From: Geoff Ruth (geoffreyruth@usa.net)
Date: Thu Nov 18 1999 - 08:11:44 PST
Message-ID: <19991118161144.29923.qmail@nwcst291.netaddress.usa.net> Date: 18 Nov 99 08:11:44 PST From: Geoff Ruth <geoffreyruth@usa.net> Subject: e-mails from Marc
I didn't realize that Marc's e-mail hadn't gone to the pinhole listserv.
Here's the message I was referring to before. At the bottom of this is a
second reply from him.
Note: 10^-26 is ten to the negative twenty-six
I like the question. The distance across a water molecule is not
significant in computing the attractive vs repulsive force at the
typical distance we place the wand except to say that there is a net
attractive force because the distance is slightly less on the oppositely
charged side than on the like charged side. If we were to make a
calculation I expect that the attractive force would equal the repulsive
force out to several decimal places but eventually we would find a
difference. This difference, though small, is still significant enough
to produce a large acceleration since the mass of the molecule is so
small (on the order of 10^-26 kg). My math is not good enough to
calculate the difference between the attractive force and the repulsive
force, but I'd be interested to see what it is. If the net accelerating
force is only 10^-25 N then we'd still see an acceleration of about 10
m/s2 for the water molecule.
a = F/m
Let's also remember that the charge on the wand is quite large, though
not large enough to produce the effect past some distance (about 10 cm
if I recall the experiment correctly). So past 10 cm the stream seems
unaffected, but closer the stream really bends. This is due to the
inverse square relation for electric charges.
The water molecule's H atoms are really very nearly exposed protons with
very little negative charge to shield them from electrical effects. It's
as if the H proton is stripped of electrons because of the very high
electronegativity difference between O and H atoms. It is this exposure
which allows hydrogen bonding to occur (which is the very nearly
covalent bonding that occurs between water molecules when the unbonded
pairs of electrons from the O atom of one water molecule interact with
the exposed proton of another water molecule). In addition, the covalent
electrons can be thought of as belonging to the oxygen atom making the
unbonded pairs push a little further still from the nucleus (e-e
repulsion) of the O atom. This distance, though small, is not
insignificant. So if we think of two exposed protons closer to the large
static negative charge combined with the extremely small mass of the
water molecule we find a large accelerating force which produces the
bent stream.
Does this make sense?
* * * * * * * * * * * * * * * *
And here's Marc's reply to what I sent out yesterday.
> I did a little algebra and came up with the following:
>
> F=(kq1q2)/d^2 for the force due to electric charges.
>
> Since kq1q2 is the same for both sides of the water
> molecule, the only difference between the two forces
> is the d^2 factor. Solving for F1/F2 I found the
> following relation:
>
> (d+x)^2
> F1=--------*F2
> d^2
>
> Where x is the distance across the water molecule.
> Since x is so small, the numerator is an approximation
> of the denominator and your intuition about the
> magnitude seems reasonable. But, as I suggested,
> because the numerator is slightly larger than the
> denominator, F1 is slightly larger than F2 and due to
> the small mass of the molecule the acceleration is
> significant.
>
> I think you need to use the binomial expansion for
> this equation to find the correct value (or do it by
> long multiplication since a calculator will
> automatically round such a small difference) but I
> don't remember how to do that. Perhaps a math
> colleague can explain it to us. Plug in some arbitrary
> number for d (say, 0.10m) and use the distance across
> the water molecule for x and see what happens. I bet
> we'd find an appreciable acceleration.
>
> -Marc
>
> P.S. I'm still thinking about your question regarding
> nonpolar molecules.
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