You make a little sidestep in meters and you measure how much

the potential changes, it's volts per meters.

It is a potential change over a distance.

So now I can write down the connection between electric

field and potential in Cartesian coordinates.

It looks much more scary than the nice way that I could write

it down up there. When I had only a function of

distance R. And so in Cartesian coordinates

we now get E equals minus, that minus sign we discussed at

length, and now we get DV DX times X roof plus DV DY times Y

roof plus DV DZ times Z roof.

And what you see here, this first term here,

including of course the minus sign, that is E of X.

And this term including the minus sign, that is E of Y.

And so on. And the fact that you see these

curled Ds it means partial derivatives.

That means when you do this derivative you keep Z and Y

constant. When you do this derivative you

do X and Z constant and so on. And so this is the Cartesian

notation for which in eighteen oh two you will learn or maybe

you already have learned we would write this E equals minus

the gradient of G. This is a vector function.

This is a scalar function. And this is just a s- a

different notation, just a matter of words,

for this mathematical recipe. And you'll get that with

eighteen in eighteen oh two if you haven't seen that yet.

So I now want a straightforward example whereby we assume a

certain dependence on X and I give you it is a given that V,

the potential, is ten to the fifth

times X. So that is a given.

And this holds between X equals zero and say ten to the minus

two meters. So it holds over a space of one

centimeter. So the potential changes

linearly with distance. What now is the electric field?

In that space? Well, the electric field I go

back to my description there. There's only a component in the

X direction. So the first derivative becomes

minus ten to the fifth times X roof and the others are zero.

So EY is zero and EZ is zero. So you may say well yeah whoa

nice mathematics but we don't see any physics.

This is more physical than you think.

Imagine that I have here a plate

which is charged, it's positive charge.

And the plate is at location X and I have another plate here,

it's say at location zero. I call this plate A and I call

this plate B and this plate is charged negatively.

So X goes into this direction. So I can put the electric field

inside here according to the recipe minus ten to the fifth

and it is in the direction of minus X roof so e- X roof is in

this direction, the electric field is in the

opposite direction, and it's the same everywhere

and that is very physical. We discussed that.

When we discussed the electric field near very large planes,

that the electric field inside was a

constant, remember, and the electric field inside

was sigma divided by epsilon zero if sigma is the surface

charge density on each of these plates.

And we argued that the electric field outside was about zero and

that the electric field outside here was about zero.

So it's extremely physical. This is exactly what you see

here. The electric field minus ten to

the fifth times e- so the magnitude of this electric field

here, the magnitude, is ten to the fifth volts per