Re: Since k varies but not G suggests an Eather


guskz@xxxxxxxxxxx wrote:
Randy Poe wrote:
guskz@xxxxxxxxxxx wrote:
Why do you say nothing is neutral? If you have equal
amounts of positive and negative charge, the net
charge is zero.


Ok, the dipole moment then aint zero, P = Qd and it's Q aint zero.
Otherwise H2O would never form?

What?

and only the field shrinks as Randy said the
residual force of dipole charges is 1/r^3

Yes, if:
(a) your + and - charges are separated so that
you have a net polarization. Just because you have
+ and - charges doesn't mean they have to separate.

(b) r is large compared to the separation between
the charges.


Ok but if "r" was the same distance as the separation then the forces
and charges would not be neutral.

What do you mean by forces being neutral?

The expression for the force from a charge Q and a charge -Q
on a charge q is F = kQq*(1/r1^2 - 1/r2^2).

If r is the same distance as the separation, then the approximation
used to convert (1/r1^2 - 1/r2^2) to 2*sin(theta)/r^3 (or whatever the
correct expression is) breaks down. Instead you can use the
exact value of (1/r1^2 - 1/r2^2).

I see "d" (P=Qd) as the related to the normal velocity of the charges

You shouldn't, since it isn't a velocity.

only and that all fields simply shrink with "d",

What?

The charges may be neutral but not the electric field

Electric fields aren't charged.

otherwise if
another charge whose "r" is almost the same as "d" would feel no
effect.

What?

But in true nature: if q = neutral = 0 then Electric field would = 0
(since E= kQq/r^2) = none sense?

With large objects at ordinary macroscopic differences, this is
usually true. The force between you and the earth is dominated by
the gravitational force. There's no net charge on you or the earth,
so no electric force between the two of you.

I think the velocity of the charges is also related to the permittivity
of space, and if so would this velocity of the charges change in a
dielectric (since light's "c" does)? ....and therefore would "d" also
change?

You can think stuff which is totally at odds with all of our experience
and theory, but that won't make it true or even plausible.

If you have two dipole charges, does the residual remain at 1/r^3 or is
it more or is it less?

It depends on how they are arranged and where you are
situated. You can have two dipoles which will give a field
that looks like one stronger dipole. You can have two
dipoles cancelling each other out so what is left is
the quadrupole moment (1/r^4 force). You could be
located near to the dipoles in which case you have
to use the exact expression, sum(kQq/r^2), not the
approximation 1/r^3.

I read that normally the quadpole is spherical thus neutral...

That's a little confused, but I can see what you read buried
in there.

1. You can have a net charge, in which case you'll see monopole
fields, E ~ 1/r^2.

2. You can have an arrangement of charges with net charge zero
but a dipole moment, in which cases at large distances E ~ 1/r^3.

3. You can have an arrangement of charges with net charge zero
and dipole moment zero but a nonzero quadropole moment,
in which cases at large distances E ~ 1/r^3.

And so forth.

maybe for
hydrogen but for other atomic mediums the arangement due to the nucleus
in any cross-section of the sphere (excluding neutrons) is:
- - - - - - - - + + + + + + + + - - - - - - - - - pole (protons
grouped together in nucleus)?

The protons are in the nucleus. I don't know what else you're
trying to say. The nucleus is very small compared to the atom,
so the protons are "grouped together". If those "---" symbols mean
you think there are electrons in the nucleus, that's wrong.

What would be the force for those (1/r^4??)

I can't figure out what "those" refers to.

--------------------------------------------------

Not really a dipole moment but a dipole wave

What?

As well for hydrogen, negative charges(e) must always be orbiting
around the heavier charge (p) ??

No, electrons are not in orbit in atoms.

... therefore creating dipole moment
pulses (a constant oribiting frequency of dipolar force as the electron
orbits the proton, therefore the dipolar moment would be a wave force
of a specific frequency and not a specific value....but

No. Not in an atom. However, it is true you could create a
situation like this with a macrosopic arrangement, and you'd
get an oscillating dipole field. I don't know what you mean by
"not a specific value". Of course it has a value.

Perhaps why this dipolar frequency in a dielectric cannot affect high
frequency EM waves (since the dipolar frequency is slower)?

There's no such dipolar frequency due to orbits since they don't
exist.

There are resonant frequencies when dipoles are free to oscillate,
such as in H2O. H2O resonates very nicely with frequencies in
the GHz, which is why microwave ovens work.

(Wikepedia says dielectric mediums (force wise) only affect static
charges and slow frequency EM waves)

That would indeed have something to do with how fast the dipoles
wiggle, but it's not connected to electrons orbiting around nuclei.

Also I don't know if this frequency would be a nice sine wave since I
believe in a previous post with you Randy long ago, that the force
varies considerably with each angle between dipoles???

Yes, "wiggles" are typically "nice sine waves".

(Forgot which
post). ....or was it that the force would be neutral if the 3rd charge
was exactly perpendicularly in between the dipoles at distance "r"?

Dipole force varies with sine of the angle (or cosine, depending on
how you define the angle). That's why a rotating dipole would give
a force which varies as a sine wave.

But an atom is not a rotating dipole.

- Randy

.

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