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.

***THE ANSWER IS NOT 9.6E5 m/s for Part A

Explanation / Answer

in horizontal electron have to travel 1.9 cm with 4.5 x 10^6 m/s.

t = x / vx = (1.9 x 10^-2 m) / (4.5 x 10^6 m/s) = 3.56 x 10^-9 sec

In vertical :

force = qE = - 1.6 x 10^-19 x 1300(j)

and a = F/m = qE/m = - 1.6 x 10^-19 x 1300(j) / (9.11 x 10^-31)

a = 2.28 x 10^14 m/s^2

vy = uy + at

vy = 0 + (2.28 x 10^14 x 3.56 x 10^-9)

vy = 0.811 x 10^6 m/s

A) magnitude of velocity = sqrt(vx^2 + vy^2)

= 4.57 x 10^6 m/s

b) direction = tan^-1(0.811/4.5) = 10.21 deg below horizontal.

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