Electric Fields and vertical deflection

[nick85]

We usually go over stuff after getting the assignments, so I have no clue how to start this question.
Figure 23N-14 shows the deflection-plate system of a conventional TV tube. The length of the plates is 3.2 cm and the electric field between the two plates is 10^6 N/C (vertically up). If the electron enters the plates with a horizontal velocity of 3.8*10^7 m/s, what is the vertical deflection y at the end of the plates?
I think F=-1.6E10 but have no clue where the direction of the electric field plays in.

 

[berkeman]

It's a trick question -- conventional TV tubes use magnetic deflection coils, not deflection plates. LOL.

Conventional TV tubes do use electrostatic lenses to help form the beam though, so I'll cut the textbook a little slack. The force from the electric field is F=qE, so the force acting on the electron is directly opposite of the E-field vector (because the charge on the electron is negative). Get it now?

 

[nick85]

Yes, but I still can't get the vertical distance for some reason knowing that the time is 8.421E-10 and using F=ma with a=(-1.6E-13)/(9.1E-31)=-1.756E17.

 

[nick85]

I was also wondering if anyone could help:
1. Two charges, -15 and +4.4 µC, are fixed in place and separated by 3.0 m.
(a) At what spot along a line through the charges is the net electric field zero? Locate this spot relative to the positive charge. Let the positive charge be at x = 0 and the negative charge be at x = 3.0 m.
(b) What would be the magnitude of the force on a charge of +23 µC placed at this spot?
I know for a that you must get E1=E2 and b is also dependent on F=kq/r^2.