Please show all work, and if written by hand written please show cleraly. An ele

Question

Please show all work, and if written by hand written please show cleraly.

An electron with kinetic energy of 2 x 10-16 J is moving to the right along the axis of a cathode-ray tube as shown below. There is an electric field E = (5 x 104 N/C) J in the region between the deflection plates. Everywhere else, E :0. Deflection plates Fluorescent screen 4cm--** 12cm 1) How far is the electron from the axis of the tube when it reaches the end of the plates? 16.086 mm Submit You currently have 1 submissions for this question. Only 5 submission are allowed You can make 4 more submissions for this question. 2) At what angle is the electron moving with respect to the axis? (Positive angle measured counterclockwise with respect to the axis) Submit You currently have 2 submissions for this question. Only 5 submission are allowed You can make 3 more submissions for this question. Your submissions: 38.66 Computed value: 38.66 Submitted: Monday, March 5 at 3:09 PM Feedback: 3) At what distance from the axis will the electron strike the fluorescent screen? cm Submit You currently have O submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question.

Explanation / Answer

1) speed along the axis v = sqrt(2KE/m) = sqrt(2*2e-16/9.1e-31) = 20965696 m/s

time taken = 0.04/20965696 = 1.908e-9 s

deflection = 0.5 at^2 = 0.5*(5e4*1.6e-19/9.1e-31)*1.908e-9^2

= 0.016 m

= 16 mm

2) Vx =  20965696 m/s

Vy = at = -(5e4*1.6e-19/9.1e-31)*1.908e-9 = -16773626 m/s

angle = arctan(Vy/Vx) = arctan(-16773626/20965696) = -38.66 degree

= 360 - 38.66 = 321 degree answer

3) new time t = (0.04+0.12)/20965696 = 7.6315e-9

distance from axis = 0.5 at^2 = 0.5*(5e4*1.6e-19/9.1e-31)*7.6315e-9^2

= 0.256 m

= 25.6 cm

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.