) This question requires a numeric or algebraic answer. In order to receive full
ID: 2104759 • Letter: #
Question
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The electron gun in a TV picture tube accelerates electrons between two parallel plates 1.20 cm apart with a 25.0 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, and then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or the potential. (a) What is the electric field between the plates? (b) What is the exit speed of an electron if its entry speed is close to zero? (Points : 20)
Explanation / Answer
The electrons accelerated by this gun are not relativistic. You need electrons well over 100 keVs in order to have significant relativistic corrections. The speed of the electrons accelerated by this gun is just
KE = 1/2 mV^2
solving for V
V = sqrt( 2 KE / m)
substituting values we get
V = sqrt(2*25E3 *1.6E-19/9.11E-31) = 9.74E+07 m/s
where I used the conversion 1 eV = 1.6E-19 Joules. The value for the speed is less than 1/3 the speed of light so a relativistic correction is really not necessary. If you insist on treating these electrons as relativistic, then you have to use the following equation for the kinetic energy
KE = mc^2 - moc^2 = mo c^2( 1/sqrt(1-V^2/c^2) - 1)
solving for V we get
V = c sqrt(1 - 1/ (KE / mo c^2 +1)^2)
substituting values
V = 3E8*sqrt(1 - 1/ (25E3*1.6E-19 / (9.11E-31*3E8*3E8) +1)^2)
V = 9.37E+07 m/s
as you can see, the relativistic value for the speed is just about the same quantity as the classical so relativistic corrections are not necessary.
ah! the electric field strength is
E = 27/1.2E-2 = 1800 kV/m
calculations are a bit approximation so check out once
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