CP Deflection in a CRT. Cathode-ray tubes (CRTs) are often found in oscilloscope
ID: 1536774 • Letter: C
Question
CP Deflection in a CRT. Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors. In Fig. P23.65 an electron with an initial speed of 6.50 times 10^6 m/s is projected along the axis midway between the deflection plates of a cathode-ray tube. The potential difference between the two plates is 22.0 V and the lower plate is the one at higher potential, (a) What is the force (magnitude and direction) on the electron when it is between the plates? (b) What is the acceleration of the electron (magnitude and direction) when acted on by the force in part (a)? (c) How far below the axis has the electron moved when it reaches the end of the plates? (d) At what angle with the axis is it moving as it leaves the plates? (e) How far below the axis will it strike the fluorescent screen S?Explanation / Answer
part a:
as lower plate is at higher potential, electric field is from lower plate to higher plate.
electric force=charge*electric field
as charge of electron is negative , direction of force is oppoiste to that of the electric field
hence force on the electron is from upper plate to lower plate.
force magnitude=1.6*10^(-19)*(22/0.02)=1.76*10^(-16) N
part b:
acceleration=force/mass
=1.76*10^(-16)/(9.1*10^(-31))=1.9341*10^14 m/s^2
acceleration direction is same as direction of force i.e. from top plate to bottom plate.
part c:
time taken to cross the plates=horizontal distance/horizontal speed
=0.06/(6.5*10^(6))=9.2308*10^(-9) seconds
then vertical distance fallen by the electron in that time
=0.5*acceleration*time^2 (as initial speed=0 in vertical direction)
=0.5*1.9341*10^14*(9.2308*10^(-9))^2=8.24*10^(-3) m
part d:
vertical velocity achieved=acceleration*time
=1.9341*10^14*9.2308*10^(-9)
=1.7853*10^6 m/s
then angle below the axis=arctan(vertical speed/horizontal speed)
=arctan(1.7853/6.5)
=15.358 degrees
part e:
time to cover 12 cm horizontally=0.12/(6.5*10^6)=1.8462*10^(-8) seconds
vertical distance fallen in that time
=vertical speed * time
=1.7853*10^6*1.8462*10^(-8)
=0.03296 m
so total distance below axis=8.24*10^(-3)+0.03296=0.0412 m
=4.12 cm
so it will strike 4.12 cm below the axis.