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22.8 continuedThe electron undergoes downward acceleration (opposite E) , and its motion parabolic while it is between the plates:S 0 L UTIO N Conceptualize This ex...

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22.8 continuedThe electron undergoes downward acceleration (opposite E) , and its motion parabolic while it is between the plates:S 0 L UTIO N Conceptualize This example differs from the preceding one because the velocity of the charged particle is initially perpendicular IO the elee- tric field lines. (In Example 22.7, the velocity of the charged particle is always parallel to the electric field lines ) As result, the electron in this example follows curved path as shown in Figure 22.21. The mo

22.8 continued The electron undergoes downward acceleration (opposite E) , and its motion parabolic while it is between the plates: S 0 L UTIO N Conceptualize This example differs from the preceding one because the velocity of the charged particle is initially perpendicular IO the elee- tric field lines. (In Example 22.7, the velocity of the charged particle is always parallel to the electric field lines ) As result, the electron in this example follows curved path as shown in Figure 22.21. The motion o the electron is the same aS that of massive particle projected horizon- mallv in gravitational field near the surface Of the Farth. Categorize The elecuon is particle field (electric). Because the elecuric field is uniform, constant elecuric force is exerted the elecuron To find the acceleration of the electron, we can model it as particle under net force. Analyze From the particle in Tield model, We know that the direc tion ofthe electric force On the electron is downward in Figure 22.21, oppesite the direction of the electric field lines: From the particle uuder nct force model, therelore , the acceleration of thc electron is downw:rd. Figure 22.21 (Example 22.8) An electron prejected horizontallv into uniform electric field producedby M charged plates Thc pattick Under nct force model Was used to develop Equation 22.H inthe cse in which the electric force O panticlee is the only force. Use this equation to evaluate the component of the acceleration of the electron: "= (1,60 * [0 C)(200 N/C) 4= 9.XIO- kg Substitute HumeTicalvales: 3X 10" m/s



Answers

Two charged metal plates in vacuum are $15 \mathrm{~cm}$ apart as drawn in Fig. 24-7. The electric field between the plates is uniform and has a strength of $E=3000 \mathrm{~N} / \mathrm{C}$. An electron $\left(q=-e, m_{e}=9.1 \times 10^{-31}\right.$ $\mathrm{kg}$ ) is released from rest at point $P$ just outside the negative plate. (a) How long will it take to reach the other plate? (b) How fast will it be going just before it hits?
The electric field lines show the force on a positive charge. (A positive charge would be repelled to the right by the positive plate and attracted to the right by the negative plate.) An electron, being negative, will experience a force in the opposite direction, toward the left, of magnitude $$ F_{E}=|q| E=\left(1.6 \times 10^{-19} \mathrm{C}\right)(3000 \mathrm{~N} / \mathrm{C})=4.8 \times 10^{-16} \mathrm{~N} $$ Because of this force, the electron experiences an acceleration toward the left given by $$ a=\frac{F_{E}}{m}=\frac{4.8 \times 10^{-16} \mathrm{~N}}{9.1 \times 10^{-31} \mathrm{~kg}}=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} $$ In the motion problem for the electron released at the negative plate and traveling to the positive plate, $$ v_{i}=0 x=0.15 \mathrm{~m} a=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} $$
(a) Fiom $x=v_{i} t+\frac{a}{a}$ $$ t=\sqrt{\frac{2 x}{a}}=\sqrt{\frac{(2)(0.15 \mathrm{~m})}{5.3 \times 10^{4^{4}} \mathrm{~m} / \mathrm{s}^{2}}}=2.4 \times 10^{-8} \mathrm{~s} $$ (b) $v=v_{i}+a t=0+\left(5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\right)\left(2.4 \times 10^{-8} \mathrm{~s}\right)=1.30 \times 10^{7}$ $\mathrm{m} / \mathrm{s}$ As you will see in Chapter 41, relativistic effects begin to become important at speeds above this. Therefore, this approach must be modified for very fast particles.

