Placeholder Image

字幕列表 影片播放

  • Prof: All right, class, I thought I'd start as

  • usual by telling you what happened last time.

  • Not the whole thing, but just the highlights so you

  • can follow what's happening today.

  • The main things I did last time were the notion of an electric

  • field, which is going to be with you from now till the end of the

  • course.

  • The idea of the electric field is that if you've got lots of

  • charges, q_1,

  • q_2, q_3,

  • instead of worrying about the force they exert on each other,

  • you ask yourself, at a generic point where

  • there's nothing, if I put 1 coulomb here,

  • what will it experience?

  • What force will it experience?

  • You compute that.

  • So in your mind, imagine a coulomb,

  • and you find the force due to q_1,

  • it looks like that.

  • q_2 may exert a force that way.

  • q_3 could be of opposite sign,

  • so maybe it will exert a force that way.

  • You add all these vectors, they add up to something.

  • That something is called the electric field at that point.

  • There is nothing there except the electric field,

  • but it's very real, because if you put something,

  • something happens to it.

  • So the electric field is everywhere.

  • The charges are in a few places.

  • Electric field is defined everywhere, except right on top

  • of the charges, which is where it probably is

  • infinite.

  • Once you know the electric field anywhere,

  • if you put another charge q,

  • a real charge, it will experience a force

  • equal to qE, because electric field was the

  • force you would have had on the 1 coulomb,

  • and if you put q coulombs, it will be qE.

  • That's the electric field.

  • So you can imagine computing it for any given distribution of

  • charges, because you know what each one of them does.

  • Then I said there's one nice way to visualize the electric

  • field, which is to draw the field lines.

  • You go to each point and you ask, if I put a charge here,

  • a positive test charge, which way will it move?

  • Then you follow that thing as it moves and you get that line

  • and you get that line and you get that line.

  • You can draw these lines.

  • The lines give you one piece of information which is very

  • obvious, namely, if you are here,

  • the force is along that line.

  • But I also pointed out to you, you get more than just the

  • direction of the force.

  • You also can understand the strength of the force.

  • The strength of the force is contained in the density of

  • these lines.

  • Now density has to be defined carefully.

  • It's not like mass per unit volume.

  • It's an area density.

  • That means if you get yourself 1 square meter,

  • you know, meter by meter piece of wood,

  • like a frame, and you hold it there,

  • perpendicular to the lines and see how many lines go through,

  • that's called the area density of lines.

  • That you can see will fall like 1/r^(2),

  • because if you draw a sphere of radius r,

  • that area is 4Πr^(2), and the same number of lines

  • are going through that as any other sphere,

  • so it will be proportional to 1/r^(2).

  • But so is the electrical field proportional to 1/r^(2),

  • and I said, let us agree that we will draw

  • 1/ε_0 lines per coulomb.

  • This is a necessity; it's just a convenience.

  • It's like saying I want to measure distances in inches and

  • centimeters for daily life.

  • You can measure them in parsecs and angstroms,

  • but you'll be dealing with nasty numbers.

  • So it's a convenience, and the convenience here is,

  • let's pick 1/ε_0

  • lines per coulomb.

  • You'll see the advantage of that, because if you ask,

  • what's the density of lines here on a sphere of radius

  • r, if there's charge q at

  • the center, I got that many lines,

  • and the area of the sphere is 4Πr^(2),

  • so the density of lines per unit area,

  • you see, is precisely equal in magnitude to the electric field

  • at that point.

  • If you had not drawn 1/ε_0,

  • but maybe 5/ε_0

  • lines, then the line density will be 5

  • times the electric field.

  • It will still represent the electric field,

  • but we don't want to simply represent it.

  • We want it to be the electric field.

  • It makes it easier.

  • Then I said, let's take a slightly more

  • complicated situation, two charges.

  • This is called a dipole, one pole and another pole,

  • and you can draw the field lines here.

  • You can see if you put a test charge, it'll run away from the

  • plus to the minus.

  • If you leave it somewhere here, it'll go like that and loop

  • around and come back, and you can calculate them.

  • This is no longer guesswork.

  • If you had enough time, I hope you all agree,

  • you can go to any point you like and find the force of

  • attraction due to this one, the force of repulsion due to

  • that one, add them up and you will get an

  • arrow that direction.

  • So one can draw these lines, and the lines tell you a story.

  • Then I said, let us find the field in an

  • analytic expression due to the dipole.

  • Yes?

  • Student: I have a question about the dipole.

  • If you put a test particle going in the positive direction

  • of the x axis, would it also _________?

  • Prof: No.

  • She's got a good point.

  • If you put somebody here, it will never loop around.

  • Can you see why?

  • Because as it goes further away, this is trying to pull it

  • back.

  • It's always going to be closer to this guy, so it's never going

  • to come back.

  • This line will go like that and this line will go like this.

  • But anything else at any other angle will loop around and come

  • back.

  • All right, now the field strength, if you want to

  • calculate it, you can use the formula for

  • E due to this one and E due to that one and add

  • them, and I did that for you.

  • I don't want to go into the details,

  • but I remind you in all cases, the electric field fell like

  • 1/r^(3), because the field of each

  • charges goes like 1/r^(2).

  • And if these were on top of each other, they will completely

  • cancel each other.

  • So the reason you have a non-zero field is thanks to

  • a.

  • Therefore the answer has to contain an a in front of

  • it, at least n the first approximation.

  • But that a, from dimensional analysis,

  • has to come with a 1/r so that the whole thing has the

  • same dimension as before.

  • That's where you get a/r^(3).

  • That a times q times 2 and so on,

  • that became the dipole moment.

  • That was true here.

  • We verified that's true here.

  • Later on, we'll verify it everywhere, because there are

  • easier ways to calculate than what I'm using.

  • Then I said, forget about the field due to

  • charges.

  • Let's look at what charges do when you put them in a field.

  • So I took two examples.

  • One was a very simple example.

  • These are two parallel plates.

  • They are not two lines; they are plates coming out of

  • the blackboard.

  • They're filled with charge and this has got charge on it.

  • This has got - charge on it.

  • And therefore, the electric field will look

  • like this, right?

  • Because if you leave a test charge, it will go away from the

  • positive, towards the negative.

  • Then I said, suppose there really is a

  • particle here with some velocity, v_0,

  • what will it do?

  • You can see that the force on this guy is going to be q

  • times E.

  • E is pointing down.

  • If you divide by the mass, that's the acceleration,

  • also pointing down, and it's constant.

  • So that's like a particle in a gravitational field,

  • except g is replaced by this number.

  • So it will just curve like that, and you can calculate the

  • trajectory.

  • The final thing I did was, what happens when you put a

  • dipole in a uniform field.

