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Electric Fields and Kinematics

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    In the figure below, an electron is shot at an initial speed of v0 = 2.25 × 106 m/s, at angle θ0 = 20.0o from an x axis. It moves through a uniform electric field = (5.04 N/C) . A screen for detecting electrons is positioned parallel to the y axis, at distance x = 3.23 m. What is (a) the x component and (b) the y component of the velocity of the electron when it hits the screen?

    2. Relevant equations

    E = F/q; F = E*q
    F = m*a; a = F/m = E*q/m
    t = Dx/Vx
    Vy = Vo + Ay*t

    3. The attempt at a solution

    It would probably be easier to look at the attached image to see my work, but I'll try explaining my logic here anyway. The x-component of the velocity is the same as the initial x-component since there is no horizontal acceleration - it turns out to be Vo*cos(20°).

    The second part, however, I've had a lot of trouble on. I've tried it multiple times so it doesn't seem to be a calculator error. My logic here is that since fields point from + to - and the magnitude of field strength at any point doesn't depend on the distance from whatever's causing the field, the electron (being negatively charged) accelerates in the -y direction. My work is shown in the attached image. I was just wondering if my logic is correct, and if so then is it just a calculation error on my part?

    Attached Files:

  2. jcsd
  3. Feb 20, 2010 #2


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    Looks OK but the numbers in the attached solution do not match the numbers that you posted in the problem statement.
    Last edited: Feb 20, 2010
  4. Feb 20, 2010 #3
    Sorry, when the answer inputted is incorrect the question generates new values to prevent you from just guessing randomly; the screenshot is from a past attempt.

    So it's most likely a calculator error I'm making when plugging in the values? I've tried it at least 5 times and double checked it, but I always seem to get part (b) wrong. Also, I've neglected gravity because it's negligible compared to the electrostatic force.
  5. Feb 21, 2010 #4


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    I did part (b) for the screen shot and I agree with you both with the method and the numbers. It is probably an error with inputting the formula that Blackboard or WebAssign or MasteringPhysics (or whatever evil thing controls this question) uses to determine whether your answer is correct. If I were you, I would take my solution to my teacher and ask him/her to show me what I did wrong. You might end up doing the entire class a favor.
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