Determine the ratio of charge to masses given by e b M equals two capital E square Developed by who? We be square. Okay. And this situation is one and it can be rewritten, as is where equals two eBay am multiplied by base pair and people. So from this equation, we can see that is where and the he's a straight line a state land who with a slow, big slope well, to e by em multiplied by this. So this is the answer for this equation. Part question. Okay, so this is the result for the part of this question now moving to the part B. So from the figure, we can find the massive immune and we select two points having values and that world former 200 multiplied by 10 to the power aid, the whole despair for me to despair and 300 world coma 600 multiplied by 10 to the power eight won't despair per meter squared which lies on the state line so we can calculate the slope of the line. Yes, slow. It plays to 600 multiplied by 10 to the power eight minus 200 multiplied by 10 to the power eight divided by 300 minus 100. So from here we get slow equals two two multiplied by 10 to the power aid world per meter is fair. So as from the part we have slope equals two to be by M molecular B B square. So from here, if I am, will be equals two Slow divide by two b square. So from here, putting values in the situation slope is too much cheaper by 10 to the power eight Walter parliamentary sphere who ended migrants will be is 0.340 Tesla. So police fear so. From here I am comes out to be 8.65 multiplied by 10 to the power eight. Who long for Katie? So this is the answer fully part B of this question. Okay, No moving to the party in which, uh, in this question we need to find out Mask. So from here, mass can be calculated as in the bar by 8.65 by 10 to the power aid, and he is equals two 1.6 to 10 to the minus 19. Too long divided by 8.65 multiplied by 10 to the power eight. So from here mass M comes out to be 1.85 multiplied by 10 to the power minus 28 pages. So this is the answer for the part of this question. Okay. Yeah. So now moving to the part C decision. And so for the party. No moving to the part C in which we have to calculate the potential and the potential is equal to email Player by D, so we can put the values to get the answer. So e is equal to two into 10 to the 45 world perimeter, and these equals to zero point double 06 needed. So from here, the is equal to 1.20 Hello, world. So this is the answer for the part C of this question. Okay, so now moving to the party in which we have to calculate the spill and the speed is given by we equals two who e capital, we develop small and and we have all the values we equals to IBM has been calculated in the part of B, so we can put that directly. Values for two multiplayer by IBM is equal to 8.65 particular by 10 to the power eight column for phasing. And we is equal to one point who ah v is equal to 400 words. Okay, 400 bowls. So from here, still weak comes out to be 8.32 multiplied by 10 to the power 5 m per second. So this is the answer for this, but the the question okay?

For the first part we're gonna use the analysts is in the text off Thompson's by an experiment which gives over m tow the acquittal. We square all right. Two times four dates. Things These days I find this gifts He's got to be equal two times over. M Dying's these grad on therefore, the graph ofthe e squared. What says the should be a straight line? As you can see, falling from the question like Waldo am expressing their sees you here. Xs being wise is God and a Mr Slow. So she opens equality don't times we squared over him for party We can find the slope using too easily read points on the graph. So using the 100 2 100 Auntie 100 600 we get the slope to be text called the slope to the end equal 600 times 10 to the five We start over them's minus 200 times Rented the car eight respectable Thanks There. So this is a lie that is now off x 300 ward minus And this gives the slope to be two times Stand to the violence point Watch me disco For this one, we can use this value that we found for Slope ever here. That means I mean, we squared over em sequel to through times 10 to the 58 one square and this gives the issue off. We buy him the equal. This values the drunk it confused with him and am here. Let's call this number dash to distinguish for this is Snow Times. Mine's one watch. Don't be scared and this gifts that issue to be coddled a 165 and the departure Katie. Which means that Mars is incredible. This value sorry. The chart of electron over this value in charge of the electron is 1.6 times and the college minus 19 column Soviet sticking the magnitude ofthe Electrica Electron judge. So the mask is control 1.35 Understand the pipe Linus during the eighties, it's on for the four part. See, we have to find a dead given the electrical bill at a distance between the plates. The electricity evidence given to be two times 10 to the five Hi, a forge for me and distance between the play It's excrement, Toby, your points. Six. Meetem exchanging Convert this meter values in this equation. I became devoted to be one point of time standard by two modes. Knowledge Can I think it's one for what C we're going to use. We're going to match the energies and I just basically eight times the potential energy. This's equal to the kind of again Auntie. So this is basically conservation off energy that we have. And this gives the Veloster two times, two times the voltage over and out on this gives the old age to be. I'm sorry. The velocity to be a point two dimes tend to the bar five meters per second. Make sure you use the valuable they'd given in the court part which is 400. Hold on hard. The voltage that began with you on your is ALS marry might vary from mine because due to the inaccuracies, maybe into the mining the slope off the graph. So if your result is close enough, the order is exact and this mining to disclose enough then

