First Name Last Name 24 marks) You have been asked to design a screen for a cath
ID: 1596579 • Letter: F
Question
First Name Last Name 24 marks) You have been asked to design a screen for a cathode ray tube television You have access to below the left ie an electrom source which shoots electron at the speed (10 between a pair of metal plates towards the A sem high screen. Theme metal plates to la this problem we t tie horizoatal dellection of electrons hat will verrie a be required to create t images A poorl on the ntillating screen light- up if an electron fall on it A uniform vertical electric field exists between the sem wide metal plate- becaste the potential difference created by the battery. You can control he strengti of this electric field f of C. ad the n of Lo 5000NA ip steps o 22 pixels required to create at the hig resolution a shape of height 3 cm. Front view of screen The parts below walk you through related Nuestions, and the steps tith which to solve this problem. Please show all work in the provided bozes or on the figures. Ensure that you eapress tnits for all numerical unswers. 1. (1 mark) on the figure below draw the trajectory of an electron as it passes through the metal plates figure below draw the trajectory of an electron as it passes through the metal plates if 2. (1 mark) on the the polarity of the plates is reversedExplanation / Answer
Given
electron speed, v = 7*10^6 m/s
height of the screen, h = 0.05 m
Range of electric field, E = -5k N/C to 5k N/C
steps, dE = 225 N/C
height of shape, h' = 0.03 m
plate width, w = 0.05 m
now, every dE electric field change will create corresponding dH change in the final location of the electorn on the scintillating screen
at electric field E
Force due to the field = qE
acceleration = qE/m ( m is mass of electron)
time spent in field, t = w/v = 0.05/7*10^6 = 7.142*10^-9 s
displacement from central axis, s = 0.5*qE*t^2/m = qE*w^2/2mv^2
for subsequent field dE
ds = qw^2dE/2mv^2
so smallest pixel width in 1d = ds = q*0.05^2*225/2m(7*10^6)^2 = 1.0091 mm
number of pixels for 3 cm height = 30 cm/1.0091 mm = 297.26 pixels
3. for the shape to be centred on the screen, the maximum displacement from the centre, s = 1.5 cm
0.015 = qE*w^2/2mv^2
E = 3344.25 V/m
4. smallest pixel size = 1.0091 mm
5. 297.26 , so apx 300 pixels
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