Problem 2.38 XYour answer is incorrect. Try again In a cathode ray tube (CRT) si
ID: 1882912 • Letter: P
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Problem 2.38 XYour answer is incorrect. Try again In a cathode ray tube (CRT) similar to those used in older television sets, a beam of electrons is steered to different places on a phosphor screen, which glows at locations hit by electrons. The CRT is evacuated, so there are few gas molecules present for the electrons to run into. Electric forces are used to accelerate electrons of mass 9 x 10-31 kg to a speed 1.3 x 107 m/s, after which they pass between positively and negatively charged metal plates that deflect the electron in the vertical direction (upward in the figure, or downward if the sign of the charges on the plates is reversed) while an electron is between the plates, it experiences a uniform vertical force 1.05 × 10-15 N, but when the electron is outside the plates there is negligible force on it. The gravitational force on the electron is negligibly small compared to the electric force in this situation. The length of the metal plates is d = 0.02 m and the phosphor screen is a distance L = 0.5 m from the metal plates. Where does the electron hit the screen? (That is, what is yf, the distance above the original line of motion?)Explanation / Answer
speed along x axis remains constant at 1.3*10^7 m/s.
force along y axis=1.05*10^(-15) N
acceleration along y axis=force/mass=1.1667*10^15 m/s^2
time taken to travel length of the plate=d/speed along x axis
=0.02/(1.3*10^7)
=1.5385*10^(-9) seconds
vertical speed achieved at the end of this time period=initial vertical speed+acceleration*time
=0+1.11667*10^15*1.5385*10^(-9)
=1.795*10^6 m/s
time taken to travel distance of length L=L/horizontal speed
=0.5/(1.3*10^7)
=3.8462*10^(-8) seconds
vertical distance travelled within this time=yf=vertical speed*time
=1.795*10^6*3.8462*10^(-8)
=0.069039 m
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