And this problem, the first part is asking us under is asking us to understand how moving charged also created. Okay, as we're moving feel so we want to use a particle cube moving with velocity V defying the position vector are hat. We want to show that what the magnetic field at the location is equal to. So in this case, we're gonna use the fact that the, um indefinitely indefinitely Small magnetic field D B is equal to mu, not over four pi r squared I ds times are so it's a a infinitely small region and an infinitely small magnetic field. We also know that our current is equal to, um are total number of charged particles in this area multiplied by the size of the area multiplied by charge, times velocity. And so for a, um, for a positive current charge, her first are for a positive charge. The direction of DS is going to be in the same direction as the and we also know that a D s is the volume occupied by the moving charges and it's equal toe one if we only have a singular charge. So in this case, that means that be must be equal to him, you know, divided by four pi times, Q v times are divided by R squared resorted to Cuba. Times are hat divided by r squared. And so we've shown that this part of the problem is true. The next part is that he got to find the magnitude of the magnetic field one millimeter to the side of a proton, moving at a given speed. To do this, we're once again going to use the equation that we just found. And we know in this case that, um, our that we're are sign of data so that would be here is going to be signed of 90 which is one since velocity is always perpendicular to magnetic field by the right hand rule. So we're gonna plug in are constants which really just include our, which is really just aren't radius. So that's tending the negative three meters that's gonna be plugged in for R squared as well as the charge on a proton. And when we do this out, we find a magnetic field of 3.2 times 10 to the negative 13 Tesla Now part C is asking us to find the force on the second proton at this point, if it's moving with the same speed in the opposite direction. So that magnetic force given by the equation Q v B we're gonna plug in the charge of a proton, which is 1.6 of him and negative 19 cool arms are velocity. Um, that are person is going to be moving at 2 10% of the seven meters per second and our magnetic field We do this, we get a total force of 1.24 times 10 to the negative. 24 students last for the problem is asked needs to find the electric force on the second proton eso. The electric force we know is just equal to um que e. So that's our constant K times queues have a first charge him. The second charge, divided by r squared. Now I tried to do the same. In this case is both of the murders the person of the charge of Proton. This is really just the same thing as q P River Port on squared que issued a constant and our is a distributor charges which is still, um one millimeters so we're gonna have 10 to the negative three meters squared and denominator and that's going to give us a final result of 2.30 times 10 to the negative 22 Newtons.

