Physics, asked by gunayasli01, 1 month ago

Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors. In (Figure 1) an electron with an initial speed of 6.80×106 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 18.0 V and the lower plate is the one at higher potential.

Attachments:

Answers

Answered by BrainlyNitya
2

Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors. In the figure(Figure 1) an electron with an initial speed of 6.50×106m/s is projected along the axis midway between the deflection plates of a cathode-ray tube. The potential difference between the two plates is 18.0 V and the lower plate is the one at higher potential.

a. What is the magnitude of the force on the electron when it is between the plates?

b. What is the magnitude of the acceleration of the electron 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?

2.0 cm Uo 0 6.0 cm >K--1 2.0 cm

Answered by shilpa85475
0

Cathode-ray tubes (CRTs) are commonly found in oscilloscopes and computer monitors:

A cathode-ray tube (CRT) is a vacuum tube containing one or more electron rifles, their beams being adjusted to display images on a phosphorescent screen.

Images can represent electric waves (oscilloscope), images (television set, computer monitor), radar targets, or other events.

In (Figure 1) an electron with an initial velocity of 6.80 × 106 m / s is projected near the center axis between the deflection plates of the cathode-ray tube. An initial electron with a speed of 7x106m / s is shown between the axis between the cathode-ray tube recording plates:

The same electric field between the plates is 1000 V / m high and high.

F = 1.6x10-16N and acceleration a = 1.76x1014m / s2.

The possible difference between the two plates is 18.0 V and the lower plate is the strongest:

With the same field between two plates ∆ V = ∆ U / q 0 = q 0 Ed / q 0 = Ed or E = -∆ V / ∆ s The potential difference depends only on the plates NOT on any shipping charge.

The work performed depends on the size of the billing process. The potential difference in electricity is far more important than the actual value of electricity.

Similar questions