Re: Since k varies but not G suggests an Eather


Randy Poe wrote:
guskz@xxxxxxxxxxx wrote:
Sue... wrote:
guskz@xxxxxxxxxxx wrote:

snip

So you're saying instead of applying the total charges in space to the
equation and using "k for space", they are varying "k" instead so as
to include the charges in the medium (as opposed to the charges outside
the medium)?

Whether charges exist ~outside the medium~ or not
does not concern us. If they are absent, their charge is
zero. If they are present, they exist in e+ e- pairs so
their value is zero.

Relativity is the word:

In a way nothing is zero nor potential and all is kinetic (always in
motion, never stops).

Paired charges are neutral and not, but without question the field
shrinks to it's smallest hypothetical value and the force will not be
neutral but at it's maximum

and if you apply energy to seperate the charges then the field is
bigger volume wise but it's force decreases.


(F_neutral = kQq/r^2 where r is the constant distance e and p remain
when neutral and Epotential = (F - F_neutral)*r^1)

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

Since nothing is neutral

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?

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.


I see "d" (P=Qd) as the related to the normal velocity of the charges
only and that all fields simply shrink with "d",
The charges may be neutral but not the electric field otherwise if
another charge whose "r" is almost the same as "d" would feel no
effect.

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


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?



same as a magnetic force.....


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...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)?

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

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

Not really a dipole moment but a dipole wave


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

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

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

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

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??? (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"?


- Randy

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