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Re: [Phys-l] Disc polarizability

Ah o.

This is the static polarizability, which means no fields in the metal. If I understand John's method, it may be similar to that of the only analytic source I've found. Furthermore, perhaps I miss read John's description and he's finding the field inside that cancels the field outside that is causing (Here we go again.) the polarization. That method, if I'm not suffering from the usual is:

AJP, v. 61 p. 12 ff. (R. Friedberg, the electrostatics and Magnetostatics of a Conducting disk)

bc

p.s. (4/3Pi)a^3 I haven't waded thru the various answers, but my impression is that those that don't include kappa zero (permitivity of mt space) means they are in esu?
p.p.s Yes I should've written, de nuovo, an E field in the plane of the disc, far from the disc, it's uniform.


John Mallinckrodt wrote:

On Feb 23, 2007, at 1:59 PM, Bernard Cleyet wrote:


In Strong's text Hull claims the polarizability of a metal disc is (2/3) the permitivity (space) times its diameter cubed. Finding this for a sphere and even an infinite cylinder is trivial (once one has seen it done). But a disc, no. Anyone know how it's done (not numerically)?


I wondered at first how the result could possibly fail to depend on the thickness since, for small thicknesses, the charge distribution should be relatively independent of thickness (leading to an induced dipole moment that is directly proportional to thickness), but then bc clarified the question saying:


my interest is w/ the E direction parallel w/ the disk, and more importantly the derivation.


I tried to get a handle on this problem by modeling the charge distribution as a line charge confined to the circumference of the disk. In that case, one can easily calculate both the dipole moment of the distribution and the field at the center of the disk and it's almost trivial to show that the ratio of the two is (pi/2) times the permittivity of free space times the diameter cubed regardless of how the charge is distributed (except for the fact that it must obey the appropriate symmetry.) This is significantly larger than the result that bc quotes.

I suppose one can take this as a proof that there is a significant amount of surface charge on the circular faces of the disk. Moreover, it seems reasonable that this distribution might reduce the polarizability by simultaneously increasing the electric field at the center of the disk AND reducing the resulting dipole moment.

John Mallinckrodt

Professor of Physics, Cal Poly Pomona
<http://www.csupomona.edu/~ajm>

and

Lead Guitarist, Out-Laws of Physics
<http://outlawsofphysics.com>

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