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1 September, 04:23

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 2.0 cm, and the electric field within the capacitor has a magnitude of 2.3 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

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Answers (2)
  1. 1 September, 07:11
    0
    7.36*10⁻¹⁵ J

    Explanation:

    From the question,

    Then kinetic energy of the electron = electric potential * charge of the electron

    Ek = q'V ... Equation 1

    Where Ek = kinetic Energy of the electron, q' = charge of an electron, V = Electric potential.

    But,

    V = E*d ... Equation 2

    Where E = Electric Field, d = distance of separation of the plates of the capacitor

    Given: E = 2.3*10⁶ V/m, d = 2.0 cm = (2/100) m = 0.02 m.

    Substitute into equation 2

    V = 2.3*10⁶ (0.02)

    V = 4.6*10⁴ V.

    Constant: q' = 1.6*10⁻¹⁹ C

    Substitute into equation 1

    Ek = 4.6*10⁴ (1.6*10⁻¹⁹)

    Ek = 7.36*10⁻¹⁵ J
  2. 1 September, 07:27
    0
    Answer: 7.37*10^-15 J

    Explanation: according to the work - energy theorem.

    The work done in moving the electron from the negative plate to the positive equals it kinetic energy.

    Mathematically, we have that

    Kinetic energy = work done = qV

    The distance traveled by the electron is simply the the distance between the plates of the capacitor (d) = 2cm = 0.02m.

    Strength of electric field (E) = 2.3*10^6 v/m

    We need to get the potential difference between the plates first, this is gotten by using the formulae below

    V = Ed

    Where V = potential difference = ?

    E = strength of electric field = 2.3*10^6 v/m

    d = distance between plates = 0.02m

    By substituting the parameters, we have that

    V = 2.3*10^6 * 0.02

    V = 0.046 * 10^6

    V = 4.6*10^4 v

    But work done = qV

    Where q = magnitude of electronic charge = 1.602*10^-19c

    Work done = 1.602*10^-19 * 4.6*10^4

    Work done = 7.37*10^-15 J
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