When the terminals of a battery are connected to two parallel conducting plates with a small gap between them, the resulting charges on the plates produce a nearly uniform electric field E between the plates. If the plates are 1.0 cm apart and are connected to a 100-volt battery as shown in figure, the field is vertically upward and has magnitude E = 1.00·10⁴ N/C.

1. If an electron (e = -1.60·10^-9 C, m = 9.11·10^-31 kg) is released from rest at the upper plate, what is its acceleration?
2. What speed and kinetic energy does it acquire while traveling 1.0 cm to the lower plate?
3. How long does it take to travel this distance?

L'accelerazione dovuta al campo elettrico sull'elettrone è: $$a=\frac{F_e}{m}=\frac{e\cdot E}{m}=\frac{1.6\cdot {10}^{-19}\cdot 1\cdot {10}^4}{9.11\cdot {10}^{-31}}\approx 1.76\cdot {10}^{15}m/s\mathrm{²}$$ Dopo 1 cm di moto la velocità è data dall'equazione cinematica: $$v=\sqrt{2as}=\sqrt{2\cdot 1.76\cdot {10}^{15}\cdot 0.01}\approx 5.9\cdot {10}^{6}m/s$$ L'energia cinetica: $$K=\frac{1}{2}m{v}^2=\frac{1}{2}\cdot 9.11\cdot {10}^{-31}\cdot {(5.9\cdot {10}^6)}^2\approx 1.6\cdot {10}^{-17}J$$

Infine il tempo per percorrere 1 cm:
$$\mathrm{\Delta }t=\frac{v}{a}=\frac{5.9\cdot {10}^6}{1.76\cdot {10}^{15}}\approx 3.4\cdot {10}^{-9}=3.4\mathit{ns}$$