The cyclotron, like many particle accelerators uses magnetic and electric fields to control and accelerate charged particles. So here we're going to take a look at the parts of the cyclotron and then get into the physics of how this all works. So there are two magnets uh circular usually uh semicircular well completely circular in shape divided each pole into two halves. Each half is called A. D. And the magnetic field is sandwiched in between the two poles. Like usual. Then there is an A. C. Voltage between the two Ds. And what happens is charged particle gets injected with very low velocity in the centre um gets accelerated by the voltage and then starts to travel in a circular path which gets bigger and bigger and bigger in radius. The more kinetic energy the particle gains. So it's a very compact arrangement because the particle moves around in a circle. Um So in order to understand how the cyclotron works, there's just a little bit of physics. Um The first is the Lorentz force on a charged particle F. Is equal to Q. V. Crosby. And here the velocity of the particle is maintaining itself perpendicular to the magnetic field. So we can just write down the magnitude of that force F equals Q. B. B. Uh The second thing we need to understand is that the circular motion that arises is really happening because that force is what's called a centripetal force that is it is always perpendicular to the velocity. Um And the force is thus holding the particle in circular motion. And so we knew needs Newton's second law. In terms of uh huh centripetal acceleration, some of the centripetal forces is equal to mm B squared over R. Now, supposedly this occurs in a vacuum. So we'll just assume that there's a single force on the particle neglect gravity since it's so weak. And if we put all that together and cancel some of the like sides, like pieces on each side, what we see is that the velocity and the radius are directly proportional to each other. As the velocity goes up, the radius of the orbit goes up and vice versa. So now we can understand why there is an exhilarating potential in between the two Ds. The idea there is that you're going to be using conservation of energy to boost up the kinetic energy of the particle and hence its kinetic energy and velocity go up. So the third piece of physics is conservation of energy. So we're going to make the kinetic energy go up and that change in kinetic energy is minus the change in electric potential, which is proportional to the charge on the object in the cyclotron and the voltage oscillating voltage amplitude. Now that oscillating voltage amplitude, usually there are two attachments. So there is a kick in the kinetic energy gain. The kinetic energy gains in two spots, one on the right hand side, and a similar thing going on on the left hand side. So you kind of have, if you want to think about it quadrants here, I'm trying to draw some plates in there. Um Yeah, actually you can just sleep the same thing connected all the way across. But you have to be careful about the phase between the two sides. So here I'm just trying to show that you get it a kinetic energy gain on the other side as well. And that can happen if the A. C. Potential is varying with just the right frequency so that there is a potential drop across the two plates on each side and in between. The potential isn't doing much at all. Okay, so let's analyze that. The oscillating voltage has to be in phase with the motion to give a boost to the the kinetic energy of the particle. So let's kind of see how that works. And so what we're going to go back and do is just slightly rewrite equation number two because we know that the angular velocity of the particle times its radius is equal to be and we can substitute that in and going back to that equation. Then we have Q. B. R over em is equal to omega times are and notice that the angular velocity does not depend on the radius at all. And this is really at the heart of how the cyclotron is able to do. What it does is if you oscillate the A. C voltage at this so called cyclotron frequency, you will be as long as you set it up correctly, you will be accelerating the particle at each time it crosses from one day to the other. Um So this is very important relationship. There are different ways to write it. For example, omega is two pi times the frequency. And so that cyclotron frequency is a product of QB over two pi am does not depend on the radius. The time it takes for the particle to go all the way around is equal to one over the frequency. And we can just invert that. So it depends what you want to do. But this is basically the design parameter that you need in order to make your cyclotron operate correctly. So has a quick finish. Let's do the case of an electron in a half a Tesla magnetic field. We know the mass of the electron and we know it's charge of course. And we can calculate that psych latin frequency using those parameters. Yeah, I don't I think I need to put those numbers in but we get about 1.4 zero times 10 to the 10th hurts. Which is in the gigahertz range. It sounds like microwaves to me. Certainly doable. Mhm. And we can ask the question, how much energy does the particle have after 100 revolutions? Well, each revolution the kinetic energy gain is the absolute value of the change in the electric potential energy. So we need to know the operating voltage. And here the operating voltage amplitude is 120 kilovolts. That's quite a substantial voltage. Um So each revolution um you're passing through the DS twice and then you have, what's that, kilovolts? So you have two times 120 times 10 to the third electron volts are 240 times tender. The third electron volts. And so after 100 revolutions rounds will call them. You just multiply that by 100 and that is a sex essentially 24 million electron volts, 24 mega electron volts of energy. Okay, so to get a feel for that, I just want to remind you that the rest energy of the electron is a half a million electron volts. So definitely usually what you do in relativity is you compare the amount of energy to the rest energy of the particle and when that energy starts to be about equal, you know, you have relativistic effects coming in. So I'll just note that you have definitely achieved relativistic energy at this point.


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