Question

An electron and a proton are each placed at rest in an electric field of 520N/C. Calculate the speed of each particle 48ns after being released?

Answer 1

So the electron final speed is `V_e = 4.4×10^6 m/s` and the
proton’s final speed is `v_p = 2.4×10^3 m/s`,
`v_e ~~ 1833 v_p`

Explanation:

The force on the electron and proton are caused by the E field causing an acceleration that is proportional the mass to respective masses. From `F = qE`, and the fact that the magnitude of the electron’s charge is `1.60×10^-19 C`, the magnitude of the force on the electron is `F = |q|E = (1 .60×10^-19 C)(520N/C) = 8.32×10^-17 N`
Now the mass of the electron is `m_e =9 .11 × 10^-31 kg`, from Newton’s 2nd Law, the magnitude of its acceleration is then:
`a_e = F/m_e =(8.32×10^-17 kg m/s^2)/(9.11×10^-31 kg)=9 .13×10^13 m/s^2`
Now the electron started from rest and reaches a final speed of `v_(e_f) = V_(e_i) + at, " with " V_(e_i) = 0` we have
`V_(e_f) = at; (9 .13×10^13 m/s^2) (48×10^-9 s) = 4.4×10^6 m/s`
Similarly for the proton that a mass of `1.67×10^-27 kg`, it will end up having an acceleration about 1000 times smaller
`a_p = F/m_p = (8.32×10^-17 kgm/s^2)/(1.67×10^-27 kg) =4 .98×10^10 m/s^2`
And then the magnitude of the velocity 48ns after being released is
`v = a_pt = (4 .98×10^10 m/s2)(48×10^-9 s) = 2.4×10^3 m/s`. ABout

So the electron final speed is `V_e = 4.4×10^6 m/s` and the
proton’s final speed is `v_p = 2.4×10^3 m/s`,
`v_e ~~ 1833 v_p`