Question

1)Two oppositely charged horizontal plates are separated by a distance d =1 cm and each has a length L =3 cm, see Fig. below. The electric field between the plates is uniform and has a magnitude E =102 N/m. An electron is projected between the plates with a horizontal initial speed of 𝑣0=106 𝑚/𝑠 as shown. Assuming this experiment is conducted in a vacuum, where will the electron strike the upper plate? Repeat when a proton replaces the electron.

Answer 1

The electric force will curve the trajectory of the electron. Its magnitude is

$$F=qE.$$

The acceleration toward the positive plate it will give the electron can be found by Newton's second law:

$$a=\frac Fm=\frac{qE}m.$$

The time it will take to pass 1/2 of the gap between the plates is

$$t=\sqrt\frac{d}{a}=\sqrt\frac{dm}{qE}=23.8\text{ ns}.$$

Meanwhile the distance the electron will make in the horizontal direction is

$$x=v_0t=0.02\text{ m}.$$

2 cm from the point where it entered the plates.

With a proton, the result will be 1 m 2 cm.