Re: Since k varies but not G suggests an Eather

guskz@xxxxxxxxxxx wrote:
Randy Poe wrote:
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?


My mistake, if a dipole moment is never zero

Any moment can be either zero or nonzero.

why does a quadrupole moment = zero when it's spherical?

If it's spherically SYMMETRIC there are no higher moments
than monopole, because higher-order moments measure
various ways in which a distribution is not symmetric.

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).

When Q and -Q are neutral,

You mean Q = 0? What else would it mean for a charge Q
to be "neutral"?

due to the constant known velocity of
charges in space

What? What "constant known velocity of charges in space"?
What velocity would that be? Charges can be not moving,
or moving at any velocity less than c.

thus "d" between them is a known constant (when
they're paired into a dipole),

Are you just trying to say "Q and -Q are located a fixed
distance d apart"?

so say "r" (of q) is = 1.5 d at first:

1. Does q orbit Q and -Q?

What?

It could be fixed, it could be in motion.

If in motion, I don't think that the motion could be a closed
orbit as that is a property of central 1/r^2 forces. The force
in this case is not 1/r^2 but something more complicated,
and it isn't central (it doesn't point toward a fixed center).

2. Does r remain at 1.5d when it returns to the same angle in it's
orbit

There's no orbit. See above.

or does it keep moving farther and farther away for the dipoles?

There's not even any guarantee it could "return to the
same angle". It might just fly by on some curved path.

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

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

Bad writing I ment: I see "d" as a constant distance which is related
(regulated) by the normal velocity of the charges in space

No such thing.

(since
velocity is an anti-attraction phenomena

What?

...the higher the velocity of
the attracting charges the bigger "d" between them).

What?

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

What?

The Electric field shrinks volume wise as the charges get closer
together?

What? What is the volume of an electric field?

The charges may be neutral but not the electric field

Electric fields aren't charged.

No but their entire value is related to the charges times the
permittivity constant: (Qxq and r^2 which is determined by the
velocity and polarities of Q &q)

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

What?

What....does charges being neutral "mean" to you?

The sentence I can't decode is "another charge whose r is almost
the same as d would feel no effect". Nothing about "neutral" there.
I have no idea what situation you're trying to describe.

It means they no longer have the same sex appeal (attraction repulsion
forces) as they once did as when they where single. This sex appeal is
the attraction/repulsion Force Field also known as the Electric field.

No clue what "they" refers to here.

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.


It was a question not a statement:

There was a statement: "I think the velocity of the charges is also
related to the permittivity of space, "

If this is a general statement, it's not true. It seems to be connected
to your belief that there is a constant natural velocity to charges,
which is also not true.

1. light and other waves slow down in dielectrics

Only on average and in the continuous approximation. In reality
light always travels at c.

therefore does the velocity of charges also slow down?

Not "therefore", but matter is slowed down when passing
through other matter, due to a host of effects which can be
lumped together as "friction".

2. I think the distance between dipoles is related to BOTH their
attraction force and velocity,

The dipoles that appear in dielectrics, which are due to
the polarizing of molecules, don't generally have a velocity.
The "attraction force" in question depends on the chemical
bonds in the molecule.

therefore if the velocity in #1 changes
therefore does the distance between dipoles also change??

I think that you have added the false beliefe "dipoles in a
dielectric move with some constant velocity" to your false
belief "charges everywhere move with characteristic constant
velocity". There's no such velocity.

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.

It's one of Sue's links, Electric Quadrupoles at:
http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elequad.html#c1

There is nothing there that says "a quadrupole is spherical".

(ellipsoidal quadpoles are non-neutral, spherical are neutral)

No, it doesn't say that. That's just an example of a charged
ellipsoidal distribution. It says that example can be represented
as a CHARGED uniform sphere and a NEUTRAL quadropole
consisting of four point charges.

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.

The "elementary quadropole" in the picture (four charges) has
net charge zero, dipole moment zero, quadropole moment
nonzero.

The ellipsoid has net charge nonzero, dipole moment zero,
quadropole moment nonzero IN THAT EXAMPLE. It's not
a general statement about charge distributions.

The spherical "component" IN THAT EXAMPLE has net
charge nonzero, dipole moment zero, quadropole moment
zero.

And so forth.

Ok but how would quadpoles (as the link above says) be 100% neutral
...doesn't that mean no "r" at all (no 1/r^2...no 1/r^3...etc)

If you have the same amount of + and - charge, your net
charge is zero.

But no, having NET charge zero is not the same as having
no moments. It doesn't mean there are no individual charges,
it means that the total charge is zero.

Something else to keep in mind is that a 1/r^3 field falls off
faster than a 1/r^2 field. So if you get very far from a distribution
with no 1/r^2 field, it does look like there's no field at all. But
as you get closer you start to perceive the nonuniformities,
the individual charges. You'll see a 1/r^3 field if there is one.
If there isn't, then you have to get closer still to see a 1/r^4
field, if there is one.

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.


no no, forgeting the neutrons...you have a sphere with protons in the
core and electrons around it

Yes.

, slice the sphere into a cross-section you
have a circle with electrons surrounding protons in the center,

Electron shells are clouds that actually penetrate the nucleus,
but yes you still have the protons confined to a tiny space in
the middle.

......then take *ANY* straight line across that circle and you have a
line of e e e e e e e p p p p p p p p e e e e e e e e e (the "- -" were
the arrangement of electrons).

Not exactly, it's more like a positive point in the middle of
a negative cloud. The nucleus is VERY small.

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

F = k Qq/r^2 where r is the LINEAR distance between charges, we were
talking about quadpoles (or more), so I took a cross section of an atom
and the most ****POPULAR**** position(orientation) for charges in space
(say only quadpoles) would be this orientiation:

I have no idea what you're saying.

e p p e (so what is the force relation for such an orientation
1/r^2 or 1/r^4 or less or more?)

I guess you're asking which moments an arbitrary atom has, all
by itself.

I'd say no net charge, no dipole moment, so only quadrupole
and higher moments.

But if the atom is ionized, then it has a net charge and so its
behavior is dominated by 1/r^2 forces.

If the atom is in a neutral molecule like H2O, then you get dipole
moments and 1/r^3 forces dominate.

(Therefore again doing my cross-section above, an eigth pole would be e
e p p p p e e

No, you don't get these poles by considering what happens
along a straight line. Multipoles describe the distribution
in 3-space.

Not really a dipole moment but a dipole wave

What?

The charges are not fixed in space,

No, this is misguided, no matter how many times you say it.

- Randy

.

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