The electron gun in an old TV picture tube accelerates electrons between two par

Question

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.0 cm apart with a 22 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.

With what speed does an electron exit the electron gun if its entry speed is close to zero? Note: The exit speed is so fast that we really need to use the theory of relativity to compute an accurate value. Your answer to part B is in the right range but a little too big.

I know the field strenght between the two plates is 2.2*10^6 but i dont know how to use that to get the answer, can someone show me the steps the the anwser please?

Explanation / Answer

V = potential difference = 22000 volts

d = distance between plates = 1 cm = 0.01 m

E = electric field = V/d = 22000/0.01 = 2.2 x 106 N/C

m = mass of electron = 9.1 x 10-31 kg

q = charge on electron = 1.6 x 10-19 C

a = acceleration = qE/m = (1.6 x 10-19) (2.2 x 106 ) / (9.1 x 10-31 ) = 3.87 x 1017 m/s2

Vi = initial velocity = 0 m/s

Vf = final velocity

x = displacement = 0.01 m

using the equation

Vf2 = Vi2 + 2 a x

Vf2 = 02 + 2 (3.87 x 1017 ) (0.01)

Vf = 8.8 x 107 m/s

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