Bill Beaty wrote:
>
> But something weird is still going on.
>
> If conductive water on the pipe's inner surface gives a second "capacitor
> plate" with an equal and opposite (positive) charge, then as this water
> evaporates and the thickness of the layer decreases... nothing should
> electrically change.
>
> If conductive water on the pipe's inner surface gives a second "capacitor
> plate" with an equal and opposite (positive) charge, then as this water
> evaporates and the thickness of the layer decreases... nothing should
> electrically change.
>
> The layer still contains a positive charge equal to the negative one on
> the outside pipe surface. As the water finally dries out, the positive
> charge should remain on the plastic surface, strongly attracted there by
> the nearby negative charge.
> The layer still contains a positive charge equal to the negative one on
> the outside pipe surface. As the water finally dries out, the positive
> charge should remain on the plastic surface, strongly attracted there by
> the nearby negative charge.
>
Bill. Wouldn't the water vapor (or the O2 which has a high electron affinity) in the air flowing over the red hot hair dryer
Bill. Wouldn't the water vapor (or the O2 which has a high electron affinity) in the air flowing over the red hot hair dryer
heating element cause the loss of electrons to the air stream?
Psychrometric Applet:
The hair dryer would gain electrons back through the power cord. No?
A cheap Van de Graaff type electrostatic generator. :-)
Frederick
