An electron with kinetic enery of 6 x 10e J is moving to the right along the axi

Question

An electron with kinetic enery of 6 x 10e J is moving to the right along the axis of a cathode-ray tube as shown below. There is an electric field-(6x 10* N/G) j in the region betwen the deflection plates. Everywhere else, E-o Deflection plates 1) How far is the electron from the axis of the tube when it reaches the end of the plates? 1.0710A7 mm Submit You currently have 5 submissions for this question. Only 10 submission are allowed. You can make 5 more submissions for this question. 2) At what angle is the electron moving with respect to the axis? (Pasitive angle measured counterclockwise with respect to the axis) Subm You currently have 1 submissions for this question. Only 10 submíssion are allowed You can make 9 more subrmissions for this question. 3) At what distance from the axis will the electron strke the fluorescent screen? 2.93 You currently have 1 subenissions for this question. Only 10 submission ore allowed You can make 9 more submislons for this question. MacBook Pro 2 3 4 5 6

Explanation / Answer

kinetic energy = 6 x 10-16 J

v = sqrt(2K/m) = sqrt(2 x 6 x 10-16/(9.11 x 10-31)) = 3.62937 x 107 m/s

as E is in +y direction, the electron will have an acceleration in -y direction,

a = qE/m = 1.602 x 10-19 x 60000/(9.11 x 10-31) = 1.055104281 x 1016

1) distance from axis while leaving plates = 0.5a(d/v)2 = 0.5 x 1.055104281 x 1016 x (0.04/3.62937 x 107)2 = 0.006408 m = 6.408 mm

2) vy = a(d/v) = 1.055104281 x 1016 x (0.04/3.62937 x 107) = 1.16285 x 107 m/s

angle = 360 - tan-1(vy/v) = 360 - tan-1(1.16285 x 107/3.62937 x 107) = 342.23o

3) in next 12 cm , distance in y direction = (0.12/v)vy

=> distance from axis = 0.006408 + 0.12vy/v = 0.006408 + 0.12(1.16285 x 107/3.62937 x 107) = 0.044856 m = 44.856 mm

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