Cathode-ray tubes (CRTs) were often found in oscilloscopes...

Cathode-ray tubes (CRTs) were often found in oscilloscopes and computer monitors. In $\textbf{Fig. P23.63}$ an electron with an initial speed of 6.50 $\times 10^6$ m$/$s is projected along the axis midway between the deflection plates of a cathode-ray tube. The potential difference between the two plates is 22.0 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?

Transcript

00:01 This question is a setup about catarrow tubes in computer monitors and then we have an electron with initial speed 6 .5 times 10 to 6 meters per second.

00:18 Okay and then the lower plate is at a higher potential so here is plus b zero.

00:25 We know that the electric field is going to be pointing up.

00:29 Okay, but the electron will experience a force downward.

00:35 Okay, because the bottom plate is at a positive potential, at a higher potential.

00:42 So, okay, so in this question, there are five parts, asking about force acceleration, the vertical deflection, the angle with the axis that it exits the plates and the vertical displacement of the electron when it tries the screen okay so there is a reference line that need to work with okay this is a horizontal line that the that the electron enters the parallel plates okay okay so in prime a we need to find a force okay so we'll be using f equals to qe okay this is the relationship between force and electric fuel and then parallel plate situations we have equals to delta v over d okay so um so the magnitude of the force skew v over d okay 1 .6 times 10 to the negative 19 v is 22 and d is 2 cm okay so 0 .02 calculate this to be 1 .76 times 10 to the negative 16 mutants.

02:10 Okay, so this is a magnitude of the force.

02:14 The direction is downward.

02:23 Okay, next in part b, you want to find an acceleration.

02:30 Okay, so using fnett equals to m.

02:37 Okay, so a equals to f, net divide by m.

02:42 Okay, put in the numbers.

02:50 To calculate this you get 1 .93 times 10 to the 14 meters per second square and then the direction is also downward.

03:10 Okay, follows the direction of net force.

03:19 Okay, in part c, you want to find a vertical displacement when it reaches at the end of the plates.

03:29 Okay, so you need to find the time taken.

03:35 For the for electron to exit the plate.

03:44 To do to find this.