Kelistrikan dan Kemagnetan

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So today no new concepts, no new ideas.

You can relax 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 look at the equation number three that the potential difference between point A and point A is obviously zero.

I 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 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 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 -- 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 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.

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, 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 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 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 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.

Pengembangan Perkuliahan

1. Buatlah sebuah Esai mengenai materi perkuliahan ini

2. Buatlah sebuah kelompok berjumlah 5 orang untuk menganalisis materi perkuliahan ini

3. Lakukan Penelitian Sederhana dengan kelompok tersebut

4. Hasilkan sebuah produk yang dapat digunakan oleh masyarakat

5. Kembangkan produk tersebut dengan senantiasa meningkatkan kualitasnya#### Staff

Visualizations:

Prof. John Belcher

Instructors:

Dr. Peter Dourmashkin

Prof. Bruce Knuteson

Prof. Gunther Roland

Prof. Bolek Wyslouch

Dr. Brian Wecht

Prof. Eric Katsavounidis

Prof. Robert Simcoe

Prof. Joseph Formaggio

Course Co-Administrators:

Dr. Peter Dourmashkin

Prof. Robert Redwine

Technical Instructors:

Andy Neely

Matthew Strafuss

Course Material:

Dr. Peter Dourmashkin

Prof. Eric Hudson

Dr. Sen-Ben Liao

#### Acknowledgements

The TEAL project is supported by The Alex and Brit d'Arbeloff Fund for Excellence in MIT Education, MIT iCampus, the Davis Educational Foundation, the National Science Foundation, the Class of 1960 Endowment for Innovation in Education, the Class of 1951 Fund for Excellence in Education, the Class of 1955 Fund for Excellence in Teaching, and the Helena Foundation. Many people have contributed to the development of the course materials. (PDF)

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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