The electron gun in a TV picture tube accelerates electrons between two parallel
ID: 2050436 • Letter: T
Question
The electron gun in a TV picture tube accelerates electrons between two parallel plates 1.3 cm apart with a 22 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.A)What is the electric field strength between the plates?
Express your answer using two significant figures.
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 using two significant figures.
Explanation / Answer
The electron gun in a TV picture tube accelerates electrons between two parallel plates 1.0 apart with a 21? The electron gun in a TV picture tube accelerates electrons between two parallel plates 1.0cm 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. has two parts part a) What is the electric field strength between the plates? Express your answer using two significant figures. = answer should be in N/C 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 using two significant figures. answer should be in m/s. Can someone please try and help me to attempt this problem, or just solve so i can prepare for my exam ...please someone who knows physics properly answer this question, thank you very much. a) E = V / d = 2.1 . 10^4 / 10^–2 = 2.1 . 10^6 N/C b) Exit energy = V . e = 2.1 . 10^4 . 1.6 . 10^–19 = 3.36 . 10^–15 J v = v 2 . energy / mass = v 2 . 3.36 . 10^–15 / 9.1 . 10^–31 = 8.59 = 8.6 . 10^7 m/s
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