An electron in a computer monitor enters midway between two parallel oppositely

Question

An electron in a computer monitor enters midway between two parallel oppositely charged plates, as shown in the figure . The initial speed of the electron is 6.20×10^7 m/s and its vertical deflection (d) is 4.70 mm.What is the magnitude of the electric field between the plates and determine the magnitude of the surface charge density on the plates in C/m^2 .
Please include all steps in solving problem Thanks so much in advance!

An electron in a computer monitor enters midway between two parallel oppositely charged plates, as shown in the figure . The initial speed of the electron is 6.2010^7 m/s and its vertical deflection (d) is 4.70 mm.What is the magnitude of the electric field between the plates and determine the magnitude of the surface charge density on the plates in C/m^2 . Please include all steps in solving problem Thanks so much in advance!

Explanation / Answer

the length L = 10 cm = 0.1 m
width W = 1 cm =0.01 m
separation of the plates d = 4.7 mm = 4.7 * 10 ^-3 m
Area of the plate A = LW = 10^-3 m^ 2
Initial speed of the electron u = 6.20 * 10 ^ 7 m / s
vertical deflection y = 4.7 * 10 ^-3 m
In horizontal direction :
distance S = L = 0.1 m
velocity u = 6.20 * 10 ^ 7 m / s
So, time t = S / u = 1.612 * 10 ^-9 s
In vertical direction :
distance y = 4.7 * 10 ^-3 m
Initial velocity U = 0
time t = 1.612 * 10 ^-9 s
from the relation y = Ut + ( 1/ 2) at^ 2
y = ( 1/ 2) at^ 2
from this accleration a = 2y / t^ 2
                                  = 3.613 * 10 ^15 m / s^ 2
From Newton,s 2nn law, the force F= ma = Eq
from this electric field E = ma / q

    here q = 1.6 * 10^-19 C (charge of electron )
           m = 9.11 * 10 ^-31 kg ( mass of electron )
plug the values we get E = 20525 N / C (b) the surface charge density = 2(0)E here 0 = 8.85*10^-12 c^2/N-m    = 363.29*10^-9 C/m^2 (b) the surface charge density = 2(0)E here 0 = 8.85*10^-12 c^2/N-m    = 363.29*10^-9 C/m^2

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