A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 45000 m/s. What will be the final speed of an electron released from rest at the negative plate?
Express your answer to two significant figures and include the appropriate units
V=

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This Post Has 4 Comments

  1. Potential energy U= k*Q*q/r
    doubling the charge Q doubles the potential energy. Since this will just be converted into kinetic energy u will double tht

  2. 2.1x10^6m/s

    Explanation:

    One electron has a charge of –1.602e-19 C

    mass of electron is 9.1e-31 kg

    mass of proton is 1.6726e−27 kg

    mass ratio is 1.6726e−27 / 9.1e-31 = 1838

    The force is constant, F

    distance is constant, d

    a = F/m

    a increases by a factor 1838, as m decreases by that factor

    a = a₀1838

    v₀² = 2a₀d

    v² = 2a₀d1838

    v²/v₀² = 2a₀d1838 / 2a₀d = 1838

    v² = 1838v₀² = 1838(45000)²

    v = 45000√1838 = 2.1e6 m/s

  3. Vf = 2,400,000 m/s

    Explanation:

    1) The only relevant force is the electrostatic force

    2) The formula for the electrostatic force is F = E×q

    Where E is the electric field and q is the magnitude of the charge.

    3) Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magntitude of the electric forces acting in both proton and electron are the same

    4) Fe = Fp (Fe stands for force on the electron and Fp stands for force on the proton).

    5) Using second law of Newton, Force = mass × acceleration

    Fe = Me × Ae (Me is mass of the electron, and Ae is acceleration of the electron)

    Fp = Mp × Ap (Mp is mass of the proton, and Ap is acceleration of the proton)

    ⇒Me × Ae = Mp × Ap

    ⇒ Ae = Mp × Ap / Me

    6) Now, state the equations for the velocity in uniformly accelerated motion:

    i) Vf² = Vo² + 2ad

    Vo² = 0 for both cases, and d is the same distance.

    ⇒ Vf² = 2ad

    ii) For the proton Vf² = 2(Ap)(d) ⇒ Ap = Vf² / (2d)

    ⇒ Ap = (55,000 m/s)² / (2d)

    iii) For the electron Vf² = 2(Ae)² (2d)

    iv) Using Ae = Mp × Ap / Me (found prevously):

    Vf² = Mp × (55,000 m/s)² / (2d) × (2d) / Me

    ⇒ Vf² = Mp × (55,000 m/s)² / Me

    Taking square root in both sides:

    ⇒ Vf = 55,000 m/s × √ [Mp / Me]

    7) These are the values for the masses of a proton and an electron:

    Mp = 1.67 × 10⁻²⁷ kg

    Me = 9.11×10⁻³¹ kg

    8) Replace and compute:

    Vf = 55,000 m/s × √ [ 1.67 × 10⁻²⁷ kg / 9.11×10⁻³¹ kg] = 2,354,841.8 m/s

    Round to two significan digits: Vf = 2,400,000 m/s

  4. The correct answer to this question is this one:

    You will need these relations

    qV = KE = 1/2mv^2

    You know the charge of a proton (the elementary charge in this case), you know its mass, and you know it's velocity. From there you can find its kinetic energy and from there you can find it's voltage. Once you know the potential difference there all you have to do is reverse the process. Multiply the charge of an electron by V and then solve for the velocity using the mass of the electron

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