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 ma gnitude 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?
1 answer:
(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.
<h3>
Acceleration of the electron </h3>
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
<h3>Speed of the electron</h3>
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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