  • Here as well, I think I was rushing near the

  • end, and even I couldn't read my stuff in the corner.

  • So I will go over that.

  • If there's something that you didn't follow,

  • then I will be happy to repeat that part for you.

  • But you should understand what the question is.

  • There is an electric field which is pointing like this,

  • as if you have two plates here, charge is here,

  • - charge is there.

  • They're producing a constant electric field in the horizontal

  • direction.

  • In that environment, I take an electric dipole whose

  • - charge and charge, q and -q,

  • happen to be oriented like that.

  • Question is, what will happen to this guy?

  • If you want, you can imagine that it's a

  • little massless stick, and one end you glue q

  • coulombs, other end you glue -q

  • coulombs, and your let it sit there.

  • What will it do?

  • First of all, it won't feel any net force,

  • because the force in this direction is q times

  • E and the force that direction is also q times

  • E, if you want,

  • it's -q times E and they cancel.

  • But that doesn't mean it won't react.

  • It will react, because you can all see

  • intuitively, it's trying to straighten this guy out and

  • applying a torque like that.

  • You follow that?

  • That's what it will do.

  • And the way to find the torque, the torque is the product of

  • the force and the distance between the point of rotation

  • and the force, and the sine of the angle

  • between them, that is to say,

  • sine of this angle.

  • What the sine of the angle does is to take the component of the

  • force perpendicular to this axis,

  • because if you resolve the force into that part and that

  • part, this part is no good for

  • rotation.

  • That's trying to stretch the dipole along its own length.

  • It's the perpendicular part that's going to rotate

  • something, so you get that times sine theta.

  • That you can write now as the vector equation

  • p x E.

  • Because p is equal to--I'm sorry.

  • I need a 2 here.

  • I forgot the 2, because this charge will have a

  • torque and that charge will have a torque and the two torques are

  • additive.

  • They are both going the same way.

  • Then 2q times a is p, and this is

  • E and the sinθ comes in the

  • cross product.

  • I'm assuming all of you know about the cross product.

  • Okay, final thing I did, which is, if you have a force,

  • you can associate with that force a potential energy.

  • Again, this is something you must have seen last time,

  • but I will remind you.

  • As long as it's not a frictional force,

  • you can say the force is connected to potential energy in

  • this following fashion.

  • Or the potential energy at x minus potential energy

  • at some starting point x_0 --

  • I'm sorry, x_0 - x is the integral of

  • the force from x_0 to

  • x.

  • This is the relation between the force, as delivered to the

  • potential, and the potential is the integral of the force.

  • For example, for a spring,

  • U is ½kx^(2),

  • and F = -kx.

  • If you go to this one, we tell you

  • U(x_0) - U(x) is equal to the

  • integral of -kx dx from x_0 to

  • x.

  • So that gives you kx_0^(2)/2

  • kx^(2)/2.

  • And by comparison, you can see U(x) =

  • ½kx^(2).

  • Actually, this is not the unique answer.

  • Do you know why?

  • Given this formula, can I immediately say this guy

  • corresponds to that, this one corresponds to that

  • one?

  • Is there some latitude here?

  • Yes?

  • Student: You can always add in a constant.

  • Prof: You can add a constant to both,

  • because if I said that, that certainly works.

  • If I add 92 to both, it still works,

  • because the 92 extra doesn't matter when you take the

  • difference.

  • So it's conventional to simply pick the constant so that the

  • formula looks simple.

  • Coming to other expression, if you had a torque,

  • which is pEsinθ,

  • you can ask--that's the torque--it's minus,

  • because it's trying to reduce the angle θ--

  • you can ask what U leads to that.

  • And you can see it's -pEcosθ.

  • See that, take the -U and take the derivative,

  • you'll get the torque.

  • But if you got two vectors, p and E and you

  • see the cosθ, I hope you guys know that you

  • can write it as a dot product.

  • So that's the end of what I did last time, okay?

  • The potential energy is proportional to the dot product

  • of p with E.

  • The torque is equal to the cross product of p with

  • E.

  • And what does pEcosθ mean

  • if you plot it as a function of angle?

  • It will look like this.

  • This is Π and this is 0.

  • That means it likes to sit here and if you deviate a little bit

  • and let it go there, it'll rattle back and forth,

  • just like a mass spring system.

  • In fact, you can very easily show, near the bottom of the

  • well, the potential energy is proportional to

  • θ^(2).

  • That's because cosθ can be written

  • as 1 − θ^(2)/2

  • θ^(4)/4!, etc.

  • for small angles.

  • If you just keep that term, you will find it looks like

  • this.

  • Not very different from U = kx^(2).

  • So what x does with forces, θ does with

  • rotations.

  • All right, so this is what we did last time.

  • Now I'm going to do the new stuff.

  • So new stuff is going to give you--I think it's useful,

  • because it tells you the level at which you should be able to

  • do calculations.

  • So here's a typical problem.

  • You have an infinite line of charge, of which I will show

  • that part, and somebody has sprinkled on it λ

  • coulombs per meter.

  • So it is not a discrete set of charges;

  • it's assumed to be continuous and it's everywhere.

  • I'm just showing a few of them, and if you cut out one meter,

  • you'll find there's λ coulombs there.

  • And you want to compute the electric field you will get due

  • to this distribution, everywhere.

  • So you want to go somewhere here and ask what's the electric

  • field.

  • That's what we're going to do.

  • Let's go here.

  • You will see why.

  • Now first of all, you've got to have an intuition

  • on which way the electric field will point.

  • You have a feeling?

  • Yes.

  • It will point here, this way.

  • Why not like that?

  • Yeah?

  • Student: >

  • Prof: Okay.

  • She said the horizontal parts will cancel.

  • That's correct.

  • Another argument from symmetry is that if anybody can give you

  • a reason why it should tilt to the left,

  • I can say, "Why don't you use the same argument to say it

  • will tilt to the right?"

  • Because this is an infinitely long wire and things look the

  • same if you look to the left and if you look to the right.

  • And the field you get should have the same property.

  • If this was a finite wire, I wouldn't say that,

  • because in a finite wire, that can be a tilt,

  • somewhere here.

  • But infinite wire, it cannot tilt to the left or

  • to the right, because each point has the same

  • symmetric situation to its right and to its left.

  • In a finite wire, it's not true.

  • Life to the left is different from the life to the right,

  • but for infinite wire, you know it cannot be biased

  • one way or the other.

  • It's got to go straight up.

  • Secondly, you can find the field here, here,

  • here or at anywhere at the same distance, you've got to get the

  • same answer.

  • Again, because if you move two inches to the right,

  • it doesn't make any difference with an infinite wire.

