See attached ouar iestnen e ca la e TAa dhap t Electrons leave the negative plat

Question

See attached

ouar iestnen e ca la e TAa dhap t Electrons leave the negative plate and accelerate toward the positive. Some pass through a hole in the positive plate and strike the the surface coated with phosphors which then glow. In between the trajectory of the electrons is steered by lesser electric fields created by other plates. The negative plate on the left would attract positive cations, and is thus called the Cathode. This is where Rays emanate in the Tube, hence the name. postively charged metal plate negatively charged metal plate Say the the plates on the left are 0.05 m apart and create an electric field of 1500 N/C. What would be the electrostatic force on an electron between the plates? N Use E notation if needed. es What would be the acceleration of the electton? m/s Starting from rest at the negative plate, what would be the electron's velocity when it arrived at the positive plate? m/s How much time would it take the electron to travel between the plates?

Explanation / Answer

1] CRT

Electric field strength = 1500 N/C

Force on an electron will then be:

F = eE = 1.6 x 10-19 x 1500 = 2.4 x 10-16 N

acceleration of the electron can be found using F = ma

where m = mass of the electron

so, acceleration of the electron will be: a = F/m = (2.4 x 10-16/ 9.1 x 10-31) = 2.637 x 1014 m/s2

now use v2 = u2 + 2aS

here, S = 0.05 m, u = 0 m/s

so, v = [2 x 2.637 x 1014 x 0.05]1/2 = 5135525.91 m/s

this is the velocity of electron at the positive plate

to get the time in which the electron reaches the positive plate, use v = u + at

5135525.91 = 0 + 2.637 x 1014 t

=> t = 1.947 x 10-8 s.

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