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Homework Help: [Solved] Velocity of a Proton in a Capacitor

  1. Mar 19, 2009 #1

    1. The problem statement, all variables and given/known data

    A proton is fired horizontally with a speed of 1.1 106 m/s through the parallel-plate capacitor shown in Figure P34.37. The capacitor's electric field is = (1.3 105 V/m, down) and the distance between the plates is 31 millimeter.
    (Image Attached)

    The proton entered the capacitor gap 18.6 milli-meters above the lower plate. Where doe the proton strike the lower plate?

    Measured in meters from the front edge of the lower plate.

    2. Relevant equations

    Kinematics Equations

    3. The attempt at a solution

    There were other questions on this problem that I already solved for. So here are other things that I know:

    The magnitude of the acceleration= 1.245E13 m/s/s

    Voltage difference= 4030 V

    Now I've set up a kinematics table (seperate for horizontal and verticle):

    Known for horizontal: Initial V= 1.1E6; A= 1.245E13
    Uknown: distance, final velocity, time

    Known for verticle: x= .0186 meters; Initial v= 0
    Uknown: final velocity, acceleration, time

    Any suggestions? I've no idea where to go from here.

    Thanks in advance!


    Attached Files:

    Last edited: Mar 19, 2009
  2. jcsd
  3. Mar 19, 2009 #2


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    Homework Helper

    Re: Velocity of a Proton in a Capacitor

    Hi Phoenixtears,

    Since the electric field is downwards, what is the direction of this acceleration? I think the answer for this will change your kinematics table below.

  4. Mar 19, 2009 #3
    Re: Velocity of a Proton in a Capacitor

    Aha! I see. That makes so much more sense.

    Becasue the field is down, the acceleration is also down. Then the horizonatal acceleration is 0 (I should have listened to my sister :-p). Thank you so much!
  5. Mar 19, 2009 #4


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    Homework Helper

    Re: Velocity of a Proton in a Capacitor

    Sure, glad to help!
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