Missing part of question
The electron beam inside an old television picture tube is 0.40 mm in diameter and carries a current of 50 μA. This electron beam impinges on the inside of the picture tube screen. 1-The electrons move with a velocity of 3.8 ×107 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 mm?
Answer:
[tex]8.21\times 10^{5} N/C[/tex]
Explanation:
From kinematics we know that
[tex]v^{2}=u^{2}+2as[/tex] where v and u are final and initial velocities respectively, s is the distance covered, a is acceleration.
We know that initial velocity is zero hence
[tex]v^{2}=2as[/tex] and making a the subject then
[tex]a=\frac {v^{2}}{2s}=\frac {(3.8\times 10^{7})^{2}}{2\times 5\times 10^{-3}}[/tex]
From Newton’s law, F=ma and also qE=ma hence [tex]E=\frac {ma}{q}=\frac {9.109\times 10^{-31}\times 1.444\times 10^{17}}{1.602\times 10^{-19}}=8.21\times 10^{5} N/C[/tex]