  • You've still got infinite wire on either side.

  • So we'll pick a typical point and calculate the field and we

  • know that answer is going to be good throughout that line.

  • So now I take this point here.

  • I want the field here.

  • I'm going to make that my origin.

  • Then I take a piece of wire of length dx.

  • dx is so small that I can treat it as a point.

  • Now the dx I've drawn is not a point, but in the end,

  • we're going to make dx arbitrarily small so that it's

  • good enough.

  • It's like a point and it is at a distance x from the

  • origin, and let's say that distance is a.

  • So let's find the field due to just this guy,

  • the shaded region.

  • Think of a charge.

  • How much charge is sitting here?

  • I hope you all agree, the charge sitting in this

  • region is just λ times dx.

  • That's just the definition.

  • If you've got that many coulombs, imagine a test charge

  • of 1 here.

  • Well, it will push it this way, and the field due to that,

  • I'm going to draw as infinitesimal,

  • so I'm going to call it dE.

  • You can call it E, but dE is to remind you,

  • it's a tiny field, due to a tiny section

  • dx.

  • Now that electric field is biased to the left,

  • but for every such section you find here,

  • I'll find a section on the other side that's precisely

  • biased to the right by the same amount.

  • Therefore the only part that's going to survive due to this

  • guy, combined with this,

  • will be the portion here, which is the vertical part of

  • that force.

  • So let me find the contribution first, only from this one,

  • then we will add the contribution for this one.

  • For that one, you just find the vertical

  • projection.

  • So how much is that?

  • Remember, the Coulomb's law for the electric field is

  • q/4Πε _0r^(2).

  • So this is the q.

  • That's the 4Πε _0.

  • r^(2) is that distance squared, which is x^(2)

  • a^(2).

  • That's really like the field of a point charge at that distance.

  • But now this is the magnitude of the electric field vector at

  • this angle, but I want the part along the y direction.

  • So I've got to take cosine of that θ.

  • You guys follow that?

  • If you took this vector, this part is

  • dEcosθ.

  • That part is dEsinθ.

  • But that angle and that angle are equal,

  • and cosθ for this triangle,

  • you can see, is a divided by

  • (a^(2) x^(2))^(½).

  • So this is the electric field in the y direction.

  • I'm going to call it dE,

  • in the y direction, due to the segment dx.

  • The total electric field is obtained by adding all the

  • dEs or adding all the contributions from all the

  • segments on this line.

  • And that goes from - infinity to infinity.

  • All right, so now it's a matter of just doing this integral.

  • So this gives me (λa/4

  • Πε _0)

  • dx/(x^(2) a^(2))^(3/2), integrated from - to infinity.

  • Now I can make life a little easier by saying that this

  • function is an even function of x.

  • That means when you change x to -x,

  • it doesn't care.

  • Therefore the contribution from a positive x region is

  • the same as the contribution from a negative x region.

  • That also makes sense in this picture here,

  • because if you look at the field I'm computing,

  • this section and this section give equal contributions in the

  • y direction.

  • But even if you did not know any of that background,

  • as a mathematician, if you see this integral,

  • you would say, "Hey, put a 0 there and

  • put a 2 here" namely,integrate over half the

  • region, because the second half is

  • giving you the same answer.

  • So you double the integral, but cut the region of

  • integration in half.

  • So at this point, you are free to look up a book,

  • if it was an exam, but maybe not even if it was an

  • exam at your level.

  • You should be able to do this integral.

  • So integrals have been around from the time of Newton and the

  • question of an integral is, find the area of some graph

  • with this particular functional form.

  • And the answer to any integral is that function whose

  • derivative is the integrand.

  • So what you have to do is guess many answers until you get the

  • right one.

  • But people have been guessing for hundreds of years,

  • and there's big tables of integrals with all the integrals

  • you want.

  • But you should still be able to do some integrals from scratch

  • and I'm going to tell you how to do this one.

  • But before you do the integral, you've got to have some idea

  • what the answer is going to look like.

  • I want you to get some feeling about this.

  • Answer depends on what, is the first question.

  • What's the answer going to depend on?

  • Student: a.

  • Prof: a, you understand?

  • Whatever this is depends only on a because 0 and

  • infinity are not going to be present as part of the answer.

  • If the lower limit was 5, it can depend on 5,

  • but it doesn't depend on any other thing, other than

  • a.

  • Then from dimension analysis, I got a length squared to the

  • 3/2, that's length cubed,

  • and a length on the top, so whole thing should look like

  • something over length squared.

  • The only length I have is a, so it's going to look

  • like 1 over a^(2) times a number.

  • Once you got the number, you're done.

  • So I'm going to do all the work now to show you that the number

  • is actually just 1.

  • This number will turn out to be 1, in which case,

  • you will find it's λ/2Πε

  • _0a.

  • Well, let's see how we get the number to be 1.

  • So does anybody know what trick you use to do this integral?

  • This is whatever, math 120 or-- yes?

  • Student: Use substitution?

  • Prof: Yes.

  • What substitution?

  • Student: x^(2) a^(2) = U.

  • Prof: That won't help you.

  • Yes?

  • Student: Can you use trigonometric substitution?

  • Prof: Yes.

  • Trigonometric substitution.

  • Which one?

  • Okay, look--no, no, I don't blame you.

  • I know the answer because I've seen it, but if I have to work

  • on it, I'll try for a while before I got it right.

  • The whole idea is, we don't like all these 3

  • _______ here.

  • We want to turn that into something nice.

  • So I'll tell you what the answer is.

  • You can all marvel at how wonderfully it works.

  • So what we are going to do is to introduce an angle

  • theta--nothing to do with the angle in the problem--so that x

  • = tanθ.

  • That means instead of going over all values of x,

  • I'll go with the suitable values of θ--

  • I'm sorry, this would not even be correct dimensionally.

  • x = atanθ.

  • You can see that every x that I want,

  • I can get by some choice of theta, because tanθ

  • goes from 0 to infinity when θ goes from 0 to

  • Π/2.

  • You cannot say x = acosθ,

  • for example.

  • You are doomed.

  • If x is acosθ,

  • the biggest x you can get is a,

  • whereas I want this x to go from 0 to infinity.

  • So when you make the change of variables,

  • you've got to make sure that for every x you want,

  • there is some θ that will do it.

  • Then the next thing you do, you say

  • dx/dθ = a times derivative of

  • tanθ which is sec^(2)θ.

  • Then you write that as dx = that.

  • What that means is an integral dx is related to an

  • integralby this factor.

  • Therefore going to the integration here,

  • I'm just doing that part, which is going to be a

  • sec^(2)θ dθ.

  • θ goes from 0 to Π/2.

