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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?

Respuesta :

oyejam

Answer:

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