Question

An electron is released between two metal plates 40.0mm apart experiences an acceleration of 1x10^12m/s^2. Determine the potential difference across the plates.

Answer 1

F=ma

E=F/e=ma/e
E=Δφ/Δx =>
Δφ =E Δx= ma Δx /e=
=9.1•10⁻³¹•10¹²•0.04/1.6•10⁻¹⁹=0.2275 V