Capacitance of Concentric Cylinders
"where the capacitance per unit length is not a function of either the charge on the system or L itself. The only parameters that matter are the radii of the cylindrical shells. This should be expected because capacitance is a feature of a system's geometry and does not depend on applied charges or potentials."
http://www.davidpace.com/physics/em-topics/capacitance-cylinders.htmhttp://www.ionizationx.com/index.php/topic,1298.msg14419.html#msg14419Chapter 4 "quote":
There are actually two principal mechanisms by which electric fields can distort
the charge distribution of a dielectric atom or molecule: stretching and rotating.
What happens to a neutral atom when it is placed in an electric field E? Your first guess
might well be: "Absolutely nothing-since the atom is not charged, the field has no effect
on it." But that is incorrect. Although the atom as a whole is electrically neutral, there is a
positively charged core (the nucleus) and a negatively charged electron cloud surrounding
it. These two regions of charge within the atom are influenced by the field: the nucleus
is pushed in the direction of the field, and the electrons the opposite way. In principle, if
the field is large enough, it can pull the atom apart completely, "ionizing" it (the substance
then becomes a conductor). With less extreme fields, however, an equilibrium is soon
established, for if the center of the electron cloud does not coincide with the nucleus, these
positive and negative charges attract one another, and this holds the atoms together. The
two opposing forces-E pulling the electrons and nucleus apart, their mutual attraction
drawing them together-reach a balance, leaving the atom polarized, with plus charge
shifted slightly one way, and minus the other. The atom now has a tiny dipole moment
p, which points in the same direction as E. Typically, this induced dipole moment is
approximately proportional to the field (as long as the latter is not too strong):
p=aE. (4.1)
The constant of proportionality a is called atomic polarizability. Its value depends on the
detailed structure of the atom in question.
For molecules the situation is not quite so simple, because frequently they polarize
more readily in some directions than others.
Problem 4.1 A hydrogen atom (with the Bohr radius of half an angstrom) is situated between
two metal plates 1 mm apart, which are connected to opposite terminals of a 500 V battery.
What fraction of the atomic radius does the separation distance d amount to, roughly? Estimate
the voltage you would need with this apparatus to ionize the atom. [Use the value of alpha in Table
4.1. Moral: The displacements we're talking about are minute, even on an atomic scale.]
For H atom!!!
E = 500V
x = 1 10^-3
Atomic polarizabilities H = 0.667 -> table 4.1 (Handbook of Chemistry and Physics, 78th ed.)
1 eV = 1.602 10^-19 C
1 debye = 3.33564 10^-30 Cm
p = dipole moment
E = electric field
E0 = 8.85 10^-12
R = bohr radius 1.0 10^-10
alpha = atomic polarizability
Problem 4.1
E =V/x =500 / 1 10^-3 =5 10^5
Table 4.1: alpha / 4 pi E0 =0.66 10^-30, so alpha = 4 pi (8.85 10^-12)(0.66 10^-30)= 7.34 10^-41
p =alpha E = e d => d =alpha E / e =(7.34 10^-41)(5 10^5)/(1.6 10^-19)=2.29 10^-16 m
d / R =(2.29 10^-16) / (0.5 10^-10) =14.6 10^-6
To ionize, say d = R. Then R = alpha E / e = alpha V / e x =>
V = R e x / alpha =(0.5 10^-10)(1.6 10^-19)(1 10^-3) / (7.34 10^-41) =1 10^8 V
So that's a lot of Volts for ionizing the H atom. When ionized, it becomes a conductor, so I think SM didn't ionize the water molecules. Only polarize the molecule (dipole moment point to the direction of the E field) thus rotating the molecule...
The dipole moment of water is 6.1 10^-30 Cm (polar molecule) but torque is N = p E in a uniform field E. If the E field is non-uniform, so that F+ does not exactly balance F-, there will be a net force on the dipole, in addition to the torque. Of course, E must change rather abruptly for there to be significant variation in the space of one molecule.
br,
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