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Surface Charge Density and high-speed proton beam

  1. Apr 12, 2007 #1
    You have a summer intern position at a laboratory that uses a high-speed proton beam. The protons exit the machine at a speed of 2.10 times 10^6 m/s, and you've been asked to design a device to stop the protons safely. You know that protons will embed themselves in a metal target, but protons traveling faster than 2.20 times 10^5 m /s emit dangerous x rays when they hit. You decide to slow the protons to an acceptable speed, then let them hit a target. You take two metal plates, space them 1.70 cm apart, then drill a small hole through the center of one plate to let the proton beam enter. The opposite plate is the target in which the protons will embed themselves.

    What are the minimum surface charge densities you need to place on each plate?

    2. Relevant equations
    Im not even too sure how to approach the problem.... I know that a parralell plate capacitor emits the same charge on each plate... Is that what these plates will be? Im guessing it could be because the charge is apparently the same on both since the question has only one answer.
    the beam goes through one of the plates and towards the other so I would assume the plate the proton goes through will be Positive charge so it will want to go towards and away from it and the Other plate a negative charge?
    I am not sure but I think kinematics is involved here, I do have a distance. so x(t)=v(x)*t, solve for t and s(t)=Vt+.5*a*t, solve for a. I know E plane=(surface charge)/(2*epsilon). and I know a=(eE/m). So I can solve for the E field to find the surface charge??
    Last edited: Apr 12, 2007
  2. jcsd
  3. Apr 12, 2007 #2
    I guess I will reinforce the hint: conservation of energy!

    You will start off with a certain kinetic energy, and you want to finish with a certain kinetic energy. What is the potential energy for a parallel plate capacitor? How does it relate to charge?
  4. Apr 13, 2007 #3
    There is a problem with the hint! Apparently we arent covering Energy for a couple of chapters and I am supposed to use Forces. We knew energy it would be really easy but since I dont and it wont be applicable on the test. I need to understand how to solve without Energy. I would be sorry about that but it was my Instructors mistake.
  5. Apr 13, 2007 #4
    You have an initial and final velocity and a distance overwhich you must do the slowing down in. A kinematic equation (the 4th one in many books) will allow you to calculate acceleration based on this input information. Knowing acceleration, and the mass of the proton, you can determine the neccessary force. The force comes from the strength of the electric field between the plates times the charge of the particle in question. Working backwards should get you the charge/area value you need.
  6. Apr 15, 2007 #5
    I am having NO LUCK. What kinematics formula specifically are you using? I have taken the velocity the electron exits , divided by the distance to get the time and then using the velocity and time using the equation, vy=vo+.5(eE/m)t^2, substituting a with the formula and also tried solving for a, nothing is working.
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