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Homework Help: Parallel plate capacitor question

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data
    The negative plate of a charged, isolated parallel plate capacitor with a vacuum gap of width d and carrying a surface charge density s is uniformly irradiated with a short pulse of ultraviolet light whose photon energy is just enough to release a photoelectron at the surface with zero kinetic energy. The radiant energy per unit area deposited by the pulse is such that a photoelectron surface charge density of exactly s is generated. Derive an expression for t_van, the time taken following the ultraviolet pulse for the electric field to vanish everywhere between the plates. Evaluate t_van for s=0.1 micro Coulomb per unit area and d=1mm

    2. Relevant equations
    None given

    3. The attempt at a solution
    I'm very confused by the question, I'm really not sure which equations to use. Presumably, the photoelectron would produce an electric field which would interfere with the two plates, but does this mean I use the equations for polarisation? Thanks in advance.
  2. jcsd
  3. Sep 23, 2011 #2

    rude man

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    1. What's the electric field set up by the photon-excited electrons in the gap d?
    2. Can you think of a relationship between h and the transit time, based on energy conservation?
  4. Sep 24, 2011 #3
    Could you apply Lorentz' law and calculate the time it would take for the charge to reach the positive plate, when the charges would cancel and, thus, the electric field would vanish? Or am I conceptualizing this wrong?
  5. Sep 24, 2011 #4


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    I think you are right. Those photoelectrons got free from the metal at the same time, but with zero kinetic energy, so they are together at the negative plate initially. Feeling the electric field of the positive charges on the other plate, they start to move towards the positive plate together, with the same acceleration and velocity. Arriving there, the charges neutralize and the electric field vanishes.
  6. Sep 24, 2011 #5

    rude man

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    I don't believe there are any magnetic effects to be considered here, so the Lorentz law is not relevant. The inductance of the gap has to be very small.

    ehild has the right idea.
  7. Sep 24, 2011 #6
    I might have the name wrong. The force on a charged particle:

    F = q(E+(vXB))

    Obviously the second term is zero. Anyway, thanks for your help, I have enough info to attempt the question at least :)
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