MosaicDave wrote:Just for fun, because I have some time before going home for supper:
Let's pretend I pour a few ounces of water in a drinking glass. Or in my green glass bottle. Which I like to drink out of, because I hate drinking out of plastic. So we can work with nice even numbers, let's call it 100ml of water.
Well after a few days the water will evaporate away and dry up, right? I think even McGinn would agree to that. But just to be clear, it's one of the things we are assuming, that the water will dry up, or "evaporate".
Let's assume as McGinn does, that it evaporates into charged drops, which then levitate out of the bottle because everything's happening within an ambient electric field.
Okay, so let's think about that:
Well the water in the bottle must be charged to begin with - otherwise McGinn's method won't work. Because glass is a really good insulator, after all. So the charge won't be going through that glass or that bottle - noway, it has to start out in there from the beginning. Aardwolf says be careful, the water out of the tap may be charged. Even though the water pipe is grounded. Well maybe it's something else. But let's just say, okay, somehow we got 100 ml of water out of the tap, into the glass, and it ended up charged.
So now those nano-droplets coming off the surface of the water, are charged, and levitating out of the glass, because there's an ambient electric field that the glass is sitting in. According to McGinn.
Okay, moving along: Let's say the ambient electric field in the room is really strong. Dave says no that's impossible, but he won't prove it to us, and we are very strict scientists, no pulling the wool over our eyes, we require proof for anything we're asked to believe. (Unless we thought of it ourselves.) So let's say, the ambient electric field is good and strong, like, as strong as it ever gets under a thunderstorm. How strong is that?
Well let's look here:
http://journals.ametsoc.org/doi/pdf/10. ... 2.0.CO%3B2
The US Weather Bureau says, they've measured fields of up to 100 V/cm. We'll work in SI units, so that's 10,000 V/m - 10^4 V/m.
So how much charge do we need, to levitate 100 ml of water in that field?
Well first, let's find the force necessary to levitate the water, in SI units. That 100 ml of water is 100 g of mass, or 0.1 kg. We are physicists, so we describe force in Newtons in SI units - and the weight of 0.1kg of mass, in the Earth's gravity, is 0.98N - let's round off and call it 1N.
So now how much charge do we need to levitate that water? F=qE, is the "electric force":
1N = q (10)4 V/m --> q = 10^-4 Coulombs (again in SI units).
So the water in the glass, begins by carrying 10^-4 Coulombs of charge. Okay.
So let's say I'm walking back from the tap, carrying that glass of water. And the charged water inside, is on the other side of the glass, from my hand. So we've got a capacitor - an old fashioned Leyden jar. I wonder what the voltage on that capacitor is?
q=CV, relates the voltage on a capacitor, to the charge it's carrying, in Coulombs. In SI units again. So we need to know the capacitance of the glass.
Well the capacitance of a glass plate capacitor, is this:
C = (K e0 A) / d
where K is the relative permittivity factor for glass; it's about 7 for normal glass;
e0, (e because I can't type epsilon) is the permittivity of vacuum, which is 8.85(10)^-12;
A is the area of the capacitor plates;
d is the thickness of the plates.
I have a pretty big drinking glass sitting here; I measure circumference and height and diameter, and come up with A = 475(10)-4 m^2.
Let's say, the thickness of the glass, is 3 mm - 3(10)-3 m.
Going to the calculator, C = 980 picofarads. Let's round it off to 1000 pF, or 1 nF, or 10^-9F.
Okay so back to q=CV:
10^-4 = 10^-9 * V
V = 10^5 Volts.
So I'm carrying that water back from the tap, in the glass that I'm carrying in my hand, and the water inside is charged to 100,000 Volts. In order for it to evaporate according to McGinn's scheme.
Maybe the ambient electric field is ten times less? Like around a weak thunderstorm?
Then we need ten times as much charge to levitate. So the water has to be charged to 1,000,000 volts, when it goes into the glass.
Hmmm....