An electron with kinetic energy of 7.0e-16 J is moving to the right along the ax

Question

An electron with kinetic energy of 7.0e-16 J is moving to the right along the axis of a cathode-ray tube as shown below. There is an electric field vector E = (1.0e4 N/C) j in the region between the deflection plates. Everywhere else, vector E = 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

Explanation / Answer

K =7*10^-16 J , charge of electron =1.6*10^-19C

From conservation of energy K = U

K =qV

7*10^-16 = (1.6*10^-19)V

V = 4375 V

E =1*10^4 N/C

(a) Ed =V

1*10^4*d = 4375

d = 0.4375 m

d = 437.5 mm

(b) tan(theta) = 43.75/12

theta = 74.55 degrees

(c) x = ((43.75)^2+(12)^2)^0.5

x = 45.37 cm

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