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Proton is shot away from inf. charged plane

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    The surface charge density on an infinite charged plane is -2.6*10^-6 C/m^2. A proton is shot straight away from the plane at 2.2*10^6 m/s. How far does the proton travel before reaching its turning point?


    2. Relevant equations
    Im very unsure of what equations to use, the only one I know is the equation for density charge. n=Q/A


    3. The attempt at a solution
    Im trying to figure out how to incorporate kinematic equations with the electric field equations. I think I may need to use the force equation for the electric field somehow.
     
  2. jcsd
  3. Sep 13, 2009 #2

    rl.bhat

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    What is the expression of electric field near an infinity charged plane?
    What is th KE of the proton near the plate?
    What is the force on the proton due to this field?
    If the proton moves a distance x before it comes to rest, what is the work done?
     
  4. Sep 13, 2009 #3
    i derived an equation for the distance.
    x= (-m*v*eps0)/(qn) n= surface charge density eps0= 8.85*10^-12

    this seemed to work, but thanks for the questions they got me to thinking!
     
  5. Sep 13, 2009 #4

    rl.bhat

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    It is not clear how you got the above equation.
    The KE of proton = 1/2*m*v^2.
     
  6. Sep 14, 2009 #5
    i used the equations
    (v-final)^2= (v-initial)^2 + 2ax
    F=qE=ma

    a= (qE)/m = -(v-initial)^2/(2x) => x= (-m(v-initial)^2)/(2qE)

    The electric field of a plane charge with surface charge density n is E=n/2(e_0)
    e_0= 8.85*10^-12

    x= -m(v-initial)^2(e_0)/qn
     
  7. Sep 14, 2009 #6

    rl.bhat

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    That is correct. Previously square was missing.
     
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