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How to find the electric field between two ideal plates?

  1. Sep 7, 2011 #1
    There are two plates that lie horizontally parallel to each other, separated by a distance of 1 cm. Each plate is 2 cm long. The question wants you to assume that the electric field between the two fields is constant, and pointing directly downwards. On the left-most edge of this plate configuration, an electron is placed 0.5 cm above the bottom plate, and has a horizontal, initial velocity of 1.46x10^6 m/s. There is no vertical component to its acceleration or velocity. The picture in question shows the trajectory of the particle moving to the right, parallel to both plates due to the initial velocity, and accelerating upwards, due to the electric field, reaching the bottom-right corner of the top plate exactly at the end of the 2 cm span of the plate. Using this information, you are to determine the magnitude of the electric field.

    To solve this, I used a kinematic equation to find the time, t, it took for the particle to reach a horizontal distance of 2 cm, the length of the plate. The equation I used was x(t) = vt, with v being the initial, horizontal velocity. Arranging the equation to solve for t, I found it to be 1.37x10^-8 seconds. Then, I set up my y-component trajectory equation as y(t) = y + 0.5at^2, where y is the initial y coordinate, and a is the acceleration due to the force exerted upon the electron by the electric field. I set y(t) = 0.01 m and y = 0.005m. By solving for t, I came to the equation a = (2(y(t)-y))/t^2, which simplifies to a = (.01)/t^2. When t found from the x-coordinate kinematic equation is substituted, I find that a is equal to 5.33x10^13 m/s^2.

    At this point, I have all of the necessary variables to solve for electric field. I set up the equation initially as E=F/q, using the definition of electric field. I replaced F with ma, and am left with E=ma/q. The mass of an electron is 9.11x10^-31 kg and the charge is 1.602x10^-19 C. When I substitute in all of my values, I am left with an electric field with magnitude 303.1 N/C.

    However, when I put my answer into Mastering Physics, I am incorrect. Could anyone out there help me figure out what I'm doing wrong? This makes sense in my head, but I am obviously not considering something. Thank you!

    Edit: added missing units
     
    Last edited: Sep 7, 2011
  2. jcsd
  3. Sep 7, 2011 #2
    I got numbers similar to you. Is it possible that the value is right but you should have a negative sign? The electron goes up so the field points down.
     
  4. Sep 7, 2011 #3
    I tried that, but the answer was still incorrect. The question only asked for magnitude, and not direction. Thank you for helping out though.
     
  5. Sep 8, 2011 #4

    ehild

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    Homework Helper
    Gold Member

    It might be the number of significant digits. The data are given with 3 digits, and you rounded the acceleration to three digits, but gave the electric field with 4 digits. The last digit has no sense.

    Never round the intermediate results, or leave more digits than it is necessary in the end result. (The intermediate rounding caused an error of 0.1.)

    The end result must not given with higher accuracy (more significant digits) as the data. Input 303 N/C.

    ehild
     
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