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What is the electric field strength inside the capacitor?

  1. Jan 26, 2016 #1
    1. The problem statement, all variables and given/known data

    An electron is launched at a 45∘ angle and a speed of 5.0×10^6 m/s from the positive plate of the parallel-plate capacitor shown in the figure (Figure 1) . The electron lands 4.0 cm away.

    a)
    What is the electric field strength inside the capacitor?

    b)
    What is the smallest possible spacing between the plates?


    e18f8d5d-d5a1-4614-9db5-ce7414952908.png
    2. Relevant equations
    F = qE
    Equations used for Projectile Motion


    3. The attempt at a solution

    For Part A

    Okay so first I thought that I may need the acceleration in the y-direction to use F = qE, so I separated the velocity into separate components.
    Vx = 5.0×10^6 m/s * cos(45)
    Vy = 5.0×10^6 m/s * sin(45)
    Then I found the time.
    0.04m = (5.0×10^6 m/s * cos(45))*t
    t = 1.131*10^-8 seconds
    Then I proceeded to finding the acceleration with the time I found by using the highest point reached (0 m/s)
    0 m/s = 5.0×10^6 m/s * sin(45) + a (1.131*10^-8 sec)
    a = -2.69 * 10^4 m/s^2

    After that I found the force on the electron using its mass, 9.10938356 × 10^-31 kilograms
    F = (9.10938356 × 10^-31 kg)*(-2.69 * 10^4 m/s^2)
    F= -2.459*10^-16 N
    Then used that to find E in F= qE using the charge of the electron 1.6*10^-19 C
    2.459*10^-16 N = (1.6*10^-19 C)*(E)
    E = -1536.57 N/C

    But it's wrong. Can someone tell me what I did wrong?


    For Part B.
    I tried calculating the amount of distance the electron traveled in the vertical direction.
    0^2 = (5.0×10^6 m/s * sin(45))^2+2(-2.69 * 10^4 m/s^2)*y
    y= 0.023 m

    But it's wrong. I think I might have misunderstood the question. How do I find the answer?


    Any help is much appreciated!
     
  2. jcsd
  3. Jan 26, 2016 #2

    TSny

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    Homework Helper
    Gold Member

    Does the electron reach max height at the same time that Δx = 4.0 cm?
     
  4. Jan 26, 2016 #3

    gneill

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    Staff: Mentor

    A possibly helpful hint (or approach to try): Consider the range equation for projectile motion.
     
  5. Jan 26, 2016 #4

    collinsmark

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    Take a moment here to reflect on what total time [itex] t [/itex] represents. Is it the time it takes for the electron to reach the highest point in its motion or the time it takes the electron to traverse the entire 4 cm?
    Take a look at the value of [itex] t [/itex] that you are using in your [itex] 0 = v_0 + at [/itex] equation. Should the value of [itex] t [/itex] used here be the time it takes for the electron to reach its highest point or the time it takes for the electron to traverse a full 4 cm?

    When the electron reaches its highest point, how far has it traveled horizontally?
     
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