1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Electric Fields and vertical deflection

  1. Jan 17, 2006 #1
    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.
  2. jcsd
  3. Jan 17, 2006 #2


    User Avatar

    Staff: Mentor

    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?
  4. Jan 17, 2006 #3
    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.
  5. Jan 17, 2006 #4
    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.
    Last edited: Jan 17, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook