An electron is launched at a 45 angle with a speed of 50106

An electron is launched at a 45° angle with a speed of 5.0×106 m/s from the positive plate of the parallel plates shown in the figure below. The electron lands 4.0 cm away. Determine the electric field that exists in the space between the plates.

Solution

Electron will undergo a projectile motion
Range , R= (v^2/a) sin 90

0.04 = (v^2/a) sin 90 => a = v^2/0.04 = (5*10^6)^2/(0.04) = a = 6.25*10^14 m/s^2

Force on electron , F= ma = (9.11* 10^-31)*6.25 x 10^14 = 5.7*10^-16 N

F= eE => E = F/e = ( 5.7*10^-16)/(1.6x10^-19) = 3.6x10^3 N/C