A proton moves at 4.50·105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60·103 N/C. Ignoring any gravitational effects, find:
  1. the time interval required for the proton to travel 5.00 cm horizontally,
  2. its vertical displacement during the time interval in which it travels 5.00 cm horizontally,
  3. the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally.


Il moto orizzontale è uniforme. Il tempo per percorrere 5 cm è: Δ t = Δ s v = 0.05 4.5 ⋅ 10 5 ≈ 1.1 ⋅ 10 − 7 s = 0.11 µs %DELTA t = {%DELTA s} over v = { 0.05 } over {4.5 cdot 10^5} approx 1.1 cdot 10^-7`s = 0.11`µs
Il moto verticale è uniformemente accelerato con accelerazione: a = F m p = e ⋅ E m p = 1.6 ⋅ 10 − 19 ⋅ 9600 1.67 ⋅ 10 − 27 ≈ 919760 m / s ² a= F over m_p = { e cdot E } over m_p = { 1.6 cdot 10^-19 cdot 9600 } over { 1.67 cdot 10^-27 } approx 919760`m/s² Lo spostamento verticale è dato dalla formula cinematica: Δ y = 1 2 a t 2 = 1 2 ⋅ 919760 ⋅ ( 1.1 ⋅ 10 − 7 ) 2 ≈ 5.7 · 10 − 3 m = 5.7 mm %DELTA y = {1 over 2} a t^2 ={1 over 2} cdot 919760 cdot left(1.1 cdot 10^-7 right)^2 approx 5.7·10^-3`m =5.7`mm
la componente orizzontale della velocità rimane la stessa: vx=4.50·105 m/s.
Quella verticale è data dalla formula cinematica: v y = a ⋅ t = 919760 ⋅ 1.1 ⋅ 10 − 7 ≈ 0.1 m / s v_y = a cdot t = 919760 cdot 1.1 cdot 10^-7 approx 0.1`m/s