Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is released from res

Question

Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, how much kinetic energy (J) has the electron gained

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Verity 5 years 2021-07-14T04:55:49+00:00 1 Answers 77 views 0

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  1. dangkhoa
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    2021-07-14T04:57:31+00:00
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    Answer:

    {\rm K} = 2.4\times 10^{-19}~J

    Explanation:

    The electric field inside a parallel plate capacitor is

    E = \frac{Q}{2\epsilon_0 A}

    where A is the area of one of the plates, and Q is the charge on the capacitor.

    The electric force on the electron is

    F = qE = \frac{qQ}{2\epsilon_0 A}

    where q is the charge of the electron.

    By definition the capacitance of the capacitor is given by

    C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60

    Plugging this identity into the force equation above gives

    F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q

    The work done by this force is equal to change in kinetic energy.

    W = Fx = (30q)(0.05) = 1.5q = K

    The charge of the electron is 1.6 \times 10^{-19}

    Therefore, the kinetic energy is 2.4\times 10^{-19}

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