Electrons are accelerated by 6.0 kV in a CRT. The screen is 30 cm wide and is 34 cm from the 2.6-cm-long deflection plates. Over what range must the horizontally deflecting electric field vary to sweep the beam fully across the screen?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. We're gonna solve for the electric field strength that's needed for this first extreme where the beam is deflected entirely to the edge of the screen. This is a top-down view and this beam is gonna scan across the width of the screen. If you were a person looking at it, your eyes would be here and you'd see just a straight line going across in your field of view. I guess this would be the front view. You would see this. There's a big computer screen there or a TV screen, and you'd see the beam do one pass right in front of you horizontally. This is the width here. So, what electric field is needed to get this beam to scan the full width? We'll solve for the electric field such that it reaches one edge and then it'll have to vary from that to the exact opposite in order to get the beam to go to the other edge. With zero electric field, the beam will go straight through to the center beam and not deflected at all. In order for the beam to get to the opposite edge up here, it'll need to encounter a field with the same strength but pointing down. The electron beam is being deflected in the opposite direction of the electric field since the electric field points in the direction of force on a positive charge and the electrons negative. The electron has to get deflected a total distance of 15 centimeters. That's half the width. These deflections gonna consist of two parts. I'm gonna do this the complicated way 'cause it's kind of, you know, sometimes more fun to do things more rigorously. We're gonna account for the deflection that occurs between these plates. And we use this letter P to represent the width of the plate, P for plate. So it's 2.6 centimeters or 0.026 meters. And so this will result to some deflection d one from the center line to a point where it is done between plates. Then after it reaches this point here, it's gonna move with a constant velocity from there to the screen because there's no electric field there to accelerate it. Whereas in the presence of the electric field between the plates, it's gonna be accelerating downwards. So its deflection from center consists of d one plus d two, and d one is the displacement during the acceleration, so that's one-half times ay times t one squared, so this is the acceleration in the y direction here times the time that it spends between the plates, t one. Plus the constant velocity that it has here which is its final vertical velocity after being accelerated by the plates multiplied by t two which is the time to get from the plate to the screen. The total displacement equals the width over two. t two is gonna be the distance x from the plates to the screen divided by the horizontal components of its velocity. We can calculate that using the fact that the kinetic energy of the electron gains when it's accelerated from the cathode and then through the hole in the anode is gonna equal the potential energy that it lost. The potential energy lost is this charge times the potential difference through which it accelerated the voltage, and its kinetic energy is one-half mVx squared, and we'll solve for Vx by multiplying both sides by two over m and then take the square root of both sides. So we have two qV over m all square rooted is the x-component of the velocity. We're doing a whole bunch of things here in order to substitute into this equation to figure out. We've found Vx now and then we're gonna have to find a bunch of other things. So we know that the vertical force that the electron experienced between the plates is the charge times the electric field strength. The vertical force, the electrostatic force we'll say is the net force whose gravity is really negligible. So that means we will replace F with mass times acceleration, and that equals qE. And we'll solve for the acceleration in the y-direction which is qE over m. Divide both sides by m there. That replaces this. So you've got this replaced and that replaced, now turn your attention to t one, it's the width of the plates divided by the horizontal speed. And we already figured out the horizontal speed. And if we're dividing it, it's the same as multiplying by its reciprocals. So we have p times the square root of m over two qV. That takes care of t one being replaced. And then let's look at Vy. Vy is the initial vertical velocity which is zero because it's moving horizontally before into the plates. Plus the acceleration that it experiences between the plates multiplied by the time it spends reaching the plates. So the acceleration is the electrostatic force divided by the mass as we know from there, and multiplied by t one which is P over Vx, which is this, P times square root m over two qV. This m here and the square root m on the top combined to make a square root m on the bottom. That's all that you can do there. Electric field strength times the width of the plate times the square root of q over two mV. Now, it's the grand substitution here, we're substituting into this equation here and so we'll replace ay with this one, qE over m. We'll replace t one with this here. We'll replace the Vy with that Ep square root q over two mV. And then this is multiplied by x, and then there's divided by Vx which is the same as multiplying the reciprocal of Vx. Multiply by square root m over two qV. That takes care of that. Then I subtracted w over two from both sides. So we have a minus w over two equals zero. At last, minusing w over two business was not important but that's what I did anyway. Then we have this line here after you simplify. This is squared which makes it P squared, then it also makes an m over two qV. The m square root m squared makes m, and so that cancels to this m. And then this q cancels to this q. This makes it two combined with this two makes a four, and we have V on the bottom and E on the top and p squared on top. Here, we have the square root m and the bottom square root m on top canceling. Likewise, the square root of qs cancel. And we have square root one over two V times square root one over two V, in the end, which makes one over two V. And it's electric field strength times with the plates times the distance x from a place to the screen on top. Minus W over two equals zero. Fractions are never friends, so let's get rid of them. Multiply everything by four V and that's the lowest common multiple of all the denominators. And we have Ep squared then 'cause the four V is cancelled there, plus 2 Ep or x. That's gonna be because four over two makes two and the Vs cancel, then minus two wV, four over two makes two, and then multiply the wV by V as well, equals zero. Then take this turn to the right-hand side making it positive two wV and factor out the electric field strength from these two terms and then divided by the bracket, the results, p squared plus px. We have electric field strength then is two wV over p squared plus px. That's two times the width of the screen 0.3 meters times the voltage, six times 10 to the three volts, divided by the width of the plates which is 0.026 meters squared, plus two times 0.026 meters times 0.34 meters. That's the distance from the plates to the screen. We get about plus or minus 1.96 times 10 to the five Coulombs. I wrote plus or minus here, because as I've mentioned before. This finds the electric field strength you need to get to this edge of the screen, and then you need to have the same magnitude but opposite sign to get the beam deflected out to the other edge. There we go.