  • Now let's look downstairs.

  • Downstairs I've got x^(2) a^(2).

  • x itself is atanθ,

  • therefore a^(2) times 1 tan^(2)θ.

  • 1 tan^(2)θ happens to be sec^(2)θ.

  • That to the power 3/2, which is what I want,

  • will give me an a^(3)sec^

  • (3)θ.

  • So what do we get?

  • You can see as promised I get a 1/a^(2) and I get integral of

  • 1/secθ, which is

  • cosθdθ from 0 to Π/2.

  • Yes.

  • And integral of cosθ is

  • sinθ from Π/2 to 0.

  • That just happens to be 1.

  • So the final answer is what I gave you here,

  • E.

  • Well, E is the vector.

  • I've just shown you the magnitude, but we've all agreed

  • what the direction is.

  • The direction is away from the wire.

  • So if you like, if you look at this wire from

  • the end, the lines will look like this.

  • If the infinite wire is coming out of the blackboard towards

  • you and you look at it this way, if you go too close,

  • you'll poke your eyes out.

  • Look from here, you'll see the lines are going

  • out radial everywhere.

  • The question is, how do the fields get weak?

  • How does it weaken with distance?

  • It weakens like 1/a.

  • That's a big of a surprise, right?

  • The field away from the wire doesn't fall like 1 over

  • distance squared, but falls like 1 over distance.

  • The reason is that every individual portion of the wire

  • has a contribution that does fall like 1 over distance

  • squared, but it is an infinite wire.

  • When you add it all up, the net answer goes like 1 over

  • the distance.

  • The field away from a wire falls like the distance from the

  • wire, on the perpendicular from the wire and there's pointing

  • away from the wire.

  • That's it.

  • Okay, so that's one calculation.

  • Then I'm going to do one more and that's going to be the end

  • of the tough calculations.

  • Second calculation is going to be an infinite sheet.

  • On the infinite sheet, the appropriate quantity is

  • called the charge density, which is coulombs per meter

  • squared.

  • That means if you cut out a tiny piece, the charge on it

  • will be sigma times the area of that piece.

  • So there is positive charge everywhere here,

  • and the number of coulombs per unit area is called sigma.

  • These are standard.

  • λ is coulombs per meter, σ is used for coulombs

  • per unit area.

  • The question is, what's going to be the electric

  • field at some point away from that plane?

  • Once again, I think we can all agree that the electric field at

  • some point from the plane will not depend on where in front of

  • the plane you are standing.

  • Are you standing here or are you standing there?

  • It doesn't matter, because it's an infinite plane.

  • If I moved 1 inch--I'll tell you why it won't matter.

  • If I moved 1 inch and the answer changed,

  • I should get the same change if I didn't move and somebody moved

  • the sheet 1 inch the other way.

  • But when I move an infinite sheet the other way by 1 inch,

  • it looks exactly the same.

  • It's got to produce exactly the same field.

  • So you can always ask, what will happen if I move to

  • the left, the same as what will happen if the sheet moves to the

  • right?

  • The sheet moving to the right looks exactly like the sheet

  • before.

  • The answer won't change, therefore the answer won't

  • change for you if you move to the left.

  • I've got infinite plane below you.

  • As long as you don't change the distance from the plane,

  • you navigate perpendicular to it, no matter where you are,

  • you will get the same answer.

  • Same answer, meaning same direction of the

  • field, same magnitude.

  • And that direction has to be perpendicular to the plane,

  • again for symmetry reasons.

  • If you tilt it in any one direction, you have no reason to

  • do it.

  • For example, if you tilt it this way,

  • I can take the infinite plane and rotate it,

  • then the tilt will be in some other direction,

  • maybe like that, but the rotated infinite plane

  • looks the same.

  • In other words, if the cause does not change,

  • the effect should not change.

  • If I can do certain things to the infinite plane that leave it

  • invariant, then I can do the same

  • transformation to the location of the point,

  • and that shouldn't have a different answer.

  • So the plane has the property that when you slide it up and

  • down parallel to itself, or twist it and turn it,

  • it looks the same, therefore the field pattern

  • should have that property.

  • Therefore the field has to be the same at all distances from

  • the plane anywhere on top of the plane, and it's going to point

  • this way.

  • But you can also find out in a minute--by the way,

  • you don't need any of the symmetry arguments.

  • You just do the calculation by brute force, it will have these

  • properties.

  • But it's good to know what to anticipate, because maybe you

  • made a mistake somewhere.

  • It's good to know some broad features.

  • So none of this is needed to calculate, even in that problem.

  • Go ahead and find the electric field not where I found it,

  • but 2 inches to the right.

  • You'll find the answer looks the same.

  • So those symmetry properties will come out of the wash,

  • but it's good for you to anticipate that,

  • and that's where you should look at the symmetry of the

  • source.

  • For example, the source was a ball of

  • charge.

  • You know if you rotate the ball, when I'm not looking to

  • rotate the ball, it's going to look the same.

  • That means the field pattern should have the property,

  • when you rotate it, it looks the same,

  • because the same cause should produce the same effect.

  • Anyway, going to this problem now, let's find the electric

  • field here.

  • Okay, now this is going to be a stretch for me to draw,

  • so I'm going to try, but you'll have to go look at

  • some textbook if you want a really nice looking picture,

  • but this is the best I can do.

  • I take a ring of radius r and thickness

  • dr.

  • I take an annulus, and I ask, what will that ring

  • do to this point?

  • So let's take a tiny part of that ring, this guy.

  • Well, for that, you just did what you did,

  • you draw the line here.

  • You'll produce a dE that looks like this.

  • What is its magnitude?

  • Magnitude is just given by Coulomb's law.

  • The q there is sigma times the tiny area,

  • dA.

  • Let's call this dA.

  • dA is the name for a small area.

  • σ times dA is the name for a small charge.

  • That charge will produce a force,

  • 1/4Πε_0, square of the distance,

  • r^(2) a^(2).

  • Finally, here is where the symmetry comes in,

  • can you see that for every section here,

  • I can find an opposite section that will cancel everything but

  • the part perpendicular to the plane?

  • So I should only keep this portion of it.

  • Namely, I should take the cosθ.

  • The cosine of that θ is the distance

  • a, just like in the other problem,

  • a/(r^(2) a^(2))^(½).

  • This is now dE.

  • If you want, you can put this following

  • symbol, dE_perp,

  • meaning perpendicular to the plane.

  • Yes?

  • Student: Do you need to multiply by 2 again,

  • because you're __________?

  • Prof: Let's be careful.

  • Her question was, should I multiply by 2,

  • because of this guy here, right?

