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Self capacitance spherical grain

  1. Nov 21, 2013 #1
    I have been given the exercise above which relates to an article we are reading. I can calculate all the results but I can't interpret them physically.
    I think my major problem is that I don't see how self capacitance is a physically measureable quantity. For a recap the self capacitance is the mutual capacitance between a given conductor and a conducting sphere of infinite radius.
    a) In the first question I am asked to find the self capacitance of spherical aluminium grain of radius R and charge Q which is just 4[itex]\pi[/itex]ε0R.
    b) In the next question I am to assume that the grain is coated on each side by two separate oxide layers which both act as plate capacitors. This is also straightforward.
    c) In this question I am asked to find how much the voltage on the grain changes if we add an electron given the two situations a) or b). The big problem for me is that I do not understand what is physically meant by the voltage on the grain. For a plate capacitor the only sense of voltage I can make is the voltage between the two plates which I don't see as a voltage on the grain? And for the self capacitance I don't know what is meant by the voltage physically either since the voltage appearing in the expression for the self capacitance as said is a voltage difference between out conductor and a sphere of infinite radius.
    So what is meant by the "voltage on the grain"? And how do we measure it physically and how does the self capacitance enter?
     

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  3. Nov 21, 2013 #2

    vela

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    When the grain is initially uncharged, you don't have to do any work to bring in a charge dq from infinity and place it on the grain. But as the grain acquires charge, you have to do work because the grain's charge Q will repel the charge dq. The potential is the work required per unit charge given the grain has a charge Q.
     
  4. Nov 21, 2013 #3
    Okay that makes sense. So the self capacitance is really a measure of the energy per charge needed to bring charge in from infinity to the grain. Now I still don't understand what is meant physically by the capacitance of the grain with respect to ground when it is coated by an oxide layer. Why does the grain have a capacitance? It is the oxide layer that acts as a capacitor not the grain. And secondly, how does ground relate to the first question. Is the potential at infinity the ground?
     
  5. Nov 21, 2013 #4

    ehild

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    First you have an isolated sphere. It can be charged, and then its surface is at some potential with respect to infinity. You certainly remember that the electric field around a charged metal sphere is kQ/r2 and the potential is kQ/r around it. At the surface of the sphere, r=R = 10 nm, the potential is kQ/R. Capacitance is defined as C=Q/U, so it is R/k=4πε0R.

    In the next situation, (read the text carefully) the grain is covered by an insulating oxide layer, and then covered by a grounded metal layer. The voltage of the grain refers to the grounded metal. It is a spherical shell capacitor. The potential difference between the inner and outer metal surfaces is U = 1/(4πκε0)(Q/r1-kQ/r2), and r2=r1+d where d is the thickness of the insulating layer. So U=Q/(4πκε0)(r2-r1)/(r1r2). r2-r1=d, and, as the insulating layer is very thin, r2≈r1=R. U=1/(4πκε0)Qd/R2.
    R^2 is related to the surface of the spherical grain, so the formula is the same as for parallel-plate capacitor.

    ehild
     
    Last edited: Nov 21, 2013
  6. Nov 22, 2013 #5
    So maybe I understood it wrong. Don't we have 2 capacitors in series? The first two pages of the attached text describe the experimental setup. I understand it like this:
    We have some kind of insulator into which we have made a hole. This hole connects one alumnium lead to the flat side of a hemispherical aluminium grain which is covered by an oxide layer. On the spherical side of the aluminium grain there is also and oxide layer and then another aluminium lead. Is this all correctly understood?
    Now in a) we have estimated the self capacitance of a free aluminium grain, that is the energy needed per charge to bring in a charge from ground an stick it onto the aluminium grain.
    In b) we then estimate how this required energy is reduced by the fact that the two oxide layers (there are 2 right?) act as a dielectric shielding the charge of the aluminium grain. Is this correctly understood? I don't think it is because I dont see why we can model anything asa plate capacitor. On the upper side of the Al-grain there is a hemispherical oxide layer and on the lower part there is a flat oxide layer (okay that is a plate capacitor, but shouldn't we also include the upper oxide layer?)
     

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  7. Nov 23, 2013 #6

    ehild

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    The paper is about a structure as capacitor - a granulated metal surface covered by an insulating layer and metal again. Fig 2 shows two capacitors in series, one between the grain and the upper lead, the other with the grain and the bottom lead, as I understand the text, but it is not clear for me... The modell in the OP - a single particle, and estimating its capacitance when it is isolated, to that, when its surface is oxidized and then covered by metal. The shape of the grain is assumed spherical - it is only a model, and you know the capacitance of a spherical capacitor

    C(isolated)=4πε0R.

    A lead is made to the grain , oxidized, and then aluminium layer deposited on the outside. That makes a spherical shell capacitor and

    C(shell) = (4πκε0)(r1r2)/(r2-r1).

    r2-r1=d << r1. So you can approximate that capacitance as

    C(shell)= (4πκε0)r2/d,

    but 4πr2=A is the surface of the grain, so the formula for the capacitance becomes

    C(shell)= (κε0)A/d,

    the same as that of a parallel plate capacitor. But is does not really matter. The author of the paper used the plate capacitor for the sake of those readers who are not familiar with the spherical shell capacitors.

    The aim of the whole modelling was to estimate the change of voltage caused by removing or adding a single electron to the charge of a grain, by assuming R=10 nm, d=1 nm and κ=8.

    U=Q/C, ΔU=ΔQ/C. C((isolated) =0.55 x 10[SUP-18[/SUP]F, C(shell) is 40 times bigger. The change of the charge by one e causes voltage change öf 290 mV on the isolated grain and about 7 mV in the grain, embedded in metal.

    ehild
     
    Last edited: Nov 23, 2013
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