Deflection in a CRT. Cathode-ray tubes (CRTs) are often...

Deflection in a CRT. Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors. In Fig. P2.. 65 an electron with an initial speed of $6.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ is projected along the axis midway between the deflection plates of a cathoderay tube. The potential difference between the two plates is 22.0 $\mathrm{V}$ and the lower plate is the one at higher potential. (a) What is the force (magnitude and direction) on the electron when it is between the plates? (b) What is the acceleration of the electron (magnitude and direction ) when acted on by the force in part (a)? (c) How far below the axis has the electron moved when it reaches the end of the plates? (d) At what angle with the axis is it moving as it leaves the plates? (e) How far below the axis will it strike the fluorescent screen S?

Key Concepts

Uniform Electric Field

A uniform electric field is one in which the electric field lines are parallel, equally spaced, and maintain a constant magnitude and direction throughout the region. In deflection systems like those in cathode-ray tubes, the field between the deflecting plates is typically uniform, and it is usually determined by the potential difference divided by the separation between the plates.

Electric Force on a Charged Particle

The electric force on a charged particle is given by the equation F = qE, where q is the charge of the particle and E is the electric field. This force acts in the direction of the electric field for a positive charge, while for a negative charge, it acts opposite to the field direction. This principle is essential in understanding how electrons, which are negatively charged, are deflected in a cathode-ray tube.

Newton's Second Law and Charge Acceleration

Newton's second law relates the net force acting on an object to its acceleration via the equation F = ma, where m represents mass and a is the acceleration. When an electron is subjected to an electric force, its acceleration can be calculated by dividing the force by its mass, linking electromagnetism with classical mechanics.

Kinematics of Uniform Acceleration

The equations of kinematics describe the motion of objects subject to constant acceleration. In the context of a cathode-ray tube, these equations are used to determine the displacement and final velocity of the electron as it moves under the influence of the uniform electric force. They allow the calculation of how far the electron deviates from its initial trajectory as well as the angle of its motion when it exits the deflection region.

Trajectory Analysis in a Cathode-Ray Tube

Analyzing the trajectory of a charged particle involves combining the effects of its initial velocity with the subsequent acceleration due to the electric field. This process often involves separating the motion into components (such as horizontal and vertical) and applying kinematic equations to find parameters like vertical deflection at the exit of the plates and the further displacement on a screen. Such analysis is fundamental in designing and understanding devices like oscilloscopes and computer monitors.

Transcript

00:01 So to answer this question, let's go over what we're given.

00:04 So the initial speed is equal to 6 .5 times 10 to the power of 6 meters per second.

00:10 The magnitude of the electric field e is equal to 1 .1 times 10 to the power 3 volts per meter.

00:16 And the charge on one electron, which we know is just standard, is equal to 1 .6 times 10 to the negative 19 cools.

00:25 So to answer part a, the magnitude of the force, which we can just use f equals qe, and since we know both these values, charge is 1 .6 times since the negative 19.

00:38 And as i said before, the magnitude of the electric field is 1 .1 times into the power 3.

00:44 So we can compute it and it's equal to 1 .7, 6 times 10 to the negative 16 mutants.

00:54 And now to move on to part b.

00:57 So the magnitude of acceleration, which we can use a is equal to f over m.

01:04 Just newton's second law and then we solve for a you get this and now we know the force from the previous equation so 1 .76 times sent to the negative 16 over the mass of the electron which is standard on 9 .1 times 10 to the negative 31 and remember this is in kg so standard units and if you compute this we're going to get that the acceleration is equal to 1 .93 times 10 to the 14 times 10 to the 14 meters per second cubed so now if you go into part c the separation of the plates has a distance of 0 .06 meters and so the required time can be calculated with this t is equal to d over v so it's just a very standard equation in physics so you should all know this equation so the distance, as i said before, is 0 .06 meters and the velocity, which we are given, is equal to 6 .5 times 10 to the power 6 meters per second.

02:24 So now if you compute this, we're going to get that the time is equal to 9 .23 times 10 to the negative 19 seconds.

02:34 So the distance moved by electron and time t.

02:37 So you can use is x.

02:39 So the distance is the distance equal one half a t squared and now since we know all these values which we computed in the previous part and in this part you can just plug them in so it's one half times acceleration which is 1 .93 times 10 to the power 19 and times time which we calculated was 9 .23 times sent to the power 9 times 10 to the power of negative 9.

03:15 Negative 9.

03:17 And this is squared.

03:19 So if you compete the central calculator, we can get that the distance moves is 8 .22 times 10 to the power of negative 3 meters.

03:33 So now the angle moves in part d.

03:37 To calculate the angle moved, all we need to do is choose this equation.

03:41 So theta is equal to the inverse time...

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