Electron gun accelerating electrons over 2.80 cm to 9.60% speed of light

[Faiza]

Q26. In the neck of the picture tube of a certain black-and-white television set, an electron gun contains two charged metallic plates 2.80 cm apart. An electric force accelerates each electron in the beam from rest to 9.60% of the speed of light over this distance. (a) Determine the kinetic energy of the electron as it leaves the electron gun. Electrons carry this energy to a phosphorescent material on the inner surface of the television screen, making it glow. For an electron passing between the plates in the electron gun, determine (b) the magnitude of the constant force acting on the electron, (c) the acceleration, and (d) the time of flight.

 

[Faiza]

i need help i dun know what to do

 

[Faiza]

im so lost on this supplementry problme and the text is of no help

 

[Faiza]

can anyone give me some hints how to solve this porblem, because I am thinking the energy of the electron wud be 1/2mv^2

 

[Faiza]

:'( ......im so lost

 

[Gopi Prashanth]

You know the mass of the electron...

Hence, you were right in thinking 1/2 mv2 is the energy of the electron and v = 0.96C where C is the speed of light...

Now you know the final velocity and the initial velocity of the electron and the distance it travels... come in Newton... equation of motion...

V2 - U2 = 2*Acc*Distance

Which will give you Acc...
Force = Mass * Acc... so there you go...

Proceed along those lines and you should be good...

 

[Faiza]

its 9.60%=9.60/100=0.096c?? right I am sure that was juss a typo??

 

[Faiza]

im going to try it

 

[Faiza]

(a) 3.7*10^-16 Joules
(b)F=1.34*10^.14 Newtons
(c)a=1.48*10^16 m/s^2
(d) time of flight=1.94*10^6 seconds
please check my answers if u have time please