Answer:
t = 2.09 10⁻³ s
Explanation:
We must solve this problem in parts, first we look for the acceleration of the electron and then the time to travel the distance
let's start with Newton's second law
∑ F = m a
the force is electric
F = q E
we substitute
q E = m a
a = [tex]\frac{q}{m} \ E[/tex]
a = [tex]\frac{1.6 \ 10^{-19}}{ 9.1 \ 10^{-31} } \ 7.8 \ 10^{-9}[/tex]
a = 1.37 10³ m / s²
now we can use kinematics
x = v₀ t + ½ a t²
indicate that rest starts v₀ = 0
x = 0 + ½ a t²
t = [tex]\sqrt{\frac{2x}{a} }[/tex]
t = [tex]\sqrt{\frac {2 \ 3 \ 10^{-3}}{ 1.37 \ 10^3} }[/tex]
t = 2.09 10⁻³ s
