An electron with a horizontal speed of 2300000 m/s and no vertical component of velocity passes through two horizontal parallel plates, as shown. The magnitude of the electric field between the plates is 160 N/C. The plates are 3 cm long. Calculate the angle to the horizontal, 0 (in degrees), that the electron would pass through the plates with. e ++++++ TITI 11 11 11 || OII 0 y 2. X

a) Calculate the acceleration in meters per second squared of the electron if the field strength is 2.50 ✕ 104 N/C.

A simple and common technique for accelerating electrons is shown in the figure below, where a uniform electric field is created between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. (a) Calculate the acceleration of the electron if the field strength is 3.45 x 104 N/C. m/s² (b) Why will the electron not be pulled back to the positive plate once it moves through the hole? O The force of gravity is too strong. O There is no field outside the plates. O The other side of the positive plate also has a negative charge.

Help me answer this question please. An electron is projected horizontally at a speed of 2.5 x 10*6 m/s between two plates. metal parallel I = 7.5 cm in length as shown in the figure below. The greatness of electric field is 130 N/C. Determined : a) Acceleration of the electron b) The time it takes to escape from the metal plates c) The final vector velocity of the electron as it escapes from the plates