Answer:
Electric field, E = 40608.75 N/C
Explanation:
It is given that,
Mass of electrons, 
Initial speed of electron, u = 0
Final speed of electrons, 
Distance traveled, s = 6.3 cm = 0.063 m
Firstly, we will find the acceleration of the electron using third equation of motion as :



Now we will find the electric field required in the tube as :



E = 40608.75 N/C
So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.
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