Question

Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 320 N/C. (a) What is the magnitude of the acceleration of the electron (in m/s2)? (b) The electron is initially at rest. What is its speed (in m/s) after 8.50 ✕ 10−9 s?

Answer 1

(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².

(b) The speed of the electron after the given time is  4.78 x 10⁵ m/s.

Acceleration of the electron

The acceleration of the electron is calculated as follows;

F = qE

ma = qE

a = qE/m

a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)

a = 5.62 x 10¹³ m/s²

Speed of the electron

v = at

v = 5.62 x 10¹³ m/s² x  8.50 x 10⁻⁹ s

v = 4.78 x 10⁵ m/s

Learn more about speed here: brainly.com/question/4931057

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