You see here the topics the way I see them, you will get three problems to --on the exam, and not all subjects can, of course, be represented on the exam. Nor can I cover all of them in fifty minutes. I will test some very basic ideas, the math will be utterly trivial, and if it becomes complicated, then you just know that you're on the wrong track.
If you get stuck, somehow, on a problem, my advice is, move on, don't stay with the problem, but move on and try some others first. There is a reason why Gauss' Law there is in red, because Gauss' Law is, of course, extremely important in the early part of the course, the closed surface integral of E dot D A is the sum of the enclosed charged, divided by epsilon zero.
And that is so important that you can be sure that there will be one problem dealing with Gauss' Law.
Now, when you have Gauss' Law problems, there's always one of three. You must have symmetry, you must have a special distribution of charges, because otherwise, Gauss' Law doesn't get you anywhere.
So we have spherical symmetry, we have cylindrical symmetry, and we have plain symmetry, and that's all there is. So you're going to get one of those three.
I will do one now, you may choose.
We're going to have a vote. One is a possibility, I do one on spherical symmetry, another one I do on cylindrical symmetry, or I do one on [pause] slab symmetry.
Who wants the spherical symmetry? Hands. Who wants the cylindrical symmetry? Way more hands.
Who wants plain? I think the cylinders have it.
But if you're clever, you can stay for the next lecture, and then you can try to get the other one.
We need a little bit of fun today.
And therefore, I want to introduce you first to something very special, which is close to my heart, it is a secret top, you're going to see it there, and that secret top, I'm going to spin, and if you're a believer in eight oh one, which you should by now, then we will --should be able to predict that that stop cannot spin for very long, there is friction with the air and friction with the surface, and so chances are it will soon fall over. We'll take a look at it later, again. So let's now start our first problem, and that is a cylindrical symmetry. Well, we have a cylinder --and here is the cylinder --
it's very long, has radius R, and I have uniform charge distribution throughout the whole cylinder, and the density is rho Coulombs per cubic meter.
Uniformly distributed through the cylinder. I want to know what the electric field inside the cylinder is and outside the cylinder. Let's first do outside the cylinder.
The gauss surface, clearly, is going to be itself a cylinder, there it goes --you can give it any random length, L, cannot have any effect on the answer --and so the end if flat, perpendicular to the axis of symmetry, and this front part is flat, and this is curved.
I give this a radius little R, and so I know that everywhere on the surface of that cylinder outside, that the electric field must be the same everywhere because the distance is the same, that's the symmetry argument.
Electric field cannot be any stronger here than it is there, if I'm on that surface.
Symmetry argument number one.
Symmetry argument number two is, given the fact that this is a cylinder, the electric field must everywhere be perpendicular to this axis, coming out --I call it radially, but, of course, it is not radially, like a sphere --it's radially coming out of this surface, always perpendicular to this axis of symmetry. Nature could not decide any other way.
That's the second symmetry argument.
One you recognize that argument, the electric flux through this flat surface and through that flat surface must be zero.
Because then, the electric field and the local D A vector, which is the perpendicular to the surface, make angles of ninety degrees with each other, because E would be like this here, but D A is in the direction of the axis of symmetry.
So no flux can, therefore, get out here and get out here.
But only through this curved surface.
But on this curved surface, if it is a positive charge, then the E vector and the D A are in the same direction, if it is a negative charge, they are in opposite directions.
Later, you can change the sign of rho, let's just make it positive for now. So if, now, I apply Gauss' Law, then I only have to take this surface into account and not these two end pieces.
And so I need to know, now, what this surface is, because E and D A are always in the same direction everywhere, thus the cosine of the angle between them is plus one.
And so what is the surface area? That is going to be L times two pi little R, and then the electric vector is everywhere, the same there, this was our symmetry argument, and that is now the charge inside this cylinder, divided by epsilon zero.
