Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors.

Question

Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors. In the figure an electron with an initial speed of 7.00×106m/s is projected along the axis midway between the deflection plates of a cathode-ray tube. The potential difference between the two plates is 21.0 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 direction of the force on the electron when it is between the plates?

c) What is the magnitude of the acceleration of the electron when acted on by the force in part (a)?

d) What is the direction of the acceleration of the electron when acted on by the force in part (a)?

e) How far below the axis has the electron moved when it reaches the end of the plates?

f) At what angle with the axis is it moving as it leaves the plates?

Explanation / Answer

let's calculate the electric field between plates = E = V/d= 21/2x 10^-2= 1050 V/m

a) F = qE = 1.6 x 10^-19 (1050) =1.68x 10^ -16 N

b)Direction: towards the bottom plate

c) accleration = F/m= 1.68x 10^ -16 N/ 9.109 x 10^-31= 0.1844 x 10^ 15 m/s^2

d)direction = towards the bottom plate

e)h = ut + 1/2 at^2---vetical dispalcement

s= ut ---horizontal displacement

6x10^-2 = 7 x10^6 (t)

t = 0.857 x 10^-8 seconds apprx

h = 0 (0.857 x 10^-8) + 1/2 ( 0.1844 x 10^ 15) ( 0.857 x 10^-8 s)^2

h = 0+0.06772 x 10^-1 m = 0.06772 x 10^-1 x 10^2 cm = 0.6772 cm ( below the vertical line)

f) angle = tan^-1 ( Vy/ Vx)

Vy = 0.1844 x 10^ 15 ( 0.857 x 10^-8) = 0.158 x 10^7 m/s

angle = (1.58 /7) =12.72 degree apprx

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