Chapter 25 Homework Due Tuesday February 27,2018 Name - Exercise 5 Deflection in

Question

Chapter 25 Homework Due Tuesday February 27,2018 Name - Exercise 5 Deflection in a CRT. Cathode- ray tubes (CRTs) are often foursd in oscilloscopes and computer monitors. In Fig, 2338 an electron with an initial speed of 65 “ 10° m/s is projected along the axis midway between the denection planes of, cathode-ray tube The uniform . electric field berween the plates has a magnitude of 1.10 10' V/m and is upward.(a) What is the force (magnitude and electron when it is berween the plates? (b) What is the acceleration of the electron (magnitude and direction) when acted on in páirt (a) (e) How far below the axis has the elkctron moving as it leaives the plates? (e) How far below the axis will i strike the fiuorescent screen S? moved when ia reaches the end of the plates? (d) At what angle with the axis is i 2.0 cm K6.0 cm 12.0 cm

Explanation / Answer

given

v = 6.5*10^6 m/s

electric field, E = 1.1*10^5 V/m

a. force on an electron moving through the plate = qE

Fe = 1.6*10^-19 * E = 17.6*10^-15 N

direction, downwards ( as electric field is upwards and electron is -vely charged)

b. accleration = F/9.1*10^-31 = 19.340659*10^15 m/s/s ( downwards)

c. by the end of the plates

let vertical distance covered by y

then

vy = a*t

where v = 0.06/t

t = 0.06/v

hence

vy = a*0.06/v = 1.78529163*10^8 m/s

hence

y = 0.5*a*(0.06/v)^2 + vy*(0.12/v) = 4.1197 m

d. when it leaves the plates, the angle of velocity with x axis is theta

tan(theta) = vy/v

theta = 87.90 deg

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