Hello. Sorry it's late. I was looking through my binder and I remembered I wanted to be the scribe for today, so here I am.

We corrected
Physics 40S Electric and Magnetic Fields Assignment
The Physics of a Cathode Ray Tube

And here are the answers:
a. PE = QE
= (1.6 x 10-19)(20 000)
= 3.2 x 10-15 J

b. Ep = Ek
Ek = 3.2 x 10-15 J
c. KE = PE
KE = ½mv22(3.2 x 10-15)
= 9.11 x 10-31 V6.4 x 10-15 = 9.11 x 10-3 1V
8.38 x 10 7m/s = V

d. T = d/v= 0.02 8.38 x 107
= 2.39 x 10-10 s

e. E = v/d
= 500 0.01
= 500 000 m
f. a = Eq/m = (500 000)(-1.6 x 10-19) 9.11 x 10-19 kg
= -8.78 x 1016 m/s2
g. V2 = V1 + at = 0 + (8.78 X 1016)(2.39 x 10 -10)
= 2.1 x 107 m/s
h. d = 1/2 at2 = ½ (8.78 x 1016)(2.39 x 10-10)
= 2.5 x 10 -3 m

We also took some notes
Cathode Ray Tube
· electrons leave the cathode and accelerate towards the anode
· the electric field, E, alone produces a force on the electrons Fe = qE causing the elections to deflect up to x
· the magnetic field, B, alone produces a force of electrons, Fb = Bqv causing the electrons to deflect down to y

1. The Find Speed V, of electrons adjust B and E until electrons hit point z
FB = Fc
Bqv = qE
V = E/B
2. To find `m` turn off E. B will deflect electrons to y where
FB = FC
Bqv/R = mv2/BqR
3. To find charge/ratio q/m = v/BR or V = E/B

Sorry if it turned out all weird. My computer's weird, I tried. I'm weird. Okay bye.