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Find the surface charge density that will stop a proton

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Problem.png


    2. The attempt at a solution

    I have the solution, but it's pretty confusing so I want to see if anyone else can walk me through step by step. The first step is pretty clear...You have to find the acceleration that the proton will experience as it moves towards the upper (positive) plate.

    F = ma
    a = F/m
    F = qE
    a = (qE)/m

    a = -eE/m

    From there what do you do? Set it up in a kinematics equation? How does the surface charge density come into play?
     
  2. jcsd
  3. Sep 30, 2011 #2
    I think you might want to include fringing effects. Consider the electric field from a uniform finite plate.
     
  4. Sep 30, 2011 #3

    gneill

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    Staff: Mentor

    Yes, you'll want to set up a kinematics equation. In particular, you're interested in the motion in the vertical direction. The surface charge density will determine the strength of the field between the plates (how?).
     
  5. Sep 30, 2011 #4

    gneill

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    Staff: Mentor

    Note: A quicker solution might be had by going via conservation of energy. Then you can skip the kinematic equation involving position, acceleration, etc..

    What's the initial kinetic energy of the proton in the vertical direction? What's the energy gained (or lost) by a proton moving through a potential difference of V volts?
     
  6. Sep 30, 2011 #5
    alright, so I need to set up a kinematics. You have to avoid the variable t, so you can use the equation:
    V_f^2 = v_i^2 + 2ad
    The final velocity will be 0

    0 = vi^2 + 2ad
    vi^2/2a = d

    Since we only care about velocity and displacement in the y direction, the equation changes to

    d = (v_i^2)sin(30)/2(-e*E/m)
    What do you do from here?
    Since you know d can't be any more than 1cm do you plug that in, and solve for...E?

    oooo man. Then from E, you can solve for the linear charge density...
    correct?
     
  7. Sep 30, 2011 #6
    I don't know how to find the energy gained when it's moving through a potential difference. Maybe in a few weeks we'll go over that?
     
  8. Sep 30, 2011 #7
    There is something misleading about the picture, how are you guaranteed the proton to land at x = W?
     
  9. Sep 30, 2011 #8
    I'm not. But I feel like I don't care about that? All I need to figure out is the charge density to keep it from hitting the upper capacitor.
     
  10. Sep 30, 2011 #9

    gneill

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    Staff: Mentor

    Here's a spoiler then: ΔE = qV

    Charge multiplied by the voltage though which the charge "falls" is the energy imparted (or extracted).

    It's the basis of the electron-volt energy unit. One eV is the energy gained by an object (particle) with the charge of an electron (an elementary charge) falling though a 1 Volt potential.

    If your proton has a certain KE in the vertical direction and its traveling against a potential difference (due to an electric field), then it will come to a halt when q*V matches the KE. Keep in mind that the units of an electric field are Volts/meter.
     
  11. Sep 30, 2011 #10

    gneill

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    Staff: Mentor

    Yes, that'll work.
     
  12. Sep 30, 2011 #11
    I did this problem and my algebraic expression is large. I have't plugged in the numbers yet...

    I also assumed the proton lands at x = W
     
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