Question

Beyond the deflection plates In this field-free region the electrons will continue to move with a perpendicular velocity component vi acquired between the deflection plates. Since the time to go the distance D is D/vo, the extra deflection occurring between the end of the deflection plates and the screen will be: m v 0 Combining (4) and (5) gives the total electron deflection eE L(L You see from (6) that the deflection of the spot on the screen is directly proportional to the strength of the deflection field, and inversely proportional to the square of the velocity of the electrons. It must follow then, that the deflection is directly proportional to the voltage applied to the deflection plates, Vd, and inversely proportional to the first power of the accelerating potential Vaccel. This can be shown by substituting equations (1) and (2) into equation (6) to obtain V L(L 2d 2 since d, L and D are constants of the apparatus: accel

please answer these two questions

A) Do these two graphs support the theory as summarized in (8)? Explain what it is about the plots that let you say this?.

B) Does (8) make sense? To answer this, discuss the physics of why the electrons deflect more or less for each of the two trials. That is, what is happening to the electron in the tube to make it deflect more when Vd is increased? What is happening to the electron in the tube to make it deflect more when Vaccel -1 is increased (means electron is deflected less when Vaccel is increased) ?

Answer 1

In electrostatic deflection, the spot is defected on the screen by applying voltage on the vertical or horizontal deflection plates. Now Let us consider a situation where an electron beam is accelerated through a potential & passes through a defection field of electric field. D rightarrow separation between deflection plates D rightarrow Distance of plates to screen l rightarrow length of plates v_d rightarrow deflection voltage v_accer rightarrow Accelerating voltage of e^- since e^- is free, its motion can be expressed by 1/2 mv^2_L = e V_acces rightarrow v^2_L = 2 e V_accel/m (1) [i.e Total = Total Electrostatic energy = qv] The acceleration of e^- a = force exerted on e^-/mass = e V_d/d/m This is so becoz Force = q times Electric Field = (e times v_d/d) Thus, time taken by e^- to move through plates is l/v_1 Therefore, deflection velocity along up - dir. [i.e t = /speed]