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Re: [Phys-L] amusing electrostatics exercise



For the numerical situation you describe, where the initial h is very small
(1e-5 m), make the rough approximation that the +Q charge on the disk is
nearly uniformly distributed, and the -Q charge is nearly all just under
the disk, nearly uniformly distributed in an area about the same as the
area of the disk. This approximate charge distribution is like that of a
parallel-plate capacitor, and given the tiny separation (1e-5 m) the fringe
field will be very small. For example, just above the center of the disk
the field is only 0.5(h/r) = 0.5(1e-5/.1) = 5e-5 times the field in the
gap. Therefore this charge distribution is consistent with the situation,
in that the field at far locations is extremely small and will have little
effect on the walls.

For larger values of h, but still small compared to the other dimensions,
the field in the gap remains approximately equal to (Q/A)/epsilon_0, which
is independent of the gap h. Therefore the potential difference is
proportional to the gap size: Delta-V = (1.5 volts)(h/1e-5).

Next I'll look at Carl's analysis....

Bruce