But, of course, the charge inside the cylinder, that's only the portion that is in this inner cylinder, and so that has also, then, length L. The cross-section here is pi r squared, so this is the volume of the volume of the charge that I have inside by Gaussian surface, I must multiply by rho, that gives in the charge, and I divide by epsilon zero.
And of course, the L cancels, as it always does, and the pi cancels here, too, and so I find that the electric field equals r squared times rho divided by two epsilon zero r, and if you want to see it vectorially, you can put an r roof there, r roof, then, would be a vector which is perpendicular to the axis --I mentioned earlier, I called that radially outwards.
So this is the electric field outside the cylinder.
R squared rho two epsilon zero r.
So it falls off as one over r. So now I want to know what it is inside the cylinder.
So now I go to r, less than equal to r.
So it's clear that what I do now, I'm going to have a Gaussian surface which, again, is a cylinder, has length L, and it has, again, two flat pieces at the end, so no flux will go through those two pieces, so my first term of Gauss' Law is going to be the same, I have L times two pi little R, because the radius of this inner circle is also r, L two pi r times the electric field, the arguments are identical --but now, there is less charge inside by Gaussian surface. Uh, the volume is L, now times pi little R squared, and then I get rho to convert it to charge, divided by epsilon zero.
I lose my L, as I always do, my pi goes, and so now I get E equals --there is an r here, and there is an R squared here, an so I only end up with one R, divide by two epsilon zero, and if you like that vector notation, you can always do this.
And of course, if rho were negative, then automatically, you see, if you put a negative charge density in here, then the E field flips over, so that's automatically taken into account both here and there.
So let's take a look at it, I'm quite happy with that.
If you substitute little R equals capital R, you are right at the surface of your cylinder, then you get the same answer in both cases.
Substitute R, capital R in here, then the magnitude of E --don't worry about the direction now --is rho capital R divided by two epsilon zero, and if you put in here for this little R, capital R, you find exactly the same answer.
So we can now make a plot of the electric field as a function of distance R, here being capital R, and here being the electric field strength. It's a linear line, zero here, it goes up to a certain maximum, and then it falls off as one over R.
And this value here is this value.
That's where little R is capital R.
It is obvious and pleasing that the electric field, on the axis itself, where little r is zero, that that electric field is zero, that is something that you could have predicted almost without any knowledge, because you have symmetry all around it, there is charge on the left, there is charge on the --one the right, there's charge on north and south, and the electric fields right at the center, of course, all pair each other [unintelligible], so you get no electric field right at the center.
If the charge, for some reason, would all be at the outer surface, if it were a solid conductor that would be the case, then the electric field would be zero everywhere inside, and this would be unchanged, assuming then, that you have the same amount of charge on the outside per unit length as you now have on the inside.
So that is cylindrical symmetry.
I am dying to take a look at my top.
I am really [pause] --and much to my shock, do I see that it stopped? It's still rotating.
So maybe I have to come to the conclusion that there is something wrong with eight oh one.
There must be a layer deeper than eight oh one --there is friction, and yet, this top doesn't come to a halt. And so that layer deeper --maybe that layer deeper is eight oh two.
Give that some thought, it may add to your sleepless nights.
We'll visit it later, because maybe it will come to a stop.
Very well. Let's now do something very different.
I have two conducting flat plates for the plane plate capacitor, and it has a certain thickness, the material is a conducting material, has a certain thickness, it's very large in size. Way bigger is the plane linearly than the separation.
And let this separation be little d, and I charge the upper plate with a positive charge so I get here surface charge density plus sigma, and here minus sigma, area is A, of both plates.
I know that in the conductor itself, there is no current, and therefore, in the conductor itself, the electric field must be zero. I have an electric field between the two plates, which can be derived using Gauss' Law, but we have already used up our time on Gauss' Law, which is sigma divided by epsilon zero. And the electric outside these two plates is very close to zero here, and also very close to zero here.
We find that from the superposition principle --because this plate, which is negatively charged, will add to the electric field pointing down, and as you perhaps remember, that that is independent of distance.