  • In fact, I should multiply by all kinds of other numbers,

  • because so far, I've found the field only due

  • to this segment here.

  • I've got to add the field due to that and that and that and so

  • on, right?

  • What will that contribution be?

  • For every one of them, this factor,

  • (r^(2) a^(2))^(3/2) is the same.

  • They all contribute to the same factor, so when I added the

  • shaded region, I'll just get the area of the

  • shaded region.

  • All these dAs, if you add them up,

  • what will I get?

  • It will be sigma over 4Πε

  • _0.

  • Now you've got to ask yourself, what's the area of an annulus

  • of radius r and thickness dr?

  • So take that annulus, take a pair of scissors and you

  • cut it, and you stretch it out like that, it's going to look

  • like this.

  • This is dr and this is 2Πr.

  • So the area of an annulus is just 2Πr dr.

  • So the sum of all these areas is 2Πr dr and then

  • I've got here (r^(2) a^(2))^(3/2).

  • But now this is the dE, due to annulus of thickness of

  • dr.

  • Then I've got to integrate over all values of r,

  • but r goes from 0 to infinity.

  • So I have here σ/2ε_0

  • times rdr, divided by (r^(2)

  • a^(2))^(3/2), 0 to infinity.

  • Student: >

  • Prof: Did I miss a pi?

  • Student: Shouldn't it be sigma over ____ pi?

  • Prof: There is a 2 pi here.

  • Student: Oh, okay, yeah.

  • Prof: 2ε_0.

  • So do you understand what I did?

  • I broke the plane into concentric rings and I took one

  • ring and looking head on at that ring,

  • I took a portion of that ring and see what field is produces.

  • And I know that even though the field due to that is at an

  • angle, the only part that's going to

  • remain is the part perpendicular to the plane,

  • because the counterpart to this one on the other side will

  • produce a similar field with the opposite angle here that will

  • cancel, so only the part perpendicular

  • will survive.

  • Then I found out that the contribution from every

  • dA had exactly these factors.

  • They all had the same r and they all had the same

  • a, so some of all the dAs,

  • all I have to add is 2Πr dr.

  • And that's the contribution from this annulus,

  • then I still have to look at annulus of every radius,

  • so that's the integral over dr.

  • Yes?

  • Student: What happened to the a _______?

  • I thought it was a over--

  • Prof: Oh, I'm sorry.

  • It's there.

  • Thank you.

  • There is an a still here.

  • Yeah, I would have caught that guy in a while,

  • but I'm always happy when you do that.

  • That's correct.

  • Okay, so now about how this integral.

  • Do you have any idea what you might do now?

  • Yes?

  • Student: Use substitution.

  • Prof: Right.

  • What substitution?

  • "Use substitution" is a pretty safe answer,

  • but you've got to go a little beyond that.

  • Student: Substitute r^(2) a^(2) for

  • the ________.

  • Prof: Yes.

  • You can do that in this problem, because there's an r

  • on the top.

  • If you didn't have the r, you couldn't do that,

  • but now you can.

  • I'll tell you how it works.

  • First of all, you can always do that

  • tanθ substitution even here.

  • It will always work.

  • The tanθ substitution,

  • if you put it here, it will still work.

  • You can go home and verify that, but I will do it a

  • different way now.

  • I will say, let w = r^(2),

  • then dw is equal to 2r dr.

  • So if I come here, I can write it as

  • aσ/ 4ε_0.

  • I borrow a 2 top and bottom to make it dw.

  • w also goes from 0 to infinity, but now I get

  • (w a^(2))^(3/2).

  • Now this is simple integral, dx/x some number

  • to some power is x^(n 1)/(n 1), but n is now -3 over

  • 2.

  • So you get aσ/ 4ε_0,

  • divided by (w a^(2))^(-½),

  • divided by -½, which is -2 on the top,

  • and that goes from infinity to 0.

  • So I'm not going to do this much slower than this.

  • This is the kind of integral that you can see right away,

  • or you can go and work out the details.

  • This is something you should do.

  • If you have trouble with such integrals,

  • then you should work harder than people who don't have

  • trouble with such integrals, because you should be able to

  • do x^(n 1)/(n 1), and know that n 1 is -½,

  • and when it comes upstairs, it becomes -2.

  • Now if you look at this integral, in the upper limit

  • omega's infinity, you get 1 over infinity,

  • which is 0.

  • The lower limit when omega is 0, you get 1/a,

  • and that will cancel the a here,

  • and you will get σ/2ε_0.

  • So that's the final answer.

  • So the electric field of this infinite plane,

  • if you look at it from the side, looks like this.

  • The σ/2ε_0.

  • So what do you notice about this one that's interesting?

  • Student: It doesn't depend on the distance.

  • Prof: It doesn't fall with distance.

  • No matter how far you go from this infinite plane,

  • the field is the same.

  • Again, each part of it makes a contribution that falls like 1

  • over distance squared.

  • As you go further and further out, you might think the field

  • should get weaker, right?

  • How could it not get weaker?

  • They're moving away from everything.

  • At least with the line charge, it didn't go weaker like

  • 1/r^(2), but it did get weaker.

  • How can you go further and further from a plane?

  • You are going further away from everybody?

  • How could it not matter?

  • Yes, any ideas?

  • For example, if you go very close to the

  • plane, what happens?

  • If you go really close to the plane what happens is,

  • the field in each section here looks like this.

  • Therefore the part that's useful is very small.

  • If you go further away, you get things like that.

  • Maybe it's a little weaker, but the part that's useful,

  • this one, is getting bigger.

  • So by magic, these tendencies cancel in the

  • end.

  • It doesn't depend on distance.

  • Now unless you do the integral, you will not know it doesn't

  • depend on the distance, because you can give arguments

  • for why it'll get weaker, arguments for why it'll get

  • stronger.

  • The fact that it'll precisely be independent of distance,

  • you have to do the calculation.

  • Yes.

  • Student: What's the negative sign?

  • Prof: Negative sign where, here?

  • Student: Yes.

  • Prof: -2 is there, but the upper limit is

  • infinity.

  • Student: Oh, okay.

  • Prof: All right.

  • Now here's the third problem, and the good news is,

  • I'm not going to solve it for you, but I'll tell you what it

  • is.

  • Here is a solid ball of charge.

  • It's got some charge density ρ coulombs per meter cubed.

  • So ρ is the standard name.

  • You use density for mass over volume and you use the same

  • symbol rho for charge per unit volume.

  • So somebody's assembled a blob of electrical charge,

  • and 1 cubic meter of that has ρ coulombs.

  • You want to find the field due to this one.

  • Now when you do a similar problem in gravitation,

  • it is generally assumed that when you're outside the sphere,

  • the whole sphere acts like a point charge with the entire

  • charge sitting at the center.

