Re: Since k varies but not G suggests an Eather


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.

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.

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.

- Randy

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