Well, provided that you are not so far away from the plates that the dimension of the plates is going to interfere.
The --if the plate is ten by ten meters, then it's fine as long as you are, say, within a few meters.
But it you go a hundred meters away, then it's not true anymore.
So if we assume that we don't go too far out, then the electric field due to this one pointing down, due to this one is pointing up, and they have equal strength, they are independent of distance, so they cancel each other here, and they cancel each other there, superposition argument. Let this point be P inside the conductor, and this point be S, and the first thing I would ask you, for instance, is what is V P minus V S? It's the potential difference over this capacitor, if you want to call it a capacitor.
That is the integral, in going from P to S, of E dot D L.
For reasons that I have never understood, your book will switch these around, and put here a minus sign, which is, of course, exactly the same thing.
And so now, we can calculate the potential difference.
If I am here, and I'm going to walk down --suppose I walk down [unintelligible] a straight line, do D L is in the same direction as E, then it's immediately obvious that this is simply E times that distance D, because E and D L are in the same direction.
So the cosine of the angle between them is plus one.
And so I find that this is E, which is, um, sigma divided by epsilon zero, times that distance D.
If I had chosen another route, I would have found the same answer, because we're dealing here with conservative fields, so the path does not matter.
So as long as you go from this plate to this plate, that integral is always E times D.
What is the potential difference between point P and point T here? V P minus V T, it's clear that that is zero, because there is no electric field here, and there is no electric field anywhere there, so the integral, obviously, is zero.
What is the capacitance of this plate? The capacitance is the charge on one plate, divided by the potential difference, which I just use the word V now, it means it is this value V P minus V S. So what is the charge on one of the plates? It doesn't matter which one you take, that is sigma times A.
That's the definition of sigma, right? It's charge per unit area.
So that's the charge on the plate, and the potential difference we just calculated, that is sigma D divided by epsilon zero, so the epsilon zero comes upstairs, and so we find, now, that it is A times epsilon zero, divided by D.
Notice it's independent of sigma, of course. Capacitance is geometry, it has nothing to do with how much charge you have on the capacitor.
I could ask you what is the electrostatic potential energy.
The electrostatic potential energy is the work that you would have to do to assemble the positive charges here and the negative charges there.
You could also look at it at the energy that it takes to create that electric field.
Same question. So this work that has to be done to assemble it is the charge on one plate times the potential difference times one-half, or --which is the same --one-half C V squared.
So, what is one-half Q V? Let's first take this one, there is one-half, Q is sigma times A, potential difference V, we have, is here, sigma D divided by epsilon zero, so this is the answer that we will find. From this one, and the answer that we find from the other one must, of course, by the same.
Let's check that. This is C, [unintelligible] is A epsilon zero divided by D, and now I must multiply by the potential difference squared, so I get a sigma squared, I get a D squared, and I get an epsilon zero squared, and these two better be the same. I have sigma sigma here, sigma squared, I have D squared divided by D, so I have only one D, I have an epsilon z-zero squared here and here, and epsilon, so I have only one over epsilon zero, so they are, indeed, the same.
Now I could ask you where is the charge located? Let's first go to the top plate, where is the charge located on the upper plate? Some of you may say, "Oh, well, maybe the charge is in the plate, somewhere here.
That cannot be.
I make a Gaussian surface all around that charge.
Gauss' Law will tell me, then, that the surface, closed surface integral of E dot D A is not zero, because there is charge inside, and if there is charge inside, that closed surface integral is not zero --but we know that the electric field must be zero everywhere, zero in the [pause] conductor, everywhere it must be zero, so the closed surface integral must be zero.
So there cannot be any charge there.
Simple argument.
Some of you may say, "OK, maybe some of the charge here is at the top surface.
Not allowed either, in this configuration.
I make myself a small pillbox, which is my Gauss surface, these are flat ends.
Electric field is zero here, electric field is zero there, sot he surface, closed surface integral must be zero because the electric field is zero everywhere, but there is charge inside the pillbox, and so Gauss' Law says that it cannot be zero.