  • But you actually have to prove that.

  • That's what took Newton a long time to prove.

  • He knew it was true but he couldn't prove it,

  • because for that, you've got to be able to do

  • integral calculus.

  • And even today, to find the field due to a

  • sphere using integration is quite difficult.

  • Think about what you have to do.

  • You want to sit somewhere here.

  • First of all, for a sphere,

  • we know the field is going to be radial.

  • It doesn't matter where you pick, everything looks the same.

  • You can decide to be horizontally here at that point.

  • Then you've got to divide the sphere into tiny pieces,

  • tiny little cubes, each with some charge rho times

  • the volume of the cube, and that will exert a force

  • like this.

  • And you've got to integrate over the volume of the sphere,

  • but each portion is at a different distance and a

  • different angle.

  • You've got to add it all up.

  • That's why it's a tough problem.

  • So to solve that tough problem, we're now going to use a very

  • powerful trick and that trick is called Gauss's law.

  • So we're going to learn today about the Gauss's law.

  • Now a prelude to that, you need a little more

  • mathematical definitions, but they're not bad.

  • I just have to tell you what the definition is.

  • Suppose I have in three dimensions a tiny little area,

  • like a snowflake, but it's flat and it's

  • rectangular, let's say.

  • I want to tell you everything about it.

  • I want you to be able to visualize the area.

  • What can I do to specify this little thing?

  • First I have to tell you how big it is.

  • If it's a tiny area, let this area be dA

  • meters squared, but that doesn't tell me the

  • orientation of this area, because that area could be like

  • this, it could be like this,

  • it could be tilted in many ways.

  • So I want to tell you it's an area in a certain plane,

  • what should I do?

  • How do I nail down the plane in which the area is located?

  • Yeah?

  • Student: Define the vector that's perpendicular to

  • that surface.

  • Prof: Define a vector normal to that surface,

  • because if you draw that vector, then there's only one

  • plane perpendicular to it.

  • Then we can follow, we can then form a vector,

  • dA.

  • It's a tiny vector whose direction is perpendicular to

  • its area and whose magnitude is the value of the area itself.

  • So areas can be associated with vectors.

  • You may not have thought about it that way, but you can by this

  • process.

  • I've told you, there's only one ambiguity even

  • now.

  • Do you know what that one-- yes?

  • Student: Which direction.

  • Prof: There are two normal's you can draw to an

  • area, right?

  • We've got an area like this, it can come out towards you or

  • go away from you.

  • Therefore simply drawing that rectangular patch is not enough

  • to nail that.

  • That's like saying, "Here is a vector."

  • That's not enough.

  • Where is the head and where is the tail?

  • That's not a vector.

  • That is a vector.

  • Similarly, this area has to be specified some more and here is

  • what you're supposed to do.

  • You take that area here, I'm just drawing it another

  • place, draw some arrows, then circulate around it in one

  • sense or another.

  • I picked a particular sense in which they're going around.

  • Then use the famous right hand rule,

  • where your fingers curl along the arrow and your thumb points

  • in some direction, that is the direction,

  • the area of the vector.

  • If the arrows are running round the opposite way,

  • then your thumb will point into the blackboard.

  • So an area like this is like a vector without a head.

  • Area like this is a signed area.

  • It's an area that's got a magnitude and unique direction.

  • So get used to the notion that a tiny planar area can be

  • represented as a vector.

  • Another way to see that is, if you took any area,

  • a rectangular area like this--square is a special case--

  • if you took two vectors A and B that form

  • the two edges, then A x B is

  • just double the area.

  • It's the fact that given two vectors,

  • I can find the third vector perpendicular to them,

  • up to a sign, is what makes a cross product

  • possible, only in three dimensions.

  • You cannot have a cross product of two vectors and four

  • dimensions because in four dimensions,

  • if I pick two vectors, they'll be two other directions

  • perpendicular to that plane.

  • Only in 3D, there's only one direction left.

  • The question is, is it in or out?

  • That you pick a sign in the cross product,

  • A x B is something that goes from A to B.

  • Or for an area, you draw arrows around the edge

  • in a certain direction.

  • So area is a vector.

  • You have to get used to that notion, along with all the other

  • vectors you know.

  • Now I'll tell you why that becomes useful.

  • So we're going to take--let's see--

  • there is a rectangular tube which has got a height h

  • and a width w, and some fluid is flowing along

  • the tube with the velocity v along the length of the

  • tube.

  • You got that?

  • It's like an air duct.

  • Stuff is going through that tube.

  • It's got a rectangular cross section.

  • The cross section area is hw.

  • If the fluid is going velocity V along the length of the

  • tube, what is the flow rate,

  • which is equal to meter cubed of stuff flowing per second.

  • I'm going to denote it by the symbol Φ.

  • If I wait 1 second and I watch all the fluid go by me,

  • past any cross section, how much stuff goes by?

  • I think you can all see, if I wait 1 second,

  • the fluid whose front was here would have advanced to here,

  • and the volume here will be v times 1,

  • because in 1 second, it goes a distance v.

  • So the flow rate will be hwv or area times

  • v.

  • That makes sense?

  • The faster it's going, the more stuff you get.

  • The bigger the cross section, the more stuff you get.

  • This doesn't depend on a rectangular cross section.

  • It can be cylindrical pipe carrying oil.

  • Again, the flow rate is area times velocity.

  • But I'm going to write this in terms of vectors,

  • because I know the velocity is a vector.

  • But now I have also learned area is a vector,

  • because area vector here, you can draw a vector

  • perpendicular to this area, and I'll draw up this

  • convention that it's area vectors along this way,

  • rather than the other way.

  • Then I can write this v⋅A.

  • Let's check that v⋅A is

  • correct.

  • v⋅A is the length of v,

  • the size of v, the size of A and cosine

  • of the angle between them.

  • Here we've got to be very careful.

  • The velocity vector is like this.

  • It's perpendicular to the plane but don't say it's cosine of 90

  • degrees.

  • The area vector is perpendicular to the area

  • itself.

  • Do you understand that?

  • When I take the dot product and I will ask for the angle between

  • the area vector and the velocity vector, that angle is 0.

  • For the area, there's a little confusion.

  • The vector representing it happens to be perpendicular to

  • the plane of the area itself.

  • So if you remember that, then

  • v⋅Ais vA times cos 0,

  • so this indeed is one way to write the flow.

  • This flow is also called a flux.

  • But now let us do the following.

  • Let's take the same problem, and I have this area here.

  • Let me now take a tilted area like this.

  • It also goes from the ceiling to the floor but still turned at

  • an angle θ.

  • So it's a bigger area than the original one.

