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Homework Help: An electron moving across capacitor problem

  1. Jan 24, 2009 #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 from the positive plate of the parallel-plate capacitor shown in the figure. The electron lands 4.0 cm away.

    What is the electric field strength inside the capacitor?

    What is the minimum spacing between the plates?


    2. Relevant equations

    F = ma
    Fon q = qE

    3. The attempt at a solution

    I am just wondering if this is correct?

    Vo = 5x10^6 m/s

    Voy = (sin45)(5x10^6 m/s) = 3.5x10^6 m/s

    t = [tex]\frac{(0.04m)(\sqrt{2})}{5x10^6 m/s}[/tex] = 11x10^-9 s

    0 = Voyt + (1/2)at^2

    a = [tex]\frac{-(3.5x10^6 m/s)(11x10^-9 s)}{(\frac{1}{2})(11x10^-9 s)^2}[/tex]

    = -6.4x10^-14 m/s^2


    =(9.11x10^-31 kg)(-6.4x10^14 m/s^2) = -5.8x10^-16 N

    Fon q = qE

    E = F/q = -5.8x10^-16 N / 1.60x10^-19 C

    = -3.6x10^3 N/C

    ymax = V0y/(2a)

    = 3.5x10^6 m/s / [(2)(-6.4x10^-14 m/s^2)]

    = 2x10^-9 m
  2. jcsd
  3. Jan 24, 2009 #2
    How did you calculate t? The time of flight is dependent on the force acting on the electron, and the electric field is not known. Try setting up equations of motion eliminating t.




    Solve for t in the second equation then substitute into the first. This gives two unknowns y and a. Find another kinematic equation that has y and a. This will give two equations with two unknowns. You will than be able to solve for y and a (hint: let y = ymax). Check your equation for ymax.
  4. Jan 24, 2009 #3
    Ok here is what I did:

    [tex]y= y_0+v_{y0}t+\frac{1}{2}at^2[/tex]



    [tex]y= y_0+v_{y0}(\frac{x}{v_{x0}})+\frac{1}{2}a(\frac{x}{v_{x0}})^2[/tex]

    [tex]v_{y}^2 = v_{y0}^2 + 2ay [/tex]

    [tex]y=\frac{v_{y}^2 - v_{y0}^2}{2a}[/tex]

    [tex]\frac{v_{y}^2 - v_{y0}^2}{2a}= y_0+v_{y0}(\frac{x}{v_{x0}})+\frac{1}{2}a(\frac{x}{v_{x0}})^2[/tex]

    [tex]a^2=\frac{-v_{y0}^2 - v_{y0}(\frac{x}{v_{x0}})}{(\frac{x}{v_{x0}})^2}[/tex]

    [tex]a^2=\frac{-(3.5x10^6 m/s)^2 - 3.5x10^6 m/s(\frac{0.04m}{3.5x10^6 m/s})}{(\frac{0.04m}{3.5x10^6 m/s})^2}[/tex]

    = [tex]9.4x10^{28} m/s^2[/tex]

    [tex]\sqrt{9.4x10^{28} m/s^2}[/tex] = [tex]3.1x10^{14} m/s^2[/tex]
    Last edited: Jan 24, 2009
  5. Jan 24, 2009 #4
    Your appoach is correct with the exception of two parts:

    [tex]v_{y}^2 = v_{y0}^2 + 2ay [/tex]

    [tex]y= y_0+v_{y0}t+\frac{1}{2}at^2[/tex]

    Is the sign of 'a' positive or negative according to the coordinates? Another clue is evaluating

    [tex]a^2=\frac{-(3.5x10^6 m/s)^2 - 3.5x10^6 m/s(\frac{0.04m}{3.5x10^6 m/s})}{(\frac{0.04m}{3.5x10^6 m/s})^2}[/tex]

    Taking the square root of each side yields an imaginary number for 'a'.
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