Since it is zero, there's no charge inside.
And so there's only one solution, nature puts all the positive charge right here at the bottom of this plate, and the negative charge right there at the top of that plate.
That's the only solution in this case. Charge cannot be anywhere else.
I can't believe it.
That thing is still running.
I want you to take a look at that.
You see that top is still happily running, so either this has to be a violation of the
conservation of energy somehow, or it is black magic --it can never be excluded in
twenty six one hundred --or, perhaps, there is some simple physics behind it.
And whatever that simple physics is, I would like you to start thinking about.
All right.
Next subject.
Again, let me come with a --with two plates, because I want to start talking about
dielectrics, and I want to [pause] massage this idea of capacitance a little further.
I have here a parallel plate capacitor.
And I put on here, positive charge, plus sigma, I call it now, sigma free, there is no
dielectric yet, but, later, there will be, so I call it sigma free, and here is minus sigma free, separation is D, surface area is A. So in the beginning, it's going to be boring, we know that the electric field here is
going in this direction, and that electric field is sigma free divided by epsilon zero.
The free goes with the sigma.
I [pause] charge it up using a power supply, and now --this is crucial --I disconnect
the power supply.
I take the leads off.
So I disconnect the power supply.
That means --and this is crucial --that whatever comes that this charge is trapped,
can never change.
No matter what we're going to do.
Power supply has been disconnected, that sigma free is trapped. I'm going to do
various things now. I'm going to change the distance between the plates, and then independently, I'm going to shove in a dielectric, we will do that separately, one at a time. And so the equations that I can now trust, and that I will be looking at, are the following. The --the free charge that I have, Q free, is obviously sigma free times the area. And that is the definition of surface charge density.
So I can trust that one.
The electric field between the plates is sigma free divided by epsilon zero, and now I get a kappa there, if we have a dielectric. The potential difference between the plates is E D.
Provided I know the E, this is the E, it's always E D.
We just had that in our previous problem. The capacitance itself, C, is the free charge on one plate divided by the potential difference between the plates, V is my potential difference, that is this V, and the electrostatic potential energy equals one-half Q free times V, but it is also one-half C V squared.
And so keep these in mind in what follows. Take a look at them, the first one is correct, second one is correct, third one is correct, that one is correct, that one I can also live with. Or you could write down, for the capacitance, if you wanted that, you can write down A times epsilon zero divided by D, times kappa.
The first thing I'm going to do with the power supply disconnected, I'm going to increase the distance D between the plates.
And I'm going to increase them by a distance --I'm going to double the distance.
So D goes up by a factor of two.
But kappa remains one.
Just air. No dielectric yet.
What happened with the electric field? E.
E can not change, because sigma free cannot change, kappa is one --there's no kappa --and if this cannot change, this cannot change. So, as I move the plates apart, there is no change in the electric field. Non-intuitive as that may be for you, the electric field remains a constant. So what happens now with the potential difference between the plates? That, now, must increase by a factor of two, because if I increase D by a factor of two, and if E is doing nothing, then V must go up by a factor of two.
So V must go up by a factor of two.
And I did a demonstration here, during one of my lectures, whereby I changed D from one millimeter to ten millimeters, and I changed the potential difference from thousand volts to ten thousand volts, you've seen it in front of your own eyes, if you were here.
So, indeed, when you separate the plates with the power supply disconnected, the potential difference goes up.
What happens with the capacitance? Well, the capacitance is Q free divided by V.
There's an R missing here.
This one doesn't change. This one goes up by a factor of two, so C must go down by a factor of two.
What happens with the electrostatic potential energy? Well, it is one-half Q free times V, Q free cannot change, V went up by a factor of two, so U must go up by a factor of two.
Remember that when I separated these plates and increased the potential difference, I told you I was doing work. U is increasing.
If I move the plates apart, I have to do that work. All right.
So this is the first part, whereby we change D.
Now, I go back to D, leave it as it was, and now I want to change kappa.
