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Coulomb's Law And Electric Field between 2 Parallel Plates

  1. Sep 17, 2013 #1
    1. The problem statement, all variables and given/known data

    A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.59cm distant from the first, in a time interval of 1.60×10−6s .

    A). Find the magnitude of the electric field.
    B). Find the speed of the proton when it strikes the negatively charged plate.

    2. Relevant equations

    F=ma
    E=F/q
    y=y_0+v_0t+1/2at^2

    3. The attempt at a solution

    Usually, to solve for the E field between two parallel plates, you use sigma/epsilon_0. However, I am not given sigma nor am I given anything about the dimensions of the plate, so I took the old kinematics route. First, I used the equation y=y_0+v_0t+1/2at^2. I said it starts from y_0=0 and it says it starts from rest, so I am left with y=1/2at^2. I plugged in the distance between the plates (.0159m) and the time (1.6*10^-6s) and solved for the proton's acceleration, which I found to be 1.24 m/s^2 in the negative y direction (towards the negative plate).

    Next, I used F=ma - using the acceleration I just found (1.24 m/s^2) and the mass of the proton (1.6*10^-27kg) and solved for F. I got 1.984*10^-27N

    Lastly, for part A, I used the equation E=F/q. Substituting the numbers in, I got that E=1.24*10^-8 N/C and it said that it was incorrect.

    For part B, i used the acceleration I found above and plugged it into v=v_0+at and got V=1.984*10^-6, which was also wrong.

    Any ideas what I did wrong?
     
  2. jcsd
  3. Sep 17, 2013 #2
    The value you obtained from acceleration seem to be off by many orders of magnitude.
     
  4. Sep 17, 2013 #3

    rude man

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    'Many' is right! :smile:
     
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