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Calculate the time it takes for an electron to travel between two parallel plates given the distance and the electric field intensity.

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Answer

t=5.69×10−13E s\mathbf{t = \sqrt{\frac{5.69 \times 10^{-13}}{E}} \text{ s}}

Step-by-step solution

5 steps

To solve this problem, we need the magnitude of the electric field intensity (EE) between the plates, which is not provided in the image. I will derive the formula for the time taken in terms of EE.

Constants used:

  • Elementary charge, e=1.602×10−19e = 1.602 \times 10^{-19} C
  • Mass of electron, me=9.109×10−31m_e = 9.109 \times 10^{-31} kg

Given:

  • Distance between plates, d=5.00 cm=0.05d = 5.00 \text{ cm} = 0.05 m
  • Initial velocity of electron, v0=0v_0 = 0 m/s (released from rest)

  1. Calculate the electrostatic force on the electron.

    The force (FF) on a charge (qq) in an electric field (EE) is given by:

    F=qEF = qE

    For an electron, q=eq = e.

    F=eEF = eE

  2. Calculate the acceleration of the electron.

    Using Newton's Second Law, F=meaF = m_e a:

    a=Fme=eEmea = \frac{F}{m_e} = \frac{eE}{m_e}

  3. Use a kinematic equation to find the time.

    Since the electron starts from rest (v0=0v_0 = 0) and moves a distance dd with constant acceleration aa:

    d=v0t+12at2d = v_0 t + \frac{1}{2}at^2 d=(0)t+12at2d = (0)t + \frac{1}{2}at^2 d=12at2d = \frac{1}{2}at^2

    Rearrange to solve for tt:

    t2=2dat^2 = \frac{2d}{a} t=2dat = \sqrt{\frac{2d}{a}}

  4. Substitute the expression for aa into the equation for tt.

    t=2deEmet = \sqrt{\frac{2d}{\frac{eE}{m_e}}} t=2dmeeEt = \sqrt{\frac{2d m_e}{eE}}

  5. Substitute the known numerical values into the formula.

    t=2(0.05 m)(9.109×10−31 kg)(1.602×10−19 C)Et = \sqrt{\frac{2(0.05 \text{ m})(9.109 \times 10^{-31} \text{ kg})}{(1.602 \times 10^{-19} \text{ C})E}} t=9.109×10−321.602×10−19E st = \sqrt{\frac{9.109 \times 10^{-32}}{1.602 \times 10^{-19} E}} \text{ s} t=5.686×10−13E st = \sqrt{\frac{5.686 \times 10^{-13}}{E}} \text{ s}

    Since the value of the electric field intensity (EE) is not provided in the question, the final answer for the time will be in terms of EE.

    t=5.69×10−13E s\mathbf{t = \sqrt{\frac{5.69 \times 10^{-13}}{E}} \text{ s}}
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