The electron gun in an old CRT television accelerates electrons between two char

Question

The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor. a. What is the electric field strength and direction between the plates? b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity] c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have? d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?

Explanation / Answer

a) electric field strength E = V/L = 25*10^3/0.012 = 2.08*10^6 v/m from anode to cathode

b) potential energy will convert in to kinetic energy of electron

e*V = (Me*v^2)/2 , v is exiting speed of electron

=> v = [2*e*V/ Me]^0.5 = [2*1.602*10^-19*25*10^3/(9.1*10^-31)]^0.5 = 9.38*10^7 m/s

c) C= 1nF , V = 25 kV

=> charge on each plate Q = C*V = 1*10^-9*25*10^3 = 25*10^-6 C

number of electron = Q/e = 25*10^-6/(1.602*10^-19) = 15.60*10^13

d) workdone = e*V = 1.602*10^-19*25*10^3 = 40.05*10^-16 J

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