Answer:
Electric field, E = 40608.75 N/C
Explanation:
It is given that,
Mass of electrons, [tex]m=9.1\times 10^{-31}\ kg[/tex]
Initial speed of electron, u = 0
Final speed of electrons, [tex]v=3\times 10^7\ m/s[/tex]
Distance traveled, s = 6.3 cm = 0.063 m
Firstly, we will find the acceleration of the electron using third equation of motion as :
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
[tex]a=\dfrac{(3\times 10^7)^2}{2\times 0.063}[/tex]
[tex]a=7.14\times 10^{15}\ m/s^2[/tex]
Now we will find the electric field required in the tube as :
[tex]ma=qE[/tex]
[tex]E=\dfrac{ma}{q}[/tex]
[tex]E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}[/tex]
E = 40608.75 N/C
So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.
