At the very rear of any traditional television picture tube (cathode ray tube, o
ID: 1688793 • Letter: A
Question
At the very rear of any traditional television picture tube (cathode ray tube, or CRT) is an electron gun, in which a stream of electrons are individually accelerated from rest through a potential difference of Vacc. Assuming a frictionless transfer of electric potential energy to the kinetic energy of the electron.a)find an expression for the electrons final speed V1 (only in terms of Vacc and any necessary physical constants).
The electron, now at this high speed V1, then passes horizontally into the region between two horizontal parallel plates. The plates have length L and are spaced a distance d apart, and have a potential difference of Vb. The electron enters exactly at the halfway point between the plates. Assume NO E-field fringing.
b) Find an expression for the magnitude of the acceleration of the electron while its between the plates.
c) what is the direction of the acceleration of the electron while it's BETWEEN the plates?
d) What is the electron's HORIZONTAL velocity component at the moment that it leaves the region between the plates?
e) What is the electron's VERTICAL velocity component at the moment that it leaves the region between the plates? Is it toward the TOP or BOTTOM of the page?
f) What is the electron's VERTICAL DISPLACEMENT at the same moment?
g) What is the maximum value that Vb can have so that the electron will NOT collide with one of the plates?
Explanation / Answer
initially the speed of electron is rest kinetic energy of electrons as they leave the gun k = ( 1 /2 ) m v_1^2 = e V v_1 ^2 = 2 e V / m where V = potential difference the speed of electrons as they leave the gun v_1 = ( 2e V_acc / m)^ (1/2) if there is no electric field between the horizontal deflecting plates the electrons enter the region between vertical deflecting plates with a speed v_x = ( 2e V_acc / m)^ (1/2)......(1) at constant up ward force the eE then acts on the elctrons and their upward acceleration is a = F / m = Ee / m a = e V_b / md.......(2) where electric field E = V_ b / d the horizontal component of velocity v_x is constant .the path of electrons region between the plates is parabolic .after electron emerg from this region their paths again becomes straight lines . time required for the electrons to travel length t = v_x / L..........(3) in this time the electron required the vertical component of velocity from equation (2) and (3) v _y = a_y t = (e V_b / md ) ( v_x /L )......(4) whenthe electrons emerging from deflecting filed their velocity makes angle with x - axis tan ? = v_y / v_x substitue the values of v_x and v_y values from equations (1) and (4) electrons vertical displacement y / D = v _y / V_x y = De V_b L / m d v_x^2 initially the speed of electron is rest kinetic energy of electrons as they leave the gun k = ( 1 /2 ) m v_1^2 = e V v_1 ^2 = 2 e V / m where V = potential difference the speed of electrons as they leave the gun v_1 = ( 2e V_acc / m)^ (1/2) if there is no electric field between the horizontal deflecting plates the electrons enter the region between vertical deflecting plates with a speed v_x = ( 2e V_acc / m)^ (1/2)......(1) at constant up ward force the eE then acts on the elctrons and their upward acceleration is a = F / m = Ee / m a = e V_b / md.......(2) where electric field E = V_ b / d the horizontal component of velocity v_x is constant .the path of electrons region between the plates is parabolic .after electron emerg from this region their paths again becomes straight lines . time required for the electrons to travel length t = v_x / L..........(3) in this time the electron required the vertical component of velocity from equation (2) and (3) v _y = a_y t = (e V_b / md ) ( v_x /L )......(4) whenthe electrons emerging from deflecting filed their velocity makes angle with x - axis tan ? = v_y / v_x substitue the values of v_x and v_y values from equations (1) and (4) electrons vertical displacement y / D = v _y / V_x y = De V_b L / m d v_x^2Related Questions
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