How Does the Work-Kinetic Energy Theorem Apply to Electron Acceleration in TVs?

[sn3t]

Homework Statement

In the neck of the picture tube of a certain black-and-white television set, an electron gun contains two charged metallic plates 2.80cm 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 material making it glow. For an electron passing between the plates in the electron gun, determine b) the magnitude of the constant electric force acting on the electron, c) the acceleration, and d) the time of flight.

Electron mass = 9.10938188 * 10^-31 kg
C = 299792458 m/s
Electron velocity = 28780075.968 m/s (.0960 * C)
Distance between plates = .028m

Homework Equations

KE = 1/2m* v^2
F= mass * acceleration

The Attempt at a Solution

a) KE = 1/2m * v^2
= 1/2(9.10938188*10^-31kg) * ( 28780075.968m/s)
= 3.77261759 * 10^-16 Joules
b) F=m*a
= (9.10938188*10^-31kg) * [ (28780075.968m/s)^2 / (.028m) ]
= 2.694722685 * 10^-14 N
c) a = F/m
= 2.694722685*10^-14N / 9.10938188*10^-31kg
= 2.9581839*10-46 m/s^2
d) ?

and I am not sure if my parts b and c are correct

 

[ehild]

b is not correct, you should divide the KE by the distance.

ehild

 

[sn3t]

ah yea you're right it's N*m then when I divide it cancels out.

w= f * d

so ke = f * d
ke/d = f?

 

[ehild]

The change of the KE is equal to the work. But initially, KE=0.

Just divide your 3.7726 * 10^-16 Joules by 0.028 m.

ehild