I'm going to move in a dielectric.
I'd like to stay working on the center board [pause] I have to change, have to --can't see this any more.
So now D is as it was before, but now kappa becomes three.
So I take dielectric and I shove it in, and sigma free is fixed, and so now, what happens with E? Well, sigma free is fixed.
If kappa, all of a sudden, becomes three, [krrk] E field goes down.
Is that surprising? No, that is not surprising, because as you move in the dielectric, the --this surface charge density is not going to change, but you are inducing now, on your dielectric, negative charge here and positive charge here as a result of that external electric field, and so that creates an induced electric field in this direction, and so, as a result of that, the net electric field goes down.
And that's what you see here, it goes down, in this case, by a factor of three.
What happens with the potential difference over the plates? Well, D wasn't changing, remember? We kept D constant now. So if E goes down by a factor of three, V must go down by a factor of three.
What happens with the capacitance, C? Well, the capacitor is free charge divided by the potential difference.
The free charge is not changing, it's trapped. The potential difference goes down by a factor of three, capacitance goes up by a factor of three. What happens with the electrostatic potential energy? Well, the electrostatic potential energy is one-half Q
V.
But Q free could not change.
V went down by a factor of three.
So U must go down by a factor of three. That means if the electrostatic potential
energy goes down, that as I move in this dielectric, that I do negative work.
If I had to push it in, you would have gone up.
So, in a way, as I move the dielectric in, it's being sucked in.
There is a force that pulls it in.
Interesting all by itself. I would like you, at home, to go to ex-through exactly the
same questions, verbatim, with one difference.
And that is, you keep the power supply connected.
Now your answers are going to be very different.
For one thing, if the power supply is connected, and if you change D --so your
power supply is connected, and you go up in D by a factor of two --the power supply is connected, [unintelligible] one thing now that cannot change throughout, and that is V. Potential difference cannot change, because the power supply is connected.
So now, if V cannot change, and you increase D by a factor of two, E, now, must go down by a factor of two. And that's very different from what happened before, when E remained constant.
So it's very, very different physics. Well, the physics is the same, but the results are
very different. And I want you do that, you have all the tools now, you can believe in those equations, and they should work for you. All right, let's visit Ohm's Law and maybe look at Kirchoff --I prefer to stay on this center board, convenient for you, and it is also convenient for me. A very simple network, keep in mind that, on an exam, all problems are extremely simple and very fundamental.
Nothing complicated.
You don't have the time for that.
I give here a problem in which I actually give numbers, on the exam you won't see
any numbers, even, not in the sense of distances, ohms, and so on, because there is
no calculator necessary.
But here, you will see some numbers, this is a battery and this battery has an EMF,
which is ten volts.
That's a given.
Plus, minus.
And here, the current is going to split into three.
There is R one, which is one ohm, R two, which is two ohms, and then we have R
three --put it a little lower --R three is three ohms, they come together here.
And here I have a resistor R four, which is four ohms, and I close the loop and go back to my battery.
Just to make it a little bit more interesting, I will introduce into this battery an internal resistance which is very small, which is oh point one ohms.
You can't remove it, it's intrinsic into that battery.
And so the first question that I would ask you in this case is, what is the total current that is going to flow? We're going to get a current I here.
Through here you get I one, through here you get I two, through here you get I three, I comes out here, I goes through here, through the fourth, come back, and I goes through the battery. So what is I? In a problem like this, there are many roads to success.
Not just one. And it's a matter of taste which one you prefer.
If I call this point A and I call this point D, then what I would do, I would ask myself the question, if this is point A, and this is point D, what resistor --which your book calls the equivalent resistor --the equivalent resistance --what resistor would I have put here, instead of these three, for the current I to be exactly the same as what it is now? So I'm going to replace these three by one resistor, which is this imaginary resistor.
As you have noticed in your book, where you undoubtedly read up on, one over R equivalent is one over R one plus one over R two plus one over R three.
