Question

7. An electron with a charge of -1.60 x 10-19 C and a mass of 9.10 x 10-31 kg passes between two charged metal plates. The electric field in the region between the plates is directed downward with a magnitude of 100. N/C (see figure below). The electron enters the uniform electric field region between the plates with an initial horizontal velocity of 3.00 x 106 m/s, and traverses a horizontal distance of 4.00 cm before exiting the plates. (a) What is the vertical deflection of the electron as it exits the uniform field region?, (b) What is the vertical component of its final velocity?, (c) At what angle does it exit? (Note: In solving this problem you can neglect the effect of the gravitational force on the electron's path. Changes in the electron's trajectory and velocity are due to the electric force acting on the electron.)

Answer 1

We can get the all parameters by using equations of motion as solved below
Charge of electron q = -1 . 60 times 10^- 19 c mass of electron m = 9 . 1 times 10^- 3 kg electric field E = 100 N/ C initial Horizontal velocity of electron v_x = 3 times 10^6 m/s Horizontal distance covered = 4 cm Time taken by electron to cover 4 cm in Horizontal direction t = x/v_x = 4 times 10^- 2/3 times 10^6 = 1 . 33 times 10^- 8 sec Vertical deflection of electron when it exist by 2 nd equation of motion s y = 4 y t + 1/2 a y t^2 sy = 1/2 e E/m t^2 u y = 0 ay = e E/m acceleration in vertical direction, m /s^2 sy