Answer: v = 2.75 * 10^7 m/s
Explanation: Since the electric field is a uniform one and the distance is small, the motion of the electron is of a constant acceleration, hence newton's laws of motion is applicable.
From the question
E=strength of electric field = 214N/c
q=magnitude of an electronic charge = 1.609* 10^-16c
m= mass of an electronic charge = 9.11*10^-31kg
v = velocity of electron.
S= distance covered = 1cm = 0.01m
a = acceleration of electron.
F = ma but F=Eq
Eq = ma.
a = Eq/m
a = 214 * 1.609*10^-16/ 9.11 * 10^-31
a = 344.32 * 10^-16/ 9.11 * 10 ^-31
a = 3. 779* 10^16 m/s²
Before the electron is accelerated, they are always not moving, hence initial velocity (u) = 0.
By using the equation of motion, we have
v² = u² + 2aS
But u = 0
v² = 2aS
v²= 2* 3.779*10^16 * 0.01
v² = 7.558 * 10^14
v = √7.558 * 10^14
v = 2.75 * 10^7 m/s.