  • How much bigger?

  • That area prime, I claim, is equal to the base

  • w, times this side.

  • This side is h/cosθ.

  • θ is the angle between these two planes.

  • It's a bigger area, but you all know that just

  • because it's a bigger area, it doesn't intercept more fluid

  • per second.

  • Any stuff crossing this guy also crosses this at the same

  • rate.

  • So how am I going to get the same rate?

  • The flux is not going to be v times A'.

  • It's going to be vA'cosθ.

  • But θ is the angle between the area vector and the

  • velocity vector, which is the same angle here.

  • So the moral of the story is, v⋅A',

  • or v⋅A in general is the flux or the flow,

  • off of any vector across an area.

  • If it's a fluid that's flowing, then v⋅A is

  • the fluid flow past that area.

  • If you need the dot product, you need the cosine of the

  • angle, because the area, if it's not perpendicular to

  • the flow, it's not useful.

  • In fact, you can take a huge area parallel to the flow and

  • nothing goes through it.

  • So area is most effective if the plane of the area is normal

  • to the fluid, or the area vector is parallel

  • to the velocity of the fluid.

  • So that's that lesson.

  • Okay, this had nothing to do with the electric field,

  • but we're going to come to the electric field now.

  • This is just a warm up.

  • Let's come to the electric field and see what's going on.

  • So I take a charge q and I draw the lines coming out of

  • it.

  • How many lines do I get crossing a sphere S?

  • Well, we know that we have agreed to draw

  • 1/ε_0 lines per coulomb,

  • so this q here, that many lines cross the

  • sphere.

  • I'm now going to relate it to something I can do with the

  • electric field as follows.

  • I'm going to say that if I go to that sphere,

  • I look at the electric field.

  • Electric field is in that direction, E.

  • And any portion of the sphere--and I want this is where

  • you got to __________ imagine--take a tiny part of the

  • sphere.

  • There's a tiny area that's got a size dA.

  • What is its direction?

  • Direction of the area vector is radial.

  • You understand?

  • The area is on the surface and normal to that is the radial

  • vector, which I always denote by e_r.

  • The electric field is equal to q/4Πε

  • _0(1/ r^(2))e

  • _r.

  • Therefore E⋅dA =

  • q/4Πε _0(1/

  • r^(2))dAe _r

  • ⋅e_r.

  • This is a dot product of the area vector with the electric

  • field vector.

  • The two unit vectors are parallel.

  • The dot product of that guy with itself is just 1.

  • So this is the number of lines crossing the tiny area,

  • because electric field numerically is equal to the line

  • density.

  • So those lines crossing this area, this is the number of

  • lines crossing that patch, which is given by

  • E⋅dA.

  • If you add up all the lines, you must add up all the

  • dAs.

  • Sum of all the dAs on a sphere of radius r is

  • just q/4Πε _0r^(2) times sum

  • of all the dAs on a sphere,

  • that is just 4Πr^(2) .

  • So you get q/ε

  • _0.

  • In other words, either you can draw the picture

  • and it's immediately obvious to you the lines crossing is

  • q/ε _0 by

  • construction, or you can remember, "Hey,

  • the electric field is a direct measure of the number of lines

  • per unit area," and the electric field times

  • area times the cosine of the angle between the area and the

  • electric field will count the lines going through a tiny area.

  • If I add them all up, I'd better get

  • q/ε _0,

  • and indeed you do.

  • So the moral of this little exercise is that

  • the surface integral, let's call it the surface

  • integral of the electric field, on a surface is equal to

  • q/ε _0 where this was

  • a sphere.

  • In other words, even if you've never heard of

  • field lines, just take the electric field and do the

  • surface integral, you get this.

  • So surface integral is a new concept.

  • You probably have not done that before and I've got to remind

  • you how it is done at least operationally.

  • If you've got a computer, you can find the surface

  • integral of anything as follows.

  • Take the surface over which you're doing an integration.

  • Divide it into tiny pieces, each is a little area

  • dA.

  • Take the dot product of the dA with the electric

  • field there, and sum over all the patches covering the sphere.

  • Then take the limit in which every patch gets vanishingly

  • small.

  • That is called the surface integral of the electric field

  • and we see that if you do that, you get the charge inside

  • divided by ε_0.

  • But that was on a sphere, and the interesting thing was,

  • the answer was independent of the radius of the sphere,

  • because the 1/r^(2) in the field canceled the

  • r^(2) in the area.

  • But it is even better than that, because it's clear to you

  • know, without any calculation,

  • that if I took some crazy surface like this,

  • the lines crossing that is also the same.

  • Because the lines leave the origin, they go radially

  • outwards.

  • They don't terminate on anything.

  • They're only supposed to terminate on other charges,

  • therefore you can count them anywhere you like.

  • You can take a census here, you can take a census there or

  • there.

  • You're always going to get the same number of lines.

  • Therefore q/ε_0

  • is also going to be equal to the line count on some

  • weird surface enclosing the charge.

  • So how am I going to count the lines on a weird surface?

  • I take the surface, divide it into little pieces,

  • but now the area vector and the electric field vector may not be

  • parallel, because it's not a sphere.

  • But it is still true that if you're trying to count the lines

  • going through, just like in the velocity,

  • you must take the dot product of these.

  • Therefore you will again find E⋅dA

  • for any surface, any closed surface,

  • is equal to q/ε_0

  • .

  • If you got lost in the mathematics, the physical

  • picture is very clear.

  • If you want to count how many lines leave the charge,

  • you can surround it with any surface you like.

  • If it's a sphere, it's very easy for you to do

  • the check.

  • If it's a crazy surface, it's harder but they all come

  • from the fact that E⋅dA,

  • count the number of lines and that's independent of the

  • surface at stake.

  • So now that it what is called Gauss's law, so I'm going to

  • write it down here.

  • What I've shown you is the following.

  • Here is some strange closed surface that's a charge q

  • inside.

  • Then E⋅dA on that surface is equal to

  • q_inside

  • _0.

  • This is not yet the theorem, but this is the case for 1

  • charge.

  • If the charge were outside, suppose the charge were here,

  • then the lines would go like this, and the total number

  • leaving the surface would actually be 0.

  • You might say, "How is that?

  • I see all these lines penetrating the surface,"

  • but you've got to remember, if you take an area vector

  • here, with the definition for a closed surface,

  • the area vector is the outward pointing normal.

  • For a closed surface, every area is a vector pointing

  • out from the closed surface.

  • Then you can see that in this surface, lines are coming out;

  • on this surface, lines are going in.

  • It's very clear what's going in is coming out,

  • because nothing is terminating.

