An electron in an oscilloscope tube is traveling horizontally with a speed of 4.

Question

An electron in an oscilloscope tube is traveling horizontally with a speed of 4.5×106 m/s . It passes through a 1.9 cm long region with an electric field pointing upward with a magnitude of 1300 N/C .

A) What is the magnitude of the electron’s velocity when it emerges from the electric field region?

Express your answer to two significant figures and include the appropriate units.

B) What is the direction of the electron’s velocity when it emerges from the electric field region?

Express your answer using two significant figures.

Explanation / Answer

E = 1300 N/C   and v = 4.5e6 m/s and ignored gravitational force

Now the e- is going to be accelerated downwards because opposites attract, alike repel
The force on the e- = F= qE = (1.602 e-19C) * (1300 NC-1 )
F= 2.08 x10^-16 N
so that force is exerted on the e-, use F=ma to find the 'a'

a= F/m = 2.08e-16N/ (9.11e-31kg)= 2.28 e14 ms-2

use the x-component velocity 4.5e6 to find the time the electron is in between the plates

t= x/v = 0.019/4.5e6 = 4.22e-9 seconds
Now v =v0+ at= 0 + 2.28e14 ( 4.22e-9)

v= 9.62 x10^5 ms-1 (downwards)

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