You know all these numbers, and so you will find our equivalent is oh point five five ohms.
Check this at home, and I hope I didn't goof up on that one.
Notice that this resistor, this equivalent resistance, is smaller than the smallest one, which is one ohm. That's obvious.
It has to be that way. Think of these as water flows.
Water flow through this one, through this one, and through this one. Remove these two.
Water is only flowing through this one. Now you add these two pipes so more water can flow, so the equivalent resistance goes down. Same with electricity.
So the equivalent resistance of parallel resistors is always lower than the smallest.
Now it's trivial to calculate the current I, I use Ohm's Law.
Ohm's Law says that potential difference that is available b-by the battery is E, it's ten volts, is now the current, the total current times all resistances along the road.
I go once around, I have here R e-equivalent, then I have R four, because the full current goes through R four, and then I have this little stinky R of I. Doesn't going to make much difference, but it's there.
And so you can find, now, what I is, because you know all other numbers, and you'll find that what I --I is my goal, and I think I found two point one five amperes, that is right. Two point one five amperes. So we know what I is.
Is this the only way? No. But it is one way, it's very effective.
So now I want to know what I one, I two, and I three is. Well, [pause] if I know the potential difference between A and D, V A minus V D, that must be, according to Ohm's Law, I one R one.
If I go this route.
But since we're dealing with conservative forces, I can also go through this path, the path doesn't matter, I must get the same potential difference.
So it's also I two times R two, and so it must also be I three times R three.
So if I only could find I two, then, of course, I would know the potential difference, out pops, immediately, I one and out pops, immediately, I three. And now I'm going to apply Kirchoff's First Rule in order to find I two.
I couldn't find I one, but I just decided on I two.
And Kirchoff's First Rule says that the closed loop integral of E dot D L --this, now, is a closed loop --is zero. In the future, you will see situations that it's not zero. Here, zero. I don't know why Kirchoff got the credit for this, this was long known before him, but nevertheless, it's called Kirchoff's First Rule.
So I go with closed path, and this is the closed path that I have decided to take.
And that closed loop integral E dot D L must be zero.
Any other path would also be zero.
I chose the one through R two, because my goal is to find I two.
Once I have I two, I want an I three follow immediately from this.
I have decided, very arbitrarily, that if I go down in potential, then I will call that --I will give that a negative sign, and if I go down in potential, I will give that a plus sign. You can reverse that.
That makes no difference, because the sum of them is going to be zero anyhow. So I will stick to my convention for now, that if I go down in potential, I will give it a minus sign, if I go up in potential, plus sign. I go, first, from A to D, through R two. The current is I two.
The resistance is R two. And I go down in potential.
So I get my first term is minus I two times R two.
I'm now at D. I started at A, I'm now at D. I went through two.
I come out here, and I go through four, R four.
I go down in potential.
The current through R four is I.
So I get minus I R four. I go up, and I see this battery in front of me, and I have to
climb up in potential.
How much do I have to climb up? That EMF, of the battery. But that dinky toy little
resistance R of I makes me down a little bit, and so I get another minus I times R of
I, and that, now, is zero. And that's one equation with only one unknown, which is I
two, because we already have I. That's the two point one five amperes.
And so you'll find, now, that I two becomes oh point six amperes.
And so you know now that V A minus V D is the potential difference, which was I two
R two, is now going to be one point two volts.
Because I two is oh point six amperes, but R two --there's a two here --is two
ohms.
So it's one point two volts. And so the one point two volts is also I one R one, and
it's also I three R three, so you get I one and you get I three.
What is the power delivered by this battery? That is the EMF times the total current
I.
We know the EMF, ten volts. The total current is two point one five amperes.
So this is twenty one point five watts.
How does that show up, that energy? Well, it comes out in the form of heat.
Heat in R four, heat in R one, R two, and R three, and a teeny, weeny little bit of
heat inside that battery because of that oh point one ohm internal resistance.
How much power comes out in resistance R two? Well, that is, of course, the
potential difference over R two, which was that V A minus V D times the current
through R two.