  • Therefore if you took a surface that did not enclose the charge,

  • this answer would be 0, but if it enclosed the charge,

  • it will be the charge inside, but it won't matter where it

  • is.

  • You can also see that you can move this charge around anywhere

  • you like, you don't change the number of lines coming out.

  • Then the most important generalization is,

  • if instead of 1 charge, I have 2 charges,

  • what do you do?

  • If you had 2 charges, let us take the total

  • E⋅dA on a surface.

  • The total electric field is the electric field due to the first

  • charge on that surface, the electric field due to the

  • second charge on the surface, thanks to superposition.

  • It's the fact that the electric field is additive over charges.

  • But this one is q_1

  • /ε _0.

  • This is q_2 /ε

  • _0.

  • Therefore, we can now write, generalizing to any number,

  • whether one draws a double integral with this thing,

  • meaning it's a closed surface.

  • You can also have an open surface, take the skin off an

  • orange and cut it in half.

  • That hemispherical skin is also a surface, but to signify it's a

  • closed surface, we do it like this.

  • On a closed surface of E⋅dA =

  • q_enclosed

  • _0, which means sum of all the

  • q's.

  • And sometimes we write this as follows.

  • Suppose inside you don't have a discrete set of charges that you

  • can count, but a continuous blob of

  • charge, and the charge has a certain density,

  • ρ is the charge density.

  • What is the q_enclosed?

  • I claim the answer is due to a volume integral of ρ

  • or any point r inside, times dxdydz inside that

  • volume.

  • You know, if you want to say how many coulombs are there,

  • well divide the volume into tiny cubes of size

  • dxdydz.

  • That times the ρ of r,

  • meaning ρ at x,

  • y and z, is the charge inside the cube.

  • You add it all up, you get the charge enclosed in

  • that funny shaped object.

  • So this is the final form of Gauss's law that we're going to

  • use.

  • I'm going to use the symbol, which is very useful.

  • This is the theorem of Gauss.

  • And S = DV.

  • DV is the boundary of V.

  • In other words, V is like a potato,

  • the skin of the potato is DV.

  • Therefore if the potato is full of charge,

  • it will emit some electric flux and the surface integral

  • electric flux over the skin of the potato is the charge

  • enclosed inside it.

  • The charge enclosed inside it is not simply some constant

  • density times volume, if the charge density varies

  • from point to point.

  • In each neighborhood inside their volume,

  • you take a tiny section, a tiny little cube of size

  • dxdydz, see how much is in there,

  • as the function of x, y and z,

  • and you do the integral over the volume.

  • Now these integrals may be hard to do, but you should know at

  • least what I'm saying.

  • This is just a way of counting the total charge inside a volume

  • when the charge is continuously distributed.

  • So I'm going to show you one application of this and we'll

  • come back to more next time.

  • And the one application I'm going to show you is to find the

  • electric field due to a solid ball, solid ball of charge.

  • So I'm going to use Gauss's theorem to do that.

  • This is true for any surface you pick.

  • It doesn't matter what surface you pick.

  • So if I want to find the electric field here,

  • I'm going to pick a surface of radius little r.

  • Let the sphere have a radius big R.

  • I want to find the field here and I'm going to use Gauss's

  • law.

  • So on the left hand side and the right hand side are two

  • different things.

  • On the right hand side, what is the charge enclosed?

  • That region.

  • Well, the charge enclosed is just some number q.

  • This whole thing is q, q spread over a sphere

  • of radius r, over

  • ε_0.

  • That's going to be the surface integral of the electric field

  • on that sphere.

  • E⋅dA on that sphere.

  • Now normally, if you knew the value of an

  • integral over a region, you cannot deduce anything but

  • the integrand, unless what?

  • There's one exception where if you know the integral of a

  • function, you can find the integrand.

  • You know when that might be?

  • Here's a function.

  • I tell you its integral from here to here,

  • but I don't show you the picture.

  • What's the integrand?

  • You don't know.

  • But if I also tell you the function is a constant,

  • and I tell you the integral, and you know the width of this

  • region, you can find the integrand.

  • So you're going to cook this up so that this whole integral is a

  • constant times the area of integration.

  • And we argue that if you've got a spherical charge density,

  • the electric field must be radial everywhere.

  • You've not proven this, but you argue that because it's

  • the only distribution with lines coming out everywhere,

  • invariant under every possible rotation of the sphere.

  • Because if you rotate the sphere, I won't know you rotated

  • it, so the field pattern cannot look different.

  • If the field pattern looked different when you rotated it,

  • then you have a problem because the cause looks the same,

  • but the effect looks different.

  • That's not allowed.

  • The only allowed this is the radially outgoing electric

  • field.

  • Therefore the electric field is radial and also constant

  • throughout the sphere, because all points on the

  • sphere are equivalent.

  • There's no reason why this is any better than that.

  • The sphere looks the same for all directions.

  • Therefore this whole integral is going to be 4Πr^(2)

  • times the electric field at that radius r,

  • because the area vector and the electric field vector are both

  • parallel, so in this dot product,

  • forget the dot product.

  • It's just E times dA.

  • There's no cosθ, just 1.

  • You can pull the E out of the integral,

  • because E is constant in magnitude on the sphere.

  • And the integral of the dA is just

  • 4Πr^(2).

  • Therefore you deduce E(r) is

  • q/4Πε _0r^(2).

  • It's a very profound result.

  • It looks very simple, but it is true for not a point

  • charge, but for the spherical distribution of charge.

  • That the field goes like that of a point charge sitting at the

  • origin, is a consequence of Gauss's law.

  • If the charge inside was not uniform,

  • suppose it's a charge q, but there's more stuff here,

  • less stuff here, this theorem would still be

  • true up to this point, but you can never deduce that

  • E is a constant on a sphere,

  • because even though a sphere is nice and symmetrical,

  • the charge distribution is not.

  • It could be big here, it could be small here.

  • You know something about the integral over a surface of a

  • varying function, then Gauss's law is no good,

  • not useful.

  • True, but not useful.

  • Gauss's law is useful only when in a given problem,

  • there's only one number you don't know.

  • That number here happens to be, what's the strength of E

  • at a radius r?

  • I know it's direction is radial, I know it's magnitude is

  • constant on the sphere by symmetry, but what is the

  • number?

  • You can trade that one number for this one number on the left

  • hand side, q/ε_0,

  • you can calculate it.

  • I'll come back and do more examples for you guys next time.

Prof: All right, class, I thought I'd start as

字幕與單字

單字即點即查 點擊單字可以查詢單字解釋

B1 中級

3.高斯定律一 (3. Gauss's Law I)

  • 90 7
    Cheng-Hong Liu 發佈於 2021 年 01 月 14 日
影片單字