An electron with kinetic energy of 4.0 x l0\'iD J is moving to the right along t

Question

An electron with kinetic energy of 4.0 x l0'iD J is moving to the right along the axis of a cathode-ray tube as shown below. There is an electric field E^ rightarrow = (5.0 x 10^4N/C) j in the region between the deflection plates. Everywhere else, E^ rightarrow = 0. Fluorescent How far is the electron from the axis of the tube when it reaches the end of the plates? At what angle is the electron moving with respect to the axis? At what distance from the axis will the electron strike the fluorescent screen?

Explanation / Answer

initial speed of electron u

so mu^2 / 2 = 4 x 10^-16

9.109 x 10^-31 x u^2 / 2 = 4 x 10^-16

u = 2.96 x 10^7 m/s

time to cross 4 cm .

t = 0.04 / (2.96 x 10^7) = 1.4 x 10^-9 sec

acc. due to field.

a = qE/m = (1.6 x 10^-19 x 5 x 10^4) / (9.109 x10^-31) = 8.78 x 10^15 m/s^2

distance traveled in y -direction = at^2 /2 = 0.0086 m

Ans = 8.6 mm ................(a)

b) angle = tan^-1(vy / vx) = tan^-1( at / u ) = 22.55 degress

c) now it have to travel 12 cm in x direction.

so tan22.55 = y / 12

y = 12 tan22.55 = 4.98 cm

total distane from axis = 4.98cm + 0.86 cm = 5.84 cm

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