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Homework Help: Guass's law & protons

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Hey , I am a new member Here . I am having some diffuculties , i did one problem and still 1 problem i can't do it coz it seems very hard as i am new to gauss's Law

    I am doing college physics II , protons are projected with initial speed of v0 = 9550m/s into a region in which a uniform electric field , E= 720N/C is present . the protons are to hit a target that lies a horizental distance of 1.27mm from the point at which the protons are lauched.
    Find the two projections angles that will result in the hit .

    b- the total duration of flight for each of thse two trajectoires.
    1zqgeo1.jpg


    3. The attempt at a solution
    I tried to do these E= K.(q)/r^2 , E.A.cos (x). well i don't have any clue....

    I hope you can help me

    Thanks
     
  2. jcsd
  3. Oct 22, 2008 #2

    LowlyPion

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    Welcome to PF.

    You can treat it like a traditional projectile range problem.
    http://www.physics.ubc.ca/~outreach/phys420/p420_00/darren/web/range/range.html

    To figure your acceleration you can use:

    [tex] \vec{F} = q * \vec{E} = m * a[/tex]
     
  4. Oct 22, 2008 #3
    Thank you so Much .

    http://www.physics.ubc.ca/~outreach/phys420/p420_00/darren/web/range/range.gif

    sorry i am not very good in physics , do i need to use this ?
    we have x= 1.27mm / so i can get the sin ?
     
  5. Oct 22, 2008 #4

    LowlyPion

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    Once you figure g you can solve for θ.
     
  6. Oct 22, 2008 #5
    g is 10 right ?
     
  7. Oct 22, 2008 #6

    LowlyPion

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    No. g = a. That is the acceleration from your E-field. It is a function of the charge of the proton and its mass. You have to figure that out.

    I said you could treat it LIKE a projectile range equation. Only you aren't concerned with gravity. You are concerned with the constant acceleration due to the E-field.
     
  8. Oct 22, 2008 #7
    :cool: aha ok Thanks , got it Now .

    Gonna solve it now , Thanks again !!
     
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