The electron gun in an old TV picture tube accelerates electrons between two par
ID: 1534547 • Letter: T
Question
The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4 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.
What is the electric field strength between the plates?
E = ______
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.
v_exit = ______
Explanation / Answer
a) E = delta_V/d
= 30*10^3/(1.4*10^-2)
= 2.14*10^6 N/C
b) use, Workdone on elctron = gain in kinetic energy
F*d = (1/2)*m*v^2
q*E*d = (1/2)*m*v^2
==> v = sqrt(2*q*E*d/m)
= sqrt(2*1.6*10^-19*2.14*10^6*1.4*10^-2/(9.11*10^-31))
= 1.02*10^8 m/s
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