The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.3 cm apart with a 21 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. Required:
a. What is the electric field strength between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero?

a. Electric field strength is 1.6*10^6 V/m = 1.6 Megavolt/meter

b. speed of an electron is 5.2*10^14m/s

Explanation:

The electric field strength is given by:

E = ΔV/d

E = electric field strength,

ΔV = potential difference,

d = plate spacin

ΔV = 21×10^3V, d = 1.3×10^-2m

a) E = V / d = (21*10^3)/(1.3×10^-2) = 1.6*10^6 V/m = 1.6 Megavolt/meter

b) Exit energy = V . e

Where 'e' is the charge of electron (1.6 * 10^(-19)

Exit energy = (21*10^3)*(1.6*10^-19) = 3.36*10−15

From the above, speed v = √2*Exit Energy/mass

Where 'm' is the mass of electron (9.1 * 10^(-31)

v = √2*energy/mass

v = √2*(3.36*10^-15)/(9.1*10^-31) = 5.2*10^14m/s

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