x c Advanced Physics ques x c Cheggcom WA ch 232 x Physics C

x c Advanced Physics ques x c Chegg.com WA ch 23-2 x Physics Capacitors (3 of x C www.webassign.net 13507624 Cathode-ray tubes (CRTs) are often found in oscilloscopes and computer monitors In the figure below an electron with an initial speed of 6.60 x 106 m/s is projected along the axis midway between the deflection plates of a cathode-ray tube. The uniform electric field between the plates has a magnitude of 1.12 x 103 v/m and is upward 2.0 em 6.0 cm 12.0 cm (a) What is the force (magnitude and direction) on the electron when it is between the plates? magnitude 1.79e-16 direction o upward downward (b) What is the acceleration of the electron (magnitude and direction) when acted on by the force in part (a)? magnitude 1.96e14 m/s direction upward O o downward (c) How far below the axis has the electron moved when it reaches the end of the plates? 88376 mm (d) At what angle with the axis is it moving as it leaves the plates? 5.096 x o below the axis (e) How far below the axis will it strike the fluorescent screen S? Cm ABP

Solution

Vertical force:

F = E*q = 1.12*10^3*1.6*10^-19 = 1.79*10^-16 N

So, acceleration:

a = F/m = 1.79*10^-16/(9.1*10^-31) = 1.97*10^14 m/s2

time taken , t = 0.06/(6.6*10^6) = 9.09*10^-9s

So, vertical displacement, s = 0.5*at^2 = 0.5*1.97*10^14*(9.09*10^-9)^2

= 8.14*10^-3 m = 8.14 mm <--------answer (c)

angle = atan((1.97*10^14*9.09*10^-9)/(6.6*10^6))

= 15.2 deg <-------answer (d)

e)

Now, time required, t\' = 0.12/(6.6*10^6) = 1.82*10^-8 s

Using the equation ,

s = u*t\' = (1.97*10^14*9.09*10^-9)*(1.82*10^-8) = 0.0326 m

So, total distance in vertical = 0.0326 + 0.00814

= 0.04074 m

= 4.07 cm <-----answer