Answer:
v = 4.8 10⁻⁴ m / s
Explanation:
To solve this exercise we can use the concepts of energy. In this case the potential energy is transformed into kinetic energy
U = K
q V = ½ m v²
v = [tex]\sqrt { \frac{2qV}{m} }[/tex]
in the exercise they indicate the value of the charge q₁ = 1 pC = 1 10⁻¹² C
let's calculate
v = [tex]\sqrt{ \frac{2 \ 1 \ 10^{-12} 120 \ 10^3}{500 ^{2} }[/tex]
v = 4.8 10⁻⁴ m / s
