Question

An electron withkinetic energy of 6.0 10-16 J is moving to the right along the axis of a cathode-ray tube as shown below. There is an electric field = (2.0 104 N/C) in the regionbetween the deflection plates. Everywhere else, = 0.

(a) How far is the electron from the axis of the tube when it reaches the end of the plates? (mm)
(b) At what angle is the electron moving with respect to the axis?
(c) At what distance from the axis will the electron strike the fluorescent screen? (cm)

Answer 1

electric field Ey =(3*104 N/C) j^
kinetic energy of the electron K.E = 4*10-16 J
charge of the electron e =1.6*10-19 C and x = 4 cm =0.04 m

a)
distance traveled by electron is
y = (-eEy)(x)2/2mv2
y = (-eEy)(x)2/(4)(mv2 /2) .......... (1)
kinetic energy of the electron K.E =(1/2)(mv2) ......... (2)
so eq (1), becomes
y = (-eEy)(x)2/(4)(K.E) ........... (3)
substitute the given data in eq (3), we get
y = -4.8*10-3 m
= -4.8 mm = -0.48 cm

b)
θ = tan-1[(-eEy)(x)/(2)(K.E)] .......... (4)
θ = -13.50

c)
vertical displacement y = (0.12 m) (tanθ)
= -2.88 cm
the electron strike the fluorescent screen at
ytotal = -0.48-2.88
= -3.36 cm (below the horizontal)