That's power, power is potential difference times current.
This is the total power delivered by the battery.
That is the total potential difference available times the total current. But of course,
R two only sees a potential difference which is one point two volts, and it has an I two which is only oh point six amperes, so this is only oh point seven two watts.
So that is the number of joules per second, in terms of heat, that is produced in R two.
I think you'll believe me when I say that the top is still running.
And the clue I will give you is that the answer lies in eight oh two.
Give it some thought, it's a very cute t-top.
All right, let's talk about kinetic energy increase due to charges that move over a potential difference.
I have two conductors, very funny in shape, but they are equally potentials, there's no current running inside the conductors.
And so conductor A is at a potential V of A, this is conductor A, potential is V of A, and this is conductor be, ahs a potential V of B, and let's assume that V A is larger than V B. If you want to change that later, that's fine.
We put the whole thing in vacuum, because I'm going to release a charge here, plus Q, and the charge will now go to B. Electric field configuration is a zoo. I don't even want to think about what it is.
One way or another, if this is vacuum, than this charge will finds it way to B. Let's say this is the routing that it takes.
It ends up on B, and the question now is, what is the speed at which it reaches B if I release it here at zero speed? So the electric field is going to do work on this charge, and the work, in going from A to B, integral A to B, is the force dot D L.
That is the electric force on that charge. Since it a conservative field, it doesn't matter what your routing is, you will always get the same answer for this.
That electric force is also the force times the electric field, at any location along the line. And so you see, you get here a Q times E D L.
But the integral of E D L is the potential difference between the two.
And so the net result, therefore, is that the work done by the electric fields, when this charge finally ends up here, that work is the charge Q times the potential difference V A minus V B, regardless of which path it chooses to go.
Let's take a practical case, we have a proton which has a mass one point seven times ten to the minus twenty seven kilograms.
The charge of the pro-proton is the same of that of the electron, but it is positive. One point six times ten to the minus nineteen Coulombs.
And let's suppose that the potential difference between A and B, this is now the potential difference, is a million volts.
Uh, let me put a delta here, because I don't want Vs on both sides. But that's the difference V A minus V B. So what is now the kinetic energy with which this proton
reaches B? So that kinetic energy must be Q times the potential difference, so that is one point six times ten to the minus nineteen, times ten to the six, so that is one point six times ten to the minus thirteen joules.
Almost no physicist would call this one point six times ten to the minus thirteen joules, but we would say, the kinetic energy of that proton is one M E V, one million electron volts.
And the reason why we do that is, an electron volt is the energy that an electron gains if it moves over a potential difference of one volt.
That's the definition of one electron volt.
Though the charge of the proton is that same of that of an electron, and it moves over a distance of one million volts, and so the energy is one million electron volts. But this is not an SI unit, so be careful.
If you work SI, you've got to use this number.
But we would say that's a proton with a kinetic energy of one M E V, one million electron volts. So what is the speed with at which it finally, then, arrives? Well, one-half M P squared is this number, this is the mass of the proton, this is the speed that the proton arrives at point B, and when you use that number, the mass, you will find that the velocity of that proton is about one point four times ten to the seven meters per second, which is about five percent of the speed of light.
In other words, we don't have to make relativistic corrections.
This answer is believable. Several students have sent me e-mail, and they have asked me for practice exams.
I am equally surprised, as you are, that the previous lecturers of eight oh two did not list on the web, on their website, exams. They didn't.
I was hoping they did, but they didn't. Professor Belcher, however, mentioned one particular practice exam to me, and when you visit the eight oh two website, you can find that practice exam, but there are no solutions on that exam. So you can discuss this exam with your instructors, with your tutors, I am available if you want to discuss it with me, I have no problems with that --it's a little difficult for me to help six hundred students, but I can help quite a few.
But my advice is that you take the study guide if you really want to do more practice, and go over some problems that have worked-out solutions.
And I wish you luck, and I'll see you next Friday.