1. Not finding help here? Sign up for a free 30min 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!

Electric Field & Capacitor Problem HELP

  1. Jul 5, 2007 #1
    Electric Field & Capacitor Problem...HELP!!

    An electron enters the lower left side of a parallel plate capacitor & exits at the upper right side making an upward curve. The initial speed of the electron is 4.00 x 10^6 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

    So I understand that F=ma=qE. I also understand that I can use kinematics to find the acceleration and I can also use the mass of an electron to figure out the force. However, how do I find q? I'm so confused!!!:confused: Please help! Thanks!!
     
  2. jcsd
  3. Jul 5, 2007 #2

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    q is the charge on an electron. Look it up in your text.
     
  4. Jul 7, 2007 #3
    q=e=1.6x10^{-19} coulombs...
     
  5. Jul 8, 2007 #4
    in the vertical direction, the electron has travelled 0.15 cm (from lower left corner to upper right corner). the initial speed in the vertical direction is 0
    => distance = a*t^2/2 (a= acceleration and t- time)
    the time is found by: t= d/v (in horizontal direction) = 2 cm/ speed
    so now you can find a, and then E.
    Good luck.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electric Field & Capacitor Problem HELP
Loading...