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Potential of a charge approaching a perfect conductor.

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data

    A point charge Q with mass M approaches a semi-infinite surface of a perfect metal and becomes trapped near the surface in a (quantum) bound state. Find the binding energy of this particle to the surface. Treat the metal classically—i.e. ignore its internal quantum levels, etc., and just consider it to be a classical perfect conductor that the point charge’s wavefunction cannot penetrate.

    2. Relevant equations

    The 1D Time-independent Schrodinger equation.

    [tex] \frac{-\hbar^{2}}{2m} \frac{d^{2}\psi}{dx^{2}} +V(x) = E\psi [/tex]

    3. The attempt at a solution

    My problem with this question mostly stems form the nature of the potential surrounding the conductor. If we assume that the charged particle is approaching the surface due to the presence of an electric field and we treat the surface as a plane then the potential is of the form:

    [tex] V(x) = QEx = \frac{Q\sigma}{\epsilon_{0}x [/tex]

    Where sigma is the surface charge of the surface.

    Or we can consider the particle to be influenced by an induced surface charge, which would take the form of a negative mirror charge (-Q) so the potential would be a simple Coulomb attraction.

    [tex] V(x)= \frac{-Q}{4\pi\epsilon_{0}r} [/tex]

    Where we would have to account for the change in separation as the charge approached the surface.

    My question is a) Which of these options seems the most reasonable. My guess is the second one but I'm not sure I feel I'm missing something.

    b) For the charge to become bound the potential needs to be "well-like" i.e. I need a repulsive component for distances very close to the surface of the conductor. I'm afraid I don't know what form this component could be. Since the wavefunction cannot penetrate the surface of the conductor I would think we could treat the potential as infinite for x<0 (putting the conductor at x=0). However, this would mean the well had no "bottom" and the charge would not bind.

    I hope that's clear. Any help resolving my confusion would be greatly appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 17, 2011 #2

    kuruman

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    I would try plan B, the image charge approach. I wouldn't worry about the lack of "bottom". An attractive delta function potential also has no bottom, yet it admits one bound state.

    I have never tried solving this problem, but that's how I would go about it for a first attempt.
     
  4. Aug 17, 2011 #3
    What is the potential energy between the charge and its image when it is at distance [itex]x[/itex] away from the boundary? Answer this classical problem first.
     
  5. Aug 20, 2011 #4
    The potential between a charge and it's image should just be the Coulomb attraction with the separation being two x where x is the distance to the boundary, I.e:

    [tex] \frac{-e^{2}}{4\pi\epsilon_{0}2x} [/tex]
     
  6. Aug 20, 2011 #5

    kuruman

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    And the next step is ....
     
  7. Aug 20, 2011 #6
    If the charge is approaching the wall at a constant velocity I would replace x with [tex] v. dt [/tex]. this makes the potential time dependent so I would need to solve the time-dependent Schrodinger equation.
     
  8. Aug 20, 2011 #7

    kuruman

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    Then you will have a vdt in the denominator of the potential. Does this look like a diff. eq. that you can solve? What if you left the x as is in the potential and put V(x) = -e2/8πε0x in the time-independent Schrodinger equation?
     
    Last edited: Aug 20, 2011
  9. Aug 21, 2011 #8
    I could probably solve the time independent version but would that account for the increased potential as the charge gets closer to the surface?
     
  10. Aug 21, 2011 #9

    ehild

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    You need to find the bound states of the charged particle near the metal plate. This comes from the solution of the time-independent Schroedinger equation with the potential [itex]V(x)=\frac{-e^{2}}{4\pi\epsilon_{0}2x}[/itex].
    The potential energy is negative so it decreases as the particle gets closer to the surface.

    ehild
     
  11. Aug 22, 2011 #10

    kuruman

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    Listen to ehild. The potential does not depend explicitly on time so there is no time-dependent Schrodinger equation to solve.
     
  12. Aug 22, 2011 #11
    If it doesn't explicitly depend on time, but may because of the motion of the particle, it still counts as time-independent because the potential itself is.

    That makes sense, I should have known when I wrote the equations out. I have a tendency to make things more complicated than they need to be.

    So just solve the Schrodinger equation for negative energies to get bound states and minimise to find the lowest?
     
  13. Aug 22, 2011 #12

    kuruman

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    Yup.
    That's the general idea. When you say "minimise to find the lowest", exactly what are you minimising? Are you going to use some variational principle? I am not criticising your method, just trying to understand how you propose to go about this.
     
  14. Aug 22, 2011 #13
    I did the math. No minimizing was needed. I got:

    [tex] -\frac{k^{2}Q^{4}m}{8\hbar^{2}} [/tex]

    Where k is the Coulomb constant.
     
    Last edited: Aug 22, 2011
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