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Re: capacitance of a disk

One thing I neglected to mention in my prevous long post about the
nitty gritty of the exact solution was what the electric field and the
surface charge density are for the problem. Since the potential function
depends only on the one coordinate u, that means that the electric
field points everywhere locally perpendicular to the surface of each u = a
spheroid. The electric field lines run parallel to the u-hat direction
of locally increasing u-value. The potential is:

V(r_vec) = V(u,v,A) = V(u) = Q*arcsin(R/u)/(4*[pi]*[epsilon]_0*R)

The only component of E_vec is its u-component E_u. Its behavior is
independent of coordinate A:

E_u(u,v,A) = E_u(u,v) = Q/(4*[pi]*[epsilon]_0*u*sqrt(u^2 - v^2))

If we have a conductor whose surface coincides with some spheroid
with u = a, then the electric field discontinuously jumps to zero as
we enter the conductor. The amount of the discontinuity gives the
strength of the surface charge density [sigma] = [epsilon]_0* /_\ E_u
on the conductor's surface. That charge density is:

[sigma](v) = Q/(4*[pi]*a*sqrt(a^2 - v^2)).

If we write [sigma] in terms of the radial distance r that a particular
part on the conducting surface is from the origin, we get:

[sigma](r) = Q/(4*[pi]*a^2*sqrt(2 - e^2 - (r/a)^2)) .

Since this formula is only for the u = a surface of the conducting
spheroid (whose eccentricity is e), we need to remember that the radial
ratio (r/a)^2 ranges over the interval: 1 - e^2 <= (r/a)^2 <= 1 . The
extremal values of the surface charge density occur on the spheroid's
"equator" where

[sigma]_max = Q/(4*[pi]*a^2*sqrt(1 - e^2))

and on the "poles" of its symmetry axis where

[sigma]_min = Q/(4*[pi]*a^2) .

If we take the limit of e --> 0 in the above formula for [sigma](r) we
get the proper uniform surface charge density for a charged conducting
sphere. Also, if we take the opposite limit of e --> 1 in that above
formula, we get the case of the flat circular disk:

[sigma](r) = Q/(4*[pi]*a^2*sqrt(1 - (r/a)^2)) .

But because this result was taken as the limit of the top and bottom
edges of the spheroid collapse to a disk this means that the surface
charge density on just one top *or* bottom surface includes 1/2 of the
total charge present. When these two surfaces merge to make a flat disk
we need to double the value of the formula and then not consider the
top and bottom anymore as having separate surface charge densites.
After doubling (and setting the disk radius R to the semi-major axis a)
we get:

[sigma](r) = Q/(2*[pi]*R^2*sqrt(1 - (r/R)^2)) .

If we integrate this (circularly symmetric) surface charge density
over the surface of the disk (but now on only one side) we obtain the
total charge on the disk, which agrees with the value Q as it should.
Note that there is a divergent 1/sqrt singularity in the value of
[sigma] when r --> R on the edge of the (zero-thickness) disk. It seems
that the charges cluster so strongly there because their mutual repulsion
tends to drive them outward as far as they can get.

BTW, in case people are having some trouble visualizing the lines of
the coordinate system and the E-field it might help if it is mentioned
that this confocal "conicoidal" coordinate system u,v,A is sort of a
distorted version of a coordinate system analogous to spherical
coordinates, and the analogy becomes ever closer at ever greater
distances from the central focal circle. The u coordinate is like the
radial coordinate and sort of measures the "distance" away from the
center. The coordinate v is sort a directional coordinate that
asymptotically acts like the polar coordinate [theta] in spherical
coordinates in that it (actually its asymyptote) sort of distinguishes
the (slope of) directions more along the z-axis from those more along
the x-y plane. Asymptotically the large distance mapping
v/R --> sin([theta]) connects sort of how v labels polar directions. Of
course, the azimuthal coordinate A is identically equal (everywhere) to
the azimuthal coordinate [phi] in spherical coordinates.

David Bowman