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!

Circuitry and then some

  1. May 20, 2006 #1
    Electrons travelling in a circular path of a radius 2m at a rate of 1.0*10^6 revolutions per second are diverting into a region between two charged plates iwth a potential difference of 40V. These plates are 3.2cm long and are seperated by 8mm. As the electrons emerge from the region between the plates, what is the amount of deflection from the original path? (If you want a diagram i'll sketch one up really quick)

    Anyways, to start..

    I tried to caluclate the velocity of the electron by using v = (2)(pi)(r) / T. I then took that velocity into the formula Vf^2 = Vi^2 + 2ad, and found out its final velocity. Before I did that though, I found out the electric field between the plates, then found the electric force using that formula. Then, using F = ma, I found the acceleration. I took everything I got into Ei = Ef, which is Kinetic Inital = Potential Final + Kinetic Final.

    I keep getting a negative "h" (which I planned to use for "d"), so I must be doing something wrong... can someone help me out? Thanks!!
  2. jcsd
  3. May 20, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hold on a minute...:wink:
    You realize that this equation must be applied separately for x and y components, right? It is really an equation for *components* of the velocity, NOT for the speed!! It should be [itex]V_{yf}^2 = V_{yi}^2 + 2 a_y (\Delta y) [/itex] and similarly for the x direction. If we use an axis where the E field is vertical and the particle has an initial veloicity to the right (along x), then in that equation V_{iy} is *zero*.

    this is true (at the condition of setting the electric potential at the initial position to be zero) BUT that will *not* help you because, again, the Vs appearing in the previous equation are *not* speed.

    The way to do this is : notice that v_x will not change since there is no acceleration along x. Find the time that it takes to get across the plate, using v_x. Now, use this time in the y direction and the value of a_y to find the total deflection, using [itex] y_f =y_0 +V_{iy} t + {1 \over 2} a_y t^2 [/itex] (notice that you may set y_0 =0 and that v_{iy} is zero).

    And btw, there is nothing wrong in getting a negative value of y_f, depending on the direction of the E field, the particle could be deflected downward. They probably want the absolute value of this anyway.


    EDIT: a typo that prevented an equation to show up was corrected
    Last edited: May 20, 2006
  4. May 20, 2006 #3
    The section in bold confuses me... what were you trying to say there?
    Also, I may be missing something, but how would you find the acceleration in the y direction? I'm a little confused from your explanation... :frown:
  5. May 20, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well, you said yourself in your first post that you had already found the acceleration! Since you know the potential difference between the plates and their separation, you can find the electric field. Then you can calculate the electric force on an electron and then the acceleration is the force over the mass. I thought that's what you said you had calculated in your first post.

    I made a typo so my equation did not show up properly. Sorry I will correct it. You will recognize the equation, it's a basic equation of Mechanics

  6. May 20, 2006 #5
    Ah, yes you're right. Some of your terms confused me I believe. Thanks for the help!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Circuitry and then some
  1. Circuitry Question (Replies: 10)