Problem 28.43 The electron gun in an old TV picture tube accelerates electrons b

Question

Problem 28.43 The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.3 cm apart with a 30 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 previous l 12 of 15 l next Part A What is the electric field strength between the plates? Express your answer to two significant figures and include the appropriate units E Value Units Submit My Answers Give U Part B 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 Express your answer to two significant figures and include the appropriate units exit Value Units Submit My Answers Give up

Explanation / Answer

A) required electric field is E = V/d = 30000/0.013 = 2.307*10^6 V/m

v = 30 kV = 30000 V
and d = 1.3 cm = 0.013 m

B) According to work energy theorem

Work done = change in kinetic energy

Change in kinetic energy = 0.5*m*vexit^2

Work done W= q*V

q is the charge of the electron = 1.6*10^-19 C

V = 30000 V

m is the mass of the electron = 9.11*10^-31 kg

then W = 1.6*10^-19*30000 = 4.8*10^-15 J

then o.5*m*vexit^2 = 4.8*10^-15

vexit = sqrt(4.8*10^-15/(0.5*9.11*10^-31)) = 1.